Wikipedia:Reference desk/Mathematics

February 1

complex integration or integration w.r.t. complex measure?

According to Cauchy's theorem the integral of an analytic function along a closed curve is zero. But an analytic function is harmonic and so the integral should be its mean value ( not necessarily zero). This apparent contradiction may have something to do with what I suggested in the title. Who can explain it nicely?DeepthiPP 05:48, 1 February 2007 (UTC)

As I understand, f(c) is the mean value of an analytic function f(z) on the circumference of any circle with center c. It is an integral with respect to the arc length (real). Look up Gauss' Mean Value Theorem. Cauchy's theorem is about complex integrals. Hope this helps. Twma 07:00, 1 February 2007 (UTC)

Take a complex integral like $\int _{\gamma }f(z)\,dz$, the dz points along the tangent to the curve. So to go in a closed curve, the $dz$ will point in one direction (in the complex plane), and eventually must point as much in the other direction to get back to the starting point. So obviously for example for any closed curve $\gamma$, $\int _{\gamma }\ \,dz=0$; and has nothing to do with any "mean value". In a related way, if you were to integrate an analytic function in a "closed curve" on the real line, you can trivially see that it is always zero, because such a curve consists of a line that goes one way from a to b, and then exactly back the other from b to a, so the integrals along these two paths must be opposites, and when you add them you get zero. --Spoon! 17:04, 1 February 2007 (UTC)

Spoon! is right, but I'll describe it another way. Suppose the closed arc has length 1, and let t be a variable that measures arc length along the arc from some starting point. If the arc is $z=z(t)$, then the usual change of variables gives

$\int _{\gamma }f(z)\,dz=\int _{0}^{1}f(z(t))z'(t)\,dt.$

Thus, Cauchy's theorem is saying (under some conditions on f) that the mean value of $f(z(t))z'(t)$ on the arc is zero. McKay 05:06, 2 February 2007 (UTC)

The line integral along a closed curve need not be zero unless the vector integrand is ir-rotational. This is the case for analytic functions. Twma 03:03, 4 February 2007 (UTC)

adding every number from 1 to 100

i heard that there is a formula like adding 1 to the end of 100 so you get 1001 and then dividing by something to go 5050 and i don't know what it is... my math tutor told me but i just don't know could someone help me out.... it's not a formula but a manipulation of the number 100 and then dividing by something...

Thanks, J- http://jdswebservice.com/xmswx

Sums of the form 1+2+3+...+n are Triangular numbers. Dugwiki 18:44, 1 February 2007 (UTC)

Ah, the arithmetic series and mathematical ingenuity! It is said that this was figured by Carl Friedrich Gauss on his early years, when an evil teacher wanted to keep his students busy for a while. Take a look at arithmetic series, summation and yes, triangular numbers — Kieff | Talk 19:53, 1 February 2007 (UTC)

The above is right - check the link - but if you want a simple way to work this out what you need to do is add the beginning and end numbers, then the next from beginning and end

eg

100+1
99+2
98+3
97+4
.
.etc
.
53+48
52+49
51+50

Note that each time the result is 101 eg 98+3=101

Also note that there are 50 or these sums

So the result is 50 times 101 = 5050 (that's the answer)

Hopefully that should explain what your teacher was on about.87.102.77.95 19:37, 1 February 2007 (UTC)

By the way thats 100x(100+1)/2 (there are fifty pairs = 100/2, and each pair adds to 101=100+1. = n(n+1)/2 when you add the number from 1 to n.87.102.77.95 19:45, 1 February 2007 (UTC)

Generalizing, the sum S of a finite arithmetic series consisting of n terms is S = n/2 (a1 + an). n/2 is 50, a1 is 1, and an is 100. Arithmetic series are things like 1 + 3 + 5 + (and so on). —The preceding unsigned comment was added by 172.146.58.73 (talk) 08:52, 26 February 2007 (UTC).

Statistics/Normal Distribution (moved here from Miscellaneous desk)

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12.If the college requires an IQ of at least 95, how many of these students will be rejected on this basis regardless of their other qualification?

First, this would be better suited to the math reference desk, and second, we don't answer homework questions. I'm certain that the correct way to solve this problem can be found in your text book. Dismas|^(talk) 13:21, 1 February 2007 (UTC)

Right, I could point you at a bunch of articles here but they will likely be more confusing than your text (not a slam to wikipedia). Just so you know, what we do here rather than perhaps do your homework for you is steer you toward references that might clear up the confusion you are having that prevents you from doing it yourself. While that is hard to diagnose "over the internet", if you give us some clue to what part you don't understand then we can help. But this does belong on the math desk. --Justanother 15:03, 1 February 2007 (UTC)

How many standard deviations below the mean is 95? What Z score does that correspond to in a normal distribution table in your textbook? What portion of the normal distribution lies below and above that Z score? How many students does that proportion imply? (Boy, do I ask a lot of questions, when all you wanted was a simple answer to the homework!). Post your answer and someone might even tell you if you got it right. Edison 19:11, 1 February 2007 (UTC)

• (Question was also posted at science Ref Desk and moved here with replies. Please do not double post.) Edison 16:19, 2 February 2007 (UTC)

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12.If the college requires an IQ of at least 95,how many of these students will be rejected on this basis regardless of their other qualification,,,,, —The preceding unsigned comment was added by 203.128.5.84 (talk) 11:42, 2 February 2007 (UTC).

The number of applicants rejected will be equal to the fourth root of the number of femtoseconds you took to write this question, integrated with respect to x over a range from minus infinity to the time it would take you to do your own homework. — QuantumEleven 13:18, 2 February 2007 (UTC)

If you did your own statistics homework, you might be able to estimate where in the roughly Gaussian distribution of class grades yours is likely to fall if you keep up this behavior. -- mattb @ 2007-02-02T13:28Z

February 2

pentiminoes

Can someone show me how to fit 12 diffrent penteminoes into a six by ten rectangle in 5 diffrent ways?

You might want to try and look at the article on pentaminoes. The question sounds like homework, so I'm not going to help you further. Oskar 07:16, 2 February 2007 (UTC)

Well, the proper article would be pentomino. — Kieff | Talk 15:51, 2 February 2007 (UTC)

Here's one possiblity as a starter:

   ***@@@@@**
   *%::::%%:*
   *%%%:%%::*
   @@*%@%*:@*
   @***@**:@@
   @@*@@@**@@

– b_jonas 19:04, 2 February 2007 (UTC)

Poincaré conjecture

The Poincaré conjecture says that if a 3space is enough like a 3sphere, then it is like a 3sphere. Um, so what does that mean? If a Poincaré-violating manifold (PVM) exists, what properties might it have that distinguish it from a 3sphere? —Tamfang 08:14, 2 February 2007 (UTC)

The conditions that make a 3-dimensional space "close enough" to a 3-sphere for the Poincaré conjecture to apply are:

• The space must be a manifold - if you look at a small enough neighbourhood of any point, it will resemble Euclidean 3-space.
• The space must be simply connected - given two points in the space, any path between these points can be continuously transformed within the space into any other path between the same two points.
• The space must be a compact space - roughly speaking, it has a finite size and does not extend to infinity in any direction.
• The space must be without boundary - roughly speaking, it doesn't have any edges where you can suddenly fall out of the space.

A Poincaré-violating manifold would have to posess all of these properties, yet still fail to be topologically equivalent to a 3-sphere. I'm not actually sure what topological property might allow you to easily distinguish such a manifold from a 3-sphere. Gandalf61 11:33, 2 February 2007 (UTC)

seeking hyperbolic cookbook

It appears that the algebraically easiest way to do stuff in hyperbolic geometry is with the hyperboloid model, in which each isometry apparently can be expressed with a matrix. But which matrix? I've read a couple of books on h.g. and plenty of websites, but found nothing that goes beyond "it can be done." Where do I find a book (or website) that will tell me:

• Given a line, what matrix expresses a unit translation along that line?
• Given a plane, what matrix expresses a reflexion in that plane?

...and like that. I might be able to work it out for myself, given free time and a better night's sleep than I've had in recent years! —Tamfang 08:47, 2 February 2007 (UTC)

If you want to experiment with the hyperbolic plane, a nice place to start is

• Stillwell, John (1992). Geometry of Surfaces. Springer-Verlag. ISBN 978-0-387-97743-0.

One reason we use so many different models is because each has its virtues. Also remember that all isometries can be composed using reflections (reflexions) only. In the hyperboloid model, a hyperbolic line corresponds to a plane of the ambient space, and any isometry must map the surface to itself. --KSmrq^T 18:23, 3 February 2007 (UTC)

Does the Stillwell book contain answers to the specific questions I ask, or only principles that I already know as in your paragraph? —Tamfang 00:41, 4 February 2007 (UTC)

triangle angle bisectors2

From Wikipedia:Reference_desk/Mathematics#triangle angle bisectors

Somebody asked "is there an expression to get each side of the triangle from given values of the angle bisectors?…86.132.237.140 21:07, 29 January 2007 (UTC)"

I've got this equation

(C2-ab)(a+b)2/c = (B2-ac)(a+c)2/b = (A2-bc)(c+b)2/a

Where capitals are bisector lengths, lower case are side lengths..

Now I could solve this.. But given that I usually fail to manipulate equations through more than two steps without making a mistake could someone check that this is correct.. Or, is there a better way to do this? Thanks87.102.4.6 11:28, 2 February 2007 (UTC)

I've just read an article in the American Mathematical Monthly of Jan 94 which shows that, given 3 arbitrary values, a unique triangle exists which has these as angle bisectors. As the easily-obtained equations giving the bisectors separately in terms of the sides were not manipulated into a vice versa form, I rather assume that this is not possible, however.86.132.238.155 15:44, 2 February 2007 (UTC)

Statistics/Normal Distribution

In a mathematics examination the average grade was 82 and the standard deviation was 5.All students with grades from 88 to 94 received a grade of B.If the grades are approximately normally distributed and 8 students received a B grade,how many students took the examination"""" 12:03, 2 February 2007 — Preceding unsigned comment added by 203.128.5.84 (talk • contribs)

For a normal distribution centered on 82 you need to know the fraction of the distribution (a value between 0 and 1 - in this case less than 1) falling between 88 and 94 when the deviation is 5. You can get the equation of the distribution when the average is 82 and the deviation is 5. Then you need to integrate the equation between 88 and 94 (or find the fraction using look up tables..) this gives you the fraction of the students getting a B. Then divide 8 by that fraction to get the total number of students...87.102.4.6 12:44, 2 February 2007 (UTC)

Note that (from Variance) the "the square root of the variance, called the standard deviation" - therefor your variance is 5x5 = 25.87.102.4.6 12:50, 2 February 2007 (UTC)

Therfor you need to integrate the below equation with respect to x between 88 and 94, with μ=82, σ=5, σ²=25.(I can't think of a simpler way - but there usually is..)87.102.4.6 12:54, 2 February 2007 (UTC)

${\frac {1}{\sigma {\sqrt {2\pi }}}}\;\exp \left(-{\frac {\left(x-\mu \right)^{2}}{2\sigma ^{2}}}\right)\!$

The form of the OP hardly suggests that the questioner is going to find integration a meaningful idea, given the essential triviality of the problem. I don't think it unreasonable to tell him/her that the key will be to use Normal tables with z=1.2 and 2.4—86.132.238.155 15:56, 2 February 2007 (UTC)

Hyperbolic geometry

I too have a simple question about hyperbolic geometry.. What lines are considered to be 'straight' in hyperbolic geometry say of a Hyperboloid surface? It seem to me that for spherical geometry great circles are the equivalent of straight lines - since they represent the longest distance.. For a hyperboloid (of one sheet)

${x^{2} \over a^{2}}+{y^{2} \over b^{2}}-{z^{2} \over c^{2}}=1$  (hyperboloid of one sheet),

are lines in the plane x/y=constant condsidered to be the equivalent of straight lines (and also parallel)(think yes)?

What about the circles formed by intersection with z=constant - are two circles formed by z=a,z=b considered to be parallel - and are they considered to be straight lines(thinks not)?

I'm using the term 'straight line' to mean the shortest distance between two points on the surface - I looked at the page and understood I think the parallel concept - but didn't get how to find a line that is the equivalent of a straight line in conventional geometry.

If you can make an answer simple I would appreciate it.87.102.4.6 12:15, 2 February 2007 (UTC)

First, it's not a hyperboloid of one sheet, it's one sheet (chosen arbitrarily) of a hyperboloid of two sheets.

?? Are you sure about this - this image shows only one sheet - that's the equation I gave? Image:HyperboloidOfOneSheet.PNG 87.102.4.6 16:20, 2 February 2007 (UTC)

True, that's the equation you gave, and I ought to have looked closer at it, rather than assuming that you had in mind the hyperboloid model of the hyperbolic plane. Since I can't tell you anything about geodesics on the hyperboloid of one sheet (other than to agree with your guess about both questions above), I'd delete my remarks below but that's bad form. —Tamfang 19:49, 2 February 2007 (UTC)

That's ok. thanks.83.100.183.48 20:05, 2 February 2007 (UTC)

In this model, hyperbolic straight lines are represented by planes passing through the origin, ax+by+cz=0, or (if you prefer) by the intersections of such planes with the hyperboloid. Note that this also describes great circles on a sphere!

(By the way, great circles are "straight" not because they are the longest possible "lines" on the sphere but because the shortest curve between two points on a sphere is a segment of a great circle.)

The lines that you describe are a valid subset of these, with c=0, but they're not parallel: they all intersect at x=y=0. A parallel set would "intersect" in a point on the circle (a line on the cone) $x^{2}+y^{2}=z^{2}$. —Tamfang 16:06, 2 February 2007 (UTC)

Given that x=y=0 doesn't have a point on the hyperboloid I described would you reconsider your last point - to be honest you've confused me a bit - are you talking about the same thing - look at the picture..87.102.4.6 16:34, 2 February 2007 (UTC)

I am not aware of any model of hyperbolic geometry based on the hyperboloid of one sheet. As Tamfang says, the model that is usually called the "hyperboloid model" is based on one sheet of a hyperboloid of two sheets. Other standard models of hyperbolic geometry are the Klein model, the Poincaré disc model, the Poincaré half-plane model and the pseudosphere. To define a model based on the hyperboloid of one sheet, you would have to define a family of curves within the surface that meet the axioms of "straight lines" in hyperbolic geometry. The curves x/y=constant within the surface could be a sub-set of this family, but there are not enough of them - there is only one curve through each point, and most pairs of points within the surface are not connected by one of these curves. The circles z=constant are certainly not candidates for "straight lines" because they are finite whereas all "straight lines" in hyperbolic geometry can be extended to infinity (as in Euclidean geometry). Gandalf61 11:07, 3 February 2007 (UTC)

Thanks.87.102.9.55 11:44, 3 February 2007 (UTC)

summation of fractions

Σ(1/n) = ?

Σ(1/n²) = ?

Σ(1/n³) = ?

—The preceding unsigned comment was added by 59.95.71.153 (talk • contribs).

Please, do your own homework. Meanwhile, take a look at summation and infinite series. — Kieff | Talk 15:47, 2 February 2007 (UTC)

For the sums to infinity see..

Σ(1/n) = ? See Harmonic series (mathematics)

Σ(1/n²) = ? See Basel problem

Σ(1/n³) = ? See Apéry's constant

whether or not this will help is another thing - also see Riemann_zeta_function#Various_properties or just Riemann_zeta_function. Good luck87.102.4.6 16:26, 2 February 2007 (UTC)

Equations

What is the best way to solve equations in fractions? Cam you please show me ways of solving it? Thank you if you have time

It would help if you could be a little more specific - take a look at the pages linked to by this link http://ww.google.co.uk/search?hl=en&q=algebra+fraction&meta= do any of the first couple of pages have the sort of thing you're asking about. - if so tell us and I'll try to give you a good point to start learning.

Also maybe this http://www.tech.plym.ac.uk/maths/resources/PDFLaTeX/alg_fracs.pdf is the sort of thing you want..

I'm assuming that you already understand basic fractions - if so this link should all amke sense http://www.mathleague.com/help/fractions/fractions.htm

Or just type in an example of the type of thing you want to solve...83.100.183.48 18:23, 2 February 2007 (UTC)

Twin Primes

The sequence 4, 6, 12, 18, 30, 42, 60, 72, 102, 108 is related to twin primes, that much I know. I just don't know the exact nature of their relation, so any help would be appreciated.

Take a look at Twin_primes#The_first_35_twin_prime_pairs - the sequence should become clear - if you can't see what it is ask for a hint...83.100.183.48 18:26, 2 February 2007 (UTC)

Just type the word sequences in a search engine search box ... you'll find The On-Line Encyclopedia of Integer Sequences [OEIS] which tells you when you paste your sequence : "A014574 Average of twin prime pairs."

Or just enter your sequence in any search box, to find that the first hit is : "Twin Primes -- from Wolfram MathWorld - The first few twin primes are n+/-1 for n==4 , 6, 12, 18" ... -- DLL ^{.. T} 11:08, 3 February 2007 (UTC)

OEIS brings up this, which i dont understand either. Can someone explain this in simpler terms? Omnipotence407 03:17, 4 February 2007 (UTC)

Try writing the twin prime pairs out in a list, and then write your sequence above it. Compare each pair in the list with the corresponding element in the sequence. It should be obvious what the connection is. Maelin (Talk | Contribs) 07:03, 4 February 2007 (UTC)

I guess its not obvious enough for me. Omnipotence407 03:19, 5 February 2007 (UTC)

Okay, here are the two lists written out.

${\begin{matrix}{\mbox{Twin prime pairs: }}&(3,5)&(5,7)&(11,13)&(17,19)&(29,31)&(41,43)&(59,61)&(71,73)&(101,103)\\{\mbox{Your sequence: }}&4&6&12&18&30&42&60&72&102\end{matrix}}$

Can you see the connection now? Maelin (Talk | Contribs) 04:02, 5 February 2007 (UTC)

Wow, yeah, my bad. I was thinking too hard. Thanks Omnipotence407 14:36, 5 February 2007 (UTC)

February 3

algebra 1

i need help on this problem:

find three solution on for this equation using -1,0,1 for x: 5x+7y=8

To do this question, you'll need to substitute those three numbers into x one at a time. So firstly, let x = -1. Therefore: 5(-1)+7y=8 which is -5+7y=8. Then you can solve for y. See this link if you need help on solving equations. Then do the same thing, but this time with x=0, then x=1 and you'll have three solutions. Hope this helps. - Akamad 22:30, 3 February 2007 (UTC)

February 4

Another triangle question

Rather similar to the angle-bisector questions above, is it possible to determine the sides of a triangle, given the perpendicular distances from the circumcentre to the sides?…81.151.247.173 00:09, 4 February 2007 (UTC)

Rectangular numbers

Please inform me what are rectangular numbers— Preceding unsigned comment added by 168.167.72.20 (talk • contribs)

The term rectangular number does not seem to have a standard definition. Looking for references, I found several sources ([1],[2],[3]) where the term is used to refer to an integer of the form n(n+1) - so the first few rectangular numbers using this definition are 2, 6, 12, 20, 30. These numbers are also called pronic numbers or oblong numbers. But other sources use rectangular number to mean a composite number, or, more rarely, a composite number that is not a square number. Gandalf61 09:46, 4 February 2007 (UTC)

Set of all integers

Can I have the set-builder notation for "x must be in the set of all integers"?

I know that it is usually represented by the letter I (or Z?) , but how can I express that x must be a member of that set e.g. for a formula?

86.146.72.220 10:55, 4 February 2007 (UTC)

The letter for integer is indeed blackboard bold Z (from the German Zahlen). The set is written as $\lbrace x:x\in \mathbb {Z} \rbrace$. For more info, see set-builder notation. Laïka 11:29, 4 February 2007 (UTC)

February 5

What is a tangent plane to z=f(x,y) at (a,b)?

My grand children just finished calculus of functions of one real variable. They can take derivatives. The tangent at a point on a curve is defined intuitively as the limiting position of secant line through the point. I am supposed to introduce functions of two variables without dot-cross products. Tangents to curves in spaces and differentiability are not in the syllabus. Question: With my hands tied, how can I convince my kids, intuitively psychologically and emotionally but need not be logically mathematically, that the plane z=f(a,b)+D_x f(a,b)(x-a)+D_y f(a,b)(y-b) is usually frequently TANGENT to the NICE surface z=f(x,y) at (a,b)? What does it mean by a tangent plane? It appears that I have only one option, that is

Shut up and accept it because I am the grand father! The appearance is an obvious extension of y=f(a)+f'(x)(x-a).

What is your alternative solutions to this teaching situation? Thank you in advance. Twma 07:06, 5 February 2007 (UTC)

Your option sounds like a great idea. Let me know how it works out! ;-)

But seriously, if it is "an obvious extension", what is so obvious about it? I claim that the core intuition is that we are always looking for a local linear approximation. For a curve, we seek a line; for a surface, a plane.

How do we find the plane? Run a vertical slice through the surface; we should see a curve (the cut through the surface) and a tangent line (the cut through the plane). Make one slice parallel the x axis, and another parallel the y axis. That should give us two independent lines through a common point, and thus determine a plane, the tangent plane.

A nice general description of a linear subspace (sometimes called a flat), such as a line or a plane, is a point plus a vector space. The vector space gives all allowed displacements from the point, and a basis for that vector space allows us to move anywhere we like in the flat.

It could help to show

• radial symmetry, like x²+y²
• reflection symmetry, like xy
• asymmetry, like x+y³
• a crease, like |x|+y²
• a point, like (x²+y²)^1/2
• a nasty point, like |x+y|+|x−y|

I see no need for cross products and "normal vectors"; the latter are not truly vectors, and fail to generalize.

Incidentally, we can attempt higher-order fits; that's what an osculating circle is. --KSmrq^T 08:32, 5 February 2007 (UTC)

Linear subspaces, flats were not available to me for explanation. Exactly two slices as you described were at my disposal. I did that and followed up with ice cream, candy and cookies. My grandchildren were happy. At the end in a subtle way, I pleased them to be good drivers (get good marks) without knowing anything about petrol chemistry (mathematical theory). Am I an acceptable grand father? Thanks. Twma 00:50, 7 February 2007 (UTC)

This is a retarded question, but...

if $a^{m}$ X $a^{n}$ = $a^{m+n}$, then $a^{m}$ X $b^{m}$ = $ab^{m}$? In other words, do I multiply numbers where the exponents(?) are equal? Is that against any mathematical laws? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:43, 5 February 2007 (UTC)

Also, in an unrelated question, what is the square root of this whole thing: ${{16^{4x}}^{2}}$? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 14:02, 5 February 2007 (UTC)

a^m x b^m = (ab)^m is true.

and the square root is 16^{2x^2} - you half the exponent - (taking logs, dividing by two - then exponentiating again..)

square root (nx) = nx/2

In fact using the first rule you supplied (which is correct) you can get

(nx/2)2 = nx/2 x nx/2 = nx/2+x/2 = nx

so:

square root (nx) = square root ((nx/2)2) = nx/2

(I get to say QED now)87.102.8.103 15:45, 5 February 2007 (UTC) Hope that helps.87.102.8.103 15:34, 5 February 2007 (UTC)

I'm not sure why you characterize the question as "retarded". In fact, I consider it a good question that brings up some valuable mathematics.

We can expand a^m×aⁿ as m copies of a followed by n more, which is m+n copies, so equal to a^m+n. But a^m×b^m expands to m copies of a followed by m copies of b, which is something completely different from m copies of a×b. We need something extra, that a×b = b×a, to complete a proof. In many important mathematical situtations, with objects more sophisticated than ordinary numbers for a and b, this commutative law of multiplication does not hold, and the proof does not go through.

Your second question requires better typesetting to be seen clearly, asking what is the square root of

$\left(16^{4x}\right)^{2}.\,\!$

This also rewards careful examination. Is it the case that the square root of a² equals a, for all a? The answer is, no. For example, −2 squares to 4, and we conventionally take the square root of 4 to be +2. We can expect this issue to arise any time more than one quantity can square to the same thing, which is common. In the given example, we should ask if 16^4x can ever be negative; if not, then the answer will be 16^4x. So, try it. --KSmrq^T 22:05, 5 February 2007 (UTC)

K Thanks very much. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:23, 6 February 2007 (UTC)

Curves of quickest descent

What precisely are bernoulli's curves of quickest descent?? — Preceding unsigned comment added by 59.183.58.204 (talk • contribs)

See our article on the brachistochrone curve. Gandalf61 15:56, 5 February 2007 (UTC)

mark up question

I have been told that when we markup costs for our work instead of just for example muliplying the cost times 1.20 for a 20% makup we should divide by 0.80. Which is correct?

Multiplying by 1.2 gives a 20% markup; dividing by 0.8 is the same as multiplying by 1.25, so gives a 25% markup. All on cost, though - if you define markup as wrt the derived figure, naturally the arithmetic will be different.…86.146.174.174 17:43, 5 February 2007 (UTC)

• Remember that dividing is the same as multiplying by one over the number. Mathmo ^Talk 20:09, 5 February 2007 (UTC)

Infinity + 1

This isn't a homework question, but it's based on something our teacher explained in class. I now have some idea why he didn't continue any farther. It's a geometric coloring problem. Take a circle of radius r, a point on that circle, and two colors. Every point on the circle must be a different color from any point on the circle that's unit distance away from it. Is it, or is it not, possible to color this circle? In some cases, where r=1 for instance, if you start coloring clockwise from a point, you return to your starting point within a finite number of steps. If this covers an odd number of points, the coloring is impossible, otherwise it's easy. What about when r is irrational, though? If you have to cover an infinite number of points before you return to the starting point, do you wind up with a contradiction, or not? Or is it like quantum mechanics, where you get both? I guess what I'm asking is, is infinity even or odd? Black Carrot 20:15, 5 February 2007 (UTC)

There's a paradox known as Thomson's lamp which addresses this issue; if you can flick a light switch an infinite number of times in a second, is it on (odd) or off (even)? I think that the current opinion is that it is both, and that defining infinity as either odd or even is meaningless (if it's odd, you can just add one to make it even and vice-versa). See also Hilbert's paradox of the Grand Hotel, which considers a similar problem. Laïka 20:58, 5 February 2007 (UTC)

You should not confuse philosophical issues with actual mathematics. The question you asked is about a concrete mathematical problem (the circle part, not the "is infinity even" part). Unless I am somehow mistaken, the answer is yes, but this may require using the axiom of choice. And no, it certainly cannot be both (in classical logic, anyway, and unless we are using an inconsistent system), though it can be independent from a given axiom system (for example, it might be independent of ZF sans choice). By the way, you meant irrational circumference, not irrational radius. -- Meni Rosenfeld (talk) 22:27, 5 February 2007 (UTC)

No, I mean an irrational radius, if I worked the formula for chord through correctly. The problem says "unit distance", not "unit distance along the circle". I think this is slightly different from Thomson's lamp and the Grand Hotel, and you're right, the divisibility of infinity may not be related either. I can't imagine, though, what "yes" even means, let alone how you got it as an answer. Could you give more detail? Black Carrot 22:46, 5 February 2007 (UTC)

What you really want to know is not whether r is rational or irrational, but whether a chord of unit length subtends a rational or irrational angle (as a fraction of the circle, not in radians). In the case of r=1, the answers are the same, but this will not be true in general.

Anyway, suppose the angle is irrational. Then you get a group action from the integers to the circle, generated by the operation "move counterclockwise by an angle with a chord length of 1", and the orbit of the action is countably infinite. From each orbit, pick one point (this is where you use the axiom of choice) and call it red; then the points one step away are black, the points two steps away are red, and so on.

Is the axiom of choice necessary to prove this result? My guess is yes, but I don't know for sure. You could prove it, if you could show, say, that the set of all red points is not Lebesgue measurable, or lacks the property of Baire (though these proofs assume the consistency of large cardinals). --Trovatore 23:16, 5 February 2007 (UTC)

Oh, actually you don't need any large cardinals to show AC is necessary if you can show the set of all red points lacks the property of Baire. If all you can show is it's not Lebesgue measurable, you need an inaccessible cardinal. --Trovatore 00:29, 6 February 2007 (UTC)

You're right, I copied the formula wrong. So, supposing it's irrational. I can pretty much understand the group action page, it seems to mean "shuffle everything in the set around." Naturally, we can say more specifically that the action consists of rotating the circle by a particular angle, and mapping everything from before to after. We chose irrational angles specifically to make the orbit infinite, so naturally the orbit is infinite. No problems so far. That's about where you lose me, though. How is it "countably" infinite? I could have sworn it would hit every point on the circle if we chose the angle/circumference relation properly, which would make it "uncountably" infinite, right? A complete line segment (and therefore the circumference of a circle) contains uncountably many points, corresponding to the real numbers on the interval [0,1). Then, you mention the axiom of choice, defined in the article as

"Let X be a set of non-empty sets. Then we can choose a single member from each set in X."

It says nothing about what that choice would entail, and reading related articles, I don't think it's supposed to. This can be applied to the current situation, as far as I can tell, by saying "if I'm trying to color this set of points, then at each and every point, there exists at least one color I can specifically choose." I kind of already knew that. So, anyway, we move on with the proof. We pick an orbit (by my reckoning, the orbit, but whatever) and pick a point in it. We then move along the orbit, coloring each point, alternating (in this case) red and black. ...Then it stops. How has that gotten me farther along? As far as I can tell, you're interpreting the axiom of choice as saying

"Let X be a set of non-empty sets. Then we can choose a single member from each set in X to satisfy any random conditions we can think of."

...which is ridiculous. But hopefully I misunderstood. Black Carrot 01:46, 6 February 2007 (UTC)

So your basic misunderstanding is that you think you can get to every point on the circle, starting from a particular point. That's not true. You can't get from x to y unless there's a finite sequence of points on the circle, the first being x and the last being y, such that successive points are distance 1 from each other. That's only countably many points, for each starting point.

That means there are in fact uncountably many orbits, not just one. To choose one point from each orbit requires AC. --Trovatore 02:11, 6 February 2007 (UTC)

The axiom of choice is not used to choose a color for every point, but rather a point for every orbit. To clarify the argument, the main observations are:

1. Two points in the same orbit can be reached (one from the other) by a sequence of chords of length 1.
2. Two points in different orbits cannot be reached (one from the other) by a sequence of chords of length 1.
3. Therefore, the selection of colors for the points of one orbit do not pose any restrictions on the selections of colors for other orbits.
4. Therefore, the problem reduces to finding a coloring for each orbit.
5. Every orbit can be put in a one-to-one correspondence with $\mathbb {Z}$ (which, by the way, means it is countable), by fixing some point and mapping it to 0, and proceeding naturally.
6. Since $\mathbb {Z}$ can be colored (black for even numbers, red for odd numbers), so can an orbit.
7. To have a coloring which applies to all points, we just need to choose, for every orbit, a correspondence between it and $\mathbb {Z}$.
8. This means that we need to consistently choose a point from any orbit.
9. This can be done using the axiom of choice.
10. To find the color of any point: Take its orbit; Take the point in the orbit given by our choice function; Take the induced correspondence with $\mathbb {Z}$; Take the integer to which our original point corresponds; Color it accordingly.
11. This is guaranteed to satisfy our requirement.

-- Meni Rosenfeld (talk) 09:36, 6 February 2007 (UTC)

I appreciate the numbering, that's a lot clearer. And I'm sorry, that makes sense as an application of the axiom of choice. I don't see that fixes the proof, but at least that wasn't the problem. Now, though I still don't quite follow how there are uncountably many countably-sized orbits, let's take that as assumed.

1. From each orbit, you choose a point. Call it Bob. (This is allowed by the axiom of choice.)
2. Proceeding clockwise at intervals of length 1, you color in the points of each orbit with alternating colors.
3. Since the points on an orbit can be put (we're assuming) into one-to-one correspondence with the natural numbers, by virtue of their countability, this is the same as coloring all the even numbers Red and all the odd numbers Black.
4. What, exactly, is the color of the point one unit counterclockwise of Bob, and what's its name in terms of the natural numbers? Or in other (and I'm hoping more precise) words, in $\mathbb {N}$, what has 1 as its successor?
5. You use, however, $\mathbb {Z}$, which I believe means the integers. This changes things a bit, in that it makes it harder to point out the problem. What, exactly, is the color one step counterclockwise of Bob, in terms of the integers, if Bob = 0? Well, -1, so that would be Black and no problem, an argument which could continue forever. However, in both cases, we're assuming that a set of numbers that's not well-known for its circularity can loop back around on itself without difficulty. I've never heard of anything that fits with that, and without some clear rule about how it loops back on itself, it can't solve the problem. It's obviously not hard to color a line, the question is whether I can color a circle. Black Carrot 20:15, 6 February 2007 (UTC)

I do indeed use the integers, $\mathbb {Z}$. It makes no sense to use the natural numbers here, since as you pointed out, you can continue either clockwise or counterclockwise. As a side note, I'll mention that $\mathbb {N}$ is sometimes taken to include 0, sometimes not. Whatever the case, the first element of the set (be that 1 or 0) clearly is not the successor of anything.

Suppose r is such that the angle between successive points is 1 (radian). So you begin with some point and mark it 0. You go counterclockwise an angle of 1, and mark that point as 1. You continue another 1, so that you are now an angular distance of 2 from the original point. Mark that point 2. Continue in this way, marking points 3,4,5, and 6. Now, if you continue another 1, you will go past the 0 point, and arrive at angle 7-2π, or roughly 0.71681. Mark that point as 7. Geometrically, this point is quite close to point 0, but this has absolutely nothing to do with our problem, since we are not imposing any restrictions on "close" points or whatever, only at point at angular distance of exactly 1. So the only way to get from one to the other is by making the complete loop of 7 steps. You can continue, marking as 8 the point at angle 1.71681, 9 at 2.71681,..., 12 at 5.71681, and then 13 at 13-4π, roughly 0.43363. Again, this point is geometrically close to the 0 point but is quite distant in terms of steps of angle 1. You can continue this forever; Since π is irrational, you will never return to the 0 point. You can get arbitrarily close to it, that's for sure - for example, the point marked 710 will be at angular distance roughly 0.00006. But you will never reach it exactly, so you will never be in a situation where you want to color it one way but have already colored it another way.

Of course, you can also go clockwise and mark as -1 the point at 5.28318, and so on. Again, you will never reach the starting point.

As already mentioned, the fact that geometrically you get arbitrarily close to 0 is irrelevant. We're only interested in exact distances, so as far as we are concerned, the set of points encountered in the process is the same as the set of integers. Of course, my thought experiment above is unnecessary - you can redefine the problem in terms of modular arithmetic, consider the equivalence classes of the relation $a\equiv b\iff \exists m,n\ \mathrm {s.t.} \ a-b+m=nL$, and so on. But that's just technical details. The essential part is that since you never reach any conflict, it is possible to have a legal coloring. -- Meni Rosenfeld (talk) 21:53, 6 February 2007 (UTC)

How do you never reach a conflict? Sure, it would take infinitely many steps to get back to your exact starting point, that's why I chose that angle, but what's wrong with that? Black Carrot 05:30, 7 February 2007 (UTC)

And in what may be a related question, does induction extend to uncountably long sequences? The article used to say it doesn't, but now it doesn't mention it, and I never found out why or even whether that's true. Black Carrot 05:30, 7 February 2007 (UTC)

You're over-complicating that aspect of it, Black Carrot. The only way there can be a conflict is if two points just one step apart are of the same color. All you need to check is that the assignment prevents that from happening. You don't need to check what happens when they're infinitely many steps apart (or even two steps, for that matter, as long as you prove you're OK for one step). --Trovatore 06:21, 7 February 2007 (UTC)

This again raises the question of whether you want to discuss this as a philosopher or as a mathematician. Philiosophy-wise, you can use words like "never" and "reach". Mathematically, these may only be used for infomral communication, and for actual arguments you should stick to words like "exists" and "for all". Consider the set of angles $A=\{x|\exists m,n\in \mathbb {Z} \ \mathrm {s.t.} \ x=m+2\pi n\}$. This is the set of all angles you can reach by steps of angle 1 (which is equivalent to a chord of length 1 for some specific radius). From the irrationality of π it trivially follows that for any $x\in A$ there exist unique $m,n\in \mathbb {Z}$ such that $x=m+2\pi n$. So we may define the color of x to be black if the unique m in its representation is even, and red if it is odd. It is then trivial to show that for all $x,y\in A$ with x and y an angle of 1 apart, they have a different color. That's the only restriction we wanted, so this is a legal coloring for A. Trying to find shortcomings in a clearly correct solution is not very helpful.

Of course, this only deals with a single orbit. We want to do the same for every orbit. This is easy, but we need the axiom of choice to ensure that every orbit has a consistent origin. This gives a coloring for the entire circle which satisfies what we were after - that any 2 points an angle of 1 apart have a different color. And this can be trivially generalized to different radii, where a different angular distance is sought.

About induction - standard induction only deals with proving properties that hold for every natural number. I'm not sure what you mean by "uncountable sequence" - if it is uncountable, it cannot be a sequence in the ordinary sense. There is, however, something called transfinite induction, which can be used to prove properties for every ordinal number, but that is more advanced. -- Meni Rosenfeld (talk) 15:34, 7 February 2007 (UTC)

Am I missing something about the problem? It seems to me that there is a fairly simple and elementary solution. As I understand the problem, an equivalent problem is to come up with a colouring of the real numbers modulo 1, such that numbers a distance δ apart are assigned different colours, where 0 < δ ≤ ½. Identifying colours with natural numbers, we are seeking a function g : [0, 1) → N such that g({x}) ≠ g({x+δ}) for all real x, where {x} = x − floor(x) denotes the fractional part of x. A solution is given by g(z) = floor(nz), where n = ceiling(1/δ).  --Lambiam^Talk 05:56, 13 February 2007 (UTC)

How can you color a point, considering that geometric points have no extent? Mr.K. (talk) 18:55, 17 February 2007 (UTC)

With my solution you actually colour the half-open intervals [k/n, (k+1)/n) for k = 0, ..., n−1, which do have extent, but only one-dimensionally. For mathematicians, to "colour" a set S simply means to define a function from S to some finite set, subject to certain restrictions. If you are an intuitionist, you believe that all functions are continuous, and so if you colour a connected subset of the real line or the plane, you cannot use more than one colour.  --Lambiam^Talk 23:11, 19 February 2007 (UTC)

Here I'll try to make this clearer to people who may not be familiar with different orders of infinity. The number of points on a circle is the same as the number of real numbers between 0 and 1, which is infinitely larger than the number of integers, as proven by Cantor's diagonal argument. So if you start at one point on the circle and keep taking steps of 1 unit, colouring each point, after an infinite number of steps it may look as if the whole circle is coloured, but actually you've only coloured a relatively very small subset of all the points on the circle. If you choose a point at random you will almost certainly find that you haven't coloured it, though if you choose any other nearby point, (even very close nearby), there will be an infinite number of points you've coloured between the two. In other words, points you have coloured can be found arbitrarily close to the point you chose at random. --Coppertwig 14:20, 22 February 2007 (UTC)

Here's an example of a point which was not coloured: the point half a unit from the first point you coloured. Proof: Suppose that point was included in the set of all points you can reach in 1-unit steps from the first point. Let k be the number of 1-unit steps needed to reach the point, and let n be the number of times you have to go around the circle to get to that point. Then $k=2n\pi +0.5$, therefore $\pi =(2k-1)/4n$, therefore $\pi$ is a rational number; but we know $\pi$ isn't a rational number, so that's a contradiction, so the supposition must be false. A similar proof can be made using any rational number instead of one-half; but rational-number steps still cover only a relatively small subset of all the real numbers that need to be coloured. --Coppertwig 14:29, 22 February 2007 (UTC)

February 6

Sterlings

How much is 4,596,411 pounds of sterling in U.S dollars.

[4] Black Carrot 00:56, 6 February 2007 (UTC)

9.04389828 million U.S. dollars, at whatever rate of exchange Google is using currently.…86.146.174.174 01:50, 6 February 2007 (UTC)

Sum of Normal distributions

It looks to me that if f(x) and g(x) are normal PDFs, then (f(x)+g(x))/2 is also a normal PDF. Is this so? What are the parameters of this new PDF? --대조 | Talk 09:22, 6 February 2007 (UTC)

Unfortunately, no. I think this is never the case (unless f and g are the same), but it is easiest to consider distributions which have very different means and very low variance. Then the average PDF will have two peaks, so the distribution will be bimodal which is clearly not normal. Compare taking the average (or any linear combination) of the variables which is normal, with the obvious mean and a variance which depends on the covariance of the variables. -- Meni Rosenfeld (talk) 09:45, 6 February 2007 (UTC)

And, in the latter case of linear combination of the variables, the pdf should be a convolution of the two original pdf. --Lemontea 01:37, 27 February 2007 (UTC)

You are, of course, right. Thank you. Once again, I am reminded that to get the right answer one must ask the right question. --대조 | Talk 14:07, 6 February 2007 (UTC)

Kurtosis

What is the difference between kurtosis, as defined in our article, and Karl Pearson's measure of kurtosis, as defined by the R package 'moments'? I realise that the former subtracts 3 while the latter does not, but is that all, or are there other differences? --NorwegianBlue ^talk 09:59, 6 February 2007 (UTC)

Found this while browsing for an answer, and I realise that the concept of kurtosis may be more complex than I thought when posting the question. Grateful for any input, though. --NorwegianBlue ^talk 11:47, 6 February 2007 (UTC)

If it takes 20 men 6 hours to dig a hole, how long will it take 10 men to dig half a hole?

Haha --AMorris (talk)●(contribs) 10:09, 6 February 2007 (UTC)

3 hours. Plus some time to figure out how they define half a hole. Since holes are usually cylindrical prisms, I would argue that half a hole would be shaped like a hemispherical prism. Depth would have to be standardized to some reference hole, so one would have to add some time to find a suitable reference hole. Manholes wouldn't be suitable because you'd need access to the sewrage system, unless one of the men was working for the city and was part of a hole-digging convention.

So what have we learned from all this? It's three hours and a bit, depending what the men work as when they're not digging holes and some thinking time. Hope that helps!

Half a hole is a hole. Proto::► 12:12, 6 February 2007 (UTC)

No, it isn't, because you have found a suitable reference hole.

Treating this as a serious question, the standard method is as follows. Work out man-hours per hole. Divide by 2 to find man-hours for half a hole. Divide by 10 to find hours it takes team of 10 to put in this many man-hours. And the result is not 3 hours. But it really depends on how you define half a hole, as everyone else has pointed out. Gandalf61 14:58, 6 February 2007 (UTC)

If you mean a hole with half the volume, then the answer is 6 hours, not 3. --Wirbelwindヴィルヴェルヴィント (talk) 18:45, 6 February 2007 (UTC)

If this is a serious question, we have a serious problem. For 20 men to take 6 hours to dig a hole, that must be one awesome hole! Is it the size of the Chunnel? Is it dug in solid granite? Are they paid by the hour? Do they spend all their time stepping on each others' toes? Recommended reading: The Mythical Man-Month. Chances are in real life the 10 men can dig half a hole much more quickly, because they'll spend less time in meetings, and they won't have to invest the 80% of effort required for the final 20% of full completion. --KSmrq^T 19:08, 6 February 2007 (UTC)

In real life I've seen 20 'men' take six hours to remove one half wheelbarrow of turf - I was one of those men.87.102.77.140 19:32, 6 February 2007 (UTC)

Whoops. So it's 6 hours and a bit, with provisos.

It's a trick question, as there's no such thing as half a hole. I've seen this before. (yes, if you used a 'reference hole', there'd be an answer, if we assume that the hole is sufficiently large for its circumference to allow 10 men to dig it simultaneously). Proto::► 15:40, 9 February 2007 (UTC)

I found this question in a puzzle book. It says that there's no such thing as half a hole, since half a hole is a hole.

Questions that pretend to be clever when they are not are annoying! If any partial progress towards the first "hole" mentioned in the question counts as "a hole", why does it "take" 20 men 6 hours to dig it? Based on the book's answer, when the men are half way into the digging (actually even much earlier), they'd have already got "a hole". It wouldn't "take" the rest of the work to "dig a hole". --72.78.237.95 21:41, 14 February 2007 (UTC)

Quadratic equations

In school, we learnt about quadratic equations:

1. Their form: ax² + bx + c = 0

2. The three methods of solving them (factorisation, completing the square and the quadratic formula)

3. How to use b² - 4ac to find the nature of the roots of a quadratic equation.

I find quadratic equations fascinating. We're going to learn more about them next year, but I don't have next year's textbook yet. So, to help me prepare for next year, and my future, could you please tell me:

1. Differentiation is to calculus as quadratic equations are to...what? Which fields of mathematics do quadratic equations play a major role in?

2. Some important applications of quadratic equations, especially in computing. — Preceding unsigned comment added by 218.186.9.3 (talk • contribs)

Sounds as if you have understood the relationship between roots and factors, and you have covered the different methods of solving quadratics. Do you understand the link between completing the square and the quadratic formula ? Can you show that they both give the same result ? Not sure if you have covered complex numbers and so complex roots yet i.e. what you do when b² - 4ac is negative - maybe you will cover that next year. To answer your Q1, quadratic equations are a special case of polynomial equations - but they come up in many different areas of mathematics, just because they are the next simplest case after linear functions. Gandalf61 12:44, 6 February 2007 (UTC)

Since x²+y²=r² is the equation of a circle - r is the radius - so for a given x , y is the solution of a simple quadratic - means that quadratic equations turn up all the time when considering circles.. and in fact ellipses, parabolas, and spheres to name a few. Quadratic equations turn up almost all the time in further mathematics - one reason being that pythagorus's theorem is quadratic in form (it uses powers of 2 eg x squared) - So you'll be seeing a lot of them.. especially in relation to triangles, angles, also movement under the effect of gravity.

For the reasons give above quadratic equations often come up in computer graphics - typically when drawing simple curves.

They also turn up in least squares statistical analysis (line fitting) which you may have or may have not done yet.

There are many more examples which I'm sure others will be able to give you.87.102.13.26 14:21, 6 February 2007 (UTC)

This all depends on which year you are in now! I can assume that you are in year 100/11 and doing GCSEs. Quadratics are in many branches of maths. they fit into differentiation/intergration, the main topic of Polynomials, Graph work and many more.MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 21:58, 12 February 2007 (UTC)

Hello, and thanks for the replies, everyone. I'm taking my O Levels this year, so I'll be studying the A Level curriculum in the next 2 years. I understand that the quadratic formula is derived from the completing the square formula. We have not covered complex roots yet, so I'd appreciate any information on them. Thanks for telling me the formulas which usually become quadratic equations when linear functions are substituted into them. Since cubic equations and quadratic equations are somewhat similar, do cubic equations also have a discriminant and formula?

Cubics do have a formula, but it is very long and messy, and you don't really need it ever, higher grade polymers (than quadratic) don't have discriminants, and with odd functions, the graph will always have at least one route. MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 19:01, 14 February 2007 (UTC)

Actually, discriminants can be defined for any polynomial (see discriminant) but it gets very complicated. For example, the cubic $ax^{3}+bx^{2}+cx+d$ has discriminant $b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd$. These higher-order discriminants are first taught to third-year undergraduates (at least where I am) so you probably don't need to know about them unless you're feeling especially keen. Algebraist 15:10, 17 February 2007 (UTC)

What's x and y

If $0,2x-0,1y=0,03$ and $0,4x+1,2y=3$, what is the values of $x$ and $y$? I've figured out that it is $0,255$ and $2,1$ respectively, but I do not think that is correct...? Lol. Please help? This is kinda a homework question, yes... ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:29, 6 February 2007 (UTC)

Multiply the first equation by -2 and add it to the second to get y. Then put your value for y back into one of the equations (your choice) to get x. –King Bee ^{(T • C)} 14:43, 6 February 2007 (UTC)

Here's a longer explanation..(try the above first it probably is a better way to learn how to do this)

The standard way to solve these linear equations is to first eliminate either x or y (it's your choice which) eg you had

0.2x - 0.1y = 0,03

adding 0.1y to both sides, the equation becomes:

0.2x = 0.03 + 0.1y  (multiply by five is the same as divide by 0.2 to give just x..)
x = 0.15 + 0.5y

The second equation was

0.4x + 1.2y = 3 (replacing x by 0.15 + 0.5y  ..)
0.4(0.15 + 0.5y) + 1.2y = 3 (expand the multiplication in the brackets)
0.6 + 0.2y + 1.2y = 3
0.6 + 1.4y = 3
1.4y = 2.4
14y = 24
y = 24/14 = 12/7 =1 5/7

You can do the same for y to get a value for x - unfortunately I can't find a simple explanation for this on wikipedia.

However you could look at http://www.themathpage.com/alg/simultaneous-equations.htm , this page is also quite clear http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-simultaneous1.pdf (neads adobe acrobat). The equations you have are called "simultaneous linear equations". There 'simultaneous' because there are two or more variables in one equation. (linear because there are no powers like x squared or x cubed..). If you want to search for other answers - be warned a lot of the pages cover examples where there are many unknowns eg 4a+7b+1.3c+0.5d = 4 ... the method is the same but might seem confusing.

Try the links or look at the method above and see if it can make sense. If you didn't understand the steps I used above then ask.87.102.13.26 14:46, 6 February 2007 (UTC)

I think I more or less understand, thanks. I had trouble multiplying the co-efficients(?) of the x en y (whatever you call that in English lol (changeable?)) so that I can add or subtract the one equation from the other in order to eliminate one of the variables, yes that's the word lol. Anyway, thanks for the help. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 18:11, 6 February 2007 (UTC)

OK good - as for multiplying the coefficients - what I'd do is first divide so that the coefficient is 1 eg:

4x+3y=15  divide by 4 gives x+0.75y = 15/4
or 0.2x + 7y = 1 divide by 0.2 gives x + 7/0.2y = 1/0.2 ie x+35y=5

and then rearrange to get the equation in terms of x eg x= 15/4 - 0.75y

then you can replace x in the other equation... (that's two steps - as you get better at it you might be able to do it in just one step)87.102.13.26 18:19, 6 February 2007 (UTC)

K cool, thanks. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 10:17, 7 February 2007 (UTC)

???

4³?

4³ is a way of writing 4 × 4 × 4 = 64. (In the same way, 4⁵ = 4 × 4 × 4 × 4 × 4, and 4ⁿ = 4 × 4 × ... × 4, n times). Darkhorse06 17:51, 6 February 2007 (UTC)

I think this was a yes or no question… Okay, I kid, but more generally, and for the glorious sake of redundancy:

$a^{n}=\underbrace {a\times \cdots \times a} _{n},$

− Twas Now ( talk • contribs • e-mail ) 06:08, 8 February 2007 (UTC)

Or you can use a calculator. — Preceding unsigned comment added by Johnnyb2963 (talk • contribs)

...which is 4 [x^y button] 3 --h2g2bob 15:42, 17 February 2007 (UTC)

Independent events

Easy question. By the definition of two independent events A and B, Pr(A ∩ B) = Pr(A)Pr(B). It can also be shown that A is independent of B′ (the complement event of B) and vice-versa. What I'd like to know is whether A′ and B′ are independent, given that A and B are independent. Thanks in advance. -- mattb @ 2007-02-06T19:31Z

Yes. If A and B are independent then, as you said, A is independent of B′. Now, using the exact same result, with B′ as the first event and A as the second, this implies the A′ is independent of B′. -- Meni Rosenfeld (talk) 19:43, 6 February 2007 (UTC)

Ah, so simple. I'm embarrassed to have missed it. Thank you. -- mattb @ 2007-02-06T20:16Z

February 7

Finding the center and radius of a circle

How do I find the center and radius of a circle when given only the coordinates of three points on the circle? Thanks, anon.

Plug the three points into the circle equation:

$\left(x-a\right)^{2}+\left(y-b\right)^{2}=r^{2}$.

and solve for a, b, and r as a system of equations. Splintercellguy 00:59, 7 February 2007 (UTC)

Or get the perpendicular bisectors of two of the lines joining the points, these will meet at the centre. The radius is the distance from this point to any of the original ones.81.154.107.174 01:10, 7 February 2007 (UTC)

Are you sure such a circle exists? If the three points are on a common line, it does not. When doing numeric computations, trouble is to be expected in situations even just approximating this, so do be careful.

Every circle in the xy plane satisfies a quadratic polynomial which is a weighted sum of

$-1,-2x,-2y,x^{2}+y^{2}.\,\!$

Call the weights a, b, c, d. Then we have a system of three homogeneous linear equations:

${\begin{aligned}0&{}=-a-2bx_{1}-2cy_{1}+d(x_{1}^{2}+y_{1}^{2})\\0&{}=-a-2bx_{2}-2cy_{2}+d(x_{2}^{2}+y_{2}^{2})\\0&{}=-a-2bx_{3}-2cy_{3}+d(x_{3}^{2}+y_{3}^{2})\end{aligned}}$

Of course, if (a,b,c,d) is a solution, so is (λa,λb,λc,λd). We can remove this freedom by requiring d to be 1, and rearrange to obtain

${\begin{aligned}a+2bx_{1}+2cy_{1}&{}=x_{1}^{2}+y_{1}^{2}\\a+2bx_{2}+2cy_{2}&{}=x_{2}^{2}+y_{2}^{2}\\a+2bx_{3}+2cy_{3}&{}=x_{3}^{2}+y_{3}^{2}\end{aligned}}$

This is simpler than it looks, because each x_i and y_j will be replaced by a given numeric coordinate; we are only solving for a, b, and c. For readers familiar with matrices, we have

${\begin{bmatrix}1&2x_{1}&2y_{1}\\1&2x_{2}&2y_{2}\\1&2x_{3}&2y_{3}\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}={\begin{bmatrix}x_{1}^{2}+y_{1}^{2}\\x_{2}^{2}+y_{2}^{2}\\x_{3}^{2}+y_{3}^{2}\end{bmatrix}}.$

• Example:

  Given

  $(17,21),\quad (9,25),\quad (-23,1)\,\!$

  Solve

  ${\begin{aligned}a+34b+42c&{}=17^{2}+21^{2}&&{}=730\\a+18b+50c&{}=9^{2}+25^{2}&&{}=706\\a+(-46)b+2c&{}=(-23)^{2}+1^{2}&&{}=530\end{aligned}}$

  Solution

  $a=620,\quad b=2,\quad c=1\,\!$

  $x^{2}+y^{2}-4x-2y=620\,\!$

  ${\begin{aligned}\mathrm {center} &{}=(b,c)&&{}=(2,1)\\\mathrm {radius} &{}={\sqrt {a+b^{2}+c^{2}}}&&{}={\sqrt {625}}=25\end{aligned}}$

The center coordinates appear directly in the solution, with the radius little extra trouble. Although this method requires that we know how to solve a system of linear equations, any method will require comparable work. This one has the advantage that it generalizes to a wide variety of curves, and to least-squares approximations. Collinearity reveals itself in lack of independence among the three linear equations (equivalently, in singularity of the matrix). --KSmrq^T 20:47, 7 February 2007 (UTC)

Airy function coefficients

Solving the Airy differential equation with a power series gives

$y(x)=\sum _{k=0}^{\infty }c_{k}x^{k}$

where

$c_{2}=0$

$c_{k}={\frac {c_{k-3}}{k(k-1)}}$

but no values for $c_{0}$ and $c_{1}$ pop out of the equation. Setting ($c_{0}$ and $c_{1}$) to (1, −1) or (1, 1) gives two solutions, but the actual Airy functions Ai and Bi use specific transcendental constants. Where do they come from? Fredrik Johansson 17:16, 7 February 2007 (UTC)

Often we choose particular constants to make other conditions fit better. Perhaps it has something to do with the Wronskian, quoted as being 1/π, which may have other implications.

As you can see in the article, function Ai is defined by a specific definite integral, which fixes all coefficients of the power series. For example,

$c_{0}={\frac {1}{\pi }}\int _{0}^{\infty }\cos \left({\frac {t^{3}}{3}}\right)\,dt.$

This fixes the numeric value at about 0.355.  --Lambiam^Talk 00:27, 10 February 2007 (UTC)

Vandalism

I just removed some (what I assumed to be minor) vandalism from Maillard reaction See http://en.wikipedia.org/w/index.php?title=Maillard_reaction&diff=105518582&oldid=105452096

Quote:

Diagram of the Maillard Reaction (English fairly obviously not the $n^{th}$ language of the website creator $\forall \,n\in \mathbb {Z} ^{+}\ll \,17$)

Can anyone tell me what it meant - or is it just gibberish? Thanks.83.100.250.165 19:35, 7 February 2007 (UTC)

It is a rude and inappropriate insult of the English language skills exhibited on the linked web site. The meaning is that, for the web site creator, English is not the native language, nor second language, nor third language, nor any language acquired significantly less than seventeenth. Ironically, anyone who could learn seventeen languages, and who can diagram and explain the Maillard reaction as chemistry, is surely far more intelligent and sophisticated than the vandal who added the insult.

Incidentally, this method of producing flavor and aroma is well-known to trained chefs; even I deliberately take advantage of it almost every time I cook. Once we used to hear that when making a beef stew, say, we should sear (brown) the chunks of meat on all sides to "seal in juices". This is scientific nonsense, as it does no such thing. Its real benefit is producing lovely rich flavor through the Maillard reaction. Vegetarians might like to try roasting onions, celery, and carrots in the oven for a comparable experience. --KSmrq^T 21:13, 7 February 2007 (UTC)

A very odd form of vandalism by all accounts...And a very long way round to give an insult.83.100.250.165 21:27, 7 February 2007 (UTC)

February 8

Do square root of -i exists?

Do square roots of -i exists? Does it have two roots? 202.168.50.40 00:39, 8 February 2007 (UTC)

$\pm {\frac {1-i}{\sqrt {2}}}.$ —David Eppstein 02:09, 8 February 2007 (UTC)

Yes. In fact, once we move up from the real numbers to the complex numbers, we finally reach a set that is algebraically closed - in fact, this property is the basis of the fundamental theorem of algebra. What this means, effectively, is that every polynomial has solutions. One upshot of this is that every number in the complex numbers has square roots, cube roots, etc, since the square roots of some number a are just the solutions to the equation $\textstyle x^{2}=a$. Maelin (Talk | Contribs) 03:13, 8 February 2007 (UTC)

e-mail of ministry education of research center

I want tokown the e-mail of ministry education of research center

Which country do you live in? - Akamad 08:36, 8 February 2007 (UTC)

Algorithm for pairing players in round-robin chess tournament

I wish to invite n players to participate in a round-robin chess tournament.

1. Is it possible to express as a function of n, the total number of ways that they can be paired for a) 1 round? b) n-1 rounds (the entire tournament)?

2. Is there an algorithm which I can use, without computer assistance, to quickly generate fixtures? Remember, in a round-robin tournament, each player plays everyone else only once. (One method I know is to factorise n, then generate fixtures for factors of n, and arrange the n players into groups of the factors.)

One way would be to have a table of n columns and n rows - for the first round pick a number randomly from the first row - this is who player one plays. Cross off that square in the array and remember to cross off the corresponding square (mark with a 1) eg if player 1 plays player 4, (ie you picked column 4 in row 1) then also cross off column 1.row 4. Repeat this for every row that does not already have a 1 in.

For the second round repeat the proceedure but obviously ignore squares that are already crossed off. Cross of each with a 2.

And repeat until all the squares are crossed out.

All you need is pen/paper (one sheet per tournament) and a method of randomly picking one square from a row.

There are other ways - if there are 6 or less players (or maybe 9) you could pick dominos randomly from out of a bag (ignore doubles) - so picking [6-3] means player 6 plays player 3. Then pick another domino - if you get for instance [4-3](player 3 has already been picked) put it back in the bag and pick again.

So you need to pick n/2 dominos per round. At the end of each round 'discard' the picked dominos and continue with the dominos remaining in the bag. You can get dominos with up to 9 dots - if you have say 6 players ignore any with more that a value 6 on either of the faces. Hope that helps.87.102.2.204 11:00, 8 February 2007 (UTC)

Have you tried the method you suggested? I used to do that, but in the last few rounds, I find it difficult to pair them and ensure everyone plays everyone else only once. That's why I thought up the "arrange players into groups of factors of n" method, which appears to be quite effective. I'm just wondering if there are methods even faster and more effecient than the "arrange players into groups of factors of n" method.

Ah yes - it does get more difficult especially towards the end - the dominos method is the other way round - as the dominos are discarded as matches are played - but in this case it might take a couple of goes to get matches at the beginning. I can't think of simpler methods - without using a computer.. Sorry.87.102.45.80 11:54, 10 February 2007 (UTC)

3. What are the algorithms computers use to quickly generate fixtures for round-robin tournaments?

There are a number of algorithms that could be used to do this - one factor is whether or not you want the selection in the first round to be random, or alternatively to always pick pairs in the same way for each round. I'm sure someone can give you more details.87.102.2.204 10:48, 8 February 2007 (UTC)

I could describe some algorhythms that would work if you wish - though they would be based on the methods already described - please say so if you want this.87.102.9.117 20:33, 10 February 2007 (UTC)

Go ahead and describe the algorithms, even if they are based on the "arrange players into groups of factors of n" method.

Oops, VERRY SORRY, the algorhythms I described "don't work" do they.. I'll take my blinkers off. I'll see if I can describe a simple one that does work, in the mean time there may may a delay...87.102.9.15 12:38, 11 February 2007 (UTC)

OK All I can come up with is a method which checks a set of n^n(n+1)/2 possible combinations until it finds a match set up for (n-1) games that isn't invalid - not very good and I'm sure you could imagine a method of setting up n(n+1)/2 loops each looping from 1 to n and then checking each combination for invalidity. Apologies for the un-thought-out gunk above.87.102.9.15 14:10, 11 February 2007 (UTC)

Here's one that should work. it uses an n x n array where array(4,3)=7 means that player 4 plays player 3 in round 7.

n=number of players
Make Array A(n,n) ; initially all values in the array are 0
x=1
Loop3: y=x+1
Loop2: A(x,y)=A(x,y)+1, A(y,x)=A(y,x)+1
IF A(x,y)<n THEN GOTO Loop4
A(x,y)=0, A(y,x)=0, y=y-1
IF y>x THEN GOTO Loop2
x=x-1, GOTO Loop3
Loop4: DO PROCEDURE "Check_Array_Validity" ; the procedure (see below) returns Q=true or false
IF Q=false THEN GOTO Loop2
y=y+1
IF y<=n THEN GOTO Loop2
x=x+1
IF x<=n THEN GOTO Loop3
DO PROCEDURE "Print results"
END.

PROCEDURE "Check_Array_Validity"
a=1
Loop1
IF A(x,a)=A(x,y) AND NOT (a=y) THEN Q=false, EXIT
IF A(a,y)=A(x,y) AND NOT (a=x) THEN Q=false, EXIT
a=a+1
IF a<=n THEN GOTO Loop1
Q=true, EXIT

PROCEDURE "Print results"
FOR r=1 to n-1
PRINT "Round Number:",r
FOR x=1 to n
FOR y=x to n
IF A(x,y)=r THEN PRINT "Player No.",x," plays ,"player No. ",y
NEXT y
NEXT x
NEXT r
EXIT

This should work - it goes through all the possible sets of matches until it finds one that is valid - Absolutely terrible when n=big I'm afraid.83.100.255.117 16:42, 12 February 2007 (UTC)

Though if you look at Round-robin_tournament#Scheduling_Algorithm it gives an algorhythm that is millions of times simpler..83.100.255.117 16:47, 12 February 2007 (UTC)

arcsin(\sqrt{2/3})

A numerical approximation of ${\hbox{arcsin}}\left({\sqrt {2/3}}\right)$ may be $0.95531661812450927816\ldots$. But is there a nice exact representation known, analogous to ${\hbox{arcsin}}\left({\sqrt {3}}/2\right)=\pi /3$.

88.74.0.242 11:14, 8 February 2007 (UTC)

Not what you want, but it equals 54.7356103° as well. − Twas Now ( talk • contribs • e-mail ) 11:47, 8 February 2007 (UTC)

This looks like it will end up an ugly number either way. − Twas Now ( talk • contribs • e-mail ) 11:50, 8 February 2007 (UTC)

As a nice english expression - how about "the angle that the edge of a tetrahedron placed on a horizontal plane makes with the horizontal" - ?87.102.2.204 11:52, 8 February 2007 (UTC) I'm guessing that this is the angle you seek - there doesn't seem an expresion for it that isn't an infinite series (I can't currently find the page that lists the angles that have 'simple' sines and cosines - there is one but I can't recall it's name - it's not in that list).83.100.158.135 15:35, 8 February 2007 (UTC)

Note if someone could supply an expression for arcsin (a/b) in terms of arcsin (a) and arcsin (b) that would help since sqrt(3)/2 and 1/sqrt(2) could be used. I'm not aware of such a thing...83.100.158.135 15:37, 8 February 2007 (UTC)

See Exact trigonometric constants - your value notable by it's absence -sorry.83.100.158.135 15:40, 8 February 2007 (UTC)

Plouffe's Inverter gives the formula $\operatorname {arcsec} {\sqrt {3}}$ which probably doesn't help you. – b_jonas 20:59, 8 February 2007 (UTC)

Hey, thanks! Ever since the Inverse Symbolic Calculator kinda died I've been wishing there were a similar thing.

I've found it in Sloane's link collection: [5]. – b_jonas 11:22, 10 February 2007 (UTC)

February 9

intercept

What does it mean "It has slope -2/3 and x-intercept of 3"?

I assume you are talking about a line y = ax + b. "It has slope −2/3" means that a = −2/3, and "it has x-intercept 3" means that y = 0 when x = 3, so that b = 2. Thus y = −2x/3 + 2. --Andreas Rejbrand 06:33, 9 February 2007 (UTC)

I'm adding a picture to help visualize.

-2/3 — This is the slope of the line. This means that on a Cartesian plane (x-y plane), that for every two points you move to the right, you must move three points down (2→ and 3↓). Or, for every two points you move left, you must move three points up (2← and 3↑).

x-intercept of 3 — This means the line crosses the y=0 line at 3. In other words, it crosses the horizontal line, the x-axis at 3. The point is (0,3).

− Twas Now ( talk • contribs • e-mail ) 03:09, 10 February 2007 (UTC)

Vectors

I sat around the other day thinking how I could apply vectors. Then I thought:

If I had a pistol and wanted to see how far the bullet went if I shot it, and my pistol could fire a bullet at 350 m/s, how far would it go?

Assumptions:

            Air resistance is negligible.
            I am firing with my pistol parallel to the ground (or at right angles).
            The terrain I am shooting out to is FLAT surface.
            The acceleration due to gravity is 9.8 metres per second per second downwards.
            I am holding the pistol 1.6m above the ground.

Am I correct in saying it would travel 1.4km before hitting the ground 4 seconds after being fired?

No, you are not. If we represent a vector V by writing $\mathbf {V} ={\underline {\mathbf {e} }}{\begin{pmatrix}x\\y\end{pmatrix}}$ where (x, y) are its coordinates, then we have that the acceleration $\mathbf {a} (t)={\underline {\mathbf {e} }}{\begin{pmatrix}0\\-g\end{pmatrix}}$ where g = 9.8 m/s². By integration, we find that the velocity $\mathbf {v} (t)={\underline {\mathbf {e} }}{\begin{pmatrix}v_{0}\\-gt\end{pmatrix}}$ where v₀ = 350 m/s. Integrating once again, we find that the position $\mathbf {r} (t)={\underline {\mathbf {e} }}{\begin{pmatrix}v_{0}t\\h-{\frac {1}{2}}gt^{2}\end{pmatrix}}$ where h = 1.6 m. The bullet reaches the ground when r_y = 0 m, i.e. when $h-{\frac {1}{2}}gt^{2}=0{\text{ m}}$. Solving for t, we get t = 0.571 ... s. Thus, $r_{x}=v_{0}t\approx 200{\text{ m}}$. --Andreas Rejbrand 22:16, 9 February 2007 (UTC)

This smells like a covert homework problem, so I'll say little. The interesting vector part of this is that we can separate the motion of the bullet into two independent components, one horizontal and one vertical. (If the viscosity of the air caused drag, could we do this?) The vertical component is pure gravity, the same as if we dropped the bullet from the given height. The horizontal component is steady inertial motion.

Recently a golf ball was hit from low Earth orbit. That is fun to analyze. --KSmrq^T 22:46, 9 February 2007 (UTC)

Here's my working:

The acceleration vector of the bullet is -9.8j, which means the velocity is -9.8tj + c At time t = 0, v = 350i ==> 350i = 0j + c ==> c = 350i ==> v = 350i + (-9.8t)j ==> the position vector is 350ti + (-4.9t^2)j + C At time t = 0, position = 1.6j ==> 1.6j = C ==> p = 350ti + (-4.9t^2)j + 1.6j ==> p = 350ti + (1.6 - 4.9t^2)j Bullet hits the ground at vertical component = 0: 4.9t^2 = 1.6 ==> 49t^2 = 16 ==> t^2 = 16 ==> t = 4 So the bullet hits the ground after 4 seconds. At time t = 4, vertical component:

350 * 4 = 1400m

Here's your error: 49t^2 = 16 ==> t^2 = 16

Can you see it? --Andreas Rejbrand 22:33, 9 February 2007 (UTC)

Yeah, kicking myself. To "KSmrq," it's not homework. I haven't done maths at school for two years. I guess you probably assumed so because the answer works out to be exactly 200m, which was purely coincidental. Also, I searched Google for the speed at which a standard 9mm pistol is fired. It told me 200-500m/s, so I took the mean. If I went out and performed this, with all those assumptions in mind, would it still be 4/7 of a second? — Preceding unsigned comment added by 138.217.32.202 (talk • contribs)

I'll say a little more then. (And please sign and timestamp your posts, even if not logged in with an account, by typing four tildes: "~~~~".)

Because the bullet is fired horizontally, its initial velocity has a zero vertical component. The acceleration of gravity depends on neither the position nor the motion of the bullet; it is constant, and purely vertical. With no gravity, the bullet would travel parallel to the flat ground forever. It hits the ground for one reason: gravity pulls it down. The time it takes to cover the vertical distance to the ground depends on its initial height, its initial vertical velocity (stated to be zero), and the acceleration due to gravity. While it falls, it travels horizontally. How far it travels horizontally depends on how fast it is moving, and on how long it takes to fall to the ground.

If we have a higher muzzle velocity, the horizontal distance increases. If we fire the pistol from higher above the ground, the horizontal distance increases. If we tilt the aim upward, we get opposing effects. Less of the muzzle velocity is horizontal, which tends to decrease the horizontal distance covered; however, some of the muzzle velocity becomes vertical, which tends to increase the time to impact because the bullet goes up before it goes down. The distance traveled varies continuously as we adjust the aim from horizontal to straight up, and so we can find an optimal angle to yield the greatest horizontal distance. You might like to try to find that angle; doing so will help you better understand these calculations without straining your nascent abilities. --KSmrq^T 05:01, 10 February 2007 (UTC)

I agree. Finding the optimal angle is an excellent exercise. It requires some (basic) calculus, though. If interested, I have actually derived this angle in a document (mechanics) at my website (the URL may be found on my user page). --Andreas Rejbrand 11:51, 10 February 2007 (UTC)

I don't really understand what you mean, but from the equation $h-{\frac {1}{2}}gt^{2}=0{\text{ m}}$ we realize that the time of the entire motion does not depend on the initial speed of the bullet. Nevertheless, the horizontal distance $r_{x}=v_{0}t$ (of course) does depend on it. --Andreas Rejbrand 23:19, 9 February 2007 (UTC)

February 10

how do you add a topic?

How do you add a topic00:39, 10 February 2007 (UTC)Swanback

If by that, you mean how do you create a new article, see: Wikipedia:Your first article. In futurem questions regarding how to use wikipedia should be asked on the help desk. - Akamad 14:31, 10 February 2007 (UTC)

Calculation of chances of spherical spawn

I have a problem.

Envision a gate in space, through which spaceships come. The gate is a sphere 2,5km in radius. At the end of this radius, distance is measured (so that two entities on opposite sides of the gate are both 0km from the gate, but 5km from each other). Now, a spaceship coming through the gate will spawn 12km away from the gatesphere, completely randomly across the 'surface' of this spawn-sphere. I have an X number of ships, for this sake let us say 10 ships, and these can fire up to 10km away, spherically of course. My question really, is how do I place these ships to have the BEST statistical coverage, with the most chance of trapping the enemy spaceship which spawns on the spawnsphere?

I can make them orbit the gatesphere at a distance of 10 or 15km. While this penalizes them - their own attack-sphere not being used to its full potential - it is not that alone, which makes me not want to solve my problem through this alternative. It is much rather that by placing my ships statically and according to formulas, I should be able to cover much more, whereas a random, uncoordinated orbit of all my ships will.. well, their attack-spheres will meet each other, overlap, and thus greatly penalize me.

Thank you for all help! 81.93.102.185 16:26, 10 February 2007 (UTC)

There's not really a rigorous way for spacing points evenly on a sphere, unless you've got a number of points that matches a particular platonic solid. Fairly good solutions can be approximated in various ways, such as starting with random points and incrementally repelling them from each other. I've heard also of starting with a spiral that wraps around a sphere and placing points evenly along that, or you could use spherical coordinates and divide things up by angle while accomodating for the smaller radius near the poles. - Rainwarrior 18:16, 10 February 2007 (UTC)

Yes, a spiral can do a reasonable job; one placement algorithm is

${\begin{aligned}z_{0}&{}={\tfrac {1}{2N}}-1;&\omega &{}=\pi \left(3-{\sqrt {5}}\right)\\k&{}=0\ldots N-1\\&&z&{}=z_{0}+{\tfrac {k}{N}}\\&&r&{}={\sqrt {1-z^{2}}}\\&&\phi &{}=\omega k\\&&p_{k}&{}=(r\cos \phi ,r\sin \phi ,z)\end{aligned}}$

Saff & Kuijlaars present another variation. An extended discussion of the topic kept by Dave Rusin covers a multitude of definitions and algorithms. --KSmrq^T 22:38, 10 February 2007 (UTC)

Question..

Can you qualify a bit about the spawnsphere - is it the surface of a sphere 12km in radius or is it a volume that is represented by a sphere 14.5km in radius minus a 9.5 radius sphere in the middle? or something different?87.102.9.117 23:29, 10 February 2007 (UTC)

First of all, thank you for the help. Platonic solids seem to be a great way to cook up some points, since I really can't make much out of that placement algorithm! :) As for the QUESTION... By spawnsphere I mean the surface of the sphere which is 14.5km in radius - just to clear that up. 81.93.102.185 13:11, 11 February 2007 (UTC)

Platonic solids are pretty to look at, but have problems in this context. We only have five: tetrahedron (4 points), octahedron (6 points), cube (8 points), icosahedron (12 points), dodecahedron (20 points). Their geometric regularity does not imply maximum spacing: twist the top of a cube 45° and move the top and bottom points towards the equator to get more separation. Splitting faces to increase the number of points disrupts regularity.

Using a computer program, we can distribute points on a sphere and nudge them with repelling forces until they stabilize. This is painfully slow for large numbers of points.

If we study the results for inspiration, we see that — apart from an inevitable randomness — we may be able to approximate the pattern with a systematic spiral. Start near the South Pole. Place a point, then rise and turn a little. The rise is chosen to assure we place all the points, stopping near the North Pole. With a good choice of turn, we get a satisfactory pattern for large numbers of points. Each point is near its predecessor and successor, but also near points above and below it. The angle in the example code is based on the golden ratio, which is known to have nice properties for breaking up a circle. Try it for a few thousand points. --KSmrq^T 22:49, 11 February 2007 (UTC)

February 11

(no new questions)

February 12

Mathematical basis of quantum physics

Quantum physics (as I understand it, at least), is dependent upon a number of results in complex analysis and linear algebra. Does anyone know what properties of the underlying "number system" (pardon my vernacular) $\mathbb {C}$ (the complex numbers) are essential to the development of these results, and thus quantum physics? I speak of things like the fact that $\mathbb {C}$ is algebraically closed (though that may not, in fact, be relevant - it just illustrates what I mean when I say "properties"). --Braveorca 05:01, 12 February 2007 (UTC)

Your question suggests the mathematics shapes the physics; actually, it's the other way around. The result of the two slit experiment is a physical fact, as is the photoelectric effect, as is quantum tunneling, and so on. It is ironic that you ask specifically about quantum mechanics, because its mathematics has a remarkable history. The physical phenomena were originally formalized in completely different ways, which later were shown to be equivalent. --KSmrq^T 07:56, 12 February 2007 (UTC)

Here's one take (scroll down to "Real vs. Complex Numbers"). Fredrik Johansson 12:07, 12 February 2007 (UTC)

How do you call 1/(1-x)?

How do you call 1/(1-x) or 1 + x + x^2 + x^3 + ...?Mr.K. (talk) 18:02, 12 February 2007 (UTC)

I don't think there is a very specific name. ${\frac {1}{1-x}}$ is an example of a rational function, and $1+x+x^{2}+x^{3}+\cdots$ (which amounts to the same value for $|x|<1$) is an example of a power series. -- Meni Rosenfeld (talk) 18:15, 12 February 2007 (UTC)

I think the better link for the infinite series is geometric progression. —David Eppstein 21:52, 12 February 2007 (UTC)

Through other sources I found that the whole thing may be called: "Gram-Schmidt Identity". Although there is a wikipedia article about Gram-Schmidt process, I didn't find more information about the identity (either here, nor somewhere else).Mr.K. (talk) 15:38, 13 February 2007 (UTC)

???

8976876876876^0.1

8976876876876^0.1 = 19.7384264 --Spoon! 21:50, 12 February 2007 (UTC)

Frequency of occurrence

Using a TI-83 Plus, is it possible to calculate (or count) the number of occurrences of an element in a list without using an third-party application or graphing? I.e., if I have {1, 1, 2, 3, 3, 3} stored to List 1 (L1), can I have the calculator count the number of times the number "3" appears? Thanks. --MZMcBride 22:40, 12 February 2007 (UTC)

If it's in a program, best way I can think of is a subroutine, comparing each element and incrementing a counter value.ST47Talk 00:21, 13 February 2007 (UTC)

Using C as the counter, E as the element input, L as the loop variable, and your L1:

PROGRAM:LSTCOUNT
:0→C
:Input "ELEMENT? ",E
:For(L,1,dim(L1))
:If L1(L)=E:Then
:C+1→C:End:End
:Disp "FREQUENCY:",C

Running the program:

prgmLSTCOUNT
ELEMENT? 3
FREQUENCY:
               3
            Done

The dim( function is in the LIST OPS menu. --jh51681 13:02, 16 February 2007 (UTC)

February 13

exchanging the variable in logarithmic cubic quations

my equation is

(1/T)=a+b(ln R)+c((ln R)*(ln R)*(ln R))

it is of the form T=f(R);

here T and R are variable. a, b , c are known values

i want above equation to be changed to R=f(T) form?? please give the detail procedure..

snehanshu tiwari bangalore india

I don't believe it possible, because of the cube of ln R. If the expression involved the log of the cube of R, that would be different, and straightforward to invert.–81.159.15.39 13:18, 13 February 2007 (UTC)

If we let x=ln(R) then we can re-arrange to get a cubic in x:

$cx^{3}+bx+(a-{\frac {1}{T}})=0$

which can then be solved (in principle - messy in practice) to find x as a function of T, say

$x=g(T)$

and then

$R=e^{x}=e^{g(T)}$. Gandalf61 14:27, 13 February 2007 (UTC)

Solving cubic equations

In school, we learnt how to solve cubic equations of the form ax³ + bx² + cx + d.

To factorise it into (x+r)(x+s)(x+t), we have to find a value for r by trial and error (with the factor theorem), then solve a quadratic equation to get the other two roots, s and t.

In the exams, the value of r will usually be 3 or less, so finding r by trial and error will not take a long time.

However, if r is large, finding r by trial and error will take a long time. Are there any faster, more systematic ways of finding r, besides trial and error? — Preceding unsigned comment added by 218.186.9.3 (talk • contribs) 09:18, 2007 February 13 (UTC)

(Please sign your posts with four tildes, "~~~~".) The roots of a cubic polynomial can be expressed in closed form, so in theory no searching is required. However, the technique is more difficult than using the quadratic formula. When the coefficients, a, b, c, and d, are real numbers (with a nonzero), the cubic is guaranteed to have a real root; but it may have only one. For example,

$x^{3}-3x^{2}+2x-6\,\!$

has only one real root.

But we can say much more. The coefficients are normalized so that the x³ term has coefficient 1. If we list them all, (1,−3,2,−6), the −6 has the largest absolute value. The Cauchy bound says that all real roots must lie between −m and +m, where m is one plus the maximum absolute value. In this example, the roots will be between −7 and 7.

We can do better. The positive coefficients before the −6 are 1 and 2, summing to 3; and before the −3 we have just 1. For an upper bound we can choose the maximum of ⁶⁄₃ and ³⁄₁, which will be 3, and again add 1. That is, the maximum real root is 4. Negating every other term (so we are evaluating the polynomial with −x), we can similarly find a lower bound, which here will be −1.

There is also a method known as Descartes' rule of signs that may help us restrict the number of positive or negative real roots. In this example it is no help with the positive roots (we will have three or one), but it tells us we cannot have a negative root.

This example polynomial has integer coefficients, and may have integer roots. If r is such a root, then it must divide the constant term, so the only possibilities here are 1, 2, 3. We have used the known bounds; and we know that 0 cannot be a root unless the constant term is 0.

We also have ways to bracket roots. If the polynomial evaluates to positive for x = a and negative for x = b, then there must be at least one root between a and b. Using a Sturm sequence, we can learn even more.

This does not exhaust our inventory of tools, but perhaps this is enough for now. Computer programs for solving cubics typically use closed form solutions. Above degree four, however, a famous theorem states that we have no closed form solutions. --KSmrq^T 11:48, 13 February 2007 (UTC)

You might want to read cubic equation. – b_jonas 14:17, 13 February 2007 (UTC)

Anyone bored?

I any of you [lovely] people have too much spare time on your hands could you give me a hand. I have the formula to work out reps until exhaustion(y) from % of max load (x), and either would like the formula rearanging, or the location of the formula the other way around, its a really nasty formula!:

$y=173.525-6.31x+(9.576\times 10^{-2})x^{2}-(6.742\times 10^{-4})x^{3}+(1.75\times 10^{-6})x^{4}$

MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 21:46, 13 February 2007 (UTC)

I can't make heads or tails out of your question. What are reps . Until what exhaustion? 202.168.50.40 22:00, 13 February 2007 (UTC)

It looks like the question is about weight training, and the equation is supposed to describe the number of repetitions the trainer can perform, as a function of the weight he is lifting (expressed as a percentage of the maximal weight he can lift). However, I cannot imagine how this formula was arrived at, or what the OP wants to do with it. -- Meni Rosenfeld (talk) 22:51, 13 February 2007 (UTC)

I did an interpolation for the pairs (50,25), (70,12), (90,5), (100,1) and arrived at something like

$102-2y-0.1y^{2}+4\times 10^{-3}y^{3}$

which gives 76% for 10 reps. Ok?--80.136.135.63 23:03, 13 February 2007 (UTC)

That's quite close, thanks! I didn't come up with it myself, it would have been easyer to rearange if I had, I got it from a sprt science website, and wanted to be able to work it backwards, if only excel had a function to work out inverse functions (or does it, maybe a qu for IT desk...), anyway many thanks for your efforts!MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 23:13, 13 February 2007 (UTC)

That $x(y)$ is of course not quite the inverse function of what you started with; if you actually want the inverse, you can get it from applying the quartic formula, but it won't be pretty: I get ${\begin{array}{l}{\frac {1}{2}}{\sqrt {{\frac {12.3344-5.0397y}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}+151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+625.7665}}\\-{\frac {1}{2}}{\sqrt {{\frac {5.0397y-12.3344}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}-151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+{\frac {425419.3114}{\sqrt {{\frac {12.3344-5.0397y}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}+151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+625.7665}}}+1251.5331}}+96.3143\end{array}}$ which gives 74.6442% for 10 reps. --Tardis 17:13, 14 February 2007 (UTC)

Please tell me that you used Mathematica or Maxima or something for that :) Doing to TeX alone would be murder! Oskar 17:42, 17 February 2007 (UTC)

Mathematica solved the quartic (with some help from me to pick a solution branch), and generated approximate TeX with TeXForm; I used Emacs to reformat the result in a WP-friendly manner: round the numbers, pretty-print the scientific notation, etc. --Tardis 16:32, 20 February 2007 (UTC)

Can someone give me a question please?

I am currently studying maths at GCSE level (first year) and am finding it relatively easy and relatively boring ("Yes miss, we know how to get a term-to-term rule. Yes, you don't need to give us 15 exercises on it, we get it already. You're not listening are you miss? You're going to give us those 15 exercises anyway aren't you miss, even the the answers are obvious. Oh look miss, you've put the answers on the board and got three wrong. Oh look, you've tried to correct it and got stuck and now you're asking us what the answer is. Well done. Oh look, now you're giving us some exercises that we don't know how to do and which neither the book nor you explain to us. It just gives us the answers without explaining why they are correct. Thanks for that."). As such, I would appreciate it if anyone can give me an equation to solve the includes a bit of everything up to the level of calculus. I've been trying to get my algebra really good, but the textbook I have doesn't really help and only gives basic questions. I am sort of looking for a question with stuff life binomial expansion, negative powers, fractional powers, rearranging divisions and multiplications, the whole lot. I most probably won't be able to solve it, but it will give me a starting point to ask questions about how to do certain sections of it. Thanks for any help you can give. P.S. I apologise for the off topic rant in between the brackets there. --80.229.152.246 22:55, 13 February 2007 (UTC)

...which is why I took my GCSE a year early and progressed on to As! Without knowing exactly what you are capable of, I darent just give you something from As (which isn't much harder than GCSE, by the way!), try looking here or here for some good qus, which come with explanations MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 23:18, 13 February 2007 (UTC)

I mean just an equation to solve that has a bit of everything in it. I don't care if I can't solve it with my knowledge, I just want to use it as set-off point to learn new things.

--80.229.152.246 17:40, 14 February 2007 (UTC)

Have you tried any of the unsolved problems in mathematics? − Twas Now ( talk • contribs • e-mail ) 05:38, 15 February 2007 (UTC)

Bearing in mind I haven't even got a GCSE in maths yet, I think attempting one of those would be pretty futile. For example, I have now taught myself the binomial theorem and would like to practice rearranging equations with negative powers, fractional powers, brackets, square roots, fractions multiplied by variable, etc. Just give me anything. I don't really mind if I can solve it or not, I would just like to know what to look up for the bits I get stuck on. --80.229.152.246 17:20, 15 February 2007 (UTC)

OK here's some things to try:

1. Find the intersection of line with a

a. sphere

b. ellipsoid

c. cone

d. plane in 3 dimensions.

2. Find the normal to this point and express it in terms of the equation of the line and the equation of the sphere/ellipsoid etc.

3. Find an expression for the closest approach of two lines in three dimensions

4. Read about integration by parts - then learn to do it.

5. Calculate the distance from the centre of a tetrahedron to a point (the tetrahedron has a known side length l)

6. Show that the differential of sin(x) is cos(x)

7. Show that the differential with respect to x of xⁿ is nx^n-1 (n is integer)

8. Find an expression/equation that differentiates to give the same equation.

I don't know if these are too hard/easy for you. But good luck.213.249.237.49 18:40, 15 February 2007 (UTC)

Thanks very much for those ideas. I'll see if I can do some of them. --80.229.152.246 18:07, 16 February 2007 (UTC)

By the way, I don't understand the first 3 questions. I don't really get what you want me to do. --80.229.152.246 18:15, 16 February 2007 (UTC)

For 1.a. Take a line in 3 dimensions eg y=4x+2, z=4y+3 (or use vectors) then find the point(s) at which it crosses through the surface of a sphere eg (x-2)^{2+(y-3)2+(z+1)2=r2 is the equation of a sphere centred on (2,3,-1) of radius r. The others are similar - you might need to look up some of the terms to do these then.87.102.7.220 14:15, 17 February 2007 (UTC)}

How to pay back infinite dollars for a finite debt

$\sum _{n=1}^{\infty }{\frac {1}{n}}=\infty$

Now this does not make sense. For example: If I borrow $1000 off you and promised to pay you back infinite dollars, it would sounds like an absolute great deal.

So year 0 : borrow $1000 from you

and year 1 : pay you back $1.00

and year 2 : pay you back $0.50

and year 3 : pay you back $0.33

and so forth. But this does not make sense at all.

I think we can solve America's national debt. 202.168.50.40 23:39, 13 February 2007 (UTC)

I would be paying back an infinite total amount of money, but it would have a finite present value. In fact, at current interest rates, you should be able to buy a promise to pay you one dollar a year, forever, for about fifty dollars. --Trovatore 00:30, 14 February 2007 (UTC)

You assume that the interest rate is greater than zero. What if the interest rate is exactly zero. Say I borrow 1000 tonnes of uranium from you and pays back in uranium. Does your argument still holds? 202.168.50.40 00:43, 14 February 2007 (UTC)

I can't imagine why you think I'd be willing to lend you uranium at an interest rate of zero. --Trovatore 01:01, 14 February 2007 (UTC)

Because usury is such a dirty word. − Twas Now ( talk • contribs • e-mail ) 05:35, 15 February 2007 (UTC)

The partial sums of the infinite series are the harmonic numbers. They increase very slowly. You need to add the first 12367 terms to even reach the value 10. To get to 100, you need about 15092688622113829863255886615135332918918094 terms, give or take a few. The universe will have evaporated long before you reach 1000.  --Lambiam^Talk 02:26, 14 February 2007 (UTC)

It's your analogy that doesn't make sense. Depending on where you live, there will be a lower bound on the amount of money that you can make up, but you will need to pay back amounts less than this lower bound, which you obviously can't do. —The preceding unsigned comment was added by 203.49.220.190 (talk) 09:29, 14 February 2007 (UTC).

If you're interested in similar situations to yours, you should check out the article on the St. Petersburg paradox Oskar 16:09, 16 February 2007 (UTC)

February 14

Clearing up a few things

I want to make sure that this:

$\lim _{a^{+}\to \infty }{\Bigg (}\sum _{a=1}^{\infty }{\tfrac {1}{a^{n}}}{\Bigg )}$

Reads: "The limit of a as it approaches infinity approximates (with increasing accuracy) the closed form of the bracketed series". Is this correct? Is there a better way of saying this?

164.11.204.51 02:35, 14 February 2007 (UTC)

It doesn't just approxmiate the right-hand side, it equals the right-hand side. --Trovatore 02:37, 14 February 2007 (UTC)

Whoops, I just noticed that you don't in fact have a right-hand side. So I'm afraid your translation is completely wrong. It shouldn't be a sentence, but a noun phrase. --Trovatore 02:38, 14 February 2007 (UTC)

Why don't I have a right hand side? You're meant to explain these things. I assume you're referring to the limit part which denotes approaching infinity from both sides. I've changed that now, does it make sense?

I would say the limit of a as it approaches infinity from the right for the function [blah blah blah]. Wait, is it supposed to be ${a^{+}\to \infty }$ or ${a\to \infty ^{+}}$? --Wirbelwindヴィルヴェルヴィント (talk) 06:32, 14 February 2007 (UTC)

It would have to be the latter. How can you approach positive infinity from the left? —The preceding unsigned comment was added by 203.49.220.190 (talk) 09:26, 14 February 2007 (UTC).

The "right-hand side" means part of the equation after the equals sign. You don't have an equals sign (except for the range of the summation, which doesn't count), so you don't have an equation; you have an expression. The translation of an expression into English is not a sentence, but a noun phrase. So the verb "approximates" doesn't belong; you shouldn't have a verb at all. (Well, you can have them in subordinate clauses, but not as a predicate.) --Trovatore 07:59, 14 February 2007 (UTC)

Actually, looking at the expression more closely, there's another confusion. This part:

${\Bigg (}\sum _{a=1}^{\infty }{\tfrac {1}{a^{n}}}{\Bigg )}$

does not depend on a at all. Did you mean to sum from n equals 1 to ∞, rather than from a equals 1 to ∞?

(Not that you can't make sense of it the way you wrote it; it's fine to take the limit as a goes to infinity of something that doesn't depend on a. It's not wrong, just kind of trivial.) --Trovatore 08:05, 14 February 2007 (UTC)

I don't know whether you're trying to find a sentence to match the formula, or a forumla to match the sentence. Either runs into problems, because neither makes sense. a can't approach positive infinity from the positive side, because it can't start out higher than infinity. It can only approach it from the negative side, and that's assumed so you don't have to write it. You can't assign a values both under the summation symbol and under the limit symbol, so you probably meant n for the summation. Also, it doesn't mean much to say that as a approaches infinity it also approaches something else, so that's probably not what you're going for with the sentence. Here's something that would make sense, and might be what you're going for: $\lim _{a\to \infty }{\Bigg (}\sum _{n=1}^{\infty }{\tfrac {1}{a^{n}}}{\Bigg )}=0$, which means "The limit of a geometric series with common ratio 1/a, as a approaches infinity, is zero." This is true. Its pieces are $\lim _{a\to \infty }{\Bigg (}\sum _{n=1}^{\infty }F(a,n){\Bigg )}=result$ where a and n are separate numbers, F(a,n) is some formula using a and n, and result is what the summation approaches as a approaches infinity. The symbol F takes a and n and does something to them, the summation takes n as input and acts on the formula next to it, the lim symbol takes a as input and acts on the summation next to it, and that must be equal to whatever's on the "right hand side" of the equals sign.Black Carrot 06:36, 24 February 2007 (UTC)

Intergral of Cosine squared X

Hi! I did the intergral of Cosine squared X or Cos²X and well... a peer of mine got me double guessing myself.. I worked it out (if correctly!) to come out to be (1/4)(sin [2x] + x) + C i was wondering if that is correct or is it just [(Sin[2x])/4] + X + C. I know I know it sounds like a silly question--Agester 03:34, 14 February 2007 (UTC)s with just the order of operations but help on this would greatly help me progress without thinking i'm a total idiot...
P.S. the difference is just the if the whole segment is over 4 or just the sine part
Thanks for input! --Agester 03:00, 14 February 2007 (UTC)

I got ${\frac {x}{2}}+{\frac {\sin(2x)}{4}}+C$

$\int \cos(x)^{2}\,dx=\int {\frac {1+\cos(2x)}{2}}\,dx={\frac {x}{2}}+{\frac {1}{2}}\int \cos(2x)\,dx={\frac {x}{2}}+{\frac {\sin(2x)}{4}}$ You can always verify by differentiation.Dlong 03:12, 14 February 2007 (UTC)

Dlong 03:12, 14 February 2007 (UTC)

hmm... i see.. that does make sense! thank you very much! That saves me from a massive headache! --Agester 03:34, 14 February 2007 (UTC)

Checking some derivatives

I'm trying to figure out whether me or the book is right on this problem:

Let $f(x)=\cot {\frac {\pi x}{10}}$ and $g(x)=5{\sqrt {x}}$

Find $(f\circ g)'(1)$

I got does not exist (since one of the factors is $-\csc ^{2}{\frac {\pi }{2}}$ which evaluates to negative one over zero), but the book has $-{\frac {\pi }{4}}$. Anyone know what's going on?

Curtmack of the Asylum 19:27, 14 February 2007 (UTC)

Yes. $-\csc ^{2}\left({\frac {\pi }{2}}\right)=-1$, not 1/0. –King Bee ^{(T • C)} 19:31, 14 February 2007 (UTC)

Oh duh, I was thinking of secant. *slaps self* Curtmack of the Asylum 21:39, 14 February 2007 (UTC)

Don't sweat it. It happens to the best of us. =) –King Bee ^{(T • C)} 21:45, 14 February 2007 (UTC)

Homework Help!!! (I know, just hear me out)

So I'm in highschool calculus and my head's about to explode. I'm trying to learn something from the examples in a book so I can do the hw, but I just don't get what happened. There's this equation in which I'm trying to separate out W in terms of t (time). F and k are constants. The original equation is:

dW

____ = dt

(F-kW)

so the book says to integrate both sides. They put the integral sign on both sides of the equation. Then they create a 1/k to the left of the integral on the left side, and counterbalance it by multiplying the numerator inside the integral by k (so it's just multiplying by a fancy form of 1, since the constant isn't affected by the integral sign). They do this because, "the differential of the denominator, d(F-kW), equals -kdW." so then it becomes integral of (-kdW)/(F-kW). Which they then say, because it's supposedly the "Integral of the reciprocal function," is ln|f-kW| But how the heck is it a reciprocal? it's essentially a derivative of a function over the function, not one over the function. How is that supposed to work? Thanks for any help, Sasha

The derivative of that function that you care about (F - kW) is -k. So it is the "derivative of the function over the function", just with a constant missing. –King Bee ^{(T • C)} 22:21, 14 February 2007 (UTC)

Basically what you did is integration by substitution, except that the substituted variable is not written out explicitly.

You have $\int {\frac {1}{F-kW}}\,dW$. You performed the subsitution $u=F-kW$; thus $du=-k\,dW$ (or $dW=-\textstyle {\frac {1}{k}}\,du$).

You then have $\int {\frac {1}{F-kW}}\,dW=-{\frac {1}{k}}\int {\frac {1}{u}}\,du=-{\frac {1}{k}}\ln |u|+C=-{\frac {1}{k}}\ln |F-kW|+C$

--Spoon! 22:40, 14 February 2007 (UTC)

Oh. Ok. So how would one perform similar mathematical magic to a new equation: integral of dM/(100-kM)? Do you multiply it by k over k again? Isn't it sort of deriv over function again? I'm having trouble really getting what the book is saying to do in these cases. And why.

And thanks for the help on that first one.

Yep, same deal. W is now M, and F is now 100. Go to town, and good luck! –King Bee ^{(T • C)} 22:29, 14 February 2007 (UTC)

So I've hit a new one, if you guys don't mind helping again. The new equation is E = RI + L(dI/dt). E, L, and R are constants. I'm trying to solve for I in terms of t. Apparently, says the answers at the back of the book, this can be rearranged and integrated into

I = (E/R)(1-e^(-Rt/L))

I can't figure this one out at all. How does one isolate the I? I tried moving the RI to the left, then multiplying by dt and fiddling around with other things, but I'm just digging myself into a whole. Is the integral of RI + L(dI/dt) equal to the integral of RI plus the integral of L(dI/dt)? Do you think this'll get me anywhere?

Try to bring it in the form (something) = dt.  --Lambiam^Talk 04:49, 15 February 2007 (UTC)

Awesome. That helped a lot. I just never know in what direction to start with this stuff.

February 15

matrices

Hey there are these weird matrices problems that need to be solved using a fast method. Things like:

1  2  3 | 4
5  6  7 | 8
9 10 11 | 12

This is like a matrix and if you know how to answer thsi please tell me, and write the method that would be dependable and fast, well to humans (i know quite a lot of ways to solve this in comp programming but requires lots and lots of doing math) fast.

Gaussian elimination is your friend.

If you're solving it by computer, it's called an "augmented matrix", and there are a number of methods for solving it quickly. If you're solving it by hand, it's called a "demented matrix", and there's no fast way to solve it. --Carnildo 20:47, 15 February 2007 (UTC)

So please tell me how people solved "demented matrices" before electronic computers were invented? What methods do they used in the 19th century? 202.168.50.40 02:27, 16 February 2007 (UTC)

They didn't. Augmented matrices are a technique for solving systems of linear equations designed to be done by a computer or programmable calculator. --Carnildo 19:27, 16 February 2007 (UTC)

Essentially the question is how to solve a system of linear equations by manual calculation, reliably and quickly. The augmented matrix

$\left[{\begin{array}{ccc|c}1&2&3&4\\5&6&7&8\\9&10&11&12\end{array}}\right]$

represents the matrix equation

${\begin{bmatrix}1&2&3\\5&6&7\\9&10&11\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}4\\8\\12\end{bmatrix}},$

which, in turn, corresponds to

${\begin{aligned}1x+2y+3z&{}=4\\5x+6y+7z&{}=8\\9x+10y+11z&{}=12.\end{aligned}}$

The standard solution techiques in numerical linear algebra factor the matrix. The most popular general method is LU decomposition with partial pivoting (and perhaps other safeguards). The singular value decomposition is more robust and flexible, but also much more expensive.

When a problem is presented as an augmented matrix, chances are you are expected to solve it using Gaussian elimination. This is a naive, more expensive, and less robust historical predecessor of LU decomposition. However, the theoretical underpinnings are the same. We employ two tools to change the form without changing the solution. First, we can multiply any row by a scalar value. Second, we can add or subtract a scalar multiple of any row to any other. To be robust, we must include a third option, exchanging two rows (pivoting).

In most cases, a series of such operations will produce

$\left[{\begin{array}{ccc|c}1&0&0&a\\0&1&0&b\\0&0&1&c\end{array}}\right],$

at which point we have revealed the solution, x = a, y = b, z = c. If we are unlucky, the last "1" (at least) will be a "0", preventing solution; but if we are really lucky, we simultaneously get c = 0, so we have a solution after all (in fact, any value of z will do).

With today's widespread availability of open-source computer algebra systems, there is rarely any practical need to solve such problems manually. The main reason to do so is to understand the fundamental ideas (which I recommend); in which case, being fast misses the point. --KSmrq^T 04:09, 16 February 2007 (UTC)

The command to do this to any matrix on your Texas Instruments graphing calculator is rref( found under a MATRIX submenu. I suggest doing your best to understand what KSmrq said even if you are in middle or high school and asking back for clarification. [Mαc Δαvιs] X (How's my driving?) ❖ 08:23, 19 February 2007 (UTC)

Regression Analysis

(question moved from Misc RD)

How would you best explain regression analysis in mathematical and practical terms? Thanks. --61.6.206.104 15:48, 15 February 2007 (UTC)

Did you read our entry on regression analysis?  --Lambiam^Talk 20:39, 17 February 2007 (UTC)

February 16

Pull tabs

What's the average mass and volume of a pull tab from an aluminum can?

^π⁄₆ (units unspecified). Go find a trivia forum and ask there. This question has no mathematical content. --KSmrq^T 10:57, 16 February 2007 (UTC)

I don't know, but here's how to find out. First, start a donation drive and get all your friends to donate their pull tabs to you. Collect maybe 1000 pull tabs. Then, weight all the pull tabs on a scale. Divide by the number of pull tabs, and that will be the average mass of one pull tab. Finding the average volume is trickier, since you haven't really defined what volume you're looking for - there's a lot of empty space in a pull tab! So are you looking for the volume of just the aluminum particles (which means I would estimate based on the calculated mass and the given density of aluminum)? -sthomson06 (Talk) 16:00, 16 February 2007 (UTC)

To find average volume fill a bucket to the brim with water, add your 1000 pull tabs to the bucket, collect the water displaced, measure its volume and divide by 1000. Running around naked crying out "Eureka" is an optional part of the procedure. Gandalf61 16:58, 16 February 2007 (UTC)

That part's optional? Shoot, if I had known it was optional it would have saved me a TON of time. :/ Dugwiki 17:16, 16 February 2007 (UTC)

Although it's an optional step that's not to say it wont affect your results. Provider uk 18:04, 20 February 2007 (UTC)

Are you sure water displacement would work? Most pull tabs I've seen are curled up on themselves at the edge, and I'm pretty sure they'd trap air. Black Carrot 07:50, 20 February 2007 (UTC)

How about if you use the average mass of a pull-tab and the density of aluminum to calculate the volume ? StuRat 00:37, 27 February 2007 (UTC)

Ellipses and hyperbolas

How can I find the equation of an ellipse, centered on a certain point, that passes through two given points?

eg, center at (0,0), passing through (1,(-10*2^(1/2))/3) and (-2, (5*5^(1/2))/3).

And how can I find the equation of a parabola, with a given vertex, that passes through two given points?

(I know this must be an very easy question but I am pretty dumb and I can't figure it out).

Thank you kind people!

It's fairly simple - I'll give an example for the parabola - the equation of the parabola I assume is y = a(x-b)²+c

Now you have three points - the vertex (the turning point at the top or bottom of the parabola), and two others.

The x value of the turning point gives b eg if the peak is at (4,7) then b=4

Now place the other two pairs of values into the equation - this gives you two linear eqautions in terms of the unknowns 'a' and 'c' - you should be able to easily solve these equations by eliminating one of the unknowns first and finding the other.

For the ellipse the method is similar - the centre of the ellipse is (0,0) therefor the equation for the ellipse (x-x1)²+b(y-y1)²=f² simplifies to (x)²+b(y)²=f² - replace values for x and y to give two linear equations with f² and b as the unknowns and solve as linear equations.87.102.20.186 11:26, 16 February 2007 (UTC)

If the ellipse center is at (x_c,y_c) translate to the origin. Further assume (as you seem to) that the ellipse is axis-aligned. Then its equation is

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.\,\!$

Let c = ¹⁄a² and d = ¹⁄b². This gives a linear equation in c and d:

$(x^{2})c+(y^{2})d=1.\,\!$

When we substitute the coordinates of the first point, (x₁,y₁), we get an equation that c and d must satisfy; likewise when we substitute the coordinates of the second point. We know how to solve a pair of simultaneous linear equations. Do so, then recover a and b using the positive square roots. Be careful; some pairs of points allow no solution.

The parabola raises a question. As before, place the vertex at the origin; but now we cannot expect incidence with most pairs of points if we insist on axis alignment and a fixed vertex. (The equation would be y = ax².) Therefore we must be free to rotate the parabola, and need a suitable formula. However, I see no way to remain with linear equations; the "aiming" of the parabola would seem to require at least a quadratic variable. Perhaps, as in the title you gave this section, you wanted a hyperbola, not a parabola? --KSmrq^T 13:09, 16 February 2007 (UTC)

Negative exponents and differential calculus help

I am currently working my way through Calculus Made Easy by Silvanus P. Thompson. I have just read the section on the power rule and am stuck on the example showing that it applies to negative exponents. In particular, the equation is set out like this.

Let $y=x^{-2}$. Then proceed as before:

$y+dy=(x+dx)^{-2}$

$y+dy=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$

I do not understand how he gets the $y+dy=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$ part from the part above. Can someone please explain it to me? Thanks. --80.229.152.246 18:47, 16 February 2007 (UTC)

I'll try. It's just an algebraic manipulation. Here:

$(x+dx)^{-2}={\frac {1}{(x+dx)^{2}}}={\frac {1}{x^{2}+2xdx+(dx)^{2}}}$ $={\frac {1}{x^{2}(1+2/x\cdot dx+(dx)^{2}/x^{2})}}$ $={\frac {1}{x^{2}(1+dx/x)^{2}}}=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$

Hope that makes sense. –King Bee ^{(T • C)} 19:20, 16 February 2007 (UTC)

$(x+dx)=x\left(1+{\frac {dx}{x}}\right)$ --Spoon! 21:28, 16 February 2007 (UTC)

Thanks for the help guys. I still don't understand why he would want it written like that though. Oh well, I think I should leave the book for a few years. Thanks very much. --80.229.152.246 09:49, 19 February 2007 (UTC)

Take a look at some precalculus topics, if you get stuck with something. Mr.K. (talk) 14:06, 19 February 2007 (UTC)

You did not ask why the author did this, but how he got there. As to the why, it should be because it is a step towards the final desired result. In general, if n is positive,

$(1+\epsilon )^{-n}=1-n\epsilon +\epsilon ^{2}U(\epsilon )\,$,

in which U(ε) goes to a definite limit as ε tends to zero. For example, if n = 2, U(ε) = (3+2ε)(1+ε)^–2. This can be used to see how (A+δ)^–n differs from A^–n for very small δ, where A ≠ 0, since

$(A+\delta )^{-n}=(A(1+{\frac {\delta }{A}}))^{-n}=A^{-n}(1+{\frac {\delta }{A}})^{-n}\,$

$=A^{-n}(1-n{\frac {\delta }{A}}+({\frac {\delta }{A}})^{2}U({\frac {\delta }{A}}))\,$

$=A^{-n}-n\delta A^{-(n+1)}+\delta ^{2}A^{-(n+2)}U({\frac {\delta }{A}}))\,$.

If δ is sufficiently small, the third term is negligeably small compared to the second term. Putting A = x and δ = dx, you get the formulas from your textbook, at least for the first line.  --Lambiam^Talk 16:46, 19 February 2007 (UTC)

February 17

homework

Kevindew 19:33, 17 February 2007 (UTC)Below is a data set. Put two more numbers in it so that 1) the median of the new data set is 6, 2) the maximum is 15, and 3) the range is 13.

4 5 4 11 8 - -

OK then. What's your question? SubSeven 20:32, 17 February 2007 (UTC)

Near the top of the page it says: "Do your own homework." That having been said, are you sure all these numbers are correct? Extending the given data set, you can achieve any two of the objectives, but not all three. The following combinations are possible:

median	maximum	range
6	15	11
6	17	13
5	15	13

It is further possible to achieve all three objectives if the data set to be extended is any of:

2 5 4 11 8 – –

4 2 4 11 8 – –

4 5 6 11 8 – –

4 6 4 11 8 – –

4 15 4 11 8 – –

 --Lambiam^Talk 20:37, 17 February 2007 (UTC)

February 18

Notation confusions with physics and mathematics

I've found that in some of the articles regarding vectors and Hilbert spaces contain the notation of vectors commonly used by physicists rather than mathematicians. I know that there are many ways to represent the same thing, and that the information is still correct, but shouldn't we adhere to one single method? It's hard to switch back and forth, especially when you're trained in pure mathematics and not physics... — Preceding unsigned comment added by Sluzzelin (talk • contribs) 07:25, 18 February 2007 (UTC)

Good question, but not one for the reference desk; this is for answering questions about mathematics, not about wikipedia. Unfortunately, I'm not sure where the right place is for this discussion. Wikipedia:WikiProject Mathematics might be a good start though. Algebraist 14:21, 18 February 2007 (UTC)

This is the right place to ask I think. The reference desks are for questions relating to improvements or otherwise to wikipedia etc.87.102.9.240 15:40, 18 February 2007 (UTC)

No, the reference desks are decidedly not for that purpose; in fact, there is no such discussion on this very long page of questions. But in the interests of economy, I can answer this one. Wikipedia doesn't enforce uniformity of spelling, or much of anything else, and there is little chance that we will try to enforce consistence of notation among disciplines. We do try to inform readers when different conventions exist. But if the Brits can deal with seeing "color" instead of "colour", and the Yanks can cope with "centre" instead of "center", surely a bright mathematician can handle the conventions of physics. Mathematics is not all that consistent within itself, as we have discovered in many an article. --KSmrq^T 15:53, 18 February 2007 (UTC)

Well, I guess the problem is that "centre" and "center" are rather close and easily discernable, but sometimes different notation isn't. A lot of times you can figure it out easily, but I still think consistency is a good idea. Perhaps on the quantum mechanics page the use of the physics notation is desirable, but it seems out of place on the mathematics page. Pure mathematics and physics really are two completely different things...

Are you talking about braket notation? – b_jonas 18 February 2007

No. Articles dealing with physics should use notation conventional to the physics community and likewise for the mathematics community.

Well, Hilbert space was an idea created for physics and is most often used in physics instead of mathematics, so I don't see why there is a problem anyway, if that information is correct. [Mαc Δαvιs] X (How's my driving?) ❖ 07:58, 19 February 2007 (UTC)

Everyone, please sign your posts with four tildes ("~~~~"), as stated at the top of this page. And those who wish to discuss this further should take the conversation to Wikipedia talk:WikiProject Mathematics. Thanks. --KSmrq^T 08:09, 19 February 2007 (UTC)

Oh sorry. Sometimes I still forget it. – b_jonas 08:26, 19 February 2007 (UTC)

February 19

Best calculus textbook

I've heard very often of Tom Apostle's Calculus I and Calculus II written in the 40s and 50s, which are very proof-heavy, well, entirely proof-based. What are your opinions on the best calculus textbooks? [Mαc Δαvιs] X (How's my driving?) ❖ 08:28, 19 February 2007 (UTC)

I can also recomend Leithold's book. Specially for beginners it is a good choice. Mr.K. (talk) 13:48, 19 February 2007 (UTC)

More information about your expectations would help. Do you want a graphic-rich easy introduction? Well-grounded proofs? Higher mathematics insight? Emphasis on engineering/physics applications? A modern revision? One of the non-standard analysis presentations?

Most texts I've seen do not impress me. Richard Courant and Fritz John do. Try a variety of books online, including those listed by Stef and at AMS. --KSmrq^T 04:27, 20 February 2007 (UTC)

simplex method

using this tool:

http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/simplex.html

I've entered in and solved:

Maximize p = 3x+y+3z subject to

• 2x+y+z <= 2
• x+2y+3z<=5
• 2x+2y+z<=6

I follow up to Tableau 2 but what doesn't make sense is in Tableau 3, column s1 and the last row. Shouldn't this number be negative from the column s1 row 2? — Preceding unsigned comment added by 75.8.72.171 (talk • contribs) 08:46, 2007 February 19 (UTC)

The given value of 1.2 is correct. This has come from a pivot on the (s2,z) element in Tableau 2, which has value 2.5; to eliminate z from the objective row (current element -1.5), 60% of the 2nd (s2) row is added to it. 60% of -0.5 added to 1.5 gives 1.2, as displayed. Someone will tell you to sign your post, can't say it bothers me.—86.132.165.26 10:25, 19 February 2007 (UTC)

(Constraints formatted, unsigned signed.) Wikipedia asks for the signature, on all talk pages; individuals like me merely remind. If you're curious why, you could read Wikipedia:Signatures. It says

• Signing your posts on talk pages and other Wikipedia discourse (but not on articles) is not only good etiquette; it also facilitates discussion by helping other users to identify the author of a particular comment, to navigate talk pages, and to address specific comments to the relevant user(s), among other things.

Coherent conversations are easier when we know who said what. (And Oleg Alexandrov would add, please remember to use the edit summary box.) --KSmrq^T 12:23, 19 February 2007 (UTC)

It's also good etiquette for the original poster to acknowledge an answer to the problem posed. —86.132.236.24 14:26, 20 February 2007 (UTC)

We don't ask for thanks, but we do appreciate it. Also note that indenting responses is standard practice; accidentally disrupting others' posts is not (both fixed). :-) --KSmrq^T 14:42, 20 February 2007 (UTC)

Math Formatting Help

If possible, could someone give me the wiki formatting for the equation F=6.6742x10^-11N*m^2/kg^2(200g*1kg)/(10cm*0.01m/cm)^2 ?

$F={\frac {6.6742\times 10^{-11}\ {\mbox{N}}\ {\mbox{m}}^{2}\ {\mbox{kg}}^{-2}\,(200\ {\mbox{g}}\times 1\ {\mbox{kg}})}{(10\ {\mbox{cm}}\times 0.01\ {\mbox{m}})^{2}}}$

Is this what you wanted? Hope I helped. Deltacom1515 03:51, 20 February 2007 (UTC)

Read up on LaTeX :) 213.48.15.234 09:57, 20 February 2007 (UTC)

WP:MATH will provide all the answers. --h2g2bob 14:11, 20 February 2007 (UTC)

February 20

Fibonacci Sequences

Alrighty, I've noticed something interesting with this generator thing. When you increase b by one (in that generator), starting at the second term you're just adding the standard Fibonacci 1,1,2,3,5...

Ex. (a,b)

(10,11)- 10 11 21 32 53 85 138

(10,12)- 10 12 22 34 56 90 146

(10,13)- 10 13 23 36 59 95 154

Is there any specific name for this or is it just there? Deltacom1515 00:37, 20 February 2007 (UTC)

OEIS kindly gives the name, Fibonacci sequence beginning 1 10.  ;) As for number 2, you get Fibonacci sequence beginning 2 10. So unless you consider these specific, I guess... No? x42bn6 Talk 01:18, 20 February 2007 (UTC)

Well, I noticed that as I increase that second number in the parenthesis, b, the number sequence starting at the second term, is increased by 1,1,2,3,5,8,13, etc. which is what I guess you would cal la normal Fibonacci sequence. I just wondered if there was a name/reason for this occurring in this fashion. Deltacom1515 02:16, 20 February 2007 (UTC)

Of course it would, and for the same reason the Fibonacci numbers exist in the first place. Each is the sum of the previous two. I'll take the second entry in your list. At 10, you've added zero to the base sequence, at 12 you've added one. To get the next number, it's (base 0 + Fib 0) + (base 1 + Fib 1), which is (base 0 + base 1) + (Fib 0 + Fib 1), which is (base 2) + (Fib 2). To get the next number, it's (base 1 + Fib 1) + (base 2 + Fib 2), which is (base 1 + base 2) + (Fib 1 + Fib 2), which is (base 3) + (Fib 3). In other words, you're doing exactly what you've described: adding up the base case as before, with the Fibonacci numbers layered on top. Black Carrot 07:40, 20 February 2007 (UTC)

This works because the Fibonacci sequence is generated by a linear recurrence relation. If F(a,b) is the Fibonacci sequence generated by initial terms a and b, then you have noticed that F(10,11)+F(0,1)=F(10,12). This is a special case of the general relationship F(a,b)+F(c,d)=F(a+c,b+d). This will not work for a non-linear relation - for example, if each term is the product of the previous two terms, instead of their sum. Gandalf61 11:31, 20 February 2007 (UTC)

primitive of 1/ln(x)

When investigating the number of primes less than x, i came across the function li(x), which is the integral from 0 to x of 1/ln(x). My computer can estimate this function, but I was wondering if a primitive existed. Does anyone know if a primitive exists or weather it does, the article on this does not seem to give an answer either way.Thepalm 06:25, 20 February 2007 (UTC)

Surely you don't mean $\int _{0}^{x}{\frac {1}{\ln x}}\ dx$? There's too many x's in there. Maybe $\int _{0}^{x}{\frac {1}{\ln t}}\ dt$? Anyway, I don't know of a primitive; something tells me such a thing is hard to find. At least, I hope it is. I'm teaching Calc II at a university right now, and my students would probably celebrate in the streets if they found an integral I couldn't do. =) –King Bee ^{(T • C)} 06:30, 20 February 2007 (UTC)

I'm not exactly sure but i think that $\int _{0}^{x}{\frac {1}{\ln t}}\ dt$ is the more correct version. Basically i was looking for a primitive of 1/ln(x). Maybe that clarifies things.Thepalm 07:08, 20 February 2007 (UTC)

Did you check the logarithmic integral function article? It has a few ways to calculate it, but not exactly what you're looking for I think. One thing that's very close that they don't mention is x/lnx. Black Carrot 07:29, 20 February 2007 (UTC)

I was wondering if there was a way to calculate it without using an infinite series, which is what most of the methods there suggest. that article does mention the x/ln(x) approximation, but the values that they give are quite different as x gets very large Thepalm 09:39, 20 February 2007 (UTC)

1/ln(x) is an example of an elementary function whose antiderivative cannot itself be expressed in terms of elementary functions. And even if you expand your list of "elementary functions" to include functions such as li(x), then there will be functions on your new list whose antiderivatives are still not on your new list - see differential Galois theory for a more precise statement of this fact. So you should be pleasantly surprised when a function does turn out to have an antiderivative that you recognise. Of course, the standard examples in introductory calculus courses are carefully chosen to disguise this inconvenient truth ! Gandalf61 11:02, 20 February 2007 (UTC)

Examples in every subject are carefully selected. I remember as a very young student being struck by the realization that answers to assigned problems always came out too cleanly to be accidental. Much later I discovered that research questions often have a different character, and may defy any answer, much less a pretty clean one. We publish the stuff that works out well, and quietly discard the rest. Worse still, we hide the usually messy path of discovery. Consider Hamilton's well-documented route to quaternions; he had fifteen years of failure before the answer suddenly became "obvious". A more modern example is the ordeal Andrew Wiles went through with Fermat's Last Theorem. :-)

Aside from deliberate selection, we also see something else, more intriguing. We can prove that most real numbers cannot be computed, most real-valued functions are wild, and so on. We can also prove that theorems can be true yet not provable. We could say that our entire mathematical playground is a set of measure zero compared to what's really out there. --KSmrq^T 14:33, 20 February 2007 (UTC)

standard integrals

1.A free electron in a metal in a state .¢=sin¶x÷a where 0<x<a or = 0 elsewhere.Determine the average value of x,<x>.Hence,find the uncertainty in position.

a mile

How many steps in a mile68.12.114.162 17:49, 20 February 2007 (UTC)

This would depend on the how big your stride is, but according to this site, the average is about 2000 steps. - Akamad 21:00, 20 February 2007 (UTC)

There are 8 furlongs to a mile. Next determine how many steps you can take in a minute. Then determine how many furlongs you can travel in a fortnight. The mathematics from here onwards is quite simple. Convert minutes to fortnight and divide the steps by the furlongs and further divide by 8. 202.168.50.40 23:29, 20 February 2007 (UTC)

.3(9)

Since, .(9) = 1 and 2.(9) = 3, does .3(9) = 4? Jac _roe 23:33, 20 February 2007 (UTC)

If you mean 3.999..., then yes. Splintercellguy 23:41, 20 February 2007 (UTC)

0.3999999999... = 0.4 202.168.50.40 00:04, 21 February 2007 (UTC)

We have a simple general procedure for converting any repeating decimal to a quotient of integers. To illustrate, take

$r=3.6{\overline {142857}}.\,\!$

The repetition involves six digits, so we multiply six times by ten, or once by 10⁶ (which is 1,000,000).

$1\,000\,000r=3614285.7{\overline {142857}}\,\!$

Since we have shifted by one complete repetition, subtraction eliminates it all.

$1\,000\,000r-r=3614282.1\,\!$

Now it is a simple matter to finish.

${\begin{aligned}999\,999r&{}={\frac {36142821}{10}}\\r&{}={\frac {36142821}{9\,999\,990}}\\&{}={\frac {253}{70}}\end{aligned}}$

The only hard part is the last step, reducing to lowest terms, which is not that hard. --KSmrq^T 11:30, 21 February 2007 (UTC)

February 21

division

6x^7/2x^2=

Having a little hw trouble there? Looks like basic algebra. Divide the coefficients and subtract the exponents. Rya Min 03:04, 21 February 2007 (UTC)

(6 / 2) x ^ (7 - 2). Whenever you multiply exponents, they're added. When you divide, it's subtracted. So x^2 * x^3 is x^5, and x^3 / x^2 is x^1 (or x). --Wirbelwindヴィルヴェルヴィント (talk) 03:06, 21 February 2007 (UTC)

Let A=x^7

Let B=x^2

6x^7/2x^2 = (6*A)/(2*B) = 3 * (A/B)

And 3 * (A/B) = 3 * (x^7/x^2)

The rest is pretty easy. 211.28.127.77 07:03, 21 February 2007 (UTC)

When two diagonizable matrices commute, are they simultaneously diagonizable?

Hello,

my question may not be that hard, but I'm really confused.

There is a theorem saying that commuting hermitian matrices are simultaneously diagonizable. I was wondering about the relevance of the matrices being hermitian.

I found a simple example... but one of the two was not diagonizable at all!

So this is my question : let$A$ and $B$ be diagonizable matrices over some field, and let $AB=BA$ . Can an invertible matrix $Q$ be found such that both $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal matrices?

My guess is yes... :| But is this true? Is it true that the matrices being hermitian isn't that important after all?!

Thank you very much, Evilbu 15:59, 21 February 2007 (UTC)

Example: $A=QD_{1}Q^{-1}$ and $B=QD_{2}Q^{-1}$ commute, simultaneously diagonalizable, and are not necessary hermitian.(Igny 18:06, 21 February 2007 (UTC))

Thank you. But the main question still remains: what about the other way around, what if they are just both diagonizable and commuting?

Yes, you are right. In fact, you can do it with more than just two; any commuting family of diagonalizable matrices is simultaneously diagonalizable; i.e., there exists a matrix Q such that for all A in the family, QAQ^-1 is diagonal. –King Bee ^{(T • C)} 18:51, 21 February 2007 (UTC)

Thanks, that's great! Do you think you can give me a hint how to prove this? I mean, I have already figured out that I can write the vectorspace as a direct sum of eigenspaces of A, and that (because of the commutating property) all of these spaces are invariant with respect to B. Any basis of eigenvectors for the restriction of B to one of these eigenspaces, would be a basis of eigenvectors of that eigenspace of A as well. But there's my problem : WHY is this restriction of B diagonizable? Thank you,Evilbu 19:35, 21 February 2007 (UTC)

Hmm...I actually cannot remember how to do it. I think you're on the right track though; I remember showing that two commuting matrices are simultaneously unitarily upper triangularizable, and I the proof of that has a lot to do with eigenspaces, as you mentioned. –King Bee ^{(T • C)} 15:23, 22 February 2007 (UTC)

Sorry about this, but I can't remember a totally elementary proof, though I'm sure there is one. Going slightly high-tech, recall (or learn ;-)) that a matrix is diagonalisable iff its minimal polynomial splits into distinct linear factors. Since the restriction of B to an eigenspace of A is a zero of the minimal polynomial of B, it has minimal polynomial dividing that of B and so it is diagonalisable. Thus your proof goes through. Algebraist 15:28, 22 February 2007 (UTC)

Eigenspaces are, indeed, key. Maybe it would help to begin with two real symmetric matrices, each with distinct eigenvalues. Then each eigenvalue has a distinct one-dimensional eigenspace, and that space is (as always) orthogonal to all the others. If A₁ and A₂ commute then their eigenspaces must align. (Discussed further below.) The general case is slightly more interesting. Of course we can have eigenvalues with multiplicity; but consider a pair of 2×2 rotation matrices. These diagonalize over the complex numbers, and always commute.

Formally, suppose A₁ maps the linear subspace E into itself; in particular, let E be the eigenspace corresponding to eigenvalue λ. Then I claim that A₂ also maps E into itself. For, let v be a vector in E; then

${\begin{aligned}A_{1}(A_{2}{\mathbf {v} })&{}=A_{2}A_{1}{\mathbf {v} }=A_{2}\lambda {\mathbf {v} }\\&{}=\lambda (A_{2}{\mathbf {v} }).\end{aligned}}$

That is, A₂v is an eigenvector with eigenvalue λ, thus in E. We can leverage this to prove simultaneous diagonalizability. --KSmrq^T 20:46, 22 February 2007 (UTC)

A riddle...

"If A is 2, B is 3 and F is 8, how high must you count to get to Heaven?"

This question was asked on this desk a few months ago, but since there was no useful answer I'm asking again. This is what I know about the riddle:

• It's NOT a trick question. There is a numerical answer that corresponds to the word "heaven".
• F is NOT supposed to be 7. F is definitely 8. It is nothing but 8 and never will be anything but 8. DON'T respond with "if F is 7" because that will just make me angry.
• The answer is not 70

This puzzle comes from an online webcomic called Jack and the correct answer appended to the site's URL should lead to a secret comic. Thanks for any help. Lovablebeautyme 21:38, 21 February 2007 (UTC)

I gave the answer before it's "70" - see below for another explanation - here is the secret page http://pholph.com/seventy/ - see it works>87.102.9.28 15:58, 22 February 2007 (UTC)

Why is it not 70? What is so special about 70? Are you absolutely positive it is not 70? (Igny 22:16, 21 February 2007 (UTC))

It could be anything. That's how riddles are. Some context might help, could you link to the actual question? One obvious solution is that each letter in Heaven is a digit in the answer. If each letter corresponds to its position in the alphabet, Heaven would be (8)(5)(1)(22)(5)(14). It doesn't, though, A=2 and B=3 and F=8, which means their positions would have to be seriously scrabbled about to make it work. If you can think of values for them, just translate from base 26 (or whatever it winds up as) to base 10 and you'll probably have the answer. Or you could just cut the knot and webcrawl through his URLs, until you find one of the form http://www.pholph.com/(somenumber). Odds are you'll find it doesn't exist. Black Carrot 22:42, 21 February 2007 (UTC)

I gave the answer to this one last time - here it is again. Count 1 through the alphabet - except on a vowel when you add 2 - thus:

A B C D E F G H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z 
2 3 4 5 7 8 9 10 12 13 14 15 16 17 19 20 21 22 23 24 26 27 28 29 30 31

So heaven = 10+7+2+27+7+17=19+34+17=19+51=70.

Now 70 = "Three score and ten" -- See: [6]

Giving the rhyme

"If A is two, B is three and F is eight,
how high must you count to get to heaven's gate?" (See now it rhymes)

Three score and ten have a man = 70 = life expectancy

This is the answer - what was wrong with my original answer?87.102.9.28 15:54, 22 February 2007 (UTC)

February 22

Different answer to an integral than that given by Mathematica

Ok, so I have this Calculus II homework problem. $\int {\cot ^{2}(2\theta )}d\theta$. When I solved it it gives me $-{\frac {1}{2}}\left[\cot(2\theta )-2\theta \right]+C$, but Mathematica's Integrator gives me $-\theta -{\frac {1}{2}}\cot(2\theta )+C$. Is my solution wrong? If not, How are these two responses equal? I guess I should sign eh?MentalPower 06:34, 22 February 2007 (UTC)

Clearly they differ, but only by one sign; check your work. I'm amazed that you could do the integral (almost) and not the comparison. --KSmrq^T 10:11, 22 February 2007 (UTC)

Yes, your answer is wrong. I assume you used the trig identity I'm thinking of, so let's start there.

$\int \cot ^{2}(2\theta )\ d\theta =\int \csc ^{2}(2\theta )-1\ d\theta =\int \csc ^{2}(2\theta )\ d\theta -\int 1\ d\theta =-{\frac {1}{2}}\cot(2\theta )-\theta +C.$

I hope that's helpful. –King Bee ^{(T • C)} 14:29, 22 February 2007 (UTC)

I doubt this is the right place to ask the question, but

Has anyone ever considered having laymen's articles on complicated subjects, especially in Math and Computers? Even the fairly noncomplicated things can be quite obtuse. I was just looking at the article on Slope Fields, a pretty straightforward Calculus concept (not that I understand it), and the wiki page didn't really clarify things at all. It just had a lot of very technical writing, and a lot of symbols. I do realize that you're all mathmaticians, and this is how such things are most directly expressed, but most of the people who look at these articles are coming at them with limited info, and they cant understand the complicated and technical jargon on most of the page. This pertains pretty much exclusively to computer and math articles with are really complicated. Science, pop culture, history, literature etc articles seem pretty straightforward, though perhaps that's due to what I know more about. At any rate, the idea would be that you could go to a page like Slope Field and instead of being expected to understand this:

"Given a system of differential equations,

   \frac{du}{dt}=f(t,u,...y,z)

           \cdots

   \frac{dy}{dt}=j(t,u,...y,z)
   \frac{dz}{dt}=k(t,u,...y,z)

the slope field is an array of slope marks in the phase space (the preceding equations imply seven dimensions, but can be any number depending on the number of relevant variables; for example, two in the case of a first-order linear ODE, as seen to the right). Each slope mark is centered at a point (t,u,...y,z) and is parallel to the vector

   \begin{pmatrix} 1 \\ f(t,u,...y,z) \\ \cdots \\ j(t,u,...y,z) \\ k(t,u,...y,z) \end{pmatrix}."

you could click on a link to a simplified version that would just explain it in English so you'd have something more tangible to work with before you jumped straight into the math. Just a thought. Didn't know where else it would go.

Clear writing is always helpful, and it doesn't need to be a simplified version of the article for clear writing to be important. Probably the article you describe could use some work. However unlike some other subjects not all topics in mathematics can be explained easily in words of one syllable: "there is no royal road". —David Eppstein 07:15, 22 February 2007 (UTC)

Wikipedia is an encyclopedia, not an instruction manual. — Preceding unsigned comment added by 203.49.242.20 (talk • contribs) 07:28, 2007 February 22 (UTC)

The plots in question are of little use in seven dimensions, and the idea is much more accessible than the article. Meanwhile, try following the references at the end. Note that we use these to get an intuition about instances of a rather complicated beast called a differential equation. Unless you understand that context, the tool will make little sense.

Physical "slope field"

Here's a physical analog. Take a bar magnet, a strip of magnetized iron with a North magnetic pole at one end and a South at the other. Physics claims the magnet creates a magnetic field, the physical embodiment of a differential equation. At every place in that field we can point in one direction as feeling North and the opposite direction as feeling South. If we take a piece of (unmagnetized) iron and shave off a pile of small iron filings, we can demonstrate. Place a very thin sheet of glass or plastic over the bar, and sprinkle the shavings on top. Give it a good shake. The little metal fibers will prefer to line up along the North-South lines they feel. This allows us to "see" the pattern of the magnetic field.

A computer program can do a similar thing numerically, using short line segments in place of iron filings, and using any differential equation in place of the magnetic field. That's the idea of a slope field. Differential equations arise frequently and naturally in many practical disciplines, including physics, biology, economics, and chemistry. Often a plot like this can be enormously helpful, especially if we believe Richard Hamming, who said "The purpose of computing is insight, not numbers." --KSmrq^T 11:18, 22 February 2007 (UTC)

Lempel-Ziv-Welch compression algorithm example on Wikipedia

Moved to Computing ref desk. Please sign your posts with --~~~~ --TotoBaggins 18:24, 23 February 2007 (UTC)

mathematical practical applications

Could mathematics calculate the date and place of a cure for cancer or any other biological disease ?

Yes, it can, but the derivation is too long to fit on this margin. – b_jonas 13:16, 23 February 2007 (UTC)

You seem to be asking if the universe is deterministic. My understanding is that even if in principle you could figure out the exact current state of a system (you can't), the randomness inherent in quantum mechanics would prevent you from making predictions based on that state. However, math does have a few practical applications besides allowing omniscience, so we should probably keep it around. --TotoBaggins 18:30, 23 February 2007 (UTC)

Since "cancer" is many different diseases, it will be cured over a long time period, perhaps a century. Early portions of the cure have already arrived, such as the new vaccine against HPV, which causes cervical cancer. StuRat 00:20, 27 February 2007 (UTC)

Size of next prime number

I'm looking for results about the size of prime numbers. Specifically, I want this:

If $p_{n}$ is the $n$th prime number then for $n>3$,

$p_{n+1}<{\sqrt {\prod _{i=1}^{n}p_{i}}}$

This is true, right? If so is this provable using not too much maths? (Not too much = half a page and no mention of Euler)

--87.194.21.177 22:22, 22 February 2007 (UTC)

Depends on what you mean by "not too much maths". Using Bertrand's postulate, we know that $p_{n-1}\geq p_{n}/2$, so for $n>4$

${\sqrt {\prod _{i=1}^{n}p_{i}}}\geq {\sqrt {2\cdot 3\cdot 5\cdot (p_{n}/2)\cdot p_{n}}}=p_{n}{\sqrt {15}}>2p_{n}>p_{n+1},$

in the last step using Betrand's postulate again.--80.136.136.113 22:32, 22 February 2007 (UTC)

February 23

How to start?

How do you start to solve the following integral? $\int {\frac {dx}{\sqrt {16+4x-2x^{2}}}}$

Thanks --MentalPower 01:43, 23 February 2007 (UTC)

You should recognize that it resembles the integral for arcsin:

$\int {\frac {1}{\sqrt {1-x^{2}}}}\,dx$

The next step is to transform the integral until it looks exactly like the known one. I would proceed by completing the square inside the square root and then performing linear changes of variables. Fredrik Johansson 02:31, 23 February 2007 (UTC)

Try a substitution of variables - I quickly tried t = 16+4x-2x^2 but didn't get too far. The square root factorises nicely into (-2)(x+2)(x-4), so that could possibly be a clue to something. --h2g2bob 16:30, 24 February 2007 (UTC)

Fredrik is right. Complete the square first, then use a change of variables. –King Bee ^{(T • C)} 17:05, 24 February 2007 (UTC)

Check math for time dilation

I was very sleepy when I did this last night so theres a good chance it is wrong. I was trying to figure out much time would go by for a person going at 99% of c from here to the Andromeda galaxy and back. Using the equations found in time dilation, I figured that about 721,520 years would go by for them (more than I expected), while about 5,114,716 years would go by for a person back on Earth. Could one of you mathies check my math real quick? Thanks much in advance. Imaninjapirate^{talk to me} 02:08, 23 February 2007 (UTC)

Well, the Andromeda galaxy is 2.5Mly away; γ for 0.99c is 7.09. The trip as viewed from Earth is not affected by relativity: it's just ${\frac {2(2.5{\text{Mly}})}{0.99c}}=5.05{\text{My}}$. The person travelling sees the distance as reduced by γ, so they count ${\frac {2(2.5{\text{Mly}})}{7.09(0.99c)}}=712{\text{ky}}$. Remember that for any actual such trip, there must be acceleration at the destination to come back: see twin paradox. --Tardis 17:08, 23 February 2007 (UTC)

Re: more than I expected. It really is depressing just how far away everything is, even at ludicrous speed. It's like a cosmic caste system where you're destined to die where you were born, give or take 100 light years.  :( --TotoBaggins 20:29, 23 February 2007 (UTC)

Exactly what I thought. Hopefully we can eventually find a way to break that barrier that is the speed of light. Right now that notion sounds ludicrous to science, but so would have most of our modern technology and science to someone only a few hundred years ago, so I suppose it could be possible sometime in the far future. Thats supposing we don't kill ourselves first. :-/ Imaninjapirate^{talk to me} 02:22, 24 February 2007 (UTC)

270 degree triangles

I was wondering, if it is possible to make a triangle in conjunction with Non-Euclidean geometry that is more than 270 degrees? — Preceding unsigned comment added by 124.191.241.62 (talk • contribs)

Sure it is. On a sphere draw the equator, then draw two lines of constant longitude from the North pole down to the equator. These lines will each meet the equator at an angle of 90 degrees, so you have two right angles in your triangle straight away. If these longitude lines meet at an obtuse angle at the North pole (say they are 0 degrees longitude and 135 degrees longitude, for example) then you have a triangle with an angle sum that exceeds 270 degrees. Gandalf61 11:33, 23 February 2007 (UTC)

Well, you can get even more... Draw a small triangle on the sphere, and then declare its "interior" being "outside" it. Now you have a triangle with a sum of angles arbitrarily close to (3×360°)–180°=900°. --CiaPan 12:34, 23 February 2007 (UTC)

"Blackboard bold 1" with subscript

In my probability courses I've come across (and used) the notation 1_{A} (where the 1 is meant to be "Blackboard Bold") to respresent a function that takes a value 1 when the event A is true and 0 when the event A is false.

I've just used it in a different course to avoid the joy of writing 0^r when dealing with the case r=0. Instead, I used 1_{r=0}, i.e. 1 if r=0 or 0 otherwise.

However, I'm aware this may not be standard notation, so my question is:

What would you call this function? Only thing I could think of was the identity function, and I know that's not right.

Many thanks. Rawling4851 12:12, 23 February 2007 (UTC)

You could call it the indicator function of the set. – b_jonas 13:12, 23 February 2007 (UTC)

Indeed, what you are describing would be the indicator function of the singleton zero. Also, I've seen some (logic) textbooks call a (usually discrete) function taking the value 1 at zero and 0 everywhere else a 0-test function. Not very original, but I guess it works. Phils 16:29, 23 February 2007 (UTC)

Also read Kronecker delta and Iverson bracket. --mglg_(talk) 17:14, 23 February 2007 (UTC)

Outside probability, it is also called the Characteristic function of the set. In probability characteristic function means something totally different of course. Algebraist 17:48, 23 February 2007 (UTC)

Our indicator function article provides an assortment of names, notations, and applications for variations of this. --KSmrq^T 18:25, 23 February 2007 (UTC)

thanks to all, guys :D Rawling4851 14:15, 24 February 2007 (UTC)

February 24

Length of Curved part of a circle

The answer I get is different to that provided in SMQ8 (Secondary Maths for Queensland) Year 8. Would you please show me where I am wrong. I have been able to work out all the others in Chap 6.5. The question and my workings follow.

Calculate the length of the curved part... In the circle there are two curved areas, opposite each other, both have an angle of 45 degrees. Therefore the curved area is two times 45 degrees of the 360 circumference. Eg 90 degrees of 360 degrees that is one quarter. Next, an arrow showing radius of 15 cm. Circumference = 2radius x pie that is 2x3.14x15 = 94.2 Curved part = 94.2 x .25

My answer 23.5. Answer in book 84 cm I would appreciate your help MathsDaphne Williams 01:40, 24 February 2007 (UTC)

My guess is that they want you to include the line segments connecting the arcs, whose combined lengths amount to four times the radius. Using R for the radius, that results in a total length of (π/2+4)R, or in a round figure 84 cm if R = 15 cm.  --Lambiam^Talk 02:47, 24 February 2007 (UTC)

If the instructions really say "the length of the curved part", then they were wrong to include straight lines in the calcs, so you should be able to argue to have your answer marked as correct. StuRat 00:13, 27 February 2007 (UTC)

Software for category theory diagrams

Hey

I want to create a wikipedia article, but I might need to draw some diagrams. What do people use to make stuff like pullback diagrams?

TIA Bgst 15:11, 24 February 2007 (UTC)

Use a (La)TeX macro package like those by François Borceux or Paul Taylor to create the diagram, and convert it (preferably) to SVG format. Apart from the last conversion step, that is how the diagram at Pushout (category theory) appears to have been created. If you get stuck, possibly its creator User:Fropuff can be of assistance.  --Lambiam^Talk 17:42, 24 February 2007 (UTC)

Fropuff has previous offered to help create diagrams. Also, in November 2006 I answered a similar question at Wikipedia talk:WikiProject Mathematics by pointing at Image:Snake lemma nat.png, the page of which documents its genesis. Most diagrams are simpler, and can use the method of Image:CategoricalPushout-02.png. --KSmrq^T 01:10, 26 February 2007 (UTC)

If you like generalized abstract nonsense, there are programs to create fake Mondrian paintings.172.146.58.73 09:10, 26 February 2007 (UTC)

February 25

Lindemann-Weierstrass Theorem

Is it possible to prove the Lindemann-Weierstrass Theorem without knowledge of the Fundamental Theorem of Algebra? Every proof I've seen usually has the statement (usually implicitly) that an integral polynomial of degree n has n complex roots.

Triangle

(Sorry I can't speak English as well.) I got a isosceles triangle with the corners ABC. At the side AC is a dot with the name X and at the side BC is a dot with the name Y. The length of AX is the same as BY. Now I must draw two circles, the first goes to the points A, Y and C and the second goes to B, X and C. Now I get another vertex (I hope that vertex is the right name) instead of in the corner C. When I draw now a line between the new vertex and the corner C, I will get a bisection of the angle ACB. Why will I get a bisection?

Think symmetry. Display the triangle with AB horizontal. The two circles are mirror images in a vertical line through C. --KSmrq^T 01:14, 26 February 2007 (UTC)

Sorry but no. I don't have the answer but it's not as simple as that. If it were an equilateral triangle then maybe, but in an isosceles the only symmetry is through point A. One of the circles has AC as a chord and intersects point y, the other has BC as a chord and intersects point x. AC is equal to AB but not to BC. Vespine 02:44, 26 February 2007 (UTC)

You added the assumption that the symmetry is through A; the problem statement did not, and so I assumed C. When you have a proof, get back to us. --KSmrq^T 17:45, 26 February 2007 (UTC)

February 26

Several proofs of the Goldbach conjecture on arXiv, how come?

Nearly all litterature, websites and articles I have read have stated that as of now there is no proof of the Goldbach conjecture. However should one continue to search for "proof goldbach conjecture" on arXiv [7] 6 supposed "proofs" turn up. Most might appreciate that these are not proofs of the the GC, but merely attempts by undergraduates. But if they are not proofs why are they still listed on the site as "proofs", a person who does not know so much about the GC might come to believe that there actually is a proof. On the other hand if one of these proofs holds up to scrutiny how does one get to know about by not reading about it in a maths journal.

Basically this question is about the workings on arXiv, who checks that the files uploaded on the page are accurate? I myself am not qualified to do so as I am only an undergraduate in maths.

Although there are certain systems in place that are designed to prevent anyone uploading anything that doesn't belong, the detailed content of files uploaded to any of the different arXiv lists are not checked by anyone, and you can sometimes find all manner of "interesting" proofs and scientific theories on there. Many researchers wait until having a paper approved for publication by a refereed journal before posting a paper there, or at least until they've submitted it to one, but not all, and many independent researchers have no such option. Take a look at Arxiv#Peer-review. Caveat lector, basically. Spiral Wave 11:51, 26 February 2007 (UTC)

What is the relationship between graph theory and topology?

Graph theory is much like a 2d topology. But there is such a thing as weighted graph theory, e.g. a problem related to travel time or fiber length. Topology doesn't seem to put much focus on "distance", as it is a geometry of position. The reason I ask is that the Konigsberg bridges and n-color theorems are covered in entry level books on graph theory *and* topology.172.146.58.73 08:58, 26 February 2007 (UTC)

Sure. Many problems in graph theory deal with weighted graphs. Some examples are finding the minimum spanning tree of a graph, finding the shortest distance between two vertices in a graph (see Dijkstra's algorithm and breadth-first search), and the famous travelling salesman problem. —Bkell (talk) 09:07, 26 February 2007 (UTC)

There is an answer at topological graph theory, but I have to say I found the description at PlanetMath more focused. 84.239.129.42 21:51, 26 February 2007 (UTC)

For the most part these are two distinct fields, with different interests and different methods. Formally, we might describe a graph using the language of topology (for example, as a CW complex), but that offers little. The Euler characteristic is important in both algebraic topology and graph theory; but considering that the former discretizes topology, the overlap is not so surprising.

Consider a tetrahedron with vertices on the surface of a sphere, projected onto the sphere. As topologists, we can consider S² itself. As graph theorists we can see the four vertices and the six edges connecting them. As algebraic topologists, we can count faces, edges, and vertices to compute an Euler characteristic, or we can compute Betti numbers via homology groups to reach the same conclusion. A topologist might ask if the space is compact or connected or path connected or the same as, say, a torus. A graph theorist might ask if the graph was complete or a tree or had multiple components or cycles. It happens that planarity forces significant constraints on a graph (see planar graph), and by puncturing the sphere we can make the graph derived from the tetrahedron planar (see Schlegel diagram). --KSmrq^T 02:31, 27 February 2007 (UTC)

February 27

Proofs and computerized searches

The question above about Goldbach's conjecture got me wondering. It seems there are a number of famous conjectures or unproven theorems which have been subjected to brute-force searches for counterexamples. For example, the Goldbach article says it has been verified for n up to around 10¹⁷. I imagine Fermat's Last Theorem received similar attention. A layman like me might be tempted to think that a search to high numbers like that makes it a pretty good bet that any further searching is likely to be unsuccessful. My question is, have any of these large-scale searches simply not gone high enough, and a solution was later found by other means? Are there any interesting mathematical relationships that only kick in for enormous values? --TotoBaggins 00:30, 27 February 2007 (UTC)

A famous historical example is Pierre de Fermat's conjecture that all numbers of the form $2^{2^{n}}+1$ are prime (see Fermat number). This is true for the first several examples, but later Euler showed that $2^{2^{5}}+1=4294967297$ could be factored as $641\times 6700417$, an amazing achievement in 1732. The next 16 Fermat numbers are also known to be composite. So this was a conjecture that proved false only for "enormous" values larger than 4294967296—not so enormous by today's standards, but 300 years ago it was astronomical. —Bkell (talk) 01:47, 27 February 2007 (UTC)

Note that most large-scale computerized searches are not run indiscriminately. Usually, "regions of the search space" (whatever that means in context) that are likely to be "interesting" are targetted. The fact that some conjecture holds for large cases that were considered in some sense likely to be problematic is often more satisfactory evidence than just the number of cases where it holds. Phils 02:31, 27 February 2007 (UTC)

This is before computers, but π(x) − Li(x) < 0 for all non-enormous (<10³¹⁶ or so) values of x, so quite possibly people believed it to be negative for all x until Littlewood proved otherwise (see Skewes' number) Algebraist 15:34, 27 February 2007 (UTC)

Good answers, thanks! --TotoBaggins 16:27, 27 February 2007 (UTC)

Also note that often one can create a computer proof by attempting to reduce the number of cases that may need to be checked to some finite number, then checking each of these cases by computer (which is fast).

It may be the case that the bounding result may be found some time later than a computer search for counterexamples, so then the bounding result essentially will establish the proof.

The next sequence

0.6180339887498949 (0.5 - 1*x + 0.5 * x^2)

0.5436890126920763 (0.5 - 1*x + 0.5 * x^3)

0.5187900636758842 (0.5 - 1*x + 0.5 * x^4)

0.5086603916420042

0.5041382583616553

0.5020170551781655

0.5009941779228898

Is there a way of getting the next sequence with the tedious effort of solving the polynomial? 202.168.50.40 02:07, 27 February 2007 (UTC)

If the polynomials are supposed to have the values to the left of them as zeroes, then the higher exponents are one off. The n-th number above is a root of the equation x + ... + xⁿ⁺¹ = 1, which is equivalent to 1 – 2x + xⁿ⁺² = 0.

For larger values of n, a way to compute these values is to start with x = 0.5, and replace x iteratively by (1 + xⁿ⁺²) / 2 until convergence. While less efficient than, for example, Newton-Raphson, the formula makes up for that in simplicity.

Another way to reduce the tedium is to delegate the computations to an automatic computer.  --Lambiam^Talk 08:24, 27 February 2007 (UTC)

And, for a rough approximation, we can notice that each term is about half as far from 0.5 as the previous term, with this relationship becoming closer with each term. Thus, we can guesstimate 0.5005 for the next term. If anyone actually solves this, please let me know how close my guesstimate came. StuRat 15:36, 27 February 2007 (UTC)

To a first approximation the n-th term equals 0.5 + 0.5ⁿ⁺³, so indeed the distance to 0.5 is roughly halved each step. The next approximation is 0.5 + 0.5ⁿ⁺³ + (n+2)×0.5²ⁿ⁺⁵, which will give you already 10 accurate decimals for n = 12. If you want to play with this, here are a few more terms to a higher precision:

 1 0.61803 39887 49894 84820 45868 34365 63811 77203 09179 80576

 2 0.54368 90126 92076 36157 08559 71801 74798 65252 03297 65098

 3 0.51879 00636 75884 22190 74538 94435 27999 88621 27809 04685

 4 0.50866 03916 42004 13646 38429 65898 41399 53244 06435 90103

 5 0.50413 82583 61655 36083 05214 03975 64928 33803 90951 87446

 6 0.50201 70551 78165 51178 05402 77860 49131 76280 09418 58461

 7 0.50099 41779 22889 83685 40843 19392 13416 42444 63056 32758

 8 0.50049 31182 86552 25605 92684 59994 20216 15720 28613 43888

 9 0.50024 54622 66794 48360 09641 13516 38045 59668 32970 76331

10 0.50012 24294 76043 26230 25101 69147 80621 05651 44971 34601

11 0.50006 11322 39058 19052 95721 38910 20517 82711 64584 78650

12 0.50003 05436 87833 37599 15252 35986 50121 37831 40657 98461

13 0.50001 52657 78675 10503 21704 66776 06039 28968 51323 26731

14 0.50000 76312 57844 59185 97067 96078 38295 32638 05704 89164

15 0.50000 38151 92125 13295 17530 07820 32593 67421 30833 67851

16 0.50000 19074 79613 29054 37344 11929 52414 93738 71876 24039

17 0.50000 09537 08879 05077 94593 43474 95695 11574 63285 05994

18 0.50000 04768 46253 40602 29374 43191 17511 31062 13753 10999

19 0.50000 02384 20966 56044 60504 21229 97561 26017 52605 83145

20 0.50000 01192 09914 83323 37916 68920 35046 84045 98586 79104

 --Lambiam^Talk 17:11, 27 February 2007 (UTC)

This would drive you nuts

Jane Smith picks up her husband everyday at exactly 6pm at the local train station. One day her husband John Smith arrived back early at 5pm. Rather than waiting an hour for his wife, he started walking back at a constant speed of 4 km/hour. Jane Smith, departed from her house at the usual time and met her husband on the road. He got into the car and she drove home and arrives back at the house 20 minutes earlier than normal.

What is Jane Smith driving speed? For this problem you must assume that Jane Smith drives at a constant speed all the time, there is no traffic on the road and there is no traffic light on the road.

Good Luck! 202.168.50.40 02:17, 27 February 2007 (UTC)

Thanks, but this is not a forum for sharing puzzles. Look elsewhere on the net if you wish to find a more appropriate place to post. --KSmrq^T 02:39, 27 February 2007 (UTC)

I probably messed up, but doing it in my head, she was driving 80 km/hr. Nevermind, that can't be right, cause I didn't use calculus, which is more or less needed in this problem. --Wirbelwindヴィルヴェルヴィント (talk) 03:07, 27 February 2007 (UTC)

This is classic puzzle. Here's how to approach it. Since Jane arrives home 20 minutes earlier than usual, she must have been 10 minutes drive time from the station when she met John. So what time did she meet John ? So how long has John been walking ? Now suppose John has walked a distance X from the station in this time. Jane can drive this same distance X in 10 minutes. So how many times faster is Jane's driving speed than John's walking speed ? And you know John's walking speed, so Jane's driving speed is ... Gandalf61 10:08, 27 February 2007 (UTC)

A measure of non-concentricity - a unit or discriptor

I'm trying to find out how to describe circular objects that are supposed to be concentric, but are not quite. How does one describe non-concentricity? What would the unit of measure be? Thanks, (4crates 05:58, 27 February 2007 (UTC))

What is appropriate as a measure depends on the purpose and context. Should the measure be scale-independent — that is, if you blow everything up by the same factor, should the outcome remain the same? Or should it follow the scale?

A simple scale-dependent measure is the distance d between the centres of the circles. A value of 0 means perfect concentricity. For physical objects, this measure has the dimension of length. One possible scale-independent measure is d / r, where r is the radius of the smaller circle. If this measure is at most 1, the centre of the larger object is inside the smaller object. Another scale-independent measure is d / (R – r), where R is the radius of the larger circle. This assumes that R > r. If this measure is at most 1, the smaller object is fully inside the larger one. These scale-independent measures are dimensionless.  --Lambiam^Talk 08:34, 27 February 2007 (UTC) (edited 17:21, 27 February 2007 (UTC))

The general area of finding appropriate descriptions for quantities like this is metrology, but Wikipedia's coverage of that area seems lacking. Lambiam's answer makes sense if the circles really are circles, but if they're lacking in concentricity they may well also be lacking in circularity. Maybe a Google search will find something more relevant? —David Eppstein 17:28, 27 February 2007 (UTC)

Squaring the circle

I understand that this is impossible with just a pair of compasses and a straight-edge, but is it possible with compasses and a ruler? How would the ruler have to be marked? Do we need the sqrt(pi) marked on the ruler, or would any two points suffice? Is it possible to produce a ruler that has a tick at exactly sqrt(pi) from the zero-tick? Vonity 23:24, 27 February 2007 (UTC)

A "straight edge" is a ruler. Or is it the other way around? Either way you cannot square a circle using only rational numbers. So what if you have the sqrt(pi) marked on your ruler? You cannot multiply two numbers on a ruler.

Pi*r*r = B*B

So B = sqrt(pi) * r

So how do you plan to "multiple" sqrt(pi) with r using your ruler? 202.168.50.40 00:52, 28 February 2007 (UTC)

A straight edge does not have any marks. The idealized straight edge in the straight-edge-and-compass construction problems allows you to construct the line between two given (i.e.: previously constructed) points, not measure distances or construct segments of a certain distance. Phils 01:15, 28 February 2007 (UTC)

Unlike trisecting the angle and doubling the cube, squaring the circle can't be done with compass and marked straightedge. I'm fairly sure that since the operations are non-linear, a pre-marked ruler won't work either: you'd need a different ruler for each circle you're working on. --Carnildo 01:51, 28 February 2007 (UTC)

A better question seems to be what exactly are the set of algebraic operations that can be constructed with a straight edge and a compass, and then in addition we had a ruler that had 2 markings: 1 marking for 1 sample unit and then 1 marking for 1 pi unit ($\pi$ as a ratio) For that matter the marking could be $e^{3\pi }+sin(4)/\pi$ units for that matter. I quote the current Compass and straightedge constructions

"The only way to construct points is as the intersection of two lines, of a line and a circle, or of two circles. Using the equations for lines and circles, one can show that the points at which they intersect lie in a quadratic extension of the smallest field F containing two points on the line, the center of the circle, and the radius of the circle. That is, they are of the form x + y√k, where x, y, and k are in F."

By adding the ratio $\pi :1$, we've now added $\pi$ to the field F, and thus ${\sqrt {(}}\pi )$ should now be constructible, unless I'm mistaken. Take this with a grain of salt. Root⁴(one) 07:43, 28 February 2007 (UTC)

Off the top of my head here. Ok, consider this argument. Using chords, you can construct the square root, right? if two chords intersect, then the product of the distances from the intersection point to the edge of the circle on one chord equals the product of the distances from the intersection point to the edge of the circle on the other chord.

Let line segment CB have 1+ pi unit length, and mark where the unit distance 1 occurs at point P. (We can do that by our assumptions here about having a marked ruler). We use the compass to bisect the 1+pi unit length and draw a circle O around it, leaving us with a 1+pi diameter circle. Now, using a compass and straight edge only we can construct a line perpendicular to line segment CB at the point P and see where it passes through O, and call those points A and A'. By some laws of Euclidean Geometry I forget, we can prove AP = A'P. By the law of chords I mentioned len(AP)*len(A'P) = 1*pi. Therefore len(AP) = ${\sqrt {(}}\pi )$

Am I off my rocker or is this correct? looks like the Power of a point article agrees with me as to what the chord law states, so maybe I'm not. Root⁴(one) 08:10, 28 February 2007 (UTC)

Finding first digit

Say, I'm doing a subtraction problem I really don't want to spend time on. Like 1235232 and 499876. Or if I was multiplying, adding, or dividing them, how could I find the digit to the far left? Thanks! [Mαc Δαvιs] X (How's my driving?) ❖ 02:14, 28 February 2007 (UTC)

What was your question again? Is it, how do I find the left most digit if I subtract 499876 from 1235232? 202.168.50.40 03:33, 28 February 2007 (UTC)

This is subtracting from left to right instead of right to left. For instance lets subtract 103 from 892. 8 - 1 = 7, but we need to check to see if we need to "borrow" from that 7. ok, 9-0, so we don't have to go any further to find the left most digit.

Generally, I think the algorithm is something as follows. If you want to do quick arithmetic, make these instructions make sense to you, don't just try to memorize them. I'm no expert at quick math, but I've come across this a time or to and occasionally try to use something like this when I'm figuring subtraction. BTW, these instructions are off the top of my head and may possibly be incorrect in some sense... (I already see one problem; this algorithm repeats indefinitely for subtraction of 100 minus 1 ... it continues to try to solve 100 minus 001 indefinitely. So it needs a bit of rehashing.)

Algorithm try 1.

1. Write each number with zeros extended to the left until both of their decimal notations are equal in length I.E. if 9359 is to be subtracted from 10072, write 9359 as 09359.
2. subtract the first two digits and make note of the subtraction call this R (this example, R = 1-0 = 1).
3. subtract the next two digits and call this S (if they exist, otherwise you're done) if S is negative, you will have to set R <- R - 1 and S<- 10 + S (Eg. if S = -4 then S becomes 10-4 = 6) Now If S is 0 or R is 0, you will have to solve the problem RSxx - yy where xx and yy are the remaining digits to be subtracted, truncating unecessary zeros as needed. Here I'm using the notation RSxx means $R*10^{(}d+1)+S*10^{d}+xx$ where d is the respective power d for digit S (sorry there's a better word for it). Otherwise, you've already found the left most digit R.

With 10072 minus 9359 we have

1. step 1 compute as above 10072 "minus" 09359
2. step 2 R <- (1 - 0) = 1
3. step 3 S <- (0 - 9) = -9, then R<-0 and S<- 1
4. solve instead 1072 - 359
5. step 1 compute as above 1072 "minus" 0359
6. step 2 R <- (1 - 0) = 1
7. step 3 S <- (0 - 3) = -3, then R<- 0 and S = 7
8. solve instead 772 - 59
9. step 1 compute as above 772 "minus" 059
10. step 2 R <- (7 -0) = 7
11. step 3 S <- 7-5 = 2
12. since S and R <> 0 left most digit is 7 and its in the hundreds place.

As you can see, the algorithm terminates before you've subtracted the whole number. But it depends upon the numbers being subtracted. Also, anytime R>1 you know that the left most digit is going to be either R or R-1 and it will have the same decimal place. Work on that a bit, maybe you can fix this algorithm's logical bug. You can probably fix it by introducing a R_old which holds the digits that may accumulate and also noting somewhere through each iteration at which decimal place you are in the subtraction.

I'm not going to work on it any further, as although I know I could fix it to where a computer could make it work easily enough, to make it human workable would take more work than I can afford at the moment.

I'm not sure if there's a good left digit finder for multiplication (division as it is taught in schools is left to right anyway, right? its been so long since I've actually done long division though.). You can always start from the left to right by moving in a criss cross pattern and stop once you can be sure you're within a certain percentage of the most accurate, correct answer, as in

345 * 678 = 3*6*10000 + (3*7 + 4*6)*1000 + (3*8 +4*7+ 5*6)*100 + (4*8 + 5*7)*10 + 5*8(1)

and evaluate that from left to right, stopping when you feel satisfied.

Root⁴(one) 07:08, 28 February 2007 (UTC)