# April 1

## applications of infinite series

1. Working out how much time it takes to do half your homework, plus a quarter of your homework, plus an eigth of your homework, plus a sixteenth of your homework ...
2. Writing a really cool-looking Taylor expansion as a means to justify various properties of an unknown function, which turns out not to be well-defined anyway. Confusing Manifestation 12:48, 1 April 2006 (UTC)
3. Writing "1 + 2 + 3 + 4 + . . . = −1/12" on a blackboard and waiting for someone uninitiated in divergent series and the zeta function to say you're wrong. Melchoir 19:54, 1 April 2006 (UTC)
You know, much as it sounds like a homework question (and if not, then merely a very strange one), I am in fact tempted to come up with 30-40 applications of variable usefulness, just for the fun of it. And I'm not even going to mention a certain Talk page debating the true meaning of infinite sums. *twitch* Confusing Manifestation 13:27, 2 April 2006 (UTC)
At the risk of sending you into further convulsions, I'd be interested to know which talk page you're talking about. -lethe talk + 14:24, 2 April 2006 (UTC)
Oh, uh, you're probably talking about Talk:Proof that 0.999... equals 1, right? -lethe talk + 14:33, 2 April 2006 (UTC)
Specifically the Arguments sub-page, which I have been taking a nice little break from. But yes, that page is bad karma. Confusing Manifestation 04:32, 4 April 2006 (UTC)

## Easy math problem

{ 10 [3(squared) + 2/4 - 5/8 ] - 9 }

I just want to know if i got the anwser right, because I was bored and I made it up... also, I don't know how to write the squared symbol, sorry. :| . --Cosmic girl 15:33, 1 April 2006 (UTC)

${\displaystyle 10(3^{2}+2/4-5/8)-9=10(9+1/2-5/8)-9=10(71/8)-9=355/4-9=319/4}$
Look at the source to see how I made it look that way, and check out Help:Formula for more help. —Keenan Pepper 15:45, 1 April 2006 (UTC)
Or if you want pretty fractions, ${\displaystyle 10(3^{2}+{2 \over 4}-{5 \over 8})-9=10(9+{1 \over 2}-{5 \over 8})-9=10({71 \over 8})-9={355 \over 4}-9={319 \over 4}.}$Keenan Pepper 15:48, 1 April 2006 (UTC)
Or even better, ${\displaystyle 10\left(3^{2}+{2 \over 4}-{5 \over 8}\right)-9=10\left(9+{1 \over 2}-{5 \over 8}\right)-9=10\left({71 \over 8}\right)-9={355 \over 4}-9={319 \over 4}.}$Ҡiff 16:25, 1 April 2006 (UTC)

Cool! I got it right! lol...(I'd be really stupid if I didn't though). thank you Keenan, and Kieff. XD. --Cosmic girl 16:00, 1 April 2006 (UTC)

And here are some ways to write "squared" without math tags (see source): 3², 3^2, 32. -- Meni Rosenfeld (talk) 11:45, 2 April 2006 (UTC)

## Tricky Cooling Problem

An object cools at a rate (in °C/min) equals (1/10) of the difference between its temperature and the surrounding air. If a room is kept at 20 °C and the temperature of the object is 28°C, what is the temperature of the object 5 minutes later?

I think the object's temperature=f(t)=8*exp(k*t) +20, where 8=temperature difference of the object and the surrrounding and 20=the surrounding temperature.

I don't understand where ${\displaystyle {\frac {1}{10}}}$ comes into play.

Patchouli 19:47, 1 April 2006 (UTC)

I think your formula is right; it just doesn't specify k. The way you've written it, k should be a negative constant related to 1/10. Melchoir 19:50, 1 April 2006 (UTC)
You've pretty much gotten it; the way you want to do this is phrase the solution in terms of a differential equation. According to the problem the way you've stated it, you should have ${\displaystyle {\frac {dT}{dt}}={\frac {1}{10}}(20-T)=2-{\frac {T}{10}}.}$ You can see by inpsection that the general solution to that equation is ${\displaystyle T(t)=C\exp ^{-t/10}+20}$, and plugging in ${\displaystyle t=0}$ gives you ${\displaystyle C=8}$.--Deville (Talk) 20:01, 1 April 2006 (UTC)
• Thus, ${\displaystyle T(5\ minutes)=8\exp ^{-1/2}+20.}$Patchouli 20:28, 1 April 2006 (UTC)

## Function problem vs. computation problem

In computational complexity theory, what's the difference between a function problem and a computation problem? Both terms seem to be used fairly similarly. The computation problem article leaves me very confused. -- Creidieki 20:59, 1 April 2006 (UTC)

I'll throw down some merge tags; they tend to drwa attention. Melchoir 21:57, 1 April 2006 (UTC)
My advice is to ignore the "computation problem" article. To me, "computation problem" is not usual terminology. Usually decision problems are contrasted with search problems, which are the same as function problems . --68.238.254.236 14:24, 7 April 2006 (UTC)

## Lost (TV series)

Hi, this is going to sound insane, but over on Lostpedia, we're working on deciphering the blacklight map on the back of the blast door, as displayed in the most recent episode "Lockdown." There's an equation we'd like to display, but Wikisyntax math markup is making my head spin. Could someone show me how to display the following on a Wikipage? Thank you! jengod 22:37, 2 April 2006 (UTC)

=2rcos[?]
=r*((sqr(5)-1)/2)
-72' = 4rcos"2(72') = r*((sqr(5)-1)/2)

(Take a look at the wikisource:)
${\displaystyle =2r\cos {?}}$
${\displaystyle =r*{\frac {{\sqrt {5}}-1}{2}}}$
${\displaystyle -72'=4r\cos {2(72')}=r*{\frac {{\sqrt {5}}-1}{2}}}$
That's my best guess at what you're after without having the image you're working from, the middle part of the third line doesn't make much sense and I had to remove the double quote mark to get TeX to render it. Take a look at Help:Formula for more detail and examples. -- AJR | Talk 23:10, 2 April 2006 (UTC)

You ROCK!

Thank you. Seriously, you're a star. The original image, enhanced can be found here. I can't make heads or tails of it, but someone broke it down as above. There are also a couple of vector-y equations on the right side of the map. Anyway, THANK YOU.

If you want to join in on the deciphering fun, check out http://lostpedia.com/wiki/Blast_Door. 71.106.0.109 01:01, 3 April 2006 (UTC)

I would read the third line in the image as ${\displaystyle -72^{\circ }=4+cos^{2}72^{\circ }=r\cdot {\frac {{\sqrt {5}}-1}{2}}}$ (although the minus sign at the start is suspect as it's emerging from outside the image.) -- AJR | Talk 12:28, 3 April 2006 (UTC)
72 degrees and golden ratio surely have something to do with pentagon. Or pentagram.  Grue  18:00, 4 April 2006 (UTC)
It's probably a pentagram, just because this is a hidden message and people love to stick those all over the place. Pentagons not quite so much :P — Ilyanep (Talk) 00:37, 5 April 2006 (UTC)
To me, that + looks more like a t, which also seems more sensible from a mathematical point of view: ${\displaystyle ...=4t\cos ^{2}72^{\circ }=...}$ Also, don't forget the backslash before the "cos" because it should not be rendered in italics, as it is not a variable. RupertMillard 10:37, 5 April 2006 (UTC)

# April 3

## Polar singularity

With reference to Euler angles, is "Polar singularity" the same thing as gimbal lock. In any case, what exactly is a polar singularity, that arises while numerically integrating differential equations involving Euler angles, and how can it be avoided? Thanks, deeptrivia (talk) 01:13, 3 April 2006 (UTC) PS: I figured out that I'll have to use quaternions and I read about them. How exactly do I make the switch from Euler angles to quaternions? deeptrivia (talk) 01:35, 3 April 2006 (UTC)

Given the Euler angles α, β and γ, the quaternion representation should be (cos γ + k sin γ)(cos β + i sin β)(cos α + k sin α). I think. Anyway, the basic idea is just to take the quaternions for the fundamental rotations around the axes by the given angles and multiply those together in the proper order. —Ilmari Karonen (talk) 19:01, 3 April 2006 (UTC)
Usually a polar singularity is the kind of thing that happens with latitude and longitude at the poles, where a latitude of 90° causes changes in longitude to have no effect. This is analogous to gimbal lock, where middle axis rotations of either 0° or 90° (depending on the axis conventions) cause the other two rotations to no longer have independent effects. Code to change any system of Euler angles to quaternions can be found at the Graphics Gems repository. However, the differential equations must also be adapted. The equation with angular velocity is
2 dq = ω q dt,    ω is the angular velocity vector
Perhaps that will suffice. --KSmrqT 20:15, 3 April 2006 (UTC)

Thanks both of you. I've been solving the same problem I had discussed in an earlier question. My differential equations look like this:

${\displaystyle \psi ''=f_{1}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$
${\displaystyle \phi ''=f_{2}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$
${\displaystyle \theta ''=f_{3}(\psi ,\phi ,\theta ,\psi ',\phi ',\theta ',s)}$

Boundary conditions:

${\displaystyle \theta (0)=\pi /2,\phi (0)=0,\psi (0)=0}$
${\displaystyle \theta '(L)\sin \phi (L)-\psi '(L)\sin \theta (L)\cos \phi (L)=M_{1}/EI_{x}}$
${\displaystyle \theta '(L)\cos \phi (L)-\psi '(L)\sin \theta (L)\sin \phi (L)=M_{2}/EI_{y}}$
${\displaystyle \phi '(L)+\psi '(L)\cos \theta (L)=M_{3}/GJ}$

where ${\displaystyle \psi ,\phi ,}$ and ${\displaystyle \theta }$ are the Euler angles. s is the independent variable w.r.t. which the differentiation is carried out. How can I convert this entire thing to quaternions? Are there any substitutions/transformations I can do? I found a source that says that if the quaternion is {b0,b1,b2,b3}, then it is related to Euler angles by:

${\displaystyle b_{0}=\cos(\theta /2)\cos(\phi +\psi ),}$
${\displaystyle b_{1}=\sin(\theta /2)\cos(\phi -\psi ),}$
${\displaystyle b_{2}=\sin(\theta /2)\sin(\phi -\psi ),}$
${\displaystyle b_{3}=\sin(\theta /2)\sin(\phi +\psi )}$

Does that mean I just have to make these substitutions into my equations? I'm not even sure how to make these substitutions because inverting these relations will be hard, and will introduce inverse trigonometric functions, which will perhaps have singularity problems of their own. I really appreciate your help! deeptrivia (talk) 03:11, 4 April 2006 (UTC)

## Deprit’s variables

What are Deprit’s variables? I couldn't find a definition anywhere. Thanks :) deeptrivia (talk) 02:38, 3 April 2006 (UTC)

I think this (click on either of the Full Refereed Article links depending on whether you want PS/PDF or scanned GIFs) might be the original paper in question. Way over my head, though. --Bth 09:11, 3 April 2006 (UTC)
Or maybe this series of papers (with a mysteriously missing number III)? --Bth 09:21, 3 April 2006 (UTC)

Thanks. I think it's the second one that I needed. You rock ! deeptrivia (talk) 03:12, 4 April 2006 (UTC)

## Formula for compound mitres

I am making a three sided pyramid out of three boards. They will be triangular and the edges will be mitred. I need a formula to calculate the angle of the mitres [based on number of sides (examp.7), angle of slope,length of base...etc.]. If possible, it would be nice to be able to put it on a spread sheet.

Can you help?

I'm not sure I understood your description. Can you clarify a bit? -- Meni Rosenfeld (talk) 15:59, 3 April 2006 (UTC)
The questioner needs to know the angle between faces of a pyramid as specified (and then, presumably, divide by two for the mitring). It's a bit unclear how regular the pyramid is. Are the three triangular boards isosceles and the same shape? If so, we should be able to give a simple enough formula. —Blotwell 01:47, 4 April 2006 (UTC)
If your interpretation is right (and I suspect it is), I'm confused by the "based on number of sides" bit in the question. Maybe he wants to generalise after making his three-sided one? --Bth 12:38, 4 April 2006 (UTC)

I believe the answer will depend on the height of the pyramids. For example, with a triangular base, there are 180° inside the base. That makes 180°/3 or 60° between the sides. In the case of an infinitely tall pyramid with a 3-sided base, there would therefore be a dihedral angle of 60°. In the case of a pyramid of zero height, the dihedral angle is 0° (or 180°). Any height in between would have dihedral angles somewhere between 60° and 180°. Do you by any chance want to always use equilateral triangles ? If so, this will give a specific height for each pyramid.

In the case of a equilateral triangle base pyramid with equilateral triangles on the sides, as well, the dihedral angle is 70.53°, and the miter cut angle is (180°-70.53°)/2 = 109.47°/2 = 54.735°. In the case of a square base pyramid with equilateral triangles on the sides, the dihedral angle is 109.47° and the miter cut angle is (180°-109.47°)/2 = 70.53°/2 = 35.265°. In the case of a regular pentagon base pyramid with equilateral triangles on the sides, the dihedral angle is 138.19° and the miter cut angle is (180°-138.19°)/2 = 41.81°/2 = 20.905°. If a regular hexagon is used as a base, then the equilateral triangles would form a flat surface, not a pyramid. A regular heptagon or higher number of sides, used as a base, with equilateral triangles, would have gaps, and not form a pyramid at all. Note that the miter cut angles between the sides and base have not been discussed here.StuRat 02:42, 5 April 2006 (UTC)

To be more specific, the base could have from three sides to eight or nine sides and tapering to a point at the top. The height dosn't matter, it could be from two inches to two km. and same with the base. I'm looking for a formula wherby entering the variables (height, width, number of sides) I can calculate the angle at which to cut the mitres. Thanks ---74.12.9.227

(I am fully aware the following formula could be simplified. I've just lost any will to do it. It is offered with no warranties, etc., though it does reproduce StuRat's results for equilateral triangle side pieces and ${\displaystyle n}$ of 3 and 4. [I also lost the will to continue testing.])
Here's what I've assumed: a regular polygon of ${\displaystyle n}$ sides of length ${\displaystyle a}$ for the base, isosceles triangles for the sides of base ${\displaystyle a}$ and whatever side length is necessary to get the required height of the pyramid, ${\displaystyle h}$.
Make a definition of ${\displaystyle \theta ={180(n-2) \over n}}$, which is simply the internal angle of the polygon in question (we could write this out in full in the equation, but it would drive us mad). This is for degrees, of course; for radians, we'd replace the 180 with a ${\displaystyle \pi }$ -- NB that Excel works natively in radians.
The mitre angle, which I'm going to call ${\displaystyle \psi }$ is then (assuming I haven't made any hideous mistakes, which I probably have):
${\displaystyle \psi ={1 \over 2}\left(180-\arccos \left({a^{2}h^{2}\cos \theta -{a^{4} \over 4}\tan {\theta \over 2}(\sin \theta -\cos \theta \tan {\theta \over 2}) \over {\sqrt {(a^{2}h^{2}+{a^{4} \over 4}\tan ^{2}{\theta \over 2})(a^{2}h^{2}+({a^{2} \over 2}\sin \theta -{a^{2} \over 2}\cos \theta \tan {\theta \over 2})^{2})}}}\right)\right)}$
Some other things which might be useful: the formula for the side length of the triangles is ${\displaystyle {\sqrt {h^{2}+{a^{2} \over 2}\left(1+\tan ^{2}{\theta \over 2}\right)}}}$ and the mitring angle for the join between the faces and the base if you had a base piece as well would be ${\displaystyle {1 \over 2}\arctan \left({2h \over a\tan {\theta \over 2}}\right)}$ (if it's free standing, you need double this, obviously).
There is almost certainly at least one mistake in the foregoing, but I hope it helps. --Bth 11:49, 8 April 2006 (UTC)

Thanks, it helps alot. ---70.50.197.146

Wait a moment, I've just realised there is an error. That big ugly arccos expression is the dihedral angle, so the mitre angle is a half of 180 degrees minus that, not just half it. I've fixed it now. (Hope you come back and check! And of course there may still be subtler problems ...) --Bth 08:56, 9 April 2006 (UTC)

Thanks for the update! ---74.12.2.9

## Approximating a parallel bezier

Here's a graphics/geometry question, although fundamentally mathematical.. What's the best way of determining the control points of a cubic Bezier, so that is approximates the curve of a different Bezier? From searching, I've found there's no exact solution. But what kind of approximation is used? --130.237.205.132 17:47, 3 April 2006 (UTC)

It can be a nasty problem; suppose the given curve has a loop. The simplest method to program is probably to treat it as a curve fit task, where points on the desired curve can be calculated from the given curve. Don't expect a single Bézier curve to suffice. --KSmrqT 20:24, 3 April 2006 (UTC)

# April 4

## product a functor?

The article on adjoint functors mentions that the diagonal functor is the left-adjoint of the product functor (this works for any limit), which gives the categorical product of two objects. This seems fishy to me. The categorical product is defined in terms of its univeral property. Thus, it is determined only up to canonical isomorphism. Thus I can't define a a functor here, since I don't have a unique choice of object. So I think this is only a pseudofunctor, not a full-fledged functor. Now, Mac Lane lists these guys as adjoint functors as well, and he makes no mention of pseudofunctors, so my question is, am I totally wrong about this? What's the deal? -lethe talk + 05:20, 4 April 2006 (UTC)

Well, at least you can say this: if a category has binary products, then it has a product functor: a real, honest functor. We can't define it uniquely, but such functors exist... and they're all adjoint to the diagonal functor on whichever side.
I'm sure this isn't news to you, but I think that's what the sources mean. Melchoir 06:06, 4 April 2006 (UTC)
Read the fine print at categorical product#Discussion; the fact that a product object is unique only up to isomorphism is not an obstacle to defining a functor. --KSmrqT 06:15, 4 April 2006 (UTC)
So I assume you're referring to the sentence "it is possible to choose the products in a compatible fashion so that the product turns into a functor CI → C". I guess whoever wrote that sentence had in mind the issue that's bothering me, and knows the solution to it. Somehow it's possible to choose particular representatives of the isomorphism classes of product objects so that the pseudofunctorial natural isomorphisms all get straightened out into honest equations. Well, I'm pleased that someone knows my woes and has an answer, but I'd be even happier if someone told me what it is. For example, do we require the axiom of choice to hold in the index category in order to be able to make this choice? -lethe talk + 13:41, 4 April 2006 (UTC)
Oh, but whatever this construction is, even if it does require the axiom of choice for general limits, I guess we wouldn't need the axiom of choice for binary products, since the index category here is finite. Right? -lethe talk + 13:53, 4 April 2006 (UTC)
I found a paper online which defines functors whose values are only defined up to isomorphism to be anafunctors, not pseudofunctors. I'm not positive that the definition is equivalent to the one I know for pseudofunctors, so now I have to think about that some more. The paper also mentions that the existence of honest functors adjoint to the diagonal functor relies on the Freyd Adjoint Functor Theorem, which does rely on the axiom of choice. I might be satisfied with that. -lethe talk + 14:30, 4 April 2006 (UTC)
That paper looks like it was written with MS Word. Hard to take such a paper seriously. -lethe talk + 17:41, 4 April 2006 (UTC)
Pardon the ignorance, but now that you mention it, which software should be used for serious papers? And for that matter, what's wrong with Word? -- Meni Rosenfeld (talk) 05:58, 5 April 2006 (UTC)
LaTeX should be used for math papers. The arXiv and many (most?) reputable journals do not accept word documents. Many journals use house formatting styles which are latex macros. In addition to not looking very nice, Word doesn't support such formatting macros. -lethe talk + 06:19, 5 April 2006 (UTC)

## palindrome

4 digit palindrome number two more than a perfect square

7227 = 85^2+2
(by manual brute force search with Windows calculator). I hope that wasn't your homework, or a particularly easy Enigma or something. --Bth 09:24, 4 April 2006 (UTC)
Or, you could have taken a calculator like casio fx-82TL or similar, start with 1026 and repeatedly calculate
${\displaystyle ({\sqrt {Ans-2}}+1)^{2}+2}$
That is, of course, unless you have any programming language handy. -- Meni Rosenfeld (talk) 09:51, 5 April 2006 (UTC)

What would the mathematical approach to the problem be?  :--) JackofOz 12:41, 5 April 2006 (UTC)

Well, you could narrow down the search with this table:
 x (1st digit = 4th digit) in x^2+2 4th digit in x^2 2nd digit in x candidates 32-44 1 9 3,7 33,37,43 45-54 2 0 0 50 55-63 3 1 1,9 59,61 64-70 4 2 \ \ 71-77 5 3 \ \ 78-83 6 4 2,8 78,82 84-89 7 5 5 85 90-94 8 6 4,6 94 95-99 9 7 \ \
Checking the candidates will reveal that only 85 produces a palindrome. -- Meni Rosenfeld (talk) 16:02, 5 April 2006 (UTC)

## what's up

what's up

A direction. <rimshot /> —Keenan Pepper 13:06, 4 April 2006 (UTC)
There is a complexity class UP, which I only mention in a vague attempt to get this section on-topic for the Maths RD. -- AJR | Talk 17:00, 4 April 2006 (UTC)
Nice. So there's no proof that UP != P or that UP != NP? —Keenan Pepper 23:58, 4 April 2006 (UTC)
The sky. —OneofThem(talk)(contribs) 17:34, 4 April 2006 (UTC)
There's UPS, trousers, scissors, but there's no single up, trouser, or scissor. How strange! – b_jonas 12:15, 5 April 2006 (UTC)
I wonder if any of the people in Seven Up ever drank 7-Up. :--) JackofOz

## too late to see answer

Help, I posted a question about house numbering here 9 days ago, and now it and its possible answer have glided off the top with now way to retrive them! No link to go back one page of answers! If you see them send it to (email excluded) as who knows, I might not even get back to see the answer to this in time!

Check the Archive. Isopropyl 23:59, 4 April 2006 (UTC)
Note that questions and answers are normally archived after a week. However, you really should check back each day to see if there are any clarifications needed. StuRat 02:29, 5 April 2006 (UTC)
Also, any answers will be in the page history. The same IP address that asked this question also created a section headed "Subject: naming points on graphs (house numbers on twisty roads)", this is the last version before it was archived. -- AJR | Talk 19:23, 5 April 2006 (UTC)

## WHAT IS 123098+1322

It is the power of the union of two disjoint sets A and B, where the power of A is 123098 and the power of B is 1322. -- Meni Rosenfeld (talk) 08:36, 5 April 2006 (UTC)

Try using a calculator, like the one built in to your computer. StuRat 08:49, 5 April 2006 (UTC)

I suspect there's more to the question than meets the I. It is the 5th of April, after all. JackofOz 08:58, 5 April 2006 (UTC)
Go see what google says. – b_jonas 12:12, 5 April 2006 (UTC)
I see nobody has stated the simplest answer there:
* * * * * * * * * *
*                 *
*   1 2 4 4 2 0   *
*                 *
* * * * * * * * * *


## The Golden Ratio

Tell me the religious and mathematical significance of the Golden Ratio.

Read it for yourself at Golden ratio. Then do your own homework. --Bth 12:54, 5 April 2006 (UTC)

The Golden Ratio is the relationship between the width and length of a rectangle which is most appealing to the eye. It is used constantly in furniture design and different aspects of architecture. The ratio is, W=L×.618. ---74.12.9.227

Explain to me why you would waste time writing that here instead of just looking at the article. —Keenan Pepper 17:23, 5 April 2006 (UTC)
Some research has suggested that it may not really be the most aesthetic, and we may just be obsessed with the ancient Greeks. Superm401 - Talk 20:27, 5 April 2006 (UTC)

## bound on number points symmetric block design

Consider a block design : ${\displaystyle 2-(v,k,\lambda )}$ such that the number of points equals the number of blocks : it is thus by definition a symmetric block design.

How can I prove that teh number of points v satisfies ${\displaystyle v\geq 4n-1}$ with ${\displaystyle n=k-\lambda }$ the so-called order of the design

It's really only a subtlety I can't figure out. I know I should consider the equation ${\displaystyle v-1={\frac {k(k-1)}{\lambda }}}$ and substitute ${\displaystyle k=\lambda +n}$

I would get the result by demanding the discriminant of the resulting quadratic equation in ${\displaystyle \lambda }$ being strictly greater than zero, but not in the exceptional case that the discriminant is exactly zero.

Any suggestions in avoiding this problem would be greatly appreciated.

## Sum of individual numbers

Consider the number 1983. For 1983 foo = 1+9+8+3 = 21

But is there a special name for summing each individual digit, and if so what is it?

The most obvious thing is that multiples of 3 always sum to a multiple of 3, but you knew that anyway. I think there is a similar effect for 9, (3*3) but not 6.

The last two digits of multiples of 4 are always a multiple of 4, (and there are probably others) but you can answer my original question. — Dunc| 20:45, 5 April 2006 (UTC)

With respect to your first question, the article digit sum seems on point, although if one carries the process on until he/she reaches a single digit (in your case, 3 [2+1]), the resultant is oftentimes referred to as a "mod sum"; the latter method, is, I think, referred to by accountants (and public practitioners of mental math, I suppose) as one in which "the nines are cast out", inasmuch as the "mod sum" of any number is also the remainder when that number is divided by 9. Joe 20:54, 5 April 2006 (UTC)

# April 6

## Matrices

What is expansion by minors?

I'm pretty sure it's another name for Laplace expansion. —Keenan Pepper 01:47, 6 April 2006 (UTC)

Thank you very much for the speedy response! :)

## STATISTICS

WHAT IS MEASURE OF DISPERSION?

See Statistical dispersion. -- Meni Rosenfeld (talk) 18:05, 6 April 2006 (UTC)

## Melting snowball

a) A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1cm3/min. At what rate is the diameter decreasing when the diameter is 10cm?. —This unsigned comment was added by 218.111.184.204 (talkcontribs) .

Fast enough that it'll be gone before you get your homework done. --Trovatore 19:43, 6 April 2006 (UTC)
So is this the way snowballs actually melt? Would not the surface area play a role, if so I would expect a quadratic term rather than just a cubic term in the homework. As to solving the homework you need to the volume of a Sphere, first, then apply some elementary algebra to find the solution. --Salix alba (talk) 20:39, 6 April 2006 (UTC)
If you start with the volume formula you're going to need calculus. Why not just evaluate the surface area of a sphere of diameter 10cm and divide? —Blotwell 03:42, 7 April 2006 (UTC)
This idealized situation is a related rates-type equation. You should relate the equations for the rates of change for volume and diameter with respect to time. As for real life, there's a billion other factors, like the ambient temperature, heat capacity of the surrounding air and the surface it's in contact with, and so forth. Isopropyl 20:51, 6 April 2006 (UTC)

# April 7

## Zero Radicals and the Transitive Property

To solve the equation of

${\displaystyle {\sqrt[{0}]{x}}=y}$

for x, you would take each side the power of zero, right? So you would get

${\displaystyle \ x=y^{0}}$

and since any number to the 0 power is 1, x=1.

So,

${\displaystyle {\sqrt[{0}]{1}}=y}$

where y is any number.

Obviously, someone could use this to prove any number is equal to any other number.

Which leads to my final question: What are the restrictions of the transitive property? Does it state when you can't use it to argue certain things (such as the simple -1=1 using square roots proof)? Have the 'rules', if you will, of the transitive property, been agreed upon? 65.31.80.100 12:01, 7 April 2006 (UTC)

I'm not sure where transitivity comes into it. If you're suggesting that we make exceptions to transitivity of = in order to avoid paradox here, that's not how we do it (I think transitivity of = is too important to give up, don't you?). As with division by zero, we simply don't allow taking 0th roots. (Evaluating ${\displaystyle {\sqrt[{0}]{x}}}$ is a lot like evaluating ${\displaystyle x^{\infty }}$, so you could also look at indeterminate form.) —Blotwell 03:37, 7 April 2006 (UTC)
To emphasize, the problem is not with transitivity of = (which always holds), but with working with operations which aren't defined. In math, you must only deal with things which are defined and follow the definition. You can in some cases use your intuition to find a plausible extension for a definition (see for example Real projective line, a structure where 1/0 is defined), but you still have to follow the definition rigorously. So your problem is dealing with ${\displaystyle 1^{\infty }}$ which isn't defined (and can't be plausibly defined, for that matter). If you'll show us the exact square root argument you have in mind, we can show you what's wrong with it too. -- Meni Rosenfeld (talk) 09:42, 7 April 2006 (UTC)
sqrt(1)=1, sqrt(1)=1-, but we obviously can't say that since 1 and -1 both solve the equation sqrt(1)=x, that they're equal.

65.31.80.100 12:01, 7 April 2006 (UTC)

Depends how you define sqrt(). Square root is usually defined to return the positive square root. See Root of unity for techniques to handle multiple solutions. --Salix alba (talk) 12:28, 7 April 2006 (UTC)
To put it differently: ${\displaystyle {\sqrt {1}}=+1}$, period. This is what follows from the definition. It's true that also ${\displaystyle (-1)^{2}=1}$ but that doesn't mean that ${\displaystyle {\sqrt {1}}=-1}$. Both +1 and -1 solve the equation ${\displaystyle x^{2}=1}$ but not the equation ${\displaystyle x={\sqrt {1}}}$. -- Meni Rosenfeld (talk) 12:40, 7 April 2006 (UTC)
Yes, taking a zeroeth root is like exponentiating by the reciprocal of zero, can't be done. I don't see where transitivity comes in though. Ozone 19:09, 7 April 2006 (UTC)
What the questioner meant was: Assuming that, as argued, :${\displaystyle {\sqrt[{0}]{1}}=y}$ for any y, then for example ${\displaystyle {\sqrt[{0}]{1}}=1}$ and ${\displaystyle {\sqrt[{0}]{1}}=2}$. By transitivity we get 1=2, a contradiction. The suggestion was that perhaps transitivity doesn't hold in such cases. What we have shown is that it is not transitivity that fails, but the equalities themselves. -- Meni Rosenfeld (talk) 17:44, 8 April 2006 (UTC)

## Concatenating two space curves

Well, I am sure this has to be simple, but my mind is all messed up right now. Suppose you have two curves, C1 defined by points (x1i,y1i,z1i) i= 1 to N1, and C2 defined by points (x2j,y2j,z2j) j = 1 to N2. How can I put the curve C2 at the end of C1 to get a smooth curve C3 with N1+N2 points. I have at my disposal a function that can numerically calculate the components of tangent, normal and binormal vectors at all these points (returning three arrays, each of size 3xN.) I guess the values of these vectors at the last point of C1 and the first point of C2 should suffice to find out the required rotation matrix for all points onC2. Any help will be greatly appreciated. deeptrivia (talk) 02:12, 7 April 2006 (UTC)

I'm not sure whether this is what you're stuck on but: if the column vectors at the last point of C1 are T1, N1, B1 and the column vectors at the first point of C2 are T2, N2, B2 then the matrix which takes one basis to the other is (T2 N2 B2) (T1 N1 B1)-1, where the notation of course means to put three column vectors in to get a 3×3 matrix. If you use unnormalized vectors then the curve will be linearly transformed in some weird way to make the parametrization line up nicely: assuming you want an actual (rigid) rotation for the curve then normalize T, N, and B to unit length before you start. And I'm too lazy to work out the right translation component. —Blotwell 03:26, 7 April 2006 (UTC)
Thanks, that's what I was doing, but didn't get the right curve. Maybe I'm making a mistake elsewhere. deeptrivia (talk) 04:35, 8 April 2006 (UTC)

## Minuscule orthodromic equals loxodromic (expressed as limit?)

On a sphere, a loxodromic arc ("H") travels in a constant direction, whereas an orthodromic arc ("ΔÔ") constantly changes direction. If one defined the loxodromic azimuth as ${\displaystyle {\overline {Az}}\,\!}$ and the orthodromic as ${\displaystyle {\widehat {Az}}\,\!}$, at infinitesimality the arc lengths and azimuths respectively equate. Would the technically proper way to express the azimuthal equity be

${\displaystyle \lim _{H\to 0}{\overline {Az}}=\lim _{\Delta {\hat {O}}\to 0}{\widehat {Az}}\,\!}$ ?  ~Kaimbridge~16:39, 7 April 2006 (UTC)
I'm confused as to why the symbol for the orthodrome has a delta and the symbol for the loxodrome doesn't. —Keenan Pepper 17:04, 7 April 2006 (UTC)
Because "ΔÔ" is the orthodromic arc segment ("angular distance") endpoint difference (ΔÔ = Ô2 - Ô1), whereas H is the loxodromic hypotenuse (here, an approximation, as cos(Latm) is the approximation of the reciprocal of the inverse Gudermannian function's divided difference): ${\displaystyle H={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!}$ AFAIK, there are no explicitly denotable endpoints to the the loxodromic arc segment ("hypotenuse")——it is a "side".  ~Kaimbridge~18:25, 7 April 2006 (UTC)
Now I'm even more confused. I thought hypotenuse meant a straight line segment, the longest side of a right triangle. A loxodrome is a curve on the surface of a sphere. Even in spherical geometry, great circles (a.k.a. orthodromes) play the role of straight lines, and loxodromes aren't straight. I have no idea what you mean by "there are no explicitly denotable endpoints to the loxodromic arc segment".
Right, the formal expression for the loxodrome is ${\displaystyle \Delta Lat\sec({\overline {Az}})\,\!}$, and since ${\displaystyle \tan({\overline {Az}})={\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\,\!}$ and sec = [1 + tan2].5, then ${\displaystyle \Delta Lat\sec({\overline {Az}})=\Delta Lat{\sqrt {1+\left({\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\right)^{2}}}={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!}$ When you look at a Mercator map, you are drawing a right triangle (The FCC expresses it this way, too: [1]).
Why can't you just say ${\displaystyle \lim _{d\to 0}}$ where d is the distance between the two points, measured any way you want? —Keenan Pepper 21:39, 7 April 2006 (UTC)
Because, since the orthodromic distance is different than the loxodromic distance (except along meridians and the equator), shouldn't the limits be different, too (perhaps ${\displaystyle \lim _{Dist_{lox}\to 0}{\overline {Az}}=\lim _{Dist_{orth}\to 0}{\widehat {Az}}\,\!}$)?  ~Kaimbridge~14:53, 8 April 2006 (UTC)

## Fourier transform of a Hilbert space

Is there such a thing as a fourier transform of a hilbert space? --HappyCamper 19:30, 7 April 2006 (UTC)

What do you mean? If you consider the space of analytic functions on ${\displaystyle \mathbb {R} ^{n}}$ and make it a hilbert space with the classical integral formula, the fourier transform is a unitary bijective operator. Evilbu 19:47, 7 April 2006 (UTC)

## Evaluating e2πi

When you evaluate e2πi, you get 1. Therefore

ln e2πi = ln 1
2πi = 0
2 = 0/πi = 0
π = 0/2i = 0
i = 0/2π = 0
How can this be true? M@$+@ Ju ~ 20:59, 7 April 2006 (UTC) Wrong. ln 1 = 0 + 2πi*k, where k is integer. ln e2πi=2πi + 2πi*k, where k is integer. The resulting sets are the same. Grue 21:08, 7 April 2006 (UTC) Unfortunately, you've made a mistake in going from the first line to the second. In general, when you have an equation of the form ${\displaystyle f(x)=f(y),}$ you cannot deduce that ${\displaystyle x=y}$. For example, if you take the equation ${\displaystyle x^{2}=y^{2}}$, it does not follow that ${\displaystyle x=y}$, for example choosing ${\displaystyle x=3}$ and ${\displaystyle y=-3}$ solves ${\displaystyle x^{2}=y^{2}.}$ The problem here is that the exponential function is not one to one, in fact it is periodic in the imaginary direction. --Deville (Talk) 21:23, 7 April 2006 (UTC) Thanks for the explanation guys. It's still funny to confuse my teachers! M@$+@ Ju ~ 22:46, 7 April 2006 (UTC)
Yeah, you'll be laughing all the way til detention...:-) --Deville (Talk) 04:28, 8 April 2006 (UTC)
Your teachers got confused on this! What's happening to our world's academic standards!! deeptrivia (talk) 04:34, 8 April 2006 (UTC)
Today's teachers were yesterday's students. If you were never taught it, you can't teach it. Sad but true. JackofOz 04:04, 9 April 2006 (UTC)

## long division

never quite got how this worked, how would you do it for:

1) 3265324 / 4124124124
2) 2312 / 312545
3) 757 / 14244
4) 141565 / 234

If you could show me how to do it step by step on these specific questions it would be very helpful, oh and show your work (: 21:20, 7 April 2006 (UTC)

The arcticle on long division has examples. I'm afraid we can't do your specific homework questions for you, though. —Keenan Pepper 21:27, 7 April 2006 (UTC)

3) 757 / 14244

is 757/14244 because it is a fraction. Perhaps you are asking about how to turn a fraction into a floating point number. Ohanian 04:52, 8 April 2006 (UTC)

Don't be a , he said quite clearly that he wanted to know about long division. Black Carrot 03:35, 9 April 2006 (UTC)

## infinitely many solutions

Hi,

I have for quite some time been trying to find a convincing proof that there are infinitely many solutions to a + b = c where c is any number. It seems possible that either a) this is a very axiomatic result which cannot actually be solved, or b) I am simply too stupid/ignorant to be able to locate the relevant proof. Since this is vital to some (hopefully) useful mathematical work I am undertaking (although I wouldn't dare call myself a competent mathematician) any help would be unbelievably appreciated.

How about if you start by assuming that there are a finite number of solutions, and try and prove that this statement is false (i.e. that our assumption was wrong - therefore there must be infinitely many solutions).Richard B 23:35, 7 April 2006 (UTC)

What set are you looking for solutions in? If a,b, and c are valued in the set {0}, then there is a unique solution. If they are valued in the group {0,1}, then there are 2 solutions if c is 0 but only 1 solution (up to ordering) if c is 1. Even more pathologically, if you assume that a and b are valued in the negative integers and look for solutions among the positive integers, there is no solution at all. Thus you see that how many solutions there are depends on what space you look for solutions in. Let's assume that you're looking for integer solutions. For each integer n, the ordered pair (n,cn) forms a unique solution. Furthermore, any solution is of this form. This is all the solutions, and there are infinitely many of them. -lethe talk + 00:09, 8 April 2006 (UTC)

That may be so, but it does not constitute a proof that there are an infinite number of solutions noting my use of the words 'any' number or indeed any mathematical set on which addition is defined. The reason this proof is so essential especially in a cryptographical sense, is that it allows us to establish with certainty that given a member of a set with certain properties, it is impossible to calculate the values of a and b from c. Can anyone think of such a proof? Thanks again.

Sorry, but this is not a mathematically meaningful question. As already pointed out, we have available myriad different kinds of "number", so a choice must be made. Typically cryptography would use natural numbers or positive integers, in which case the number of solutions is finite. Sadly, I must agree with your self-assessment of not being a competent mathematician, and serious cryptography research requires such compentence. As an amateur you can still have fun, but it is unrealistic to expect your work to be of much interest to professionals. --KSmrqT 02:46, 8 April 2006 (UTC)
As my examples show, there can be no proof of infinitely many solutions in a general additive number system, because it's not true in a general additive number system; there are number systems for which the equation has a finite number of solutions. Generally things that are not true do not have proofs. -lethe talk + 03:14, 8 April 2006 (UTC)

KSmrq, actually I was intending this proof to be a model for a proof for finite permutation groups, which are, i know, finite. However, all i wanted is a proof that covered say real numbers, and I'm pretty sure that it would go something like:

Set a to half of c and b to half of c. Then, add and take a number to each, until a and b reach their upper and lower limit. However, since you can always add and take a number to a and b, it is impossible for a and b to reach their upper and lower limits, and thus no such limits exist. Vice versa works also, with decrementing a and b incrementing. Therefore a and b have no upper or lower limits, rendering them infinite.

I know the way I have expressed this may not please you 'professional' types out there, but considering that you failed to produce even a rudimentary attempt, and instead stated the *blatantly* obvious, then proceeded to belittle someone, I'd say that any attempt is better than your pathetic contribution. I don't know what rock you come from under, but you mustn't be very proud of it, because you seem to be getting your self-esteem from the internet, of all places. Either help or crawl back home.

I merely agreed (in part) with your own description of yourself. You, however, have violated Wikipedia's core rule of no personal attacks in your remarks towards me. Don't do it again; you may be blocked for such behavior.
Since the reals include the natural numbers, a proof by induction would trivally show a solution for every natural number a. Namely, let b equal c and let a equal 0. Now increment a and decrement b. Or do you also want a proof that the natural numbers have infinite cardinality?
Ask smart questions if you expect good answers. --KSmrqT 12:20, 8 April 2006 (UTC)
Wait, what do you mean, "failed to produce even a rudimentary attempt"? I gave you a proof that there were infinitely many solutions over the integers, and showed by counterexample that there is no proof in general? What more do you want? -lethe talk + 04:47, 8 April 2006 (UTC)

Yes, that tirade wasn't in any way pointed at your post, which i only just read enough to comprehend that it was exactly what I wanted in the first place (i read the bits about the more restricted cases and just assumed that the post was going to go on and say exactly what KSmrq did say) and I convey my greatest thanks for it. I think that what KSmrq said was based solely on miscommunication, but even if what i said was dead wrong, I really do not appreciate the insult. Incidentally, would you accept my proof as valid?

Your proof looks pretty much like another form of my proof. So I think it's OK. I do have one complaint though: now you're considering real-valued solutions, and there are actually uncountably many solutions over the reals, which is a lot more than just countably infinitely many. By listing them one at a time in increments, you restrict yourself to only countably many solutions. You can make sure to get all uncountably many solutions if you model the proof more closely to mine: note that for each real number x, (x,cx) constitute a unique solution to the equation. Now the solutions are indexed by a real variable, instead of an incrementing index, which allows you to catch all the solutions. -lethe talk + 05:59, 8 April 2006 (UTC)

Thanks a lot, greatly appreciated.

# April 8

## improvement

how can i improve my mathematics? your help might save my life! thx

Practice, practice, and more practice.
Probably not what you wanted to hear, but it's true. --Bth 11:53, 8 April 2006 (UTC)

## Form an equation

3 and -3

Thankyou

Um... what equation? What do 3 and -3 have to do with anything? —Keenan Pepper 17:52, 8 April 2006 (UTC)
I guess the request was to find an equation the solutions of which are 3 and -3. And in that case, the equation
${\displaystyle x^{10}-45x^{8}+810x^{6}-7290x^{4}+32805x^{2}-59049=0}$
Will satisfy the homework assignment's requirements. -- Meni Rosenfeld (talk) 17:58, 8 April 2006 (UTC)
So will

${\displaystyle 5919012181389927685417441689600000000x-9447709684208047354981782650880000000x^{3}+}$${\displaystyle 4337015638473273668425522182881280000x^{5}-906380005918141132650786081964032000x^{7}+}$${\displaystyle 105315135918687298508885950223794176x^{9}-7607858213674594456495183800161280x^{11}+}$${\displaystyle 366661782054884005855608205864192x^{13}-12365589876694504346988698445440x^{15}+}$${\displaystyle 301748325708943677229642930528x^{17}-5456090765994828963719786980x^{19}+74326386672885754888959569x^{21}-}$${\displaystyle 771182311331381631254950x^{23}+6130120665016658846445x^{25}-}$${\displaystyle 37368696356054464800x^{27}+173944259366417394x^{29}-}$${\displaystyle 611675276741620x^{31}+1593528150578x^{33}-2975110060x^{35}+}$${\displaystyle 3757117x^{37}-2870x^{39}+x^{41}=0}$

w00t--Deville (Talk) 19:18, 8 April 2006 (UTC)
Sorry, but we're looking for an equation the solutions of which are +3 and -3. Equations with additional solutions aren't good. Since this one has 2 roots right and 39 wrong, you won't get more than 4.9 (out of 100) for this answer. -- Meni Rosenfeld (talk) 06:25, 9 April 2006 (UTC)

3 + -3 = 0. -lethe talk + 18:08, 8 April 2006 (UTC)

How about the lowly quadratic (x-3)(x+3)=0? Am I missing something here? —The preceding unsigned comment was added by 84.234.167.166 (talkcontribs) .

Shhh! —Keenan Pepper 01:37, 10 April 2006 (UTC)

People from Britannica randomly add these articles to create anarchy in wikipedia...

## Mathematics/measurement

What is the smallest possible measurement of time? 204.112.201.7 20:02, 8 April 2006 (UTC)

5.391 × 10−44 seconds (see Planck time for more information). —Ruud 20:16, 8 April 2006 (UTC)
Not that this has anything to do with mathematics, but whether or not Planck time is the smallest time you can measure, or what it even means to measure time, is a rather subtle, and perhaps model-depentent, question. To measure something smaller than the Planck time would require a clock with higher frequency, thus greater than Planck mass. String theory probably predicts such particles (it predicts an infinite tower of particles starting at nearly Planck mass), but there are probably a lot of subtleties to address before you could use such a particle as a clock. Anyway, most "time" measurements of nuclear decays are actually measurements of the dispersion of energy. I guess with modern clocks, you can measure things as small as one electron decay of Cesium, since that's how we build our atomic clocks. -lethe talk + 20:39, 8 April 2006 (UTC)
What? Planck mass is a quite large amount. From our article on planck mass: "Unlike most of the other Planck units, the Planck mass is on a scale more or less conceivable to humans, as it is roughly the mass of some fleas."Ҡiff 08:53, 9 April 2006 (UTC)
Yeah, so? What's your point? -lethe talk + 18:39, 9 April 2006 (UTC)
You said "it predicts an infinite tower of particles starting at nearly Planck mass". It predicts particles with mass closer to Planck mass? That's very massive for a single particle. Doesn't sound right. ☢ Ҡiff 04:46, 11 April 2006 (UTC)

## symmetric power and alternating power commute?

If

${\displaystyle \sum ^{p}V\,\!}$

is the pth symmetric power of an n-dimensional vector space, and

${\displaystyle \bigwedge ^{q}V}$

is the exterior power or alternating power, do I have a canonical isomorphism

${\displaystyle \bigwedge ^{p}\sum ^{q}V=\sum ^{q}\bigwedge ^{p}V?}$

I think I ought to, but I'm having a hard time writing down such an isomorphism. -lethe talk + 20:14, 8 April 2006 (UTC)

No, if my arithmetic is right. I have
${\displaystyle dim\bigwedge ^{2}\sum ^{2}F^{2}=3,\quad dim\sum ^{2}\bigwedge ^{2}F^{2}=1.}$
Melchoir 22:56, 8 April 2006 (UTC)
Whoa, well I guess that puts a nail in that coffin. Thank you. -lethe talk + 17:57, 9 April 2006 (UTC)

# April 9

## Existence of characteristic functions for decision problems

Is considering decision problems in lieu of computational problems sound? The relevant articles use an argument that deciding if x is in language L is equivalent to computing the characteristic function and checking if output is 1. But the characteristic function is defined in terms of the decision problem.

I can't seem to understand why this is sound? Further, if a characteristic function doesn't exist without resorting to using the decision problem, what good is it in showing their equivalence?

Do I understand the proof correctly?

Admittedly I don't know much about the particular field, but looking at the generalities I'm not sure what the problem is here. Having two equivalent formulations of the same problem can be very helpful, both for understanding the nature of the underlying concepts and for actually finding solutions. (Closer to my understanding, I'd give the example of quantum mechanics where the same problem can be represented in quite different ways that can be proved equivalent.) The fact that the different representations are not independent of one another is only to be expected, surely? --Bth 12:01, 10 April 2006 (UTC)
I don't know what you mean by "soundness", but a characteristic function (a special case of a function) is just an another way of characterizing a set. It is obvious that, as a way to model "problems", function problems are at least as expressive as decision problems. The relevant question to ask to is: given a class C of function problems, is there a function f in C that cannot be reduced to a decision problem under a reduction that preserves membership in C? I believe the answer depends on C. --64.236.170.228 14:11, 13 April 2006 (UTC)

## Is this algorithm for calculating log2(x) correct/

I apologizes for the computer language python, but is the fundamental algorithm for calculating log2(x) aka logarithm base 2 correct?

The numeric value of the binary logarithm of a positive real number can easily be calculated using only the addition, subtraction, multiplication and division arithmetic operators. Here is a sourcecode in python which produces the value.

#!/usr/bin/python

from __future__ import division

def log2(X):
epsilon = 0.000000000001
integer_value=0
while X < 1:
integer_value = integer_value - 1
X = X * 2
while X >= 2:
integer_value = integer_value + 1
X = X / 2
decfrac = 0.0
partial = 0.5
X=X*X
while partial > epsilon:
if X >= 2:
decfrac = decfrac + partial
X = X / 2
partial = partial / 2
X=X*X
return (integer_value + decfrac)

if __name__ == '__main__':
value = 4.5
print "     X  =",value
print "LOG2(X) =",log2(value)

# Sample output
#
#    $python log2.py # X = 4.5 # LOG2(X) = 2.16992500144 #  Can't comment on your coding, but the algorithm looks fine. Preamble determines the integer part of the logarithm, and leaves you with X between 1 and 2, so log2(X) is between 0 and 1. Squaring X doubles log2(X), which you can think of as a shift left operation on the binary representation of log2(X). The algorithm then tests whether X^2 is >=2, which is testing whether the integer part of the left-shifted logarithm is 1. If the integer part of the logarithm is 1 it throws this away (by dividing X^2 by 2), and then repeats. Essentially it is building up the logarithm one binary place at a time. There is nothing special about base 2 - you could use the same algorithm to calculate logarithms to base N by replacing the X>=2 tests by X>=N and replacing X=X/2 by X=X/N. Gandalf61 09:08, 10 April 2006 (UTC) Are you sure this algorithm also works for any base? Say base 3. What should the value of partial be 1/3 , I don't think the algorithm works for base 3. I shall test it by running it and check with my scientific calculator. Ohanian 10:32, 10 April 2006 (UTC) I'm sorry but I having trouble getting this to work for base 3. I even tried cubing it aka X=X*X*X but it gives the wrong answer. Ohanian 10:52, 10 April 2006 (UTC) I think I got log3(X) and log4(X) to work now. Here are the results #!/usr/bin/python from __future__ import division def log3(X): epsilon = 0.000000000001 integer_value=0 while X < 1: integer_value = integer_value - 1 X = X * 3 while X >= 3: integer_value = integer_value + 1 X = X / 3 decfrac = 0.0 partial = 1/3.0 X=X*X*X while partial > epsilon: while X >= 3: decfrac = decfrac + partial X = X / 3 partial = partial / 3 X=X*X*X return (integer_value + decfrac) def log4(X): epsilon = 0.000000000001 integer_value=0 while X < 1: integer_value = integer_value - 1 X = X * 4 while X >= 4: integer_value = integer_value + 1 X = X / 4 decfrac = 0.0 partial = 1/4.0 X=X*X*X*X while partial > epsilon: while X >= 4: decfrac = decfrac + partial X = X / 4 partial = partial / 4 X=X*X*X*X return (integer_value + decfrac) if __name__ == '__main__': value = 4.5 print " X =",value print "LOG3(X) =",log3(value) print "LOG4(X) =",log4(value) #$ python log34.py
#       X  = 4.5
#  LOG3(X) = 1.36907024643
#  LOG4(X) = 1.08496250072

Ohanian 11:21, 10 April 2006 (UTC)


So, presumably we can generalise to a function giving logN(X):

#!/usr/bin/python

from __future__ import division

def log(N,X):
epsilon = 0.000000000001
integer_value=0
while X < 1:
integer_value = integer_value - 1
X = X * N
while X >= N:
integer_value = integer_value + 1
X = X / N
decfrac = 0.0
partial = 1/float(N)
X=X**N
while partial > epsilon:
while X >= N:
decfrac = decfrac + partial
X = X / N
partial = partial / N
X=X**N
return (integer_value + decfrac)


though as I don't do Python my syntax may be wrong in places ... Would it work for non-integer N? --Bth 11:49, 10 April 2006 (UTC)

It should, in principle. Of course, to compute XN for non-integer N you need to use logarithms in the first place, rather defeating the point of it. It'd be much easier to use the identity logNX = log2X / log2N instead. —Ilmari Karonen (talk) 13:48, 10 April 2006 (UTC)
There is no need to raise X to the power of N. Just leave the lines X=X*X, start partial at 0.5 and divide it by 2 at each repeat. This adds one binary place of accuracy to the logarithm for each repeat for any value of N. In theory you could add one decimal place of accuracy at each repeat instead by calculating X^10, and working out which integer powers of 10 this falls between, but this makes the algorithm much more complicated, and you will probably loose accuracy in practice. Computers like to calculate in powers of two ! Gandalf61 15:20, 10 April 2006 (UTC)
Let's not forget that we're working with computers here. Floating-point numbers are already stored in sign-mantissa-exponent format, so you can read off the exponent without doing any computation. Unless this is a homework problem, just call frexp and be done with it. Melchoir 18:24, 10 April 2006 (UTC)
Personally, I was just interested in general in the applicability of the algorithm. I was stating it in Python 'cos of the overall context. I'd never suggest actually using such a process to calculate a log to arbitrary base -- it's much easier to use the relevant properties of logs (as Ilmari pointed out) with whatever intrinsic log function the language has (which will almost certainly be faster than some user-defined function). --Bth 10:40, 11 April 2006 (UTC)
Gandalf61, you are a bucking genius! A true genius. I followed your advice and it actually works. Here is the latest version of the logarithm algorithm. I'm very happy because when I started, I only had the algorithm for log2(x) but now not only I have an algorithm for any base, I even had an algorithm for a non-integer base as well (such as 2.71828). Thank you very much. Ohanian 23:24, 10 April 2006 (UTC)
#!/usr/bin/python

from __future__ import division

def log(N,X):
epsilon = 0.000000000001
integer_value=0
while X < 1:
integer_value = integer_value - 1
X = X * N
while X >= N:
integer_value = integer_value + 1
X = X / N
decfrac = 0.0
partial = 0.5
X=X*X
while partial > epsilon:
while X >= N:
decfrac = decfrac + partial
X = X / N
partial = partial / 2
X=X*X
return (integer_value + decfrac)

if __name__ == '__main__':
value = 45.7
print "     X  =",value
print "LOG6(X) =",log(6,value)
print "  LN(X) =",log(2.718281828,value)

#  SAMPLE OUTPUT
#    $python log.py # X = 45.7 # LOG6(X) = 2.13315367578 # LN(X) = 3.82209829854  # April 10 ## Anyone know how do you derive the distance formula (for neutral geometry) I checked the article but it simply said it was derived from the Pythagorean theorem. Don't worry its not homework I'm just curious... I'm not sure what you're asking for here. There's a proof of the Pythagorean theorem onm its page. One way to understand the generalisation from 2 dimensions is to see it as repeated applications of the rule. If you have a point ${\displaystyle (x,y,z)}$ then you can draw a triangle with the hypotenuse going from the origin to the point (so its length will be the distance of the point from the origin, which we'll call ${\displaystyle r}$), one side going from the point to meet the ${\displaystyle xy}$ plane at a right angle (of length ${\displaystyle z}$, obviously) and a third side which for now we'll say has length ${\displaystyle l}$. So ${\displaystyle r={\sqrt {l^{2}+z^{2}}}}$ by the Pythagoras formula. Now ${\displaystyle l}$ is fairly obviously the hypotenuse of a right-angled triangle in the ${\displaystyle xy}$ plane with sides of length ${\displaystyle x}$ and ${\displaystyle y}$, so ${\displaystyle l={\sqrt {x^{2}+y^{2}}}}$. Plugging this into the earlier formula, ${\displaystyle r={\sqrt {\left({\sqrt {x^{2}+y^{2}}}\right)^{2}+z^{2}}}={\sqrt {x^{2}+y^{2}+z^{2}}}}$. Is that what you wanted, or have I missed the point entirely? --Bth 09:37, 10 April 2006 (UTC) Having read the article in more depth, I think you're actually wondering about how we get to the ${\displaystyle {\sqrt {(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}}}$ formula for two points ${\displaystyle (x_{0},y_{0}),(x_{1},y_{1})}$ in 2 dimensions. Basically, this relies on the fact that the x and y directions are at right angles to each other. Thus you can take the distance between the two points in each direction as the two other sides of a right-angled triangle and calculate the distance (which is the hypotenuse) by Pythagoras. Here's a noddy ASCII diagram (you'll have to pretend the diagonal really is a diagonal):  (x1,y1) 1 /\ . . | . . | . . (y1-y0) . _ | 0 . . . . . . .|. \/ <----(x1-x0)----> (x0,y0)  --Bth 09:49, 10 April 2006 (UTC) Thanks Bth...but isn't that the same as sqauring the two parts of the slope? i.e if the slope is 7 over nine, the sum of those squares would be the same thing right? The same quantities are involved, but not combined in the same way. In general, ${\displaystyle (x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}\neq \left({y_{1}-y_{0} \over x_{1}-x_{0}}\right)^{2}}$. Think about a similar triangle twice as big (labelling the distances as ${\displaystyle x=x_{1}-x_{0}}$ and ${\displaystyle y=y_{1}-y_{0}}$ for compactness). The slope would be the same -- ${\displaystyle {2y \over 2x}={y \over x}}$ -- but the distance would be doubled -- ${\displaystyle {\sqrt {(2x)^{2}+(2y)^{2}}}={\sqrt {4x^{2}+4y^{2}}}={\sqrt {4(x^{2}+y^{2})}}={\sqrt {4}}{\sqrt {x^{2}+y^{2}}}=2{\sqrt {x^{2}+y^{2}}}}$. --Bth 08:11, 11 April 2006 (UTC) No, I mean that the 2 components are the same in distance (as you said) so if you sqaured the two components of slope and added them, then took the root of that, would not that be the same thing? Sorry, with you now. Yes, it would. But note that you can in general take the slope of a straight line using any two points along it, whereas to get the distance of a line segment between two points you obviously have to use those specific two points. --Bth 07:27, 12 April 2006 (UTC) Oh yeah, didn't think of it that way...anyway thanks for the help. ## Range and 0 ok, with working out the range, range=highest - lowest, does the lowest include zero. e.g. if the numbers are 1, 5, 7, 8, 0, 10 would the range be 10-1 = 9  or  would it be 10-0 = 0 i really dont understand and usually i'm pretty good at this whole maths thing. thanks, skye Yep, when calculating range you always consider all the scores. And if there were negative numbers in there too, you'd count them. One interesting thing to note though is that if you're drawing a box plot, there may be occasions where you ignore certain outlying values, but that's a bit past what you're asking. Confusing Manifestation 12:49, 10 April 2006 (UTC) Thanks heaps, got lab report due tomoz and was stressin heaps, thanks sooooooooo much, cya, skye I would say it depends on what you are measuring. Is zero really a valid answer or not ? If you were timing how long it takes a ball to drop a certain distance and got a zero answer, something is obviously wrong, so discard this answer (and redo the test, if possible). However, if you are counting the number of seeds that sprouted in a sample, then zero is a perfectly valid answer and should be included. StuRat 20:35, 11 April 2006 (UTC) ## Permutations Hi I am trying unsuccessfully to make permutations for a golf trip for 20 players. Over 4 rounds, played in groups of 4, the players should rotate so that none of them plays together more than once. I am sure that there is an easy answer, but we can't find it. Please help. Thank you Label the players a-t. For the first round group them abcd, efgh, ..., qrst. In the next round a,e,..,q stays in their group, b,f,..,r rotates one group, c,g,...,s rotates two groups and d,h,..,t rotates three groups. This gives you a round-robin tournament of five rounds. This should work in general, when (players/4) is a prime. Writing it out, the tournament looks like this: abcd|efgh|ijkl|mnop|qrst arol|ebsp|ifct|mjgd|qnkh angt|erkd|iboh|mfsl|qjcp ajsh|encl|irgp|mbkt|qfod afkp|ejot|insd|mrch|qbgl  I think bridge players call this a rainbow movement. Rasmus (talk) 14:14, 10 April 2006 (UTC) ## Undefined, DNE(do not exist), & No Solution When talking about a limit, equation, solution, or stuff like that, what is the different between using undefined, DNE, and no solution? And when should I use which one? —Preceding unsigned comment added by 72.59.31.118 (talkcontribs) Some expressions (like ${\displaystyle 1 \over 0}$) are undefined. It may also be indeterminate (like ${\displaystyle 0^{0}}$). Equations can have solutions or not (in fact, they can have any number of solutions: 0, 1, 6, or various kinds of infinite numbers of solutions). Certain concepts like the limit of a sequence, the sum of an infinite series, or the derivative of a function at a point, are not always defined; however, instead of saying that ${\displaystyle \lim _{x\rightarrow \infty }\sin x}$ and ${\displaystyle f'(0)}$ for ${\displaystyle f(x):=\left|x\right|}$ are undefined, we typically say that they do not exist, because (in contrast to ${\displaystyle 1 \over 0}$) there is a way to evaluate the expression -- it just doesn't work when you try. (However, functions like ${\displaystyle f'(x)}$ are often said to be "undefined at 0" or so; there is some overlap.) --Tardis 20:39, 10 April 2006 (UTC) Context and care are our guides. Depending on how a question is posed, and what is assumed, any one of these might apply. However, we have some archetypal examples. • A limit either does or does not exist. • An equation either does or does not have at least one solution. • The meaning of an expression either is or is not defined. • The truth of a logical statement either is or is not decidable. • A result either is or is not computable. (You didn't mention decidability nor computability, but they will turn up.) Now consider √2. We can ask what it means as a rational number; its meaning is undefined. We can ask about the existence of a rational number x such that x squares to 2; such a rational does not exist. We can ask about a rational solution to the equation 0 = x2−2; there is no solution. ## Solving Rational Equations Will you explain to me how to do this problem? I'm doing a makeup assignment from being gone, so I missed the lesson. Also, it can help me do the others once I know how to figure these out. Ok here's the problem 2 over (x-2) squared = 1-1 over x-2 Your help would be greatly appreciated. Also, sorry for the equation form, I don't know how to do all of the math symbols on the computer! Thanks alot! Kirsten (aka 65.101.68.120) You need not use fancy mathematics layout, but please use parentheses. Do you mean: • ${\displaystyle {\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}\,\!}$ • ${\displaystyle \left({\frac {2}{x-2}}\right)^{2}=1-{\frac {1}{x-2}}\,\!}$ Assuming the first, write 2/((x−2)^2) = 1 − 1/(x−2). --KSmrqT 03:45, 11 April 2006 (UTC) Hey, try multiplying out the brackets from underneath so you don't have any awkward divisors. Then multiply out the brackets to get loads of terms of x and clean up! It helps that the 1/(x-2) cancels. Anand 20:43, 10 April 2006 (UTC) I'm guessing you mean ${\displaystyle {\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}}$? (See Help:Formula.) This problem can be reduced to a polynomial equation by writing both sides in the standard rational function form: ${\displaystyle {\frac {2}{(x-2)^{2}}}={\frac {x-2}{x-2}}+{\frac {1}{x-2}}={\frac {x-2+1}{x-2}}}$. Then multiply by the LCM of the denominators. You should get a simple quadratic equation to solve. --Tardis 20:49, 10 April 2006 (UTC) Hi, thank you so much for your help, I'm sorry, but I still don't get it :( do you think that you can explain it step by step? sorry, it's just not clicking with me. thanks again kirsten Consider an equation, A = B. If c is not zero, then we can multiply both sides by c without changing the truth of the equality: cA = cB. The equation you have written has denominators that get in the way of seeing a solution. In particular, the left-hand side has a denominator of (x−2)2. Assuming x is not 2, the quantity (x−2)2 is not zero; thus we can use it as c to eliminate the denominator. ${\displaystyle {\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}\,\!}$ becomes ${\displaystyle 2=(x-2)^{2}-(x-2)\,\!}$. The right-hand side expands to x2−4x+4 − x+2, or simply x2−5x+6. We can also add any c we like (whether zero or not) to both sides of an equation. Here, if we add −2 to both sides we reduce the left-hand side to zero, leaving standard quadratic equation form for the right-hand side. ${\displaystyle 0=x^{2}-5x+4\,\!}$ Such an equation can be solved by the method known as "completing the square", or equivalently, by using the quadratic formula. Is that explicit enough? --KSmrqT 04:05, 11 April 2006 (UTC) But don't forget: Since the original equation has (x - 2) in the denominator, x must be different than 2. If you get x = 2 as a solution to the final equation, it is ignored (you don't include it as a solution to the problem). -- Meni Rosenfeld (talk) 06:56, 11 April 2006 (UTC) Er, is the question not as it obviously appears (isolating x) or am I suffering a brain cramp?: ${\displaystyle {\frac {2}{(x-2)^{2}}}\neq 1-{\frac {1}{x-2}}}$ (except when x = 1) and ${\displaystyle 1-{\frac {1}{x-2}}={\frac {x-2}{x-2}}-{\frac {1}{x-2}}={\frac {x-2-1}{x-2}}\neq {\frac {x-2}{x-2}}+{\frac {1}{x-2}}={\frac {x-2+1}{x-2}}}$!?! ~Kaimbridge~18:54, 12 April 2006 (UTC) Brain cramp. Try x = 4. Also notice that what was requested was a general approach to such problems, and feedback suggested the directions should be elementary. --KSmrqT 19:12, 12 April 2006 (UTC) ## Math e+26??????? I have a number--> 1.4190792748769801e+26 --> and i want to know what the e+26 part of it means. Do i move the decimal 26 places? help please 1.4e+26 = 1.4 * 100000000000000000000000000 Hint: count the zeros Ohanian 23:48, 10 April 2006 (UTC) See Scientific notation. Melchoir 01:39, 11 April 2006 (UTC) # April 11 ## Trig integration Hi, I have a set of parametric eqations defined as ${\displaystyle x=3\,cos\,(t),y=9\,sin\,(2t)\,}$ where ${\displaystyle 0\leq t<2\pi }$ and I have to find the cartesian equation in terms of y. So far I've got ${\displaystyle y^{2}\,=\,(9\,sin\,(2t))^{2}}$ ${\displaystyle =\,81\,sin^{2}\,(2t)}$ ${\displaystyle =\,81\,(1-cos^{2}\,(2t))}$ ${\displaystyle =\,81-81\,cos^{2}\,(2t)}$ but I cant get any further. I've got a feeling that ${\displaystyle cos\,(2t)=2\,cos^{2}\,(t)-1}$ comes into it somewhere. Can anyone help? Whoops - must remember to sign Anand 14:11, 11 April 2006 (UTC) ${\displaystyle \cos(t)={x \over 3}(Equation1)}$ Now, from the equation for y: ${\displaystyle y=18\sin(t)\cos(t)\,}$ (by ${\displaystyle \sin(2x)=2\sin(x)\cos(x)\,}$) ${\displaystyle y=6x\sin(t)\,}$ (substitite in (1)) ${\displaystyle y^{2}=36x^{2}\sin ^{2}(t)\,}$ (square it) ${\displaystyle y^{2}=36x^{2}(1-\cos ^{2}(t))\,}$ (pythagorean trigonometric identity) ${\displaystyle y^{2}=36x^{2}(1-{x^{2} \over 9})\,}$ (substitite in (1)) ${\displaystyle y^{2}=x^{2}(36-4x^{2})\,}$ (simplify) RupertMillard 15:02, 11 April 2006 (UTC) Thanks Rupert! Anand 15:18, 11 April 2006 (UTC) ## Fun math class Quick question for ya'll. I'm an engineering student, and have taken all the required math courses (calc up to diff eq). I'm going to be solving PDEs until I die, I know that already. But what's a "fun" math course to take? Something theoretical? Analysis? Probability? I'd like a broad knowledge of some basic stuff. Even though it does not pertain to my major, I plan on taking relativity next year for some kicks. Thanks in advance! Isopropyl 14:32, 11 April 2006 (UTC) My personal favorite was "History of Math," although it was only a one month class. Abstract algebra was cool, and I regret not taking any geometry. (also, "y'all" not "ya'll"). --LarryMac 14:48, 11 April 2006 (UTC) Additional thought after a couple hours ... if you really want broad knowledge, find some electives in art or literature or something completely away from engineering or math. It's good to learn how to use all the parts of your brain <g>. --LarryMac 19:30, 11 April 2006 (UTC) I'm actually an urban planning minor, which is pretty far away from biological engineering. Even art has its place in BE; we're doing some stuff in synthetic biology where we take pictures using E. coli. Interesting stuff. Isopropyl 20:14, 11 April 2006 (UTC) Probability and/or statistics are useful for anybody, in my opinion. I also took combinatorics, which I found thoroughly enjoyable and which is sort of probability-ish. — Lomn Talk 14:55, 11 April 2006 (UTC) Ooh, take topology. It'll change the way you think, or your money back! Melchoir 19:37, 11 April 2006 (UTC) Should I be able to understand intro level combinatorics or toplogy with my math background? I've only had three semesters of calculus. Isopropyl 20:16, 11 April 2006 (UTC) I should also admit that while I remember how to use variation of parameters (of all things!), there's a lot of things I've forgotten. In the fall, it'll be more than a year since I last took math. Isopropyl 20:19, 11 April 2006 (UTC) Three semesters of calculus should be enough for both combinatorics and topology. Combinatorics should be no problem, it mostly uses counting. Topology uses a lot of set theory and proof writing techniques. You don't need prerequisite coursework to take the course, but if it's your first foray into abstract mathematics, you may find yourself overwhelmed. Many students are recommended to take a transition course into logic/set theory/proof writing before they take something like topology. But like I say, this is a recommendation only, not a requirement. You can pick it up as you go along, if you're careful. The easiest abstract mathematics course is probably considered linear algebra, with abstract algebra in second place. I might recommend one of those for you instead of topology, depending on how good a math student I thought you were. Furthermore, it is my opinion that first semester topology is kind of dry, you don't get to study things like the classification of surfaces, homotopy theory, homology theory, knot theory, until a semester or more, usually. General topology is cool too, but I think those other subjects are the really cool parts. If you wanted to do topology, I would recommend you try to do it for more than just a semester. -lethe talk + 20:42, 11 April 2006 (UTC) I endorse everything lethe says, but I want to say that I took an essentially one-semester course on general topology that reached some very interesting material. However, that particular course had high prerequisites and a truly exceptional professor. I can easily envision a deadly-boring class if it's taught badly... Perhaps you ought to ask someone in your math department you can trust for advice? Melchoir 21:05, 11 April 2006 (UTC) Oh man, I had to slog through a whole semester doing point-set topology, and it was my second year at uni, and for a while, I was quite flustered. My two buddies were totally mystified, and ended up dropping their math majors largely because of this course. Later semesters were awesome though, and I later even came to have an appreciation for point-set topology. But I think you could pass through a lot of that stuff and get to something interesting in a single semester. As you say, depends on the course, the prof, the uni. -lethe talk + 21:19, 11 April 2006 (UTC) (edit conflict) I can't speak for combinatorics, but topology can be a great course because it often starts from scratch; it's mathematics without the numbers. For example, Munkres assumes almost no background at all! On the other hand, topological concepts are often taught alongside advanced analysis or set theory. Ultimately it's up to your particular school whether or not they want to offer a course on topology that will interest you. Melchoir 20:45, 11 April 2006 (UTC) I have another idea for a fun math class for you. Differential geometry. Often, a first course in differential geometry is simply a continuation of stuff you learned in vector calculus, the differential geometry of curves and surfaces in R3. Thus it fits your prerequisites perfectly. It's a nice springboard into Riemannian geometry, should you decide to take that at a later date, or general relativity. And you'll learn the Gauss-Bonnet theorem, which may whet your appetite for a topology course. -lethe talk + 21:14, 11 April 2006 (UTC) I would like to thank you all for your recommendations. However, it seems that I need to take Analysis 1 at my school before I can take anything that's not engineering-oriented applied math (probablity for example). I'll look into your suggestions at a later date. Thanks though! Isopropyl 22:01, 11 April 2006 (UTC) ## Sinusoidal Graphs Transformations What is the best way to approach something like y = -4sin (2x + π/4)+ 5? No it's not for homework, any example would be great (anything with a reflection, vertical shift, horizontal shift and all of that together so it's approaching the highest degree of difficulty). Also I was wondering how do you know whether you should graph the x-axis in radians or degrees? Thanks C-c-c-c 21:22, 11 April 2006 (UTC) I assume it's graphing, so you could take these points into consideration: It has an amplitude of 4 Because it has a negative sign, it means it's going the opposite way of a sine graph There is a 2x in the sine function, which means it has a period of π radians It approaches π radians ${\displaystyle {\frac {\pi }{4}}}$ before a usual such graph It has a minimum value of -4+5=1, and maximum value of 4+5=9 The hardest part here could simply be the shift. Therefore a quick sketch of ${\displaystyle -4\sin {2x}+5}$ should help. You can then transpose the graph to the left ${\displaystyle {\frac {\pi }{4}}}$ radians. You should use radians, not degrees, because there is a ${\displaystyle {\frac {\pi }{4}}}$ within the sine function. 00:57, 12 April 2006 (UTC) Thanks a lot! C-c-c-c 01:52, 12 April 2006 (UTC) Watch out! If you want to talk about shifts, you need to talk about adjustments made to ${\displaystyle x}$, not ${\displaystyle 2x}$. As such, you need to write it as ${\displaystyle -4\sin(2(x+\pi /8))+5}$. Then you can see that it is in fact only shifted by ${\displaystyle \pi /8}$. The other points given by x42bn6 are all accurate. I might add, though, that properly functions like sine and cosine are functions of numbers, not of angles, and that numbers and angles are only equivalent when the angles are stated in radians. The degree symbol (°) should always be interpreted as simply denoting the quantity ${\displaystyle \pi /180}$ as a multiplier. --71.38.221.214 03:44, 12 April 2006 (UTC) Whoops. :( 01:06, 13 April 2006 (UTC) ## Consecutive Integers Relatively Prime plus new triangle I know, but cannot prove, that two consecutive integers are always going to be relatively prime. Does anybody have a proof? Let the integers be n and n+1. Suppose p divides n, so that n = kp for some integer k. Then n+1 = kp+1, which means that for p to divide n+1 it must also divide 1. Clear? --KSmrqT 08:23, 12 April 2006 (UTC) Also, my teacher gave us the following puzzle: 12 1112 3112 211213  And asked us to find the next row (which is 312213). He then said that these triangles can start with any numbers (except 0), and asked us if they will all eventually repeat, or whether some will continue to grow. (A side problem which I have already solved, what is the shortest set of starting numbers that will repeat immediately). This is an example of one of the "visual" series. Row n simply states the numbers in the n-1 row. Let's take the second row. There are three (3) ones (1) and one (1) two (2). This is reflected in row three. Isopropyl 21:54, 11 April 2006 (UTC) Nevermind. I think you already understand the concept and are asking a question that I don't understand. Isopropyl 22:02, 11 April 2006 (UTC) The answers to your teacher's questions concerning the puzzle, and more, are found under Look-and-say sequence. Melchoir 22:23, 11 April 2006 (UTC) As for consecutive integers, check out the article Coprime. You could argue from the definition or exploit one of the theorems listed under "Properties". Melchoir 22:29, 11 April 2006 (UTC) The Euclidean algorithm terminates after one step. —Keenan Pepper 03:42, 12 April 2006 (UTC) It really isn't the Look-and-say sequence. Here is an example of the way that my teacher's sequence repeats with 12 and 34 as starting numbers: 12 34 1112 1314 3112 211314 211213 31121314 312213 41122314 212223 31221324 114213 21322314 31121314 etc. 41122314 31221324 21322314 21322314 etc.  Whereas the Look-and-say sequence goes as follows: 1 11 21 1211 111221 312211 13112221 1113213211 etc.  My question is now, for any starting numbers, will the sequence eventually fall into a loop as shown above for 12 and 34? Yes, it will. Here is a sketch of a proof: 1. There are two ways for a "step" to grow in length: Fill in a missing number (ie. x1y3 => z1v2w3), or not to have any missing numbers and have all odd position equal (ie. x1x2x3 leads to a 4 in the next step (unless x>3), which leads to ${\displaystyle x_{1}1x_{2}2x_{3}314}$). 2. If the sequence is 2*4 or longer, then after the third step, the only way to have ${\displaystyle x1x2x3...x_{n}n}$ is for x=1.(Since from the second step all steps are on the form ${\displaystyle x_{1}1x_{2}2x_{3}3...x_{n}n}$, so if a step was x1x2...xn, x>=3, the previous step would have to be xn>=3n long, yet only contain numbers from 1 to n. If it was 2122...2n the previous step would have to be ${\displaystyle x_{1}1x_{2}2x_{3}3...x_{n}n}$ with ${\displaystyle {x_{1},...x_{n}}={1,...n}}$ (perhaps a permutation), but then the step before that would have to be of length (n+1)n/2>2n (since n>=4) long yet only contain numbers from 1 to n.) 3. If a step (except for 12) have no "missing numbers", the successor cannot have all odd numbers equal to 1. If you combine these, you can show that the length of the steps are limited. But if the length of the steps is limited, the size of the numbers in a step must be limited, too (by the length and by any initial numbers). So there is a finite possible "steps" for a given starting number, so the sequence must repeat after some time. Note: the above assumes that multiple digit numbers are treated as a single entity, ie. starting with 9999999999 gives (10)9 => 191(10) etc. If you instead want 9999999999 => 109 => 101119 you will have to change the proof a bit, but I think it will still hold. Rasmus (talk) 08:36, 12 April 2006 (UTC) Ah, so it isn't. I didn't notice that 3112 -> 211213 step, where your teacher diverges from Conway. Melchoir 19:53, 12 April 2006 (UTC) It is based on the Look-and-say sequence method, plus sorting digits : instead of 132112 it goes 211213. Now the rules for evolution may differ, but do they differ slightly or much ? Also, the Look-and-say sequence article seems to postulate that there is only one sequence ... I'd say it is just an example, and Mathworld is more accurate. --DLL 20:34, 12 April 2006 (UTC) Apparently, given Rasmus' reasoning, they differ drastically. As for whether there is a One True Look-and-say sequence, I'd say that's more semantics and history than mathematics. Melchoir 23:48, 12 April 2006 (UTC) ## math tell each gcf you would use to simplify the fractions. I don't really understand your question. Do you have any specific set of fractions? The question is asking you to give the number that you'd divide the numerator (the top) and denominator (the bottom) by to get a faction in its simplest form (ie with no factors in common between the two). So, if you had 4/6, the answer would be "2" because 4/6 = 2/3 when you cancel the common factor of 2 between the top and the bottom. Or if you had 40/100 the answer would be "20" because that's the factor that reduces it to 2/5. But no, we're not actually going to do the homework for you. --Bth 09:23, 12 April 2006 (UTC) PS Does the question really say "tell" rather than "state" or something? That sounds horribly informal to me for a question. What part of the world are you in? # April 12 ## Vectors in EMW Is there an easier practical illustration of the common terms we use in Electromagnetic Theory like Gradient,Curl,Divergence? "Easier" is a comparative term; compared to what? We have articles on gradient, curl, and divergence, and the first two give basic examples. Or do a web search; tutorials are common. --KSmrqT 10:56, 12 April 2006 (UTC) This was asked over at the science desk, too. Naughty, naughty, 210.212.194.215. --Bth 11:13, 12 April 2006 (UTC) You might want to look at the book referenced at vector calculus by H. M. Schey; it's specifically written for such practical illustration. --Tardis 17:00, 12 April 2006 (UTC) ## Calculate the distance between the two points on the earth what is the simple low to calculate the distance between the two points on the earth by using geographic Coordinate See great-circle distance. --Bth 13:53, 12 April 2006 (UTC) ## Statistics, cat ownership, and elections Say there is an election in which 100 people vote, and the result is split 50-50. If it was noticed that 20 of the people in favor of the measure owned cats, while 5 of the people against it owned cats, how would I determine if cat ownership was statistically likely to have influenced the outcome of the vote? Say the vote is run again, with 100 different people, this time including 30 cat owners. Assuming cat ownership is significant, and that there are no other influencing factors, how would I determine the odds of the majority being in favor? --Serie 21:46, 12 April 2006 (UTC) These questions make a fundamental error in statistics: correlation is not causality. Knowing a correlation between cat ownership and voting may help predict votes, but that says nothing about "influence". --KSmrqT 09:01, 13 April 2006 (UTC) Also, your sample size is quite small, so error would be huge in any calculations. Any group of 20 people are not likely to be evenly distributed as 10 voting for and 10 against. StuRat 09:39, 13 April 2006 (UTC) No matter what the size of the group, it's not likely to be that perfectly evenly distributed in reality. But it's a valid problem-solving technique to assume something (however unlikely) is true and ask 'what if' questions. JackofOz 11:54, 13 April 2006 (UTC) No, in this case the question is semantically malformed. The statistics cannot determine influence, only correlation. Suppose the measure reads "Be it resolved that cats make excellent pets." Although the voters who own cats are likely to agree, that is not at all the same as saying the ownership influenced the vote. More likely, the attitude that would cause someone to choose a cat as a pet would also lead to a vote in favor. That is, the correlation stems from a common cause (attitude), not from one event (ownership) causing the other (favorable vote). It is far more important to know this limitation of statistics than to know how to calculate various statistical quantities. As for the latter, try the article on Pearson's chi-square test. --KSmrqT 12:40, 13 April 2006 (UTC) # April 13 ## Unit Circle Are there any tricks to remembering the unit circle? C-c-c-c 00:34, 13 April 2006 (UTC) I do not understand the question. Does Unit circle help? Melchoir 00:43, 13 April 2006 (UTC) When trying to remember specific cos, tan, sin, sec, csc, and cot values for certain angles (30, 45, 60, 90, 135...etc) C-c-c-c 01:08, 13 April 2006 (UTC) Ah! Well, my strategy is: visualize, exploit patterns, and avoid actual memorization. Well, except for a few basics. Try starting with sin(30°)=1/2 and sin(45°)=cos(45°)=1/sqrt(2); these should be automatic, and if you want to reassure yourself, just draw the appropriate triangles. I guess cos(30°)=sqrt(3)/2 is useful too; you can get it from the sine with the Pythagorean trigonometric identity. 90° should also be automatic, as well as 0°; just visualize them to figure out where the 0s and 1s go. As for 60° and 135° and so on, these can be related back to other facts using the identities listed at Trigonometric identities#Periodicity, symmetry, and shifts; the rules can be justified visually. Finally, values of all the other functions (tan, sec, csc, and cot) are not worth remembering. If they come up, you can just express them in terms of simpler functions as at Trigonometric identities#Definitions. Melchoir 01:25, 13 April 2006 (UTC) Thanks! C-c-c-c 01:54, 13 April 2006 (UTC) Here's a mnemonic: ${\displaystyle \sin 0^{\circ }={\frac {\sqrt {0}}{2}}}$, ${\displaystyle \sin 30^{\circ }={\frac {\sqrt {1}}{2}}}$, ${\displaystyle \sin 45^{\circ }={\frac {\sqrt {2}}{2}}}$, ${\displaystyle \sin 60^{\circ }={\frac {\sqrt {3}}{2}}}$, ${\displaystyle \sin 90^{\circ }={\frac {\sqrt {4}}{2}}}$. Then either use a visual interpretation or the mnemonic ASTC ("all students take calculus" / "another stupid trig class") to remember which functions are positive in which quadrants (all, sin and csc, tan and cot, cos and sec). --Geoffrey 13:49, 13 April 2006 (UTC) ## Digits of an Irrational Number I'm doing some computer programming and I need some way to compute the digits of an irrational number (specifically, a square root or something like pi or e anything will do). But I need more digits than C or VB can provide with a regular classification, which I think is in the double, 15 digits?? (also I need the program to do it itself, no cheating off the web to get the digits). Is there an easy way without any severe progamming to calculate digits? I was thinking the only way would be to program a new function that does multiplication and adding longhand, storing each subsequent number from a series like Ramanujan's (History_of_numerical_approximations_of_π) and adding them longhand, digit by digit, carrying the 10, 100, 1000 etc. I found this for pi: Bailey-Borwein-Plouffe_formula but I don't think it is applicable. As far as I can tell, the method to compute 16^(n-k) mod(8k+1) described in Exponentiation by squaring still requires that my data will be able to store 16^(n-k). And what about that second term? I don't know enough (or have forgotten enough) calculus to determine what the infitine sum of 1/(16*(8k+1)) from k = n+1 to infinity will equal...but it probably won't affect the rounding of the integer which I assume is the nth digit? any ideas would be appreciated. Thanks, -Snpoj 03:42, 13 April 2006 (UTC) I didn't follow the links you provide, so I can't comment on those algorithms, but it's certainly true that if you want many digits of a number, then you need to code a data structure that can hold the digits. The blindingly obvious thing to do is make an array that holds all your digits (or bits, they'd be faster), and code a multiplication subroutine. You can use the algorithm you learned in 1st grade for multiplication (carry the 1). This algorithm is easy, no "severe programming" required. However, it takes O(n2) time, where n is the number of digits. There is an algorithm to do multiplication in O(n) time, which you might be interested in if you're going to keep a lot of digits, since calculating π to all those digits may take quite a lot of multiplications. This algorithm is described in Knuth's The Art of Computer Programming. So anyway, once you have code that knows how to multiply as many digits as you want, then you have to decide an algorithm for generating those digits. I understand that Newton's method for finding square roots converges quite fast; the number of accurate digits doubles at each iteration, so you should get a lot of digits quite fast. As for which algorithms are fast for calculating π, I know nothing. -lethe talk + 04:36, 13 April 2006 (UTC) According to Arbitrary-precision arithmetic, most systems have libraries for arbitrarily many digit numbers. For example, you could use GNU Multi-Precision Library. Then you don't have to code your own multiplication. -lethe talk + 04:45, 13 April 2006 (UTC) Yeah, I guess the multiplication subfunction would eventually degenerate into the addition one I mentioned. Anyway, thanks for the suggestions and the links. I'll check those out. Thanks -Snpoj 00:05, 14 April 2006 (UTC)  ? # timer = 1 (on) ? \p 2000 realprecision = 2003 significant digits (2000 digits displayed) ? sqrt(Pi) time = 74 ms. %5 = 1.772453850905516027298167483341145182797549456122387128213807789852911284591032181374950656738544665416226 823624282570666236152865724422602525093709602787068462037698653105122849925173028950826228932095379267962800174 639015351479720516700190185234018585446974494912640313921775525906216405419332500906398407613733477475153433667 989789365851836408795451165161738760059067393431791332809854846248184902054654852195613251561647467515042738761 056107996127107210060372044483672365296613708094323498831668424213845709609120420427785778068694766570005218305 685125413396636944654181510716693883321942929357062268865224420542149948049920756486398874838505930640218214029 285811233064978945203621149078962287389403245978198513134871266512506293260044656382109675026812496930595420461 560761952217391525070207792758099054332900662223067614469661248188743069978835205061464443854185307973574257179 185635959749959952263849242203889103966406447293972841345043002140564233433039261756134176336320017037654163476 320669276541812835762490326904508485320134192435989730871193799482938730111262561658818884785977875963761363218 634246546641333954355703201522654193952186030497310513829498439659165614245955421226615102478536098095510395600 789402188099613382854025016800745802729119366425192820510001936350073914643295493433951928853735459200563766502 880540575532123189009126322819150914980836695624483100852221923973646324842863261145766932425371577377894414090 544573595351225626391080239236909732127905807617134603914574791879794124850218445145811341888880413220955332184 646709727491028565262707845453262227848800982385836300754950954764062377083388357225436621567481327668384244972 420874516161833205077991480184666814236693651902845463857614827857037774388376297479982737705431583682410998683 228503805526355369722293133805264428410372312043967004307612454138311792278275363715598398376884537027842985707 0905112238405367790133854145853162080730431380697399874366931660138170792720560419548828580630931116362970478678140270  PARI/GP computer algebra system --GangofOne 06:23, 13 April 2006 (UTC) Interesting. I must take a look at the code to see how they do that. -Snpoj 00:05, 14 April 2006 (UTC) ## Period of a (Sine or Cosine) Function I know that Period = 2π/|b| where b is the horizontal stretch or compression. So for example, 2SinΘ would be 2π/1 = 2π or 360 degrees. Sin2Θ would be 2π/2 = π or 180 degrees. Now, if I switch this formula around I can have |b| = 2π/period. That means that b must always be positive, something I'm not quite sure of. Now that I think about it I don't think we've ever had a negative b in our tranformation graphing. Does it not exist or is that too complicated for my level (high school)? Thanks. C-c-c-c 04:41, 13 April 2006 (UTC) Well |b| = 2π/period, but the | | around b denotes the absolute value, which is there so that the period will come out positive; thus for a given period, b could be either +|b| or -|b|. If b in sin(bΘ) is negative, then all you've done is taken sin(|b|Θ) and flipped it backwards. As it happens sin(-x) = -sin(x) for all x, so you can always rewrite sin(bΘ) with a positive b by flipping the sign. Hope that helps. -- SCZenz 06:54, 13 April 2006 (UTC) ## Publishing Proofs I came up with a proof of the angle bisector theorem (I'm not sure if its' original or not, but I haven't encountered it anywhere else). I'm trying to see if anybody will publish it, but I can't find any appropriate journals. I know that Wikipedia has an article on mathematical journals, and the two big English ones I see are Annals of Mathematics and the Journal of the American Mathematical Society, but the papers in these journals are extraordinarily complex; they're dozens of pages long, use computers, cite references, etc. Could anyone suggest a mathematics journals that would publish such a simple, geometric proof? Thanks --JianLi 18:43, 13 April 2006 (UTC) I presume you've read the angle bisector theorem article? Professional scientists tend to be very unenthused about submissions to their journals from amateurs. This might seem harsh, but the simple fact is that 99.9% of the time, such submissions turn out to be either a) wrong, or b) already known. If you're not a working mathematician, you'll almost certainly need help from one to get something published in a mathematics journal. One suggestion: go find a grad student whose work is related to geometry, and show them your idea. If they agree you've got something, they might then take it to their professors. Alternatively, just put up a webpage outlining your proof (taking as much care as you can), and add a note saying that you think that this is a new proof of the theorem, that you're unsure of its correctness, and feedback would be appreciated, and link to it from somewhere where Google will find it. Before too long, somebody will find it, and if you're not a crank (tip: don't make your page look like http://timecube.com), you stand a pretty good chance of useful feedback. In my own research work, I once came across somebody's (admittedly, an academic's) hobby mathematics posted on their personal webpages, and found that it was relevant to my own "serious" research. :--Robert Merkel 09:01, 14 April 2006 (UTC) haha, funny website. Yes, I've read teh angle bisector article and looked at the proofs. my proof doesn't use any trig functions, which is why i think it might be original -JianLi ## Most Monumental Discovery in Recent History Just curious, what's the biggest discovery in the field of mathematics in recent history (excluding prime numbers, etc.)? Modern mathematics is pretty large and specialized, so I find it hard to be abreast of major discoveries in all fields, and therefore hard to have an opinion about biggest discoveries overall. Let me just mention some big discoveries which happened in the last couple decades: the classification of finite simple groups, including the discovery of some new sporadic groups; the discovery of nonstandard smooth structures on four dimensional Euclidean space by Milnor et al.; the discovery that to every Chern form is a knot invariant, by Witten. And of course everyone's heard of the the proof by Wiles of Fermat's last theorem and the proof by Perelmen of Thurston's geometrization conjecture. Anyway, those are some of the most important mathematical results (I hesitate to use the word "discovery" for mathematical results) in recent times, where recent means probably the last coupla decades. -lethe talk + 23:45, 13 April 2006 (UTC) Yes, "discovery" is an uncertain term in mathematics. Did Hamilton discover quaternions or invent them? Fermat conjectured that an+bn = cn had no nonzero integer solutions (a,b,c) for all integers n > 2. Only recently was Wiles able to prove the conjecture, in the context of a many-year assault on the much deeper Taniyama-Shimura theorem. So is Fermat's Last Theorem a recent discovery? Certainly its proof is. However, the discovery part of Wiles' proof lies more in the methods and insights developed along the way, and less in finally being able to assert that the theorem is proved. Contrast that with the discovery of non-Euclidean geometries, showing that the parallel postulate is independent of Euclid's other axioms. Gödel's incompleteness result surely qualifies as a discovery, being completely unexpected. We might also view fractals, strange attractors, and chaos theory as discoveries. I'd say non-standard analysis qualifies, though it's had less impact that one might expect. Or what about the demonstration that topos theory can replace set theory as a foundation for logic and mathematics? Which brings us to the discovery/invention of categories; and before that, well, name your favorite: scheme, homology theory, manifold, group, quaternion, complex number, integers, …. Also, keep in mind that only with the passage of time can we see which ideas have the biggest impact. --KSmrqT 00:35, 14 April 2006 (UTC) I'd tell you, but it isn't published, and this reference desk is too small to contain it. (ducks) Melchoir 00:02, 14 April 2006 (UTC) I dunno whether it counts as "recent", but the discovery of NP-complete problems in 1971 is one of the most influential. Proving a problem is NP-hard, by reduction to an NP-complete problem, is a standard way of showing that a problem is very difficult, and such proofs are often quite routine and appear in hundreds (thousands?) of CS papers. Furthermore, the follow-on question it raised - the (P = NP question) has exercised the minds of theoretical computer scientists (who are essentially mathematicians) ever since Cook's paper in 1971. See Complexity classes P and NP. --Robert Merkel 08:45, 14 April 2006 (UTC) Well, I've recently seen an inverse Laplace transform that doesn't require the use of imaginary numbers (and especially imaginary contour integrals), although it does require taking arbitrarily high derivatives of a function. Probably not that huge a discovery, though. Confusing Manifestation 13:16, 14 April 2006 (UTC) The Four color theorem was proved in 1976. Chuck 19:55, 14 April 2006 (UTC) # April 14 # April 15 # April 16 ## radical of square ${\displaystyle \pm x={\sqrt {x^{2}}}}$ OR ${\displaystyle x={\sqrt {x^{2}}}}$? I believe it is the second, but my friend disagrees. -Mystaker1 03:08, 16 April 2006 (UTC) ${\displaystyle |x|={\sqrt {x^{2}}}}$ HTH. -lethe talk + 03:45, 16 April 2006 (UTC) Thank you! -Mystaker1 03:51, 16 April 2006 (UTC) Err... lemme elaborate on that answer then. There are indeed two different numbers which square to x, for any nonzero x. However, the square root function is defined to only return the positive choice. Thus, whether or not x is negative, its square is positive, and the square root is the positive number. This is the same as the absolute value function. This choice is a bit unnatural, and causes some problems for complex numbers, but nevermind that. -lethe talk + 03:55, 16 April 2006 (UTC) I understand. Grazi. -Mystaker1 04:17, 16 April 2006 (UTC) So who wins the argument? -lethe talk + 04:24, 16 April 2006 (UTC) I'd say the friend. ${\displaystyle x={\sqrt {x^{2}}}}$ is untrue no matter how you look at it; ${\displaystyle \pm x={\sqrt {x^{2}}}}$ is true if you interpret it as "√x2 is eiher +x or -x". In this sense, this is true for complexes as well - as opposed to |x|. -- Meni Rosenfeld (talk) 09:40, 16 April 2006 (UTC) lethe was right. The ${\displaystyle {\sqrt {\;\;}}}$ in ${\displaystyle {\sqrt {x}}}$ denotes a function. When x is a non-negative real number, ${\displaystyle {\sqrt {x}}}$ is by definition the non-negative value y that satisfies ${\displaystyle y^{2}=x}$. --68.238.254.236 11:57, 16 April 2006 (UTC) We know lethe was right. Lethe is always right. Why did you feel the need to reinstate his argument? In case you didn't understand my comment - Of course √x2 is non-negative, but x can be negative. -- Meni Rosenfeld (talk) 14:16, 16 April 2006 (UTC) ## infinitesimal comprehension Also, I'm not sure if I understand infinitesimals. Can you tell me whether this equation is solved correctly or not? -Mystaker1 03:31, 16 April 2006 (UTC) I've replied on the page. -lethe talk + 03:49, 16 April 2006 (UTC) ## Is this log 2? We were doing some stuff on Hilbert matrices in applied maths (really just MATLAB practice) and I was playing around and discovered that ${\displaystyle \lim _{n\to \infty }{\sum _{j=1}^{n}{\frac {1}{n+j-1}}}\approx \log 2}$. Are they exactly equal? And could anyone give me a hint on proving it? (Just for fun - this isn't homework or anything.) --대조 | Talk 11:54, 16 April 2006 (UTC) Yes, the limit is ${\displaystyle \ln 2}$. Here is an outline of a proof. Your sum is (more or less) ${\displaystyle H_{2n}-H_{n}}$ where ${\displaystyle H_{n}}$ is the nth harmonic number. Now ${\displaystyle H_{n}=\ln(n)+\gamma +O(1/n)}$ ${\displaystyle \Rightarrow H_{2n}=\ln(2n)+\gamma +O(1/n)}$ ${\displaystyle \Rightarrow H_{2n}=\ln(2)+\ln(n)+\gamma +O(1/n)}$ ${\displaystyle \Rightarrow H_{2n}-H_{n}=\ln(2)+O(1/n)}$ ${\displaystyle \Rightarrow \lim _{n\to \infty }H_{2n}-H_{n}=\ln(2).}$ Gandalf61 13:46, 16 April 2006 (UTC) A different way to solve it is to use the Riemann sum ${\displaystyle \lim _{n\to \infty }{\sum _{j=1}^{n}{\frac {1}{n+j-1}}}=\lim _{n\to \infty }{\frac {1}{n}}{\sum _{j=1}^{n}{\frac {1}{1+j/n-1/n}}}}$ ${\displaystyle =\lim _{n\to \infty }{\frac {1}{n}}{\sum _{j=0}^{n-1}{\frac {1}{1+j/n}}}=\int _{0}^{1}{\frac {1}{1+x}}dx=\ln 2}$ (Igny 15:23, 17 April 2006 (UTC)) ## simple algebra inequality Dear Math Ref. desk: I have a really simple question that is really dumb but I have been struggling with it. It is a SAT II Math IC question. I am not posting a homework. Thank you for your help. 5. What are all values of x for which (4−x)2 ≥ (x−2) (A) x ≥ −3 (B) −5 ≤ x ≤ 0 (C) −3 ≤ x ≤ 2 (D) x ≤ −3 or x ≥ 2 (E) −2 ≤ x ≤ 3 The given answer is (C) and the difficulty level is 3. I would like to have some procedure also. I tried to solve it myself but I think I am missing something. -- Kushal [e-mail address not entered] The preceeding question was added by Kushal one Kushal one 18:11, 16 April 2006 (UTC). Please visit my talk page to resolve any suspicion of trolling by me. I previously used to make (very few) comments under the name "My IP address is not permanent." and variants. Thank you for taking yout time once again. A good way to solve inequalities is to solve the corresponding equality, in this case (4−x)2 = (x−2), and then determine the intervals in which the equality holds by inspection. —Keenan Pepper 19:22, 16 April 2006 (UTC) [edit conflict] Well, I'll give a sketch of the solution, if you don't understand it, state which step you are having trouble with: a) Expand the square and simplify to get x2 - 9x + 18 ≥ 0 b) Solve the quadratic to get (x - 3)(x - 6) ≥ 0 c) The solution is x ≥ 6 or x ≤ 3. Since this is different from the answer you give, you may also want to check that you have typed the equation correctly. The same method works anyway. -- Meni Rosenfeld (talk) 19:23, 16 April 2006 (UTC) It might be helpful to expand the left-hand side and bring the right-hand side over, producing ${\displaystyle x^{2}-9x+18\geq 0.\,\!}$ Now the left-hand side describes a parabola with arms up (as x goes to ±∞ the value goes to +∞). Thus the answer must be of the form "xa or xb", which only (D) has. But this is clearly wrong, because when x is zero the inequality is satisfied, 16 ≥ −2; and when x is 4 it is unsatisfied, 0 ≥⃒ 2. In fact, the parabola crosses zero at the roots of the polynomial, 3 and 6. So you have given us a mistaken problem. The correct problem statement uses (4−x2) ≥ (x−2); try applying this same kind of analysis to it. --KSmrqT 19:45, 16 April 2006 (UTC) Correction: Thank you, KSmrq. The parentheses were clearly wrongly put. Signs of a clumsy student... Please read the above question as KSmrq put it. I assure you, however, I did not make that mistake on paper. Thank you I will see your guidance and try to do the problem again. I still think there is something I am missing... Kushal One STILL HAVE TROUBLE What are you having trouble with? The steps are the same as with the original equation posted. -- Meni Rosenfeld (talk) 15:56, 18 April 2006 (UTC) Please do not use ALL CAPS; it's the typographical equivalent to shouting. Please do not repeat a question; it is rude and unhelpful. Please do add details under the original section if followup is necessary. But this question has been answered completely, so far as I can see. The essential points are: 1. Bring the content to one side, forming an inequality with zero. 2. Look at the sign of the highest degree term, which we assume quadratic. 3. Solve for the roots (zero-crossings). Do yourself a favor and learn to ask smart questions. --KSmrqT 17:42, 18 April 2006 (UTC) Another way to solve this type of multiple choice question is to plug in each answer and see which one works. In this case, since the answers range from -5 to 3, you could create a table with each integer between -6 and 4 and try all those values in the equation and record which ones work. Then you could infer the correct answer from the table. StuRat 08:01, 19 April 2006 (UTC) ## group theory prove that if N is a normal subgroup of G and H is any complex of G then NH=HN Isn't that pretty much the definition of a normal subgroup? —Keenan Pepper 20:01, 16 April 2006 (UTC) What do you mean by a "complex" of a group? Dysprosia 06:30, 18 April 2006 (UTC) ## Frequency Distribution of Numerals I read some where that given a long list of multi-digit (random?) numbers, the number 1 will occur as the first digit far more often than 10% (I think it was somewher around 30%...). Can anyone tell me more about this? SigmaEpsilonΣΕ 21:24, 16 April 2006 (UTC) See Benford's law. --KSmrqT 22:46, 16 April 2006 (UTC) Thanks! SigmaEpsilonΣΕ 19:03, 17 April 2006 (UTC) # April 17 ## Accounting Need assistance with this question, not clear on how to put the monthly summary together to get the total material payments for January. Please email me at e-mail removed; see the note at the top of the page. Many thanks The Denver Corporation has forecast the following sales for the first seven months of the year: January$9,000 May $10,000 February 12,000 June 16,000 March 14,000 April 20,000 July 18,000 Monthly material purchases are set equal to 30 percent of forecasted sales for the next month. Of the total material costs, 40 percent are paid in the month of purchase and 60 percent in the following month. Labor costs will run$4,000 per month, and fixed overhead is $2,000 per month. Interest payments on the debt will be$3,000 for both March and June. Finally, the Denver sales force will receive a 1.5 percent commission on total sales for the first six months of the year, to be paid on June 30.

Prepare a monthly summary of cash payments for the six-month period from January through June. (Note: Compute prior December purchases to help get total material payments for January.)

This has nothing to do with mathematics. Ohanian 04:49, 17 April 2006 (UTC)

I disagree; it's essentially a word problem, with dollar signs. The questioner just wants the math worked out. (I agree that most accounting questions -- say, where to put goodwill on a balance sheet, or what accrual means, would belong on the Misc. page.)
Anyways, for this company, for each month n, materials are bought based on sales projections for month n+1. They are paid for, 40% in month n and 60% in month n+1. So in January (which is month n+1 using my notation), the total materials payment is 40% of the January purchase, and 60% of the December purchase.
To get this, we need to calculate what the December and January purchases are. Luckily, this company is looking ahead, so we know that the December purchase is 30% of the January sales, and the January purchase is 30% of the February sales.
Putting this together, we get 0.4*0.3*(February sales) + 0.6*0.3*January sales) = January purchases. --ByeByeBaby 23:31, 17 April 2006 (UTC)

I am submitting this with all respect to the respected RD sirs and madams. Please take it with a pinch of salt.

Dear ‘Aklepard’:

Month Forecasted sales Monthly material payment payment same month Payment next month Labor Fixed overhead Interest payment Commission

December -- 3000 1200 -- 4000 2000 0 0

January 10000 3600 1440 1800 4000 2000 0 0

February 12000 4200 1680 2160 4000 2000 0 0

March 14000 6000 2440 2520 4000 2000 3000 0

April 20000 3000 1200 3600 4000 2000 0 0

May 10000 4800 1920 1800 4000 2000 3000 0

June 16000 5400 2160 2880 4000 2000 0 3000

July 18000 these details here are not required for the present calculation..........................................................................................................................................

[NB: ‘--’‘means ‘cannot calculate’ and ‘0’ means no ‘payment’ necessary.]

Add columns ‘payment same month’ and ‘Payment next month’ to get ‘total material payments’ for that month. FYI, the January figure is USD 1440 +USD 1800 =USD 3240

To get total value of monthly value of cash payments, add columns 3, 4, 5, 6, 7, 8, and 9.

If you are still confused, post your confusion on the reference desk.

Yours truly,

Kushal Hada [contact info removed but ca be found on my talk page.]

PS It is not considered a good idea to post your e-mail address on the RD. I hope you understand.

PPS I apologize in advance if I offended you in any way.

-- Kushal one 00:25, 18 April 2006 (UTC)

## Amen

So there is an article about Fibonacci Poems at the NYTimes [2]. They introduce the concept of Fibonacci numbers by saying, "Readers of the blockbuster best-selling "Da Vinci Code," of course, may recognize the Fibonacci sequence as the key to one of the first clues left for the novel's hero and heroine" which just shows you how idiotically innumerate the general public is. But anyway, that's not my point. I have a question about this:

Let P be a set of points in general position in the plane. Amen.
The last line, said Mr. Venkatasubramanian, is an inside joke in geometry.


Can anybody explain this joke to me? I googled it, and found it again at [3], but with no explanation. JianLi 01:41, 17 April 2006 (UTC)

I think the joke is simply that very many geometry problems begin that way, so it's little more than a formulaic invocation, like "in the name of the father and the son and the holy ghost" in Catholicism, for example. This says it's called the "Sacred Sausage Invocation". Geometers are weird. —Keenan Pepper 04:17, 17 April 2006 (UTC)
The link now has a comment saying SoCG (suggesting the pronunciation "sausage") is a Computational Geometry conference. I am not familiar with the "joke" as such, but the association with computational geometry is important. In pure mathematics it seems perfectly reasonable to solve problems without "accidental" alignments. Thus three points do not form a line or a right triangle, two lines are not parallel, that kind of thing; hence, "general position". In computational practice, points frequently align in such ways, wreaking havoc with the algorithms and forcing a great deal of ugly special case handling. So for a computational geometer, the statement has two meanings: "Let … general position" is a standard form of mathematical postulate, but it can also be read as a request to some deity that the problem inputs actually satisfy this postulate — hence the "amen". --KSmrqT 13:05, 17 April 2006 (UTC)
Hmm, interesting. thanks for your responses! -JianLi 00:35, 21 April 2006 (UTC)

## Complex coeficients in a quadratic equation

Can anyone, please, tell me the method of how to resolve these? Thanks in advance.

I don't understand the question. Melchoir 20:22, 17 April 2006 (UTC)
Well, that's covered by Quadratic equation. Melchoir 00:21, 18 April 2006 (UTC)

## Mathematical Function

Afraid the link wont work therefore it is named after this man

http://img133.imageshack.us/my.php?image=6b7ld.gif

Please Can You tell me the name of this Mathematical Function?

Its Similar to a wave its pink and black.

See Benoît Mandelbrot, and the Mandelbrot set. --KSmrqT 13:08, 17 April 2006 (UTC)

## mathematical formulas on a sphere

Which article should I look to search for mathematical formulas in non-euclidian universes, that is not a infinite plane.

More specific, i want to be able to calculate things like shorter distance between two points, pythagorean theorem and point of intersection of two line segments, but all of those in the surface of a sphere.

Thanks --Alexandre Van de Sande 20:03, 17 April 2006 (UTC)

Try Spherical geometry, particularly the links at the bottom. Melchoir 20:20, 17 April 2006 (UTC)
There is no Pythagoras theorem on a sphere. Spherical "line segments" are great circle arcs, and have two points of intersection where the line of intersection of the two circle planes pierces the sphere. For a unit sphere centered at the origin, coordinates of surface points act as unit vectors, and the dot product of two is cosine of the angle between them (automatically the smaller of the two choices); angles are equal to chord lengths, thus shorter distance. --KSmrqT 23:12, 17 April 2006 (UTC)

## program to draw lines

I have a mac and a pc, a big interest in mathematics and some basic experience on some programming languages. I want to program an application that map networks, wich basically draws sets of lines among a growing point set according to some basic rules.

which program do you recommend, something with a low learning curve. Thanks

If you know at least the basics of matrix arithmetic, I would say you should look at Matlab. It's powerful, not too hard to learn the basics of, and is very good at drawing lots of cool graphs and stuff. Confusing Manifestation 00:50, 18 April 2006 (UTC)
Mathematica does graph theory. Dysprosia 06:26, 18 April 2006 (UTC)
For mapping networks, a good candidate comes from a telephone company! Try GraphViz. --KSmrqT 15:42, 18 April 2006 (UTC)

# April 18

## longitude circumphrences

I'm looking for the equation for the circumphrance of a certain longitude; for example, the circumphrance of a smaller circle that's 45 degrees above the exact middle of a spere. Tell me if I need to be clearer than that. Jonathan talk 19:31, 18 April 2006 (UTC)

I think you mean latitude. The circumference C of the circle at latitude φ on a sphere of radius R is
${\displaystyle C=2\pi R\cos \phi .}$
Of course, cos 45°=1/sqrt 2. Melchoir 20:25, 18 April 2006 (UTC)
Could you be a little more specific about that formula? I understand the 2 pi R part but as far as I know a cosine isn't something you multiply with. Jonathan talk 21:05, 18 April 2006 (UTC)
You're right, let me try again with a clearer style:
${\displaystyle C=2\pi R\cdot \cos(\phi ).\,}$
As in, you take the cosine of the latitude, and then you multiply that with the circumference of the equator, 2 pi R. Melchoir 21:13, 18 April 2006 (UTC)
For the Earth, the equator is roughly 24,000 miles in circumference, or about 1,000 miles per time zone. The 2πR in the above formula can be replaced by the circumference. (Earth's circumference is also about 40,000 km, but I don't know a memory trick for that.) --KSmrqT 23:06, 18 April 2006 (UTC)

Are you sure about that? I tried that formula and kept getting negative numbers. (For instance, 2 times pi times 10 times the cosine of 10 equals: -52.7204.) Jonathan talk 00:29, 19 April 2006 (UTC)

Try 10°, not 10 radians. (Multiply angle by π/180, which is 0.0174532925 approximately.) --KSmrqT 01:23, 19 April 2006 (UTC)
Thanks! I tried it and it worked. Jonathan talk 23:52, 19 April 2006 (UTC)

## Terminator line and cylindrical projections

I'm trying to find out the function to draw the terminator line on a plate carrée projection of Earth's surface. Thing is, I have a map with the continents and everything, and I want to generate the terminator line of the given period of the year and use it to create a shadow layer over the map. The time of the day part I can figure by myself. Oh, and it doesn't have to be utterly precise, just good enough.

Anyway, this proved to be a tricker than I thought. As you probably know, Earth axis has a tilt (23.45°), so the terminator isn't a simple meridian line, but a curve, and depending of the time of the year this curve changes. Can anyone help? ☢ Ҡiff 20:46, 18 April 2006 (UTC)

If you know your Java, this dude has an open-source applet (the source code is a link at the bottom) that does exactly what you're looking for. Or, you could simply skip the whole thing and buy a classy leather-framed original Geochron, right from the manufacturers. Plug it in, set the time and voila! Of course, the one the geomatics department at my school had was broken, but at least it broke when it said the sun was up, so we weren't plunged into perpetual darkness. --ByeByeBaby 04:27, 19 April 2006 (UTC)

## Mathematics

A circle is drawn. Two points are marked on its sides. We join the 2 points. Thus, we divide the circle into two parts. Into how many parts can the circle be divided for "n" no. of points?. Find the formula

Depends on what you're allowed to do with the points. It's not clear. Melchoir 21:05, 18 April 2006 (UTC)

I worked it out on paper and it looks to me like your formula is:

${\displaystyle 2^{n-1}}$

Because when I used 2 points I got 2 section; 3 points = 4 sections; 4 points = 8 sections; 5 points = 16 sections... Jonathan talk 21:13, 18 April 2006 (UTC)

This is a well-known example showing how this kind of "obvious" conjecture can be wrong. Using 6 points, you only get 31 sections. (@Melchoir: Choose n points on a circle in general position, draw all ${\displaystyle {n \choose 2}}$ connecting segments.)--gwaihir 21:30, 18 April 2006 (UTC)
ah. Very interesting... I found a reference here. D'ya suppose it's an article-worthy topic? Melchoir 21:38, 18 April 2006 (UTC)
This series can also be found using Pascal's Triangle:
                         1                  | = 1              = 1
1  1               | = 1+1            = 2
1  2  1            | = 1+2+1          = 4
1  3  3 1          | = 1+3+3+1        = 8
1  4  6 4 1        | = 1+4+6+4+1      = 16
1|  5 10 10 5 1        | = 5+10+10+5+1    = 31
1  6| 15 20 15 6 1        | = 15+20+15+6+1   = 57
1  7 21| 35 35 21 7 1        | = 35+35+21+7+1   = 99
1 8 28 56| 70 56 28 8 1        | = 70+56+28+8+1   = 163


If we add all the numbers on the right hand side of the line per row, we get 1, 2, 4, 8, 16, 31, 57, 99, 163. --Alexs letterbox 23:06, 18 April 2006 (UTC)

See also Ronald Graham, Donald Knuth, Oren Patashnik, Concrete Mathematics chapter 1.2. – b_jonas 20:56, 23 April 2006 (UTC)
This question certainly deserve an article on its own, it's in no way simple, and was even used as the question for the high-end entry test(test for exceptional iq, or some name like that) on International High IQ Society a few years before!(PS:Now it became level5 puzzle4) Anyway, one method to solve it is by Euler's formula V-E+F=1 for plane figure. Find the general formula for both V and E(including intersecting segments), and you'll got it. --Lemontea 08:19, 25 April 2006 (UTC)
Update: Shot it! Rush over to Dividing a circle into areas and see the proof. --Lemontea 10:41, 28 April 2006 (UTC)

## Curious Puzzle

I'm not quite sure how to ask this, so bear with me. I'm looking for a field of mathematics/analysis that I've never heard of or run across, that I can't think of a name for, but that I'm certain must exist somewhere. I'm trying to solve a problem that involves taking a set of objects (one-digit numbers), arranging combinations of them within a structure, and finding a particular combination that satisfies certain criteria. It would have to be able to deal with a situation where 1)There are too many possibilities to try every one (unlike Anagrams), and 2)There's no obvious series of deductions that solves it one piece at a time (unlike Sudoku). A bit like a Magic Square, really, though apparently there are formulas that solve most of those. Any ideas? Black Carrot 21:07, 18 April 2006 (UTC)

Combinatorics? It depends on what sort of criteria you have. With certain criteria, you can get a linear algebra problem... Dysprosia 03:17, 19 April 2006 (UTC)
Possibly. Do you know of any sources on how that works? The article isn't very helpful. The basic problem is this: choosing two series of one-digit numbers such that the many-digit numbers they form multiply to a predetermined number, using the normal multiplication algorithm (multiplying each pair of digits, tacking on the right number of zeros, and adding them up). The resultant puzzle isn't all that different from magic squares and the like (especially if you format it right, with the pair of numbers along the axes of a table and the products in the cells), and it seems like there should be some method for working it out. Black Carrot 23:43, 20 April 2006 (UTC)

Well, if a0,...,an is your first sequence, note that (in base 10) the integer they form is a010n+...+10an-1+an, then do the same for b0,...,bm which is your second sequence. Multiply the two sums, collect, and compare coefficients. This should solve the problem. Dysprosia 14:33, 24 April 2006 (UTC)
I'm not quite sure I follow you. Black Carrot 01:43, 25 April 2006 (UTC)
Oh, I think I might have misunderstood you. Do you want to see how many different products a given number has? Or do you want to see if a product of two numbers gives you a given number? Dysprosia 02:00, 25 April 2006 (UTC)

# April 19

## Tricky Java Program

Hi, I have a Java program to do that I just can't crack.

These are the instructions given:

Create a class called Sentence with the methods as listed below. This class has one String field called aSentence that holds several words separated by spaces. Your client class used to test these methods called RunSentence should use file input. Use the file in the pickup folder called “sentence.txt”

//create a constructor method, a mutator method and an accessor method

//purpose: the method should return the number of words in the sentence

public int numWords( )

So, the question asks for the number of words. I thought about this and assumed: number of words is number of spaces plus 1. For example "This is a sentence", has three spaces, and 4 words, and so on ( I don't think we have to worry about two spaces in between, just started strings not so long ago).

aSentence is my string here, n keeps track of the spaces and I've set n to 1, because if anything is written there will always be something (ie. one word). I have no syntax erros but the debugger isn't working for me so I have no idea what's wrong with it. Int index gives me an integer value of how long the string is. I've set position to 0 so that it counts from the first index. The repition statement is position == index, so once I get position to equal index, I'm at the very end of the string.

In the first part of the for loop, position is 0, and it checks for where " " is found. So then position becomes 4, and it adds one to n because it found a space, therefore a word. Next it does it pretty much the same thing, but adds one to the that 4, which then becomes 5. After it does aSentence.indexOf ( " ", 5) it becomes 7, where 7 is the next space. Then it adds 1 to that 7 and starts again. I have it add this one to stop it counting the space twice, if that makes any sense to anyone.

  public int numWords()
{
int n = 1; //counts number of spaces

int index = aSentence.length()-1;
for ( int position = 0; position == index; position = position + 1)
{
position = (aSentence.indexOf ( " ", position));
if ( position != -1)
{

n = n +1;
}

position= (aSentence.indexOf( " ", position+1));
if ( position != -1)
{

n = n +1;
}
}

return n;


}

C-c-c-c 06:05, 19 April 2006 (UTC)

Sorry that i don't know java, but I think that the "position == index" should be "position != index" because otherwise it will terminate immediately unless the length was one?
it typical to use position <= index. I don't think you need to repeate the indexOf test, just once should do. You may find it easier to use String.charAt(int pos) which will let you loop through the string testing one character at a time. --Salix alba (talk) 07:22, 19 April 2006 (UTC)

I see several problems with this code. For one, you shouldn't change your position variable inside a block of code whose termination rests on position reaching the length. For example, it looks like if you feed that code a string with no spaces, it will set position to -1 on the first pass, and never terminate. If all you want to do is count the number of spaces, then why don't you just do

n=0;
for (int i=0;aSentence.length()-1;i++)
if (aSentence[i]==' ')
n++;


-lethe talk + 07:39, 19 April 2006 (UTC)

For the record, this assignment is horribly designed. There's no point in having a class with one member variable that just analyzes its member variable. Much better would be, in some appropriate class, to just have
public class StringUtilities { // name arbitrary; would be combined with other such functions
public static int countWords(String s) {
int ret=0;
boolean wordstart=true;
for(int i=0;i<s.length();++i) {
if(s.charAt(i)==' ') wordstart=true;
else if(wordstart) {++ret; wordstart=false;}
}
return ret;
}
}


This handles multiple spaces properly, as well as spaces at the beginning or end of the string. It's still a very naïve definition of a "word" counter, though. --Tardis 19:21, 19 April 2006 (UTC)

However the assignment offers an object oriented solution and yours is a procedure oriented solution. Ideally you would want to sub-class String to add the countWords method, but string is final so thats not possible. The assigment design does follow well established design patterns a fully fleged OO solution would also need to add a number of accessor methods to allow other string methods to be used. This way sentance would be a self contained enterty without the need for utility classes hanging around. --Salix alba (talk) 20:01, 19 April 2006 (UTC)
I'm not sure what you want is a "proper" object-oriented solution either. The notion of counting words in a string is not restricted to some subset of strings, WordCountableString extends String or so. Certainly some strings cannot be said to contain words, for instance "+!@#($!*(%@@^" or "42", depending on your definition of "word". But because it's entirely fair to say that such strings have 0 words, the function ${\displaystyle f(s)}$ that gives the number of words in ${\displaystyle s}$ is well-defined for any ${\displaystyle s}$, given a suitable unambiguous definition of "words" and boundaries between them. So what we'd really want is some sort of aspect-oriented programming approach where we embued all Strings with the capacity to report on their own word counts. What you're suggesting (aside from duplicating String functions on a "fake subclass") would seem to naturally extend to having a FirstLetterMayBeUppercaseString class which supports the boolean beginsWithUpper() function. And then why not (supposing multiple inheritance, while we're invalidly subclassing String anyway) have WordCountableStringPerhapsWithInitialUppercaseLetter that extends both of these? I don't think that the notion of defining arbitrary functions on Strings is possible to fit into the usual OOP model. Thus my simple suggestion of a utility function, since that's what's being requested (a function). What might be useful as a OOP kind of solution would be, say, a WordCountingReader extends FilterReader with the obvious semantics. Then you're adding behavior to a well-defined object, rather than simply defining a function of a well-defined object. Seems like a better fit to me. (Sorry this is so long, and again if I'm misunderstanding you!) --Tardis 21:54, 19 April 2006 (UTC) Is it just me, or does it feel like this question and subsequent discussion would find a better home on a different reference desk? Are we now to expect all computer science questions to migrate here? --KSmrqT 23:04, 19 April 2006 (UTC) Indeed we are. See Wikipedia talk:Reference desk#Science/Mathematics subcategory renaming. Basically, the science desk was getting a lot of traffic, the mathematics desk not so much, so we decided to try moving all the computer stuff. It makes more sense to me for computer science to be grouped together with mathematics anyway. If it doesn't work out we can always change it back. —Keenan Pepper 00:21, 20 April 2006 (UTC) ## Simultaneous Rational Equations solve for x and y: m/x +n/y = a p/x +q/y = 0  I see from the wikisource that you meant those to be separate equations (not one equality among three quantities, one of which is 0), so I fixed it. The easiest thing to do is solve the second equation for, say, x, and plug it into the other equation. Then you have one equation in one variable, which can probably be reduced to a polynomial or so. --Tardis 19:21, 19 April 2006 (UTC) If either x or y are zero, the equations are meaningless, so assume they're nonzero. Multiply both equations by xy to clear denominators, leaving a total degree 2 equation and a linear equation. Eliminating y from the latter reduces the former to a quadratic in x. Solve as usual, then back-substitute. --KSmrqT 21:25, 19 April 2006 (UTC) solve for x and y: m/x +n/y = a p/x +q/y = 0  becomes solve for U and V: mU + nV = a pU + qV = 0  x = 1/U and y = 1/V very easy. Ohanian 02:22, 20 April 2006 (UTC) ## Bracket nomenclature Numerous papers (such as this) explaining Rand index (for which there's no Wikipedia article) use a bracket nomenclature similar to that of column vectors ${\displaystyle {4 \choose 2}}$ but with normal parentheses () rather than square brackets []. I can't understand what this bracket nomenclature is referring to. The worked example is: ${\displaystyle {2 \choose 2}+{4 \choose 2}=7}$ Can anyone explain to me how this answer is achieved? I'd be most grateful!! UkPaolo/talk 19:52, 19 April 2006 (UTC) See binomial coefficient. Fredrik Johansson 19:59, 19 April 2006 (UTC) thank you! Now just need to figure out where ${\displaystyle {2 \choose 2}}$ and ${\displaystyle {4 \choose 2}}$ came from.... :o) UkPaolo/talk 20:14, 19 April 2006 (UTC) Seems like we need an article on the Rand index, stub now created. --Salix alba (talk) 20:11, 19 April 2006 (UTC) would be nice... UkPaolo/talk 20:14, 19 April 2006 (UTC) # April 20 ## Geometry Question A bridge needs to be suspended over a river. The bridge will be one mile in length. If a (one mile + 2 feet) cable is used as the suspension cable, what is the largest type of animal that will be able to cross under the suspension cable at the mid-point of the bridge? Choices: • a) an ant • b) a mouse • c) a guinea pig • d) a groundhog • e) a small dog • f) a goat • g) Danny DeVito • h) an average human • i) Yao Ming • j) a really tall giraffe • k) a really tall giraffe on Yao Ming's shoulders How do I solve this? Thank you for your help. --User:Archie44 10:20, 20 April 2006 (UTC) Approximate that the suspension bridge is just an isoceles triangle, with base of one mile and sloping sides with total length 1 mile + 2 feets. Use Pythagoras' theorem to see how high the tip of the triangle is and use that as an estimate of how tall an animal can fit under it. I think that's what they're asking you, at least (you may be surprised). Confusing Manifestation 12:25, 20 April 2006 (UTC) It depends on the ratio of weight between the cable and the deck, but if the deck is much heavier than the cable, the cable will follow a parabola. So if we don't use the approximation, we have to examine the arc-length of the parabola: ${\displaystyle f(a,l)=\int _{0}^{l}{\sqrt {1+(2ax)^{2}}}dx}$ and find the solution to f(a, 0.5 mile)=(0.5 mile + 1 feet). Unless I made a mistake (very possible!), the result is very different from the approximated answer - I doubt that an ant could pass under that. Rasmus (talk) 13:53, 20 April 2006 (UTC) Well, using a whole bunch of programs and stuff, I have solved said function and found a height of 62 feet. Likewise, I could be wrong. For reference, I found the integral using The Integrator, solved it using DeadLine, and converted the units using Google. The triangular approximation gives 72 ft, which isn't a huge difference (particularly in terms of what the question actually asks). Confusing Manifestation 18:11, 20 April 2006 (UTC) That sounds more sensible - I was quite surprised about the difference. Redoing the calculations, I get 62.9 feet. I think I might have taken a as the result the first time rather than a*(0.5 mile)2. Sorry about that. Rasmus (talk) 20:14, 20 April 2006 (UTC) The question makes no sense. Skim through the pictures of suspension bridges and think about how the deck relates to the cable. So the cable is lowest at the middle, and we can calculate the curve of the cable; so what? No matter what the cable does, its least height above the deck is an independent design parameter. --KSmrqT 14:52, 20 April 2006 (UTC) I think you should think of a bridge with only one pillar, like this. Rasmus (talk) 14:58, 20 April 2006 (UTC) Yes, that makes more sense. A careful question would still explicitly state not only that design, but also that the ends are at the level of the deck. --KSmrqT 22:38, 20 April 2006 (UTC) I don't think it makes a difference whether it has one central tower or one tower at each end, only that the lowest point of the cable is level with the deck, if we're modelling the cables as an isoceles triangle. The only difference is that with two towers the "base" of the triangle is an imaginary line between the tower tops, whereas with a single central tower the base is the deck of the bridge. (The engineer in me is screaming out that this "two towers" design wouldn't work as they would topple inwards due to the lack of any support on the outside of the bridge - i.e. you need to have /|\/|\ not just |\/| - but then that engineer is also pointing out that from a structural POV the entire setup is ludicrous. But you can ignore everything in these brackets since it's a geometry question not a structures questions.) -- AJR | Talk 00:57, 21 April 2006 (UTC) See catenary for a solution. (Igny 20:47, 20 April 2006 (UTC)) Actually, no. Without loading the curve is a catenary, but the weight of the deck makes it a parabola. See this discussion. --KSmrqT 22:38, 20 April 2006 (UTC) I would say it couldn't support any weight, because to use a 5282 foot cable to span a 5280 foot gap, you would need to exert an enormous amount of tension to prevent it from sagging more in the middle. This tension would be so high that any materials we have would fail under such a stress. StuRat 07:28, 21 April 2006 (UTC) ## Physics question I'm not sure if this comes under maths or science, but I'll try it here: sand is falling onto a conveyor belt at a rate of 0.2kg/s. The conveyor belt is moving at a rate of 3m/s. There are four questions: 1. What force is needed to keep the belt moving at this speed? 2. At what rate is work being done by the force? 3. What is the rate of change of kinetic energy? 4. Why are the answers to 2 and 3 different? These are fairly straightforward. Since force is ${\displaystyle {\frac {dmv}{dt}}}$, the answer to the first part is clearly 0.6N. For the second, we see that 0.6N is applied over 3m every second, so the rate of work is 1.8W. For the third, there's 0.2kg of new mass every second, which will add ${\displaystyle {\frac {1}{2}}(0.2)3^{2}}$ of kinetic energy every second, i.e. 0.9W. I can't for the life of me explain 4, though---where the extra energy every second is going, if not being stored as kinetic energy. I've considered examining the centres-of-mass (consider the 0.2kg being dropped at one point - it takes 0.9J of work to spread that out over 3m, and another 0.9J to accelerate it to 3m/s), but this falls down because you get the spreading for "free" (I think). This isn't homework. Any help definitely appreciated. --83.147.171.12 10:57, 20 April 2006 (UTC) Your problem is in the answer to question two. For easier visualization, assume the sand falls in large clumps of 0.2 kg a piece, one clump falling each second. The acceleration from 0 to 3 m/s will not be instantaneous, let's assume the force on the belt is constant, so each sandclump accelerates to 3 m/s in exactly one second. In this time it will have moved 0.5*3m/s2*(1s)2=1.5 m. So the work applied is only 0.6N * 1.5m = 0.9W, similar to the answer in 3. The answer remains the same if you let the sand fall continously and let the acceleration become near-instantaneous, but I will let you work that out for yourself. Rasmus (talk) 12:46, 20 April 2006 (UTC) That's what I thought, too, except the answers provided with the problem set correspond to those I've given for the first three questions (there was no answer given for the fourth). And it's not like it's physically impossible to have the conveyor belt instantaneously (or as close to it as makes no difference) each grain of sand to 3m/s. At worst, the acceleration happens over a very short time. Thus it does travel 3m, or very close to it. So I don't think this is the answer. --83.147.171.12 14:12, 20 April 2006 (UTC) If a grain of mass m accelerates to 3m/s in a short time, Δt, assuming constant acceleration, the acceleration will be (3 m/s)/Δt and the force applied during this time will be m*(3 m/s)/Δt. During the acceleration it will have moved 0.5*(3m/s/Δt)*(Δt)2 = 1.5 m/s * Δt. So the work applied during this time is m*(3 m/s)/Δt * 1.5 m/s * Δt = m * 4.5 m2/s2. Each second we accelerate 0.2kg/m grains, so the work done each second is 0.2 kg * 4.5 m2/s2 = 0.9J. Feel free to let Δt approach zero :-) Rasmus (talk) 14:39, 20 April 2006 (UTC) The key here is interpreting question 2: Work is being done by which force? There are three forces working here: 1) Friction on the sand ((which accelerates it), 2) friction on the belt (which attemps to decelerate it), and 3) a force done on the belt by an engine (or whatever) to keep it moving at a constant speed. 1 and 2 are action-reaction, so are both 0.6N. 1 does a power of 0.9W, as calculated by Rasmus. 2 does a power of -1.8W - it's working on the belt, moving at a constant 3 m/s. 3, to keep the constant speed, does a power of 1.8W. The question probably refers to force 3, explaining the 1.8W. and for #4 - the remaining 0.9W is wasted as heat. Every time force acts between bodies moving relative to each other (the sand and belt in this case), energy is lost as heat, proportional to the force (0.6N here) and relative velocity (average 1.5 m/s here). Thus 0.9W is wasted. This is similar to what happens when you try to accelrate your car with the clutch half-way. -- Meni Rosenfeld (talk) 15:17, 21 April 2006 (UTC) ## factorize integers in polynomial time? Do we have an article that covers how to factorize integers in polynomial time, or whether this is possible, or the repercussions on the entire computer security systems we use if this mathematical breakthrough happened, or if somebody is attempting to solve this problem?--Sonjaaa 14:36, 20 April 2006 (UTC) The most relevant articles are probably Integer factorization, Shor's algorithm and Complexity classes P and NP. I expect security systems will migrate to Elliptic curve cryptography (or perhaps a cryptosystem based on yet another hard problem) relatively quickly once integer factorization of 2048 bit numbers becomes computionably feasible (whether classicly or using quantum algorithms). Most of the standards for the infrastructure already allows many other cryptosystems than RSA. Rasmus (talk) 14:51, 20 April 2006 (UTC) To the best of my knowledge, no published algorithm factors integers in polynomial time, despite deep, sustained, and well-funded interest for cryptography and other reasons. In academia, the existence of such a breakthough would be big news; in other circles, a big secret. What Ralph Merkle, Whitfield Diffie, and Martin Hellman described ("one-way trap-door functions"), the basis for public key cryptography, could take advantage of many different alternatives, so long as one way is easy and the other hard. So theoretically, we could simply move on to a new choice. Practically, there would be a span of time during which many secrets would be vulnerable prematurely. People who take cryptography seriously plan conservatively, but the average consumer of cryptography chooses passwords that can be guessed with a dictionary or a little personal information. Fast integer factorization is not the biggest risk! --KSmrqT 15:18, 20 April 2006 (UTC) The links above do an excellent job discussing the problem. I can say with confidence that people are attempting (and have attempted) to solve this problem. As those articles state, there is no publically known algorithm that factors integers in polynomial time. However, many of the underlying assumptions in cryptography are related but not always equivalent to integer factoring, e.g. RSA problem. It should be stressed that integer factorization is sufficient to break RSA but not necessary, i.e. RSA may be insecure while integer factorization remains hard. You might also browse through the Discrete logarithm for another number-theoretic problem used in cryptography. ## ID Card size What is the logic behind the size of ID Card as 85.6 mm x 53.98 mm ? —Preceding unsigned comment added by 202.70.64.15 (talkcontribs) It is an international standard, the size is defined as ID-1 in ISO 7810. What the logic is behind standardizing that size, I'm not sure. UkPaolo/talk 16:15, 20 April 2006 (UTC) Divide 85.60 by 53.98 to give 1.585 this is close to ln(3)/ln(2) which is related to the Sierpinski triangle. --Salix alba (talk) 18:54, 20 April 2006 (UTC) Thank You both but i meant to ask why that size is choosen, for example the A series paper size is obtained from an A0 size paper with an area of 1 sq.m. and aspect ratio of 2^0.5. There may be no reasoning behind choosing the dimensions as above. Not all the ISO measurements need to have such a reasoning behind it. Dysprosia 05:03, 23 April 2006 (UTC) The ratio of the sides is related to the golden mean', thought by the Greeks to be the most pleasing ratio of sides in a rectangle See http://goldennumber.net/creditcard.htm ## Vector & set size nomenclature A quick question regarding vectors and sets. Is the length of a vector ${\displaystyle \mathbf {a} }$ written ${\displaystyle |\mathbf {a} |}$ or ${\displaystyle ||\mathbf {a} ||}$, or can it be either? Which is most commonly used? What about a set ${\displaystyle b}$, does that have size ${\displaystyle |b|}$ or ${\displaystyle ||b||}$ (or either)? I'd be grateful if someone could clarify this for me! UkPaolo/talk 18:20, 20 April 2006 (UTC) For a vector, usually the latter, but sometimes the former. For a set, the former. I've never seen the latter. (But then I voted for ruler and compass over compass and straightedge.) — Arthur Rubin | (talk) 19:00, 20 April 2006 (UTC) I agree with Arthur. -lethe talk + 19:37, 20 April 2006 (UTC) I agree with Arthur as well, but with a little clarification: I have mostly seen ${\displaystyle |\mathbf {a} |}$ used for the length of vectors in ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle \mathbb {C} ^{n}}$ as a generalisation of absolute value in ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$, particularly in undergraduate university courses, while ${\displaystyle ||\mathbf {a} ||}$ is more used once you're looking at more general normed vector spaces. Confusing Manifestation 19:55, 20 April 2006 (UTC) The single bar notation, |·|, is used for absolute value of a real or complex number. It is also used for many situations where a size or count is needed, such as the size of a set or group. It may also denote the determinant of a matrix. The double bar notation, ‖·‖, usually denotes a norm, with the Euclidean norm being the common length of a vector. It is used frequently in linear algebra and matrix theory, but not much elsewhere. When necessary, a subscript is used to indicate which norm is desired; for example, ‖MF is the Frobenius norm of matrix M. To put both notations to work, we can define the Lp norm of a vector x = (x1,…xn) as ${\displaystyle ||\mathbf {x} ||_{p}=\left(\sum _{i=1}^{n}|x_{i}|^{p}\right)^{\frac {1}{p}}.\,\!}$ The Euclidean norm is the case p = 2, implied if there is no subscript. --KSmrqT 20:03, 20 April 2006 (UTC) Also, when there are matrix norms involved, single bar notation is used for vector norms and double bar for matrix norms. Grue 07:41, 21 April 2006 (UTC) Maybe it should be mentioned that ${\displaystyle \|\mathbf {a} \|}$ (\|) looks better than ${\displaystyle ||\mathbf {a} ||}$ (||).--gwaihir 08:20, 21 April 2006 (UTC) Thanks to you all for having taken the time to reply! That clarifies things a lot :o) UkPaolo/talk 09:30, 21 April 2006 (UTC) On the other hand, the phyics books (at least those I read) tend to use bold text (or bold italics) for vectors and italics for their lenghts, like r2 = r2. Conscious 19:19, 26 April 2006 (UTC) ## Using a program without inserting the CD every time I have several computer programs that I installed on my computer - but every time I try to use these programs it say I have to have the CD in the drive before it will run. Is there a way to completely install these CD's and then point the program to the full install rather than the CD drive? Thanks for you help in advance Maria For individual programs, there are often no-CD cracks available. Use of these cracks may not be legal in all countries, to obtain the crack may require trawling through distasteful websites, and installation may require some technical knowledge. The cracks may be infected with viruses or spyware. A more general solution involved copying the entire CD or DVD to your hard drive and mounting the image in a disk image emulator. The article on that subject gives several examples. There are many such programs; some are commercial, others are free. They don't work with all programs as some programs specifically detect such disk image emulators and refuse to run. Sometimes you can load a patch to the emulator to avoid such detection... it can get complicated. I don't know about the legality of running such an emulator; I'm not a lawyer but I can't see why it would be illegal. Please read the two articles I've linked to above, but if you want further help you may be better finding a forum devoted to discussing such matters, since I don't think we can provide detailed advice from this reference desk page.-gadfium 00:31, 21 April 2006 (UTC) This happens for two reasons. (1) An astronomy program, say, with a huge catalog of known stars may keep hundreds of megabytes on a CD rather than encumber a hard drive. These days, hard drives are cheaply available in sizes much larger than CDs, so there's less reason to do this. (2) The author may be concerned about software piracy, and read special hidden information on the CD as an attempt to be sure this is an authentic original version of the program. Such steps make it difficult to create even a legal backup of purchased software, something most parents would like to do with their children's games. In this case, a quiet war is escalating between the copy protection schemes and methods to circumvent them. The article on copy protection may be helpful. --KSmrqT 01:35, 21 April 2006 (UTC) # April 21 ## RSA and prime numbers Please can someone explain to a non-mathematician why prime numbers are needed to create keys for RSA cryptosystems. Wouldn't any sufficiently large number be o.k.? There are two reasons. 1. Security. It is easier to factorize a number that is a product of many smaller numbers than a number that is only a product of two large ones (at least with some methods). There are also other attacks on RSA that work better if the modulus has small factors. 2. Practical reason: Some of the mathematics involved in the cryptosystem relies on the numbers being prime. Basically, if p and q aren't prime you can't rely on the decrypt-operation being the inverse of the encrypt-operation. For a workable example, let p=3, q=8. Then the modulus is pq=24 and the totient is (p-1)(q-1)=16. 3 is relatively prime to 16, so we can choose that as the public key and 11 is the inverse modulo 16, so that will be our private key. Let us transmit the message "5". We encrypt: 5^3(mod 24)=5, and decrypt: 13^11(mod 24)=5. That worked! But now let us transmit "4": Encryption: 4^3 (mod 24)=16, decryption:16^11(mod 24)=16. That didn't work! You can generalize the RSA-system to using more than two primes (which is "sort of" the same as just using arbitrary large numbers). That is called multi-prime RSA. Rasmus (talk) 12:53, 21 April 2006 (UTC) The core idea behind the RSA version of public key cryptography are that we have a public key/private key pair based on prime factorization. Previous cryptography depended on a strictly private key, which caused a practical problem: how to securely transmit the key to wherever it was needed. The ability to have public keys revolutionized cryptography. Number theory is used to construct a pair such that the public key can be used to encrypt, the private key can be used to decrypt, but knowing the public key does not make it possible to decrypt nor to discover the private key. One piece of the public key is the product of two large prime numbers. It is relatively easy to multiply p and q to get their product n; but if p, q, and n are sufficiently large, it is extremely timeconsuming to factor n to recover p and q. The technical details of the encryption and decryption don't actually require n to be a product of two large primes, but the security of the overall scheme relies upon it. Any integer n greater than 1 is a product of some number of prime factors. The methods used to find the factors typically find small factors quickly. So a very large n like 2187250724783011924372502227117621365353169430893212436425770606409952999199375923223513177023053824, even though it has 100 digits, can be factored almost instantly, because it is a product of 2s. But the number 2269868715257204734612142004082495122481735606947863309410992763455491328649178167436710530263617567, which also has 100 digits, is the product of two random 50-digit primes, and considerably harder to factor. --KSmrqT 20:22, 21 April 2006 (UTC) While the fact that the first number equals 2330 is obvious to a computer, the point might be made even clearer by choosing an example that is more easily factored by humans, such as, say, 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000. That's also a 100-digit number, but I doubt anyone here has any trouble factoring that. —Ilmari Karonen (talk) 22:48, 21 April 2006 (UTC) True, except for the Lewis Carroll problem: • "Can you do Addition?" the White Queen asked. "What's one and one and one and one and one and one and one and one and one and one?" • "I don't know," said Alice. "I lost count." • "She can't do Addition," the Red Queen interrupted. It's a quotation that always makes me smile. :-) --KSmrqT 00:43, 22 April 2006 (UTC) Thanks for your help guys, I still don't think I understand. Probably because I stopped learning maths aged 14 (now 29). Please can you recommend a good primer on this?EmperorMoo 20:53, 21 April 2006 (UTC) This is the good primer, along with the RSA article and its links. Can you condense your confusion into a clear followup question? --KSmrqT 00:33, 22 April 2006 (UTC) Thanks everyone, I think I'm going to have to read more maths background before I'm able to understand. EmperorMoo 12:48, 23 April 2006 (UTC) One thing I think that wasn't obvious in the above explanations that I wish to add is that somebody who knows all the prime factors of n (one part of the public key) can easily calculate the private key, but the cipher works with just knowing n and not the factors. This is why you have to make n hard to factor and also why you need large primes. Jeltz talk 14:54, 24 April 2006 (UTC) And perhaps it is because RSA use modular arithmetic and rely on some very slightly advanced theorem in this field, for example, fermat's little theorem. You may want to have an in-depth understanding of this before thinking about RSA. Try searching on wikibooks for modular arithmetic also. --Lemontea 02:49, 28 April 2006 (UTC) ## variational methods O genius Wikipedians, I pray to thee to give a helping hand to a fellow bemused user. This is a homework question, I apologize - but I'm not asking for the answer (as it is given), but more so the best way about going about the question in order to get the answer. If possible pls put this into proper Maths syntax, and give me a link to how to write in maths syntax: A person wishes to walk from (0,0) to (1,1) in the x-y plane in the shortest possible time. The nature of the ground is such that the max. speed of walking is given by "v=v0(1 + αy)" where v0 and α are constants. Show that the path taken is the integral between 0 and 1 of √(1 + (y′)2) / (v0(1 + αy)). The next bit of the question is "Show that the path taken should be an arc of a circle with centre on the line y = −1/α". Please give me a nudge in the right direction. All help is completely appreciated. --Dangherous 13:31, 21 April 2006 (UTC) You want to minimize the time. So given a path, let's find its time. If you know velocity and distance, then time = distance/velocity. For a curved path, this becomes ${\displaystyle dt={\frac {ds}{v}}.}$ Then the total time is ${\displaystyle T=\int {\frac {\sqrt {1+y'^{2}}}{v_{0}(1+\alpha y)}}\,dx}$ simply by substituting the arc length formula and the velocity. To solve for the path, write the Euler-Lagrange equations for this quantity. For help on marking up math formulas, see meta:Help:Formula. -lethe talk + 14:46, 21 April 2006 (UTC) # April 22 ## Finding a given mean I am having trouble with a specific problem from my statistics homework. I'd appreciate any help given. Suppose you own a basketball team. Suppose also that in the NBA the average salary of players is$5,000,000. The lowest salary you can offer is $385,000. You plan to sign 15 players. Construct a set of 15 data elements that have a mean of$5,000,000 spending the least amount of money possible.

Is there a formulaic way to solve such a problem? Or is the only way to fiddle around with various numbers? --Impaciente 02:25, 22 April 2006 (UTC)

Is that really the exact statement of the problem? It reads as if you always have to spend exactly $75,000,000 (to get a mean of 5 million among 15 people) but have many ways of doing so, the easiest being giving 5 million to each. Kusma (討論) 02:46, 22 April 2006 (UTC) (after edit conflict) You want 15 numbers, none less than 385,000, with a mean of 5 million. I don't understand the bit about spending the least amount of money possible, since the total will always be 15*5 million = 75 million. To make it really simple, you could have 15 numbers, each exactly 5 million, or you could offer 1 million to each of 14 players, and 75-14 million = 61 million to one player.-gadfium 02:48, 22 April 2006 (UTC) Yes, that was the wording of the problem. Come to think of it, it sounds sort of self-explanatory. But thanks a lot, to both of you. For future reference, anytime that a mean is given, the sum of the elements is also (implicitly) given as well? --Impaciente 02:54, 22 April 2006 (UTC) So long as you know the number of elements, yes.-gadfium 03:34, 22 April 2006 (UTC) ## Geometric Naming Conventions What does the "kis" particle mean in the names of the following geometric figures (and others)? I've noticed that all the figures in question have triangular faces, but I don't understand what the number (tria, tetra etc) represents. --72.140.146.246 03:34, 22 April 2006 (UTC) They're from the Greek words for "three times", "four times", etc. In an n-akis polyhedron, each of the faces is replaced by n congruent isoceles triangles. For example, a pentakis dodecahedron is a dodecahedron with each pentagonal face replaced by five triangles. —Keenan Pepper 04:16, 22 April 2006 (UTC) The Greek triakis means thrice, triple, three times, 3-fold; similarly for the others. Consider the triakis octahedron; each triangular face of the original octahedron is multiplied by raising a center point and creating a face for each edge, three altogether. Thus the triakis octahedron is a "tripled" octahedron, with 24 faces instead of 8. Similarly, the pentakis dodecahedron multiplies each original pentagonal face of a dodecahedron by raising a center point and creating five new faces. With so many polyhedra to name, it's a challenge inventing systematic, meaningful, and distinct choices, nevermind memorable! --KSmrqT 05:06, 22 April 2006 (UTC) Thank you. This is exactly what I wanted to know. But what about "disdyakis"? --72.140.146.246 12:55, 22 April 2006 (UTC) I believe that's "double double", but check the articles for confirmation. --KSmrqT 22:33, 22 April 2006 (UTC) Thanks. I did some investigation, and found that the disdyakis dodecahedron can be formed from a rhombic dodecahedron by quartering each face and raising the center point (as tetrakis). It was the rhombic part that got me. --72.140.146.246 20:56, 24 April 2006 (UTC) See http://en.wikipedia.org/wiki?title=Talk:Catalan_solid and http://en.wikipedia.org/wiki/Conway_polyhedron_notation ## possiblity of 10 events occurring (No question.) The probability of ten independent events with identical probability occurring is P^10, where P is the probability (from 0.0 to 1.0) of each individual event occurring. If the probability of each event is different, multiply all those probabilities to get the answer. StuRat 20:42, 22 April 2006 (UTC) That's assuming the events are independent of eachother. --BluePlatypus 20:57, 22 April 2006 (UTC) Hence my inclusion of the word "independent", LOL. StuRat 22:10, 22 April 2006 (UTC) Try Poisson distribution. linas 00:28, 23 April 2006 (UTC) The question is, may events be independent of each other? Luthinya 16:57, 29 April 2006 (UTC) # April 23 ## Physics: A Dynamics Problem A 20 kg box rests on a table. What is the weight of the box and the normal force acting on it? b) A 10 kg box is placed on top of the 20 kg box. Determine the normal force that the table exerts on the 20 kg box and the normal force that the 20 kg box exerts on the 10 kg box. First I did: a)  W = mg W = 20(9.81) W = 196.2 N (DOWN)   ΣFy = ma  Since there is no movement in the vertical direction, Fy = 0  FN - FG = 0 FN = FG FN = 196.2 N (UP)  This part was easy, but my question concerns the second part. I did: b)  ΣFy = ma  This equals zero again...and this where I'm unsure (FN2 = 10 kg object, FN1 20kg object)  FN2? + FN1 - FG = 0 FN1 = FG - FN2 FN1 = 9.91(20) - 9.81(10) FN1 = 98.1 N (UP)  And the second part of part b...  ΣF = ma FN - FG = 0 FN = FG FN = 10(9.81) FN = 98.1 N (UP)  I'm not sure about this, that's what I have up there so if anyone can point me in the correct direction or explain to me why I'm right (I think I am, but I am not sure, and I need to be sure of the concept) it would be greatly appreciated. Thanks C-c-c-c 04:02, 23 April 2006 (UTC) For the second part, I assume you meant 98.1 N, which is correct. For the first part, there are two forces pushing the 20 kg box down, its own weight, and the weight of the box on top of it. The normal force from the table must cancel both of those. You can think of it as if the two together were a single 30 kg box. —Keenan Pepper 04:18, 23 April 2006 (UTC) Right, I changed that. So, wait, have I done it right? I'm starting to think I haven't since I didn't include the FG for the 10kg box in the first part of question b. Instead should it be:  ΣFy = ma FN1 + FN2 - FG1 -FG2 = 0 FN1 = FG1 + FG2 - FN2 FN1 = 9.81(20) + 9.81(10) - 9.81(10) FN1 = 196.2N (UP)  This doesn't seem right either, since it's the same as up top.... Help, utterly confused! C-c-c-c 04:26, 23 April 2006 (UTC) FN2 is acting on the 10 kg box, not the 20 kg box. —Keenan Pepper 04:38, 23 April 2006 (UTC) Isn't that what I have there? I have FN2 = 9.81(10) C-c-c-c 04:44, 23 April 2006 (UTC) Yes, but you have FN1 + FN2 - FG1 - FG2 = 0. Why should the sum of those forces be zero if they are not all acting on the same object? —Keenan Pepper 04:53, 23 April 2006 (UTC) But isn't that an entire system, can't you do it like that too? C-c-c-c 04:55, 23 April 2006 (UTC) If you're considering both boxes together as a 30 kg object, then the weight is 30 kg * 9.8 m/s2 and you wouldn't include the normal force FN2 because that's internal to the object. The top box is pushing down just as hard as the bottom box is pushing up. —Keenan Pepper 05:47, 23 April 2006 (UTC) Oh, I didn't know you didn't include internal forces. Okay that makes sense, thank you so much for you taking the time to going back and forth and helping me, I appreciate it! Thanks a bunch again. C-c-c-c 06:08, 23 April 2006 (UTC) ## ISBN Checksum question After explaining the process of determining the checksum of an ISBN, the ISBN article says, "Since 11 is a prime number, this scheme ensures that a single error (in the form of an altered digit or two transposed digits) can always be detected." How does 11's being prime affect its usefulness for checksumming? 139.55.22.138 19:54, 23 April 2006 (UTC) If the modulus were a composite number, then some multipliers could be factors of the modulus, in which case some single-digit errors would not be detected. For example, if the modulus were 10 and the 5th digit were changed by 2, then that digit's contribution to the checksum would be changed by 2 * 5 = 10, so the total checksum would be the same. If the modulus is prime, that guarantees that none of the multipliers are factors. —Keenan Pepper 20:04, 23 April 2006 (UTC) ## Stochastic Processes How can I compute E[Max(X,Y)] (Mathematical Expectation)?! ‍‍‍‍‍Armandeh 22:07, 23 April 2006 (UTC) Given what? —Keenan Pepper 15:32, 24 April 2006 (UTC) With some assumptions, it is ${\displaystyle \int _{-\infty }^{\infty }t(f(t)G(t)+g(t)F(t))dt}$ -- Meni Rosenfeld (talk) 18:27, 24 April 2006 (UTC) Just to clarify and fix typos (hoping that I don't offend by overwriting Keenan's edit-conflicting post that asked for them): the lowercase letters are probability density functions, and the uppercase letters are cumulative distribution functions. f/F go with X, and g/G with Y (or vice versa, of course). The assumptions are that X and Y are real and independent, unless I'm mistaken and you need more than those. Of course, the stuff in parentheses there is useful outside this formula as well, since it's just the PDF of ${\displaystyle T:=\max(X,Y)}$. If X and Y are codependent, you have to consider a PDF ${\displaystyle f(x,y)}$, and then you can calculate ${\displaystyle E[T]=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }t(x,y)f(x,y)\,dy\,dx}$. To get a more fundamental result -- the PDF for T in this case -- note that ${\displaystyle \max(x,y)=t{\mbox{ iff }}x=t\land y\leq t\lor y=t\land x\leq t}$, so (neglecting infinitessimal probabilities that will clearly contribute nothing) ${\displaystyle h(t):=P(T\in [t,t+dt])=P(x\in [t,t+dt])P(y. For the case of ${\displaystyle f(x,y)=f(x)g(y)}$, one recovers the ${\displaystyle f(t)G(t)+g(t)F(t)}$ above. --Tardis 21:18, 24 April 2006 (UTC) A good assumption to add is that f and g exist to begin with, that is, F and G are differentiable. Depending on how strict our definition of the integral is, we can weaken this condition somewhat. -- Meni Rosenfeld (talk) 15:07, 26 April 2006 (UTC) # April 24 ## Vinogradov's version of the big O notation in TEX Does anyone know what the control sequence for Vinogradov's version of the big O notation is? The closest I can find is ${\displaystyle \ll }$, but the right "<" should be cut off where the left one ends. Often found in books on analytic number theory. Mon4 00:16, 24 April 2006 (UTC) I think it is \lll as defined in the mathabx package. Unfortunately, the AMS packages also define \lll but for a different symbol (the same as ${\displaystyle \ll }$ but with three smaller-than symbols), so you'll need to do something special if you want use both packages. -- Jitse Niesen (talk) 02:05, 24 April 2006 (UTC) If ${\displaystyle <\!\!\!<\!\!\!<}$'s not the exact symbol you want, look at how \lll is defined in the mathabx package and if it's an appropriate definition, copy it over to your TeX file. Dysprosia 02:36, 24 April 2006 (UTC) Perhaps the Unicode character "" (U+2AA1, LessLess, [DOUBLE NESTED LESS-THAN]) is what you seek. (Shown here at twice life size.) Your fonts may not include higher code points like this, so here's a link with an image. Of course, convincing your TeX to use it is another matter. --KSmrqT 04:21, 24 April 2006 (UTC) This is the symbol I'm looking for. Indeed, it is not properly displayed in my browser. What font does render this character correctly? And, does anybody have an idea on how to use it on Wikipedia within a [itex] formula? Mon4 11:18, 24 April 2006 (UTC) As it's not exactly standard notation, you probably shouldn't use it on Wikipedia. Use ${\displaystyle f\in O(g(n))}$ instead. Dysprosia 14:26, 24 April 2006 (UTC) It isn't non-standard but rather archaic. It used to be used common (Hardy and Wright use it in their number theory textbook for example) and is still used by some authors. However, Big-O notation is more common now (at least in math written in English) Dysprosia is correct. JoshuaZ 14:34, 24 April 2006 (UTC) I have seen ${\displaystyle \ll }$ being used instead of the little o notation, I guess it is a matter of choice where the right < ends (similar to ${\displaystyle \rho /\varrho }$).--gwaihir 08:06, 24 April 2006 (UTC) ## Programming Languages Hello. I'm a beginning programmer. Two years ago, I tried to learn Visual C++ .NET. That was really, really hard. I wasn't able to do anything besides the tutorial in the book, so I quit. About four months ago, I picked up Liberty BASIC. I am able to do a lot and understand the language. I really enjoy it and am doing a lot of fun stuff. I wondered if anyone had any suggestions on other languages to try next or any suggestions about a good progression of languages for a learning programmer. Any stories about what you did, what you wish you did, or just simply any advice you have would be great. Thanks for your help. --Think Fast 01:23, 24 April 2006 (UTC) Python. It's simple but powerful, it runs on all major platforms, and there's good free documentation at http://docs.python.org/Keenan Pepper 02:04, 24 April 2006 (UTC) Shift attention from programming languages to algorithms and data structures. A warmly recommended text is Cormen, Leiserson, Rivest, Stein, Introduction to Algorithms, 2/e, ISBN 0262032937. --KSmrqT 09:24, 24 April 2006 (UTC) A lot depends on what you want to do. Different programing languages are better at different tasks. Do you want to write Microsoft Windows programs with graphical user interfaces try revisiting Visual C++ you may understand it better, now. Java is not too bad (debatable). Web programming Perl, Python, PHP all do good job on the server side, javascript on the client side. Wikipedia is done using PHP. Ajax is the current buzzword. Databases MySQL a very different way of thinking about programming, very useful skill. In terms of your programming development getting some Object-oriented programming under your belt is a good idea. You might like to look at Slashdot where the pros and cons of different languages are endlessly debated. --Salix alba (talk) 10:12, 24 April 2006 (UTC) ## Oil API Gravity Can you please explain how this formula was formulated? (141.5/SG at 60degF)-131.5 this formual is used to calculate the API gravity of cude oil. thanks From the API gravity article, the formula appears to be an arbitrary definition. Its motivation is due to a historical error in measuring instruments. Melchoir 23:56, 24 April 2006 (UTC) thank you Melchoir....... anyone has more?? ## variational methods revisisted O genius Wikipedians, I pray to thee to give a helping hand to a fellow bemused user. This is a homework question, I apologize - but I'm not asking for the answer (as it is given), but more so the best way about going about the question in order to get the answer. If possible pls put this into proper Maths syntax, and give me a link to how to write in maths syntax: A person wishes to walk from (0,0) to (1,1) in the x-y plane in the shortest possible time. The nature of the ground is such that the max. speed of walking is given by "v=v0(1 + αy)" where v0 and α are constants. Show that the path taken is the integral between 0 and 1 of √(1 + (y′)2) / (v0(1 + αy)). The next bit of the question is "Show that the path taken should be an arc of a circle with centre on the line y = −1/α". Please give me a nudge in the right direction. All help is completely appreciated. --Dangherous 13:31, 21 April 2006 (UTC) You want to minimize the time. So given a path, let's find its time. If you know velocity and distance, then time = distance/velocity. For a curved path, this becomes ${\displaystyle dt={\frac {ds}{v}}.}$ Then the total time is ${\displaystyle T=\int {\frac {\sqrt {1+y'^{2}}}{v_{0}(1+\alpha y)}}\,dx}$ simply by substituting the arc length formula and the velocity. To solve for the path, write the Euler-Lagrange equations for this quantity. For help on marking up math formulas, see meta:Help:Formula. -lethe talk + 14:46, 21 April 2006 (UTC) Following on from this, how would I show that the path taken should be an arc of a circle with centre on the line y = -1/α? When I've tried this, I've done about 5 pages of working out, which is far too much I'm sure. What's the quickest way to get to the answer? Thanks awfully. --Dangherous 11:26, 24 April 2006 (UTC) The Euler-Lagrange equation for ${\displaystyle y(x)}$ is ${\displaystyle -{\frac {\alpha {\sqrt {1+y^{\prime 2}}}}{v_{0}(1+\alpha y)^{2}}}-{\frac {d}{dx}}{\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}=0}$ with the boundary conditions ${\displaystyle y(0)=0,y(1)=1}$. Multiply by ${\displaystyle {\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}}$, ${\displaystyle -{\frac {\alpha y'}{v_{0}^{2}(1+\alpha y)^{3}}}-{\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}{\frac {d}{dx}}{\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}}$ ${\displaystyle {\frac {d}{dx}}{\frac {1}{v_{0}^{2}(1+\alpha y)^{2}}}-{\frac {d}{dx}}\left({\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}\right)^{2}=0}$ Integrate, ${\displaystyle {\frac {1}{v_{0}^{2}(1+\alpha y)^{2}}}-\left({\frac {y'}{{\sqrt {1+y^{\prime 2}}}v_{0}(1+\alpha y)}}\right)^{2}=C}$ Can anyone check my math and finish it from here? (Igny 12:59, 24 April 2006 (UTC)) I looked through your steps and didn't see any mistakes. The resulting equation looks depressingly unlike the equation for a circle, though I'm not up for wading into it at the moment. -lethe talk + 01:21, 25 April 2006 (UTC) Thanks for checking the steps so far. The previous equation is equivalent to ${\displaystyle {\frac {y^{\prime 2}}{1+y^{\prime 2}}}+c(1/\alpha +y)^{2})=1}$ where ${\displaystyle c=C\alpha ^{2}v_{0}^{2}}$. Then solving it for ${\displaystyle y'}$ results in ${\displaystyle {\frac {{\sqrt {c}}(1/\alpha +y)y'}{\sqrt {1-c(1/\alpha +y)^{2}}}}=1}$ Denote ${\displaystyle u=1-c(1/\alpha +y)^{2}}$ and get ${\displaystyle {\frac {u'}{2{\sqrt {cu}}}}=1}$ Integrate, ${\displaystyle -{\frac {\sqrt {u}}{\sqrt {c}}}=x+c_{1}}$, therefore ${\displaystyle 1-c(1/\alpha +y)^{2}=c(x+c_{1})^{2}}$ This looks like a circle. The unknown constants can be found from the boundary conditions on y. The center is on the line ${\displaystyle y=-1/\alpha }$. Again I am not 100% sure I got all the steps right.(Igny 02:45, 25 April 2006 (UTC)) ## Latex to Microsoft Equation Editor I have a huge number of equations in Latex, and I need to make a presentation in Powerpoint. Is there a way to convert equations from Latex to the equation editor format used in Powerpoint? I seriously hope there is! Thanks :) deeptrivia (talk) 14:06, 24 April 2006 (UTC) PS: I know about Texpoint, but it doesn't seem to work for me. deeptrivia (talk) 14:10, 24 April 2006 (UTC) Why don't you take screenshots and snip out the images of equations? Or perhaps use Texvc to possibly do the job? Equation Editor is quite intensely horrible and cannot handle everything that TeX can, so you're bound to run into problems. Dysprosia 14:23, 24 April 2006 (UTC) A great way to make presentations from Latex is Ppower4 (see http://www-sp.iti.informatik.tu-darmstadt.de/software/ppower4/) They also look a lot like Powerpoint. Texpoint isn't working for me either. It works for my advisor, so I am confused. moink 02:17, 25 April 2006 (UTC) Another approach is to use TeX4ht to convert to OpenOffice format (which stores the equations in MathML). OpenOffice can then export the document in MicroSoft formats, although it also includes a presentation package of its own (Impress). -- Avenue 10:32, 26 April 2006 (UTC) ## Simultaneous equation Solve this simultaneous equation? Find the value for a + t if: 4a=8t-12 a=9-t You may use this equation to help you: a=2t-3 easy then, since a=2t-3 and a=9-t, then 2t-3=9-t, so 3t = 12, so t= 12/3 = 4. with this knowledge, 4a=8t-12=36-12=24, so a=24/4 =6 thus t=4, a=6. hope this helps! UkPaolo/talk 21:14, 24 April 2006 (UTC) This reeks of being a homework question, about which the directions at the top of this page have comments both the asker and the answerer should read. --KSmrqT 21:26, 24 April 2006 (UTC) Yes, it does seem like a homework question. I have read the directions at the top of this page, but I was assuming good faith and helping nonetheless, since simultaneous equations was clearly something the poster was having difficulty with, and I thought a worked example might help. UkPaolo/talk 07:34, 25 April 2006 (UTC) Um, 8*4 isn't 36, so the answerer technically didn't help with the homework. Of course, the silly thing is that the last equation is simply the first equation divided by 4; why is it even given as part of the homework? --Tardis 21:30, 24 April 2006 (UTC) Yes, I made a mistake, as you pointed out 8*4 isn't 36, and a is thus in fact 5 (as could have been worked out easier as a=9-t, so a=9-4 =5). Still, you got the idea... (and also learnt a valuable lesson to always check stuff!) UkPaolo/talk 07:34, 25 April 2006 (UTC) The web can solve it for you. http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/SysEq.htm Gauss-Jordan_elimination or System_of_linear_equations Ohanian 23:24, 24 April 2006 (UTC) It pays to actually read the question. It asked to solve the equation not for "a", or "t", but for "a+t". That can be simply obtained by ignoring the first equation and looking at the second. If a=9-t, then a+t = 9. QED. If you were asked this question in an exam, and then went off on a tangent working out the individual values of a and t, you would be marked down (and particularly if you worked out that a=6 when it actually equals 5.) JackofOz 01:13, 25 April 2006 (UTC) What kind of stupid exam question would ask for a+t? And if you know a and you know t then logically you would know a+t, it sounds like an exam question purely for the sake of an exam question without any real use in real life. Ohanian 03:09, 25 April 2006 (UTC) Exam questions aren't motivated by real life? Noooooo! I... I won't believe it! Let me out! Melchoir 03:12, 25 April 2006 (UTC) One that was testing your ability to identify which bits of information are important and which unimportant. Which does make more sense of the "you may use the fact that you can divide the first equation by four to help you" bit, if it's just to obfuscate the fact that the first equation is completely irrelevant. --Bth 07:27, 25 April 2006 (UTC) Heh. Actually, I presumed that the poster just used + in place of an &, since it seemed unlikely that you'd ever want to solve the equations for a+t. Incidentally, I don't know about elsewhere in the world, but in British exams you certainly wouldn't loose marks if you worked out (correctly) the individual values of a and t and then added them, even if there was an easier way to finding the answer. UkPaolo/talk 07:34, 25 April 2006 (UTC) The poster said "the value for a+t", not "the values of a+t". And because this was a mathematical question, using a + sign if they really meant "and" would be obviously very ambiguous, so I discount that possibility. I think he definitely wanted one answer, not two. The use of the term "simultaneous equation" was a red herring, probably deliberately put there to see if the student was awake.JackofOz 09:07, 25 April 2006 (UTC) Seems the British educational system has gone the same way as other western systems - deskilling students about answering the question that was asked, rather than what they assume the questioner must have meant. JackofOz 09:07, 25 April 2006 (UTC) How so? If it asks for a+t and you answer 9, correctly, you should surely get full marks howsoever you achieved that answer. UkPaolo/talk 09:27, 25 April 2006 (UTC) Maybe I've just become a crusty, hard-hearted curmudgeon (not the Nirvana one, the other one). JackofOz 09:37, 25 April 2006 (UTC) Note that even if the student isn't marked down for doing a problem "the hard way", it still takes longer that way so could cost them on a timed test. StuRat 12:54, 25 April 2006 (UTC) • I don't see why asking for a+t has any less real-life use than asking for a and t. There can very well be real-life problems where you have equations describing a and t, and what you ultimately need to solve the problem is a+t. • Asking for a+t or the like is probably characteristic of SAT-style exams (though I don't know about SAT itself, only Israel's psychometric exam), in an attmpt to trick the student. Of course, failing to notice the object of the question in such multiple-choice exams is deadliest. -- Meni Rosenfeld (talk) 15:26, 26 April 2006 (UTC) # April 25 ## What is this surface? Does the surface with equation z = xy have a name? I'm curious because it looks a little like a hyperbolic paraboloid, and the equation is similar to the reciprocal function xy = k (ie y = k/x), which is a hyperbola. --72.140.146.246 01:00, 25 April 2006 (UTC) There is a name for each class of quadric surface. And you're right about this one. It is a hyperbolic paraboloid, if you rotate your coordinates about the z-axis, you'll get z=x2y2. -lethe talk + 01:20, 25 April 2006 (UTC) Rearrange into implicit form xyz = 0, and notice that x, y, and z form a multivariate polynomial of maximum total degree 2. So we know immediately that this is a quadric, a quadratic surface. Holding x constant we get lines on the surface as y varies, and holding y constant we get lines as x varies; so this is a doubly ruled surface. We could call it a bilinear surface. Slicing it with the plane y = x gives a parabola, z = x2. All of this supports the identification of this as a hyperbolic paraboloid. The canonical equation for such given in the article is ${\displaystyle \left({\frac {X}{a}}\right)^{2}-\left({\frac {Y}{b}}\right)^{2}+2Z=0.}$ We have already observed that yx produces a parabola (cupped upward); so does y+x (cupped downward). So we make the substitution X = yx, Y = y+x, Z = z/2, and let a = b = 2. (As lethe says, we rotate 90° around the Z axis. Essentially, we are diagonalizing the bilinear form of the quadratic xy.) The canonical equation transforms into −xy+z = 0, which yields the original formula. --KSmrqT 02:42, 25 April 2006 (UTC) ## Names for ${\displaystyle Z_{p}^{*}}$ The group theory I did in my computer science courses is getting amazingly rusty. I know that ${\displaystyle Z_{p}^{*}}$ is the group of all of the numbers from 1 to (p - 1) under multiplication, and has a lot of nice properties, but I can't remember the other names for it. Could someone point me to the article on that group? I'll be happy to create some redirects from Zp* or Z_p^*, to make finding the article easier. Thanks! -- Creidieki 01:01, 25 April 2006 (UTC) Multiplicative group of integers modulo n treats the general case where n doesn't have to be prime. I don't think there's an article specializing on the prime case. Melchoir 01:20, 25 April 2006 (UTC) If n is prime, it's simply the cyclic group of order n−1, right? All cyclic groups of a given order are isomorphic. So, the appropriate article would seem to be cyclic group. —Keenan Pepper 02:37, 25 April 2006 (UTC) So the group that Multiplicative group of integers modulo n is talking about is just ${\displaystyle Z_{n}^{*}}$ in other notation? They don't use that notation in the article, but I can certainly add it. -- Creidieki 03:54, 25 April 2006 (UTC) Yep. The article doesn't use the exact notation, but it does use ${\displaystyle \mathbb {Z} /n\mathbb {Z} ^{*}}$, which I consider a trivial variant of the notation you prefer. But yes, it's not bad to mention alternative notations, so I encourage you to add a mention of alternatives. -lethe talk + 04:05, 25 April 2006 (UTC) ## Relearning How to Walk (Metaphorically Speaking). Long, long ago, when the Internet was still young and lived under the name of Sanders (or at the very least with a sign out front reading "COMMERCIAL INTERESTS WILL"), I obtained undergraduate degrees in computer science and applied mathematics. In the case of the math degree, I'd already completed some advanced coursework in high school, so I entered college taking courses in linear algebra and differential equations. None of that is intended to impress anyone; in fact, just the opposite, because after I wasted the next ten years trying in vain to convince myself that a lucrative career in the computer industry was the route to happiness and financial success, I finally gave up (just wasn't good at programming for a living -- I'm too social). Before long I found myself tutoring at the local community college, and I've discovered that while I'm very good at helping students learn pre-algebra, algebra, and geometry, when the topic turns to more advanced matters, like trig or calculus, I start to falter... and pretty much everything I learned in college I've now forgotten. Diff EQ's, Linear, abstract, even advanced calc -- gone from, or buried somewhere in, my brain. Now I'm considering teaching math for a living, preferably at the high school level, and I have a feeling I'm going to have to pursue an advanced math degree at some point. But I don't have any of my old textbooks, which makes it harder to re-learn all that stuff. And that brings me to my question: besides the Wikibooks and the stuff in this portal, which strike me as neither complete (yet) nor conveniently set up for structured learning (yet), where can I go online, or what books should I buy? If I need to start investing in textbooks again (my wallet trembles at the idea), which ones do you recommend? Thanks for any advice you can offer! --Jay (Histrion) (talkcontribs) 20:02, 25 April 2006 (UTC) Several suggestions; they may or may not help. (1) Visit a bookstore that (also) sells used textbooks, preferably one that serves more than one college. Near (or shortly) after the end of the academic year, superseded editions of textbooks may be sold at firesale prices. For your purpose, it doesn't matter whether you're getting the latest edition. (2) Go to a bookstore with a good collection of math books, and check out the books reprinted by Dover, but make sure that the material, terminology, and notations are reasonably modern. (3) Collect freely distributed lecture notes and tutorials from the Internet. Try a combination of search terms like "www.math" "edu" "lecture notes". That should give you links to lecture notes on a web page hosted by a math department. With some luck, you may be able to find something on the topics you're interested in. --68.238.254.236 01:24, 26 April 2006 (UTC) The Web has become the electronic equivalent of a large, strange, wonderful public library. Some links: If you are interested in specific topics, or general culture questions, ask for more focused guidance. --KSmrqT 03:51, 26 April 2006 (UTC) Hi, i'm currently doing a maths degree (although i'm guessing on the other side of the atlantic) and i find that the schaum's outline series are very helpful and very cheap. they are appaulingly edited but all things considered they're superb. # April 26 ## Integration based on constant endpoints vs. varying intermediate points Let's say Q = arctan(.5*tan(P)). Generally speaking, ${\displaystyle \int _{P_{f}}^{P_{c}}F'(P)dP\quad \neq \quad \int _{Q_{f}}^{Q_{c}}F'(Q)dQ\,\!}$ However, if P equals either 0 or 90°, then Q = P. Now consider ${\displaystyle \int _{0}^{90^{\circ }}F'(P)dP\,\!}$ and ${\displaystyle \int _{0}^{90^{\circ }}F'(Q)dQ\,\!}$ If Pm = 45°, Qm ≈ 26.565°. If you wanted to integrate between Qf and Qc, but using even spacing (i.e., so that Qm = .5*[Qf + Qc]), how would you define/describe it? Does either the integral or interval have a special name? ~Kaimbridge~12:47, 26 April 2006 (UTC) It is somewhat similar to the substitution ${\displaystyle x=2y}$ in ${\displaystyle \int _{0}^{\infty }f(x)dx=\int _{0}^{\infty }2f(2y)dy}$ (Igny 15:18, 26 April 2006 (UTC)) I think you might be confused about the definition of an integral. Assuming we're talking about Riemann integrals, in order for a function to be Riemann integrable, every sequence of Riemann sums must converge to the same limit. If you sometimes get different limits depending on the partitions of the interval you take, then the function is not integrable. Therefore it doesn't matter if you use "even spacing" or uneven spacing, because the limits must be the same for the integral to exist. Please forgive me if I have misinterpreted your question. —Keenan Pepper 17:10, 26 April 2006 (UTC) I think I know what the question is. One of the functions -- either P or Q, it doesn't matter which -- in some sense runs through its early values more quickly than the other. You could change the sample spacing in the Riemann integral while holding the nominal widths of the bins constant, but then it wouldn't be a Riemann integral any longer. (As Keenan Pepper points out, the true Riemann integral is constructed so that it doesn't care about spacing.) A more elegant solution is to simply weight those intermediate points in order to compensate for the speed at which they're being traversed. This technique does have a name: Integration by substitution, and the "weight" is the Jacobian. Melchoir 20:40, 26 April 2006 (UTC) I'm not sure, but I believe the first example in Integration by substitution is what I'm talking about. The thing is, I'm not questioning how you do it, but rather how you distinguish it notationally. Consider these possibilities: First, let ${\displaystyle \cos(\propto _{tn})={\frac {TN-1}{UT-1}}\,\!}$ and ${\displaystyle {\begin{matrix}Q(P)_{tn}&=&Q(P_{1})+\cos(\propto _{tn})(Q(P_{ut})-Q(P_{1})),\\=Q_{tn}&=&\qquad Q_{1}+\cos(\propto _{tn})(Q_{ut}-Q_{1});\qquad \qquad \\Q(P_{tn})&=&Q(P_{1}+cos(\propto _{tn})(P_{ut}-P_{1}));\qquad \quad \end{matrix}}\,\!}$ ---------------------- • ${\displaystyle \int _{Q_{1}}^{Q_{ut}}F'(Q)dQ=\sum _{TN=1}^{UT=\infty }{\frac {F'(Q_{tn})}{UT}}(Q_{ut}-Q_{1});\,\!}$ • ${\displaystyle \int _{P_{1}}^{P_{ut}}F'(Q(P))dP=\sum _{TN=1}^{UT=\infty }{\frac {F'(Q(P_{tn}))}{UT}}(P_{ut}-P_{1});\,\!}$ • ${\displaystyle \int _{P_{1}}^{P_{ut}}(F\circ Q)'(P)dP=\int _{P_{1}}^{P_{ut}}F'(Q(P))Q'(P)dP,\,\!}$ ${\displaystyle =\sum _{TN=1}^{UT=\infty }{\frac {F'(Q(P_{tn}))Q'(P_{tn})}{UT}}(P_{ut}-P_{1});\,\!}$ • ${\displaystyle \int _{Q(P_{1})}^{Q(P_{ut})}F'(Q(P))dQ(P)=\int _{Q(P_{1})}^{Q(P_{ut})}F'(Q(P)){\frac {dQ(P)}{\Delta P}}\int _{P_{1}}^{P_{ut}}Q'(P)dP,\,\!}$ ${\displaystyle =\sum _{TN=1}^{UT=\infty }{\frac {F'(Q(P)_{tn})}{UT}}(Q(P_{ut})-Q(P_{1})),\,\!}$ ${\displaystyle =\left(\sum _{TN=1}^{UT=\infty }{\frac {F'(Q(P)_{tn})}{UT}}\right)\sum _{TN=1}^{UT}{\frac {Q'(P_{tn})}{UT}}(P_{ut}-P_{1});\,\!}$ The first integral looks like a normal integral. However, if you've identified Q as a function of P, Q(P), have some integrands with direct equivalencies (e.g., F'(Q) = F'(Q(P)) = G'(P)) and other integrands that aren't equivalent, but their antiderivatives are (e.g., F(Q) = F(Q(P)) = H(P), but F'(Q) ≠ H'(P)), it would seem that ${\displaystyle \int _{P_{1}}^{P_{ut}}F'(Q(P))dP=\int _{P_{1}}^{P_{ut}}G'(P)dP,\,\!}$ but ${\displaystyle \int _{Q_{1}}^{Q_{ut}}F'(Q)dQ=\int _{Q(P_{1})}^{Q(P_{ut})}F'(Q(P))dQ(P)=\int _{P_{1}}^{P_{ut}}H'(P)dP.\,\!}$ So, going back to my original endpoints, where Q1 = P1 = 0 and Qut = Put = 90°, you have, ${\displaystyle \int _{0}^{90^{\circ }}G'(P)dP=\int _{0}^{90^{\circ }}F'(Q(P))dP,\,\!}$ and ${\displaystyle \int _{0}^{90^{\circ }}H'(P)dP=\int _{Q(0)}^{Q(90^{\circ })}F'(Q(P))dQ(P)=\int _{0}^{90^{\circ }}F'(Q)dQ.\,\!}$ It just seems too easy to confuse, even using "dQ(P)", as ${\displaystyle \int _{Q(0)}^{Q(90^{\circ })}\,\!}$ suggests that the midpoint is Q(45°) when it is actually Q(63.435°)! Given that there is special notation for such things as "composite functions" and "path integrals" (${\displaystyle \oint _{C}}$), it would seem that something as potentially ambiguous as this would too (and, since Q(P) = arctan(.5*tan(P)) it goes the other way, too: P(Q)= arctan(2*tan(Q))!). FYI, this isn't just idle hypothetics——"P" is geodetic and "Q" is reduced latitude and they are used in those nasty things known as elliptic integrals! P=) ~Kaimbridge~21:59, 29 April 2006 (UTC) I have no idea what all those expressions meant. Do you have a question? Melchoir 22:09, 29 April 2006 (UTC) # April 27 ## .gif versus .jpg Which type of image has a better resolution and is sharper?Patchouli 05:06, 27 April 2006 (UTC) Neither, as you would discover by a quick glance at the JPEG and GIF articles. In fact, this question is meaningless for most image formats. The JPEG format condenses photographic material better by using lossy compression, throwing away data that the eye notices less. The GIF format is a poor choice for photographs, and even for graphics it is inferior to PNG in compression, color control, and other desirable features. Resolution can be whatever you want. Consult the article on image file formats and the graphics file format summary. Graphics uploaded to Wikipedia now use SVG as well, for numerous benefits. --KSmrqT 05:40, 27 April 2006 (UTC) Thank you.Patchouli 06:51, 27 April 2006 (UTC) I would argue that gif is sharper, since it has no lossy compression that blurs edges. If you need full color then go with PNG. Obviously this is not the whole story, JPEG can do a far better job of compressing. Also, GIF can only be sharper for originals that have 256 colors or less. Any more means that the GIF must include dithering. So GIF can be a good choice for drawn graphics, things containing text etc. JPEG is rotten for text. PNG does not enjoy universal support, so your improved quality comes with it the knowledge that fewer people will see the picture. (I cannot see the PNG graphic in the top left corner of Wikipedia pages because it is a PNG, the Windows IE uses ActiveX to display PNG, and I have turned this off for security reasons). Notinasnaid 08:19, 3 May 2006 (UTC) Given that PNG has been a W3C recommendation for about a decade, if one's browser does not properly support it one might consider switching browsers, if one has that option. (Mozilla Firefox and Opera are solid choices; there's a slim chance IE7 will be another.) Or should we also avoid CDs and DVDs, sticking with vinyl LPs and VHS tapes? --KSmrqT 10:12, 3 May 2006 (UTC) ## affine Varities and grobner bases.. What are affine varieties and how do they relate with grobner bases? Is there any other type(s) of varieties apart from affines if there what are they and how do they differ from the affine varieties? —The preceding unsigned comment was added by Nkomali (talkcontribs) . You raise two related topics, neither of which is trivial. Back when the world was simpler (say, the nineteenth century), mathematicians investigated the relationship of geometry to algebra, where geometry was projective and algebra meant the zeros of homogeneous polynomials over the complex numbers. Some polynomials, like x2w2, split into factors (here xw and x+w). Others, like x2+y2w2, do not. (Notice that polynomials in multiple variables are quite different from polynomials in a single variable in this regard.) When a polynomial factors, the zeros of the full polynomial are the union of the zeros of the factors; so we focus attention on the irreducible polynomials, those without factors. The geometric manifestation of an irreducible polynomial is a variety. Usually a variety would be a projective variety; affine varieties are the pieces we stitch together (like a manifold) to cover all the parts of the projective variety. These days algebraic geometry has wandered far from its humble and intuitive beginnings to varieties defined in highly abstract ways with much broader application, using concepts like "scheme" and allowing different kinds of fields besides the real or complex numbers. Already in the classic work we notice quickly that the zeros of a polynomial are also zeros of all of its multiples, and so we shift algebraic attention from a single polynomial to an "ideal" in the ring of polynomials. It is not especially difficult to write a polynomial in a unique canonical form, but to do so we must order the variables and then order the terms. Thus we might order x before y and higher degree terms before lower, so that x2+xy+y2 is canonical while its reverse is not. Writing a canonical form for an ideal is more challenging, and it also depends on this kind of ordering. Given an ordering on terms, a Gröbner basis for an ideal is a route to a unique canonical form consisting of a list of polynomials. It has properties which are of great help in computations. --KSmrqT 09:50, 27 April 2006 (UTC) ## sine and sine hyperbolic If sine can be defined within a triangle,how about sine hyperbolic?? —The preceding unsigned comment was added by 210.212.194.215 (talkcontribs) . The sine and cosine make up the coordinates of a point of a circle. Draw some line segments to the origin, and you get the triangle you're talking about, and the argument of the functions is an angle of the triangle. The hyperbolic sine and cosine make up the coordinates of a point of a hyperbola (hence the name!). If you draw the line segments to the origin, you only get your boring triangle again, though. In this case, the argument of the functions isn't an angle of the triangle, unless you're willing to consider a funny version of angle (if you endow your space with a funny metric (just putting a minus sign in the Pythagorean theorem will give you one way to do it), then the notion of angle gets replaced by something called rapidity, and you could say that the rapidity of your triangle is the argument of the functions. In this funny space, the hyperbola is a circle). To sum up, sine is the angle in a Euclidean triangle, hyperbolic sine is the angle in a non-Euclidean (hyperbolic) triangle. -lethe talk + 10:57, 27 April 2006 (UTC) ## Easy way in Matlab In Matlab, is there a loop-free way to create an nxn matrix with diagonal and upper-triangular entries equal to 1, and lower-triangular entries equal to 0? eg. for n = 3: [ 1 1 1; 0 1 1; 0 0 1]  Confusing Manifestation 12:45, 27 April 2006 (UTC) toeplitz([1 zeros(1,n-1)], ones(n,1)); -- Jitse Niesen (talk) 13:39, 27 April 2006 (UTC) Either of the following work in octave, so I suppose they should work in matlab too.  repmat(1:n,n,1)'<=repmat(1:n,n,1), triu(ones(n)),  b_jonas 21:11, 27 April 2006 (UTC) Other, wierder ways:  cumsum(eye(n), 2), fliplr(hankel(ones(1,n))), inv(eye(n) - diag(ones(1, n - 1), 1)), !!(eye(n) + diag(ones(1, n - 1), 1))^n,  b_jonas 21:28, 27 April 2006 (UTC) Thanks for all of those suggestions. I'll use Jitse's version just because at a quick glance it seems to be fairly simple, although it's nice to know the alternatives exist. (First rule of Matlab: there's almost always at least one loop-free way to do something.) Confusing Manifestation 00:31, 28 April 2006 (UTC) ## Topology. Is the continuos image of a complete metric space necessarily complete? (Prove or give counter example) Nkomali. All the Cauchy sequences still converge, right? This shouldn't be that hard to prove... —Keenan Pepper 14:06, 27 April 2006 (UTC) No, I think it isn't, and I've just seen a nice example today. Let's take ${\displaystyle \mathbb {R} }$ with the usual metrics, and map it to the open interval ${\displaystyle (-\pi /2,\pi /2)}$ with the ${\displaystyle \arctan }$ function. This is continuous, but the image isn't complete. (I think it's true the other way: the continuous inverse image of a complete metric space is complete.)b_jonas 20:57, 27 April 2006 (UTC) My comment in parenthesis was stupid. – b_jonas 14:36, 28 April 2006 (UTC) Stereographic projection of the complex plane (without point at infinity) onto the Riemann sphere (without northpole) is another counter example (of similar kind, though). Mon4 00:57, 30 April 2006 (UTC) ## Groebner bases What are the properties and applications of groebner bases. --NkomaliNkomali 14:07, 27 April 2006 (UTC) We are not a substitute for your brain and effort. Read the article (and the response to your earlier question). --KSmrqT 18:03, 27 April 2006 (UTC) ## Topology Show that the spaces l power alpha, the set of all bounded sequences aaaa9 in R or C with norm xn = sup{\xn\}, is a normed linear space. Show also that l power alpha is a complete normed linear space with the transition invariant d(x,y) = sup subscript j in N absolute value of the difference between epsilon j and mu of j.--Nkomali 14:26, 27 April 2006 (UTC) Read the top of this page about homework. --KSmrqT 19:14, 27 April 2006 (UTC) ## Chaos Theory hi, can anyone explain to me what does the 'unpredictable, yet deterministic' behaviour of things in chaos theory means?... I don't get it :|, how can something be 'unpredictable yet deterministic'? If we find it's deterministic, aren't we somehow making it predictable?.--Cosmic girl 15:55, 27 April 2006 (UTC) A chaotic system exhibits sensitivity to initial conditions to an extreme degree. Although the system is deterministic, in the sense that the laws that govern the evolution of the system are known exactly, small differences in initial conditions will lead to huge differences in long-term behaviour. Since there is a practical limit to the accuracy to which initial conditions can be known (and, in computer systems, a limit to the accuracy with which calculatuons can be carried out) this means that such systems are, in practice, unpredictable. See butterfly effect for a longer explanation. Gandalf61 16:21, 27 April 2006 (UTC) Thank you! :) --Cosmic girl 19:41, 27 April 2006 (UTC) With perfect mathematical precision, deterministic does imply predictable. Even without it we can say a great deal about the behavior of a chaotic system. So, yes, this language is more provocative than impeccable. Still, it conveys an essential germ of insight in contrasting the behavior of a chaotic system to more familiar ones. Consider the trajectory of a cannonball. Tip the cannon up a little and the ball lands in a slightly different place. Turn it to the side, and again the flight path and landing spot change only slightly. This is the predictability we expect of Newtonian dynamics, F = ma and all that. It applies to celestial mechanics, such as the orbits of planets, as well as to Earthly cannon balls. We can say with great confidence where Mars or Jupiter will be 2000 years from now. But long unsuspected, lurking within is a radically different possibility. In some planetary systems we have no ability to make longterm predictions, because small changes in the present scatter the distant future over a vast range of possible positions. Perhaps our measurements of the current position are microscopically in error; perhaps a comet wanders by; perhaps the mass of its star(s) is not exactly what we thought. A chaotic system spreads and folds the trajectories again and again, like a mixing machine, amplifying even the tiniest initial uncertainty. --KSmrqT 19:46, 27 April 2006 (UTC) There is another thing - even if we know exactly the initial condition, it can take an astonishing amount of computational effort to actually determine the outcome. By "astonishing" I don't mean "I'll let my PC work on it for several hours", but rather - even if all the computers on earth join forces to calculate it, and work an amount of time greater than the 15 billion or so years that the universe is believed to exist - we'll be nowhere near a solution. That's one way of interpreting "unpredictable". As an example, suppose I give you a 10000-digit number which is a product of 2 large primes (known only to me), and ask you what the prime factors are. This problem is deterministic - you know the number, so you can calculate its factors. You don't need any additional information. But the computation is in practice of the scale I mentioned above. So, theoretically you can determine what the factors are, but you have no practical way of actually predicting them. -- Meni Rosenfeld (talk) 17:31, 28 April 2006 (UTC) The folks at JPL have some remarkable software for celestial mechanics, including relativistic effects; but even they cannot compute with infinite precision. Long before we exhaust time we will corrupt the initial trajectory with accumulated calculation error. --KSmrqT 07:23, 3 May 2006 (UTC) Hi! The butterfly effect states that a butterfly wing flap, there, may give a tornado here. Nobody dares to ignore the image, become that of the sensitivity to initial conditions paradigm. You can exhibit models, but you can never show the right butterfly. Therefore, no experience in that domain is reproductible. There was, long ago, other domains that would give models but no reproductible experiences : some are called pseudoscience now. Research goes on, provoking either adhesion or derision. We should no more adhere without thinking, but we may take a different look at that. --DLL 17:55, 28 April 2006 (UTC) I would make a comment on Gandalf's observation above: I think the weather is probably not deterministic. That's because the initial conditions cannot, even in principle, be known precisely; according to the Copenhagen interpretation of quantum mechanics, they're not even precisely well-defined. Though this initial indeterminacy is very slight, any change at all in the initial conditions is amplified geometrically with the passage of time. Thus if you wait long enough, the question "will it rain on such-and-such a date in Dallas, Texas?" becomes literally undetermined, not just unpredictable in practice. --Trovatore 00:04, 29 April 2006 (UTC) It is a distracting cheat to introduce quantum mechanics. One of the intriguing facts about chaos is that it can occur strictly within that paragon of determinism, Newtonian physics. --KSmrqT 07:23, 3 May 2006 (UTC) Sorry to edit this in archive, but I think the point should be addressed: Yes, chaos can occur in deterministic settings. My point was that the example introduced was the weather, which is not deterministic. Or at least I believe it is probably non-deterministic, for the reasons I've explained. --Trovatore 21:23, 5 May 2006 (UTC) ## Charateristic functions If I have a random variable Z = aX + bY, what is the characteristic function of Z in terms of the pdfs of X and Y? --HappyCamper 16:34, 27 April 2006 (UTC) If X and Y are independent, then ${\displaystyle \chi _{Z}(s)=\chi _{X}(as)\chi _{Y}(bs)}$ or am I thinking of something completely different? — Arthur Rubin | (talk) 16:48, 27 April 2006 (UTC) See characteristic function (probability theory). It specifically gives your example. If you want it in terms of the PDFs of X and Y rather than their characteristic functions, you'll need to first get the characteristics using the integrals also given on that page. --Tardis 17:35, 27 April 2006 (UTC) # April 28 ## WATER'S DEPTH I HAVE A RECTANGULAR TANK MEASURING 55 METRES BY 48 METRES CONTAINING 33 LITERS OF WATER. HELP ME FIND THE DEPTH OF THE WATER. —Preceding unsigned comment added by 208.131.187.81 (talkcontribs) 02:22, 28 April 2006 First, don't write in all captical letter, this is considered shouting and rude. Second, it is obvious that this is a homework problem, please read the instruction on the top of this page. Simply put, Volume of rectangular prism is length times width times height. Just plug in the numbers, and divide to get the answer. Besure to use consistent unit. --Lemontea 02:57, 28 April 2006 (UTC) Convert the volume from units in litres to units in metre cube and the answer is as plain as daylight. Ohanian 09:15, 28 April 2006 (UTC) I thought yards & gallons were still in use in some parts of the world. Do they like plain daylight ? --DLL 17:45, 28 April 2006 (UTC) 1 gallon is 231 cubic inches. The rest is easy. Chuck 18:55, 28 April 2006 (UTC) Bloody little. I'd call that "moisture" rather than "depth of water". —Ilmari Karonen (talk) 19:37, 30 April 2006 (UTC) Do you perhaps mean centimeters (cm)? You could always convert to chains and hogsheads.... Skittle 23:24, 1 May 2006 (UTC) ## Trig Identities How would you go about doing the question below... ${\displaystyle {\frac {2\sin ^{2}\theta -1}{\sin \theta \cos \theta }}={\tan \theta -\cot \theta }}$ All I could think of was changing 2sin²θ-1 into -2cos²θ + 1. I can't change sinθcosθ (at least I don't think so). I tried the right hand side next, and I got: ${\displaystyle {\frac {\sin ^{2}\theta -\cos ^{2}\theta }{\cos \theta \sin \theta }}}$ By the way, I keep on getting Failed to parse (lexing error) whenever I try to properly write a fraction, I just copied and pasted one from this page and changed it to this question but it still doesn't work, sorry about that. Thanks C-c-c-c 03:34, 28 April 2006 (UTC) It parses now. Also, you're close. Compare the numerators of your two fractions --Deville (Talk) 03:43, 28 April 2006 (UTC) Mhm, well I've actually located a little thing on the formula sheet, and noticed that 2cos²θ -1 = cos²θ - sin²θ(yes I should have looked earlier, but it had it as Cos2θ = 2cos²θ -1, Cos2θ = cos²θ - sin²θ and I didn't link cos²θ - sin²θ and 2cos²θ-1 together until then.) I have -2cos²θ +1 so that means it should be -(cos²θ - sin²θ) so it should be sin²θ - cos²θ. Yep it works, thanks a lot. --- C-c-c-c 04:02, 28 April 2006 (UTC) Hey, just remember that all you ever need to know is$sin^2 + cos^2 = 1\$, and the angle-addition formulas... everything else comes from these. --Deville (Talk) 15:18, 29 April 2006 (UTC)
Try using the definitions of tan and cot. Does that help? Confusing Manifestation 04:09, 28 April 2006 (UTC)

## Retire

How much money do I need to retire? And what % return can I expect to get from investing?

If I can achive X% Average return, does that mean I can always take out X% of the princaple each year? 12.183.203.184 03:52, 28 April 2006 (UTC)


How to calculate various interest rate problem
----------------------------------------------


extremely long post cut by me. For full version, see this diff. -lethe talk + 09:25, 28 April 2006 (UTC)

Note that taxes and inflation also need to be subtracted from the interest rate each year to find your true gain (or loss). StuRat 17:28, 29 April 2006 (UTC)

## Duplex Numbers

What are duplex numbers? Ohanian 09:06, 28 April 2006 (UTC)

The first page or so of this document has some background of duplex numbers. It looks like they're a parallel to the complex numbers or quaternions. Confusing Manifestation 12:42, 28 April 2006 (UTC)
The same concept is covered by split-complex number.--gwaihir 12:19, 3 May 2006 (UTC)

## Hex, square and triangle grids

Some board games or maps use squares (4 exits from every cell, or 8 if you include diagonals, but the latter are not the same distance away as the former), some use hexes (6 equidistant exits), some use triangles. Is it possible for a map to use other shapes, and allow other numbers of possible equal exists besides 4, 6 or 3? Is it possible to have with 8, for instance? Or why is it impossible? The shape doesn't have to be symmetrical, but ideally the exit points should be as equidistant as possible from the centre of one to the centre of another.--Sonjaaa 10:28, 28 April 2006 (UTC)

You are looking for List of uniform planar tilings (the main article is tessellation, the current Mathematics Collaboration of the Week). There are only the three regular tilings that you mention, but there are some semiregular ones, too. Rasmus (talk) 11:25, 28 April 2006 (UTC)
If you start putting together congruent regular polygons in a repeating pattern, the overall fate of the surface is determined by the value of 1/(number of edges per polygon) + 1/(number of polygons meeting at each corner). If that's equal to 1/2, you get a tiling of the plane. Only three pairs of unit fractions fractions add up to 1/2: 1/6 + 1/3, 1/4 + 1/4, and 1/3 + 1/6. These correspond to tilings of the plane by hexagons, squares, and triangles. If the value is less than 1/2, you get a tiling of the sphere (a.k.a. a Platonic solid). If the value is greater than 1/2, you get a tiling of the hyperbolic plane. —Keenan Pepper 13:05, 28 April 2006 (UTC)
The life game believes that a square has 8 neighbours, including diagonally opposites. Hexes cant' choose (the diagonal is just between two other hexes) ; and triangles are back to hexes, but with more density. So we have 3, 4, 6 and 8 on a plane (according to OEIS : very round numbers ?!) --82.227.17.30 17:37, 28 April 2006 (UTC) (DLL)
Displaying 1-10 of 1281 results found. Wow, useful. —Keenan Pepper 17:58, 28 April 2006 (UTC)
It's in the 10 first. P.S. I feel dizzy now I found a life game played on a semiregular pentagonal tessellation. --DLL 19:52, 28 April 2006 (UTC)

## Geometry question

Dear Wikipedians. I'll be very pleased if I'll the solution for the following geometrical rider: ABC is a triangle in which angle B = 2 angle C. D is a point on BC such that AD bisects angle BAC and AB=CD. Prove that angle BAC=72 degrees.

Thank you very much. User Akshitapatel

This sounds like a homework question, but it's an interesting one--took me quite a while to work out--so I'll give an outline which I hope will lead you in the right direction.
• Given that ${\displaystyle \angle ABC=2\angle BCA}$, and ${\displaystyle \angle ABC+\angle BCA+\angle CAB=180^{\circ }}$ (assuming Euclidean geometry!) we can define everything in terms of a single angle. Let ${\displaystyle \angle BCA=2\theta }$ (using ${\displaystyle 2\theta }$ rather than ${\displaystyle \theta }$ avoids some ugly fractions later on, but is not strictly necessary).
• Determine ${\displaystyle \angle ABC}$, ${\displaystyle \angle ADB}$, ${\displaystyle \angle BCA}$, and ${\displaystyle \angle CAD}$ in terms of ${\displaystyle \theta }$.
• Apply the sine rule to determine a relationship between ${\displaystyle AD}$ and ${\displaystyle CD}$, and another between ${\displaystyle AD}$ and ${\displaystyle AB}$.
• Substitute ${\displaystyle CD=AB}$, and rearrange both equations from the previous step to generate two different functions of ${\displaystyle \theta }$, both equal to ${\displaystyle {\frac {AD}{AB}}}$. Set them equal to each other. (It may also help to apply the fact, at this point, that ${\displaystyle \sin(90^{\circ }-x)=\cos x}$, but it is not strictly necessary.)
• At this point there are two approaches, depending on how rigorous you need to be:
• Easy, but non-rigorous approach: See if you can find a value of ${\displaystyle \theta }$ which makes the equation true. It's not too hard to do by inspection with the right insight. (And it's even easier since you know what ${\displaystyle \theta }$ is supposed to work out to.)
• Hard, but rigorous approach: substitute the trigonometric functions you have in your equation of ${\displaystyle 2\theta }$, ${\displaystyle 3\theta }$, and ${\displaystyle 4\theta }$ with the equivalent values in terms of trigonometric functions of ${\displaystyle \theta }$. Solve for ${\displaystyle \sin \theta }$, and then for ${\displaystyle \theta }$. This involves a fifth-order equation in ${\displaystyle \sin \theta }$, but it has a quadratic factor.
It's possible there's a geometric approach to this proof which doesn't involve trigonometric functions, but I wasn't able to find it. Chuck 20:43, 28 April 2006 (UTC)
Found it! Actually it's the famous classical Pentagon or Pentagram. Anyway, the geometric approach goes below:
For simplicity, call the angle BAD as b, and call angle ACD as a. Now the first and most important step, Find a point on AC such that BE bisect angle ABC. Join BE as well as DE.
Because angle EBC = angle ECB = a, triangle BEC is isosceles and so BE = EC.
We also know that AB = CD, and that EB = EC, and that angle ABE = angle DCE. Hence, by SAS, triangle ABE and DCE are congruent.
Consequently, AE = DE, so angle ADE = angle EAD = b. Also angle EDC = angle EAB = 2b. So in summary, angle ADC = angle ADE + angle EDC = b+2b=3b.
But by exterior angle of triangle, angle ADC = angle BAD + angle ABD = b+2a. So 3b=b+2a, and therefore a=b.
The rest should be obvious enough for you to see. --Lemontea 01:32, 29 April 2006 (UTC)
The data are : (1) B = 2C ; (2) AD bissects BAC ; (3) AB = CD ; (4) A = 72° (to be proven).
Let (4) be true. (4) => (5) B+C = 108° ; (5) and (1) => (6) C = B/2 = 36° ; (2) and (6) => (7) now we have thoses triangle measures : ABD (36°, 72°, 36°) and ADC ( 36°, 108°, 36°). Then both are isosceles (golden triangles) and remind of pentagram angle measures.
(7) isoscele means that => (8) AB = AD and AD = DC ; since (3) it is true. --DLL 07:50, 29 April 2006 (UTC)
Let me clarify the above comment's purpose: it demostrates why the figure is part of the pentagram, and shows that there is no contradiction for the assertion(that A=72 degrees). But it isn't a solution, just an, err, exploration, if you ask me. It doesn't show (4) is the only possible solution, in particular, see Affirming the consequent and Logical fallacy. Thanks. --Lemontea 09:01, 29 April 2006 (UTC)

## Sin x into Sec x

Hello fellow Wikipedians. Today I had the unfortunate task of writing a trig test (but I don't think it went too bad for that matter), and the question below stumped me.

Use only sec x:

${\displaystyle {\frac {\sin x}{\sin 2x}}}$

I didn't know what to do, and I looked at all the formulas on the formula sheet to see how I could convert sin x into anything, but alas, nothing. The question below has to be in terms of sec x, that's the challenging part. I spent a good deal towards the end trying to think this through, and all I could come up was converting sin x into 1/cscx, but that didn't help either. Also, I converted sin 2x into 2sinxcosx (edited) I believe or something similar, I don't know it off the top of my head. Any thoughts on this would be greatly appreciated. --- C-c-c-c 22:52, 28 April 2006 (UTC)

See Trigonometric identities#Double-angle formulae. You appear to have confused sin(2x) with cos(2x). Melchoir 22:58, 28 April 2006 (UTC)

I have an OT question—I've just recently administered my first final exam as an instructor in Canada, and have come across this usage of "write an exam", which confuses me a little. If the student writes the exam, just what is it the instructor does, when creating it? Isn't that also "writing", or is there a different verb used? --Trovatore 23:55, 28 April 2006 (UTC)

Right, never mind, I did confuse it, but I think it was 2sinxcosx I had on the bottom, whatever it was, I got it from the identity sin 2x on my formula sheet. --- C-c-c-c 02:44, 29 April 2006 (UTC)

Sorry, I may not have expressed myself clearly. My confusion has nothing to do with the mathematics, but with this usage of the phrase "write an exam". To me, an American, writing an exam means devising questions, TeXing them up, printing them out, and copying them for the hapless victims^W^W eager students. But in Canada, I find that it's the students who "write the exam" (what in the States we would call "taking an exam"). So if it's the students who wrote my exam, then what exactly did I do to it? --Trovatore 02:48, 29 April 2006 (UTC)
Mhm true, the people often around me say "write an exam", (here in Canada) or take an exam, both are interchangable in the high school setting at least. I would assume that actually creating an exam would be referred to "making up the exam", though now as I reread that it sound goofy, but maybe adding more detail to the end of the phrase, "I'm writing an exam...for my students". I can't think of a different verb you'd use though. --- C-c-c-c 02:57, 29 April 2006 (UTC)
How about "set an exam"? Mon4 11:27, 29 April 2006 (UTC)
Is that what's used? To my ear that sounds like deciding when an exam will be held, how much it's worth, maybe what it will cover. But this could be a difference between Canadian and American usage, I guess. That would bring my count of such differences up to, I don't know, five or so. Canadian speech, customs, way of life, are a lot more similar to the American ones than Canadians like to think, but that sometimes makes it all the more disconcerting when you run into one. --Trovatore 14:41, 29 April 2006 (UTC)
Oh, you wacky Canadians!!!!  :) Seriously, my experience is that in American English usage, the instructor "writes", or "writes up" the exam, and the students "take" it. In British English usage, the instructor more commonly "sets" or "designs" the exam, and the students "write" it. Actually it makes a little more sense, as I don't "write" exams, I "TeX" them, and the students are the only ones doing actual writing... --Deville (Talk) 15:17, 29 April 2006 (UTC)
In New Zealand, an instructor sets an exam, and students sit it. I don't think I've ever heard the term "write an exam" before seeing this.-gadfium 20:29, 29 April 2006 (UTC)
That would be true of Australia as well. JackofOz 06:47, 2 May 2006 (UTC)
The students sit it because they can't stand it. --LambiamTalk 00:17, 5 May 2006 (UTC)

So, you got as far as:

${\displaystyle {\frac {\sin x}{\sin 2x}}={\frac {\sin x}{2\sin x\cos x}}}$

What did you do next? Melchoir 04:23, 29 April 2006 (UTC)

Sinx in the numerator cancels out with the sinx at the bottom which leaves you with 1/(2cosx). Since 1/cosx = secx, 1/(2cosx)=0.5secx. Hope that helped.
Oh, so that's how you would do it. Thanks C-c-c-c 05:00, 30 April 2006 (UTC)

## Sine, cosine, tangent

Today we started learning trig in my geometry class (our teacher calls it plain-vanilla trig, as it does get harder) and we learned the simplest of rules. We took notes, started on homework with extra class time, etc. When I got home to finish, I found some one had taken my notes and homework (while it is entirely possible I just lost them on the floor at school somewhere, I will stick with them being stolen). What I need to do is just make sure I'm sure. Sine=Opposite side/Hypotenuse; Cosine=Adjacent side/Hypotenuse; Tangent=opposite side/Adjacent side; Area= (1/2)(angle a)(angle b)(sinC). I pretty sure about all but the last one. If any corrections are needed please help me. Thanks. schyler 00:05, 29 April 2006 (UTC)

I don't understand what the last one is even saying, but the rest look fine! There's plenty of information at Trigonometry and related articles if you want to verify any other formulas. Melchoir 00:53, 29 April 2006 (UTC)
Actually... you probably mean the formula at Triangle#Using trigonometry, where a and b are side lengths, not angles. Melchoir 00:57, 29 April 2006 (UTC)
Have you been taught the popular trigonometry mnemonic "SOHCAHTOA" for the basic formulae? --Bth 08:23, 29 April 2006 (UTC)
I've always preferred "sex on holidays can advance happiness to outrageous amplitudes" - more memorable, and less chance of mixing up sohcahtoa and sahcohtoa. Confusing Manifestation 14:59, 29 April 2006 (UTC)
I prefer "Some Old Hippy, Caught Another Hippy, Tripping On Acid". StuRat 17:24, 29 April 2006 (UTC)
We always used "The Old Arab Sat On His Camel And Humped", which takes a different order. I think the main reason we used it was because our teacher was embarrassed by it, since it was clearly not quite ~right, and she tried to make it a little better by using 'hoped' instead of 'humped', which clearly missed the point of why it wasn't ~right and spoiled the pun at the same time. Bet she wished she never made that poster. (PS, humped as in sulked) Skittle 23:38, 1 May 2006 (UTC)
I think you must mean area = side a * side b * sin gamma / 2. – b_jonas 18:10, 29 April 2006 (UTC)

Thank you. Now that you mention it, my teacher did tell us this story about an indian princess named Sohcahtoa. I forgot. Also, that is the formula I was thinking of. Again, thanks. schyler 01:32, 30 April 2006 (UTC)

When my physics teacher (Mr. Rippetoe) related that story, he claimed that Sohcahtoa was related to his own Indian ancestors, the Rippetoas. Black Carrot 22:09, 4 May 2006 (UTC)

# April 29

## The Basic rules of Differential Calculus

I have been looking through the proofs of the rules of differential calculus, and I understand how we find d/dx(x^n)=nx^(n-1) (something I was taught to take for granted), as well as the quotient rule and d/dx(sin x) and d/dx(cos x). However, I reached a stumbling block with the proof of the Chain rule. From the first line:

${\displaystyle g(x+\delta )-g(x)=\delta g'(x)+\epsilon (\delta )\,}$ where ${\displaystyle {\frac {\epsilon (\delta )}{\delta }}\to 0\,}$ as ${\displaystyle \delta \to 0.}$

I assume this was reached by:

${\displaystyle g'(x)={\frac {g(x+\delta )-g(x)}{\delta }}}$
${\displaystyle \delta g'(x)=g(x+\delta )-g(x)}$

Where did the epsilon(delta) come from?

Also, in the proof of the product rule (which is more clearly laid out), it is said that:

${\displaystyle g(x+\Delta x)h(x+\Delta x)-g(x)h(x)=g(x)(h(x+\Delta x)-h(x))+h(x+\Delta x)(g(x+\Delta x)-g(x)),\,}$

I assume this is done with some sort of factorisation, but nothing I can see. The two terms on the left have no common factor. What am I missing?

Thanks in advance. --Alexs letterbox 05:00, 29 April 2006 (UTC)

For your first question, the epsilon comes from replacing an equation that only holds in the limit with a straightup equation. If lim f(x) = lim g(x) as xa, I cannot conclude that f(x) = g(x), but I can say that f and g are approximately equal. That is, that f(x) = g(x) + ε, with lim ε =0 as xa. Proof: let ε = f(x) – g(x). In the equation you cite, on one side of the equation you have the fraction [g(x+δ) – g(x)]/δ and on the other you have g'(x). Your mistake is assuming these are equal. They're not equal, they're only approximately equal, up to a small error which vanishes in the limit that δ → 0.
For your second question, I think the easiest way to see that the two sides are equal is to distribute out the right-hand side. Once you do that, you'll realize that the way to procede from the left-hand side to the right-hand side was not to just factorize, but to first add zero in the form of 0=g(x)h(xx)–g(x)h(xx). Then you factorize, then you get the equation. -lethe talk + 09:15, 29 April 2006 (UTC)
Thanks. --Alexs letterbox 23:15, 29 April 2006 (UTC)
Just looking through the working again (I'm nearly there!), once again in the Chain rule, how does one get from f(g(x)+δg'(x)+ε(δ))-f(g(x)) to (δg'(x)+ε(δ))f'(g(x))+ɳ(δg'(x)+ε(δ))? What justifies taking the δg'(x)+ε(δ) out of f? --Alexs letterbox 11:00, 30 April 2006 (UTC)
For function f we have a similar equation as for g. Using all different variables to avoid mixups, it is: f(z+ω) – f(z) = ωf'(z) + η(ω). Now simply plug in g(x) for z and δg'(x)+ε(δ) for ω. --LambiamTalk 14:08, 30 April 2006 (UTC)
Edit conflict: The equation h(x+ζ) = h(x) + h'(x)ζ does not hold exactly, but only in the limit ζ→0. In other words, h(x+ζ) = h(x) + h'(x)ζ + η, for some η for which η/ζ → 0 as ζ→0. Substitute fg for h, and δg'(x)+ε(δ) for ζ, and denote the error η for this choice as αδ, and you have your equation. -lethe talk + 14:21, 30 April 2006 (UTC)

## Difference between Absolute Graph and Relative Graph

Hi,

I would like to know the difference between Absolute Graph and Relative Graph? I have searched all over the internet and i havent found any information. Its not even in my college notes. Any help would be appreciated as i have no idea what absolute and relative graphs are!

I've never heard of them, and Wikipedia has nothing. (See Graph for common usages.) If I were you, I would be really suspicious of whoever told me about these things! Melchoir 08:16, 29 April 2006 (UTC)

Its for my assignment for Quantitative Maths. I am stumped at it too!

The two most likely looking results I dug up from Google are:
Something to do with electronic engineering, and measures of components' response, or something.
A page about investment which talks about "absolute" and "relative" graphs in the context (I think) of looking simply at a stock's performance (absolute) or its value divided by its value on some chosen reference date ("relative").
The second one seems like it might possibly be the general idea, but it seems too simple really. --Bth 08:57, 29 April 2006 (UTC)

Hey thanks alot. Atleast that gave me some idea. Maybe i will be able to expand on it but any new info is always welcome. thanks!

If you gave us some context, we might be able to make good educated guesses. What course is this for? -lethe talk + 09:19, 29 April 2006 (UTC)

Quantitative Maths. I have my asked my instructor and hopefully she will respond quickly. Its an online course.

I suppose any graph of an index, like the Dow Jones Industrial Index, would be a relative graph. That is, the actual values each day are meaningless, it's only the change from day to day that's significant. In a normal (absolute) graph, the individual values do have some meaning, like the price of an individual stock each day. StuRat 17:08, 29 April 2006 (UTC)

## Newton

Hi is the Newton who created the Newton-Rapson Method (for finding roots of an equation) Sirr Isac Newton. (the apple one)

Yes. See Newton's method#History. Melchoir 21:15, 29 April 2006 (UTC)

# April 30

## Computer Audio Recording

Hello, everyone. I'm interested in making my own "book on tape" type of project, except on a CD which is made with my computer. Does anyone know what I would need to do this and what software I could use, preferably downloadable for free from the internet from a site such as tucows? Thanks. --Think Fast 01:13, 30 April 2006 (UTC)

—The preceding unsigned comment was added by 213.42.2.28 (talkcontribs) 08:26, April 30, 2006 (UTC).

## Negative exponential distribution

Say shoppers enter a shop at a mean rate of 4 per minute. This can be modeled as an exponential distribution parameter 0.25. How do you show that the difference in times between successive shoppers entering is distributed exponentially as well? Is this to do with Exponential distribution#Memorylessness? 12:53, 30 April 2006 (UTC)

Yes, that is what the "memorylessness" gives you. Just after a shopper entered at time t is no different from any other moment, including time 0. LambiamTalk 13:24, 30 April 2006 (UTC)
Er, I don't think that's strictly speaking true. If your model is "an exponential distribution with parameter 0.25", then by definition you're only modelling the time till the first customer. So you haven't said anything about any subsequent customer. However, if you also assume that shoppers occur independently, then you can use the memoryless property (which tells you how much longer to expect to wait for a shopper if one hasn't occurred yet—read the article—not what you want) together with that assumption (which tells you that the time you expect to wait is the same whether a shopper has occurred or not) to prove that the shoppers collectively form a Poisson process, and hence in particular you can use the strong Markov property to prove the required result. —Blotwell 22:07, 30 April 2006 (UTC)
I agree, that is entirely correct. What I was trying to say is that if you assume the process has no memory (which I took to mean that whatever happened or did not happen in the past was irrelevant for the expected value of future events) the maths needed to show that the time differences have negative exponential distribution proceeds essentially as sketched under Exponential distribution#Memorylessness (even though this covers a different question, namely the expected time to a single event, and not an ongoing process). I thought that that was what was being asked. This was somewhat cryptically expressed as 'that is what the "memorylessness" gives you'. I should perhaps have said: 'that is what the assumption of "memorylessness" gives you'. LambiamTalk 23:51, 30 April 2006 (UTC)
Ugh, I misread the question. It is modelled by X~Po(4) where X is the number of shoppers per minute. Can you then assume that Y~exp(0.25) from X?
And then, how on Earth do you prove that the difference in times is distributed as a negative exponential distribution?
I'm pretty sure I don't have to learn about this for my exam, but seeing this makes me annoyed... 03:23, 1 May 2006 (UTC)
The solution is implied by the previous answers. Did you look up the Poisson process article? The basic observation that may help you make the bridge is that if it takes time τ for the first event to happen after time t, then there were zero events in the interval [t,t + τ). So if x is the random variable standing for how long it takes for the first event after time t to happen, then Prob(x < τ) = Prob(#{events in the interval [t,t + τ)} = 0). I hope this helps. LambiamTalk 23:56, 1 May 2006 (UTC)
Yeah, figured it out, though by not that article, though that would've helped a lot. 07:30, 2 May 2006 (UTC)

## Ellipse-like shapes

Are there any formal names for 3-focus, 4-focus, etc. ellipse-like shapes?

Tuvwxyz 18:53, 30 April 2006 (UTC)

Perhaps if we had a definition, or even an illustration, we could be of more help. --KSmrqT 19:02, 1 May 2006 (UTC)
I can provide a fully-rigorous definition, despite not knowing the answer. As an ellipse with the usual 2 foci is the locus of all points such that the sum of the distances between the respective foci and the point is some constant, the natural analog is simply to increase the number of distances to sum to that constant. For example, with foci (0,1), (0,-1) and (1000,0) with a total distance of 2010, the shape would be very narrow near the third focus, but around the origin would more nearly resemble a 500-radius circle. (A contour plot of ${\displaystyle \sum _{i}d({\vec {f_{i}}},{\vec {x}})}$ shows how these look -- in particular, mine looks like a circle mated to a square-ish corner. I can't seem to get MediaWiki to like mine, though.) Note that it's meaningful for the constant to be so small that none of the foci are even within the curve! --Tardis 22:37, 1 May 2006 (UTC)

## daffy definitions from gfeometry

i need help find it some definitions for word like PI,ratio ect where i can look?

Well, Wikipedia's a good place to start. Try Pi and Ratio. —Keenan Pepper 22:05, 30 April 2006 (UTC)