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April 3

Polar singularity

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

Given the Euler angles α, β and γ, the quaternion representation should be (cos γ + k sin γ)(cos β + i sin β)(cos α + k sin α). I think. Anyway, the basic idea is just to take the quaternions for the fundamental rotations around the axes by the given angles and multiply those together in the proper order. —Ilmari Karonen (talk) 19:01, 3 April 2006 (UTC)

Usually a polar singularity is the kind of thing that happens with latitude and longitude at the poles, where a latitude of 90° causes changes in longitude to have no effect. This is analogous to gimbal lock, where middle axis rotations of either 0° or 90° (depending on the axis conventions) cause the other two rotations to no longer have independent effects. Code to change any system of Euler angles to quaternions can be found at the Graphics Gems repository. However, the differential equations must also be adapted. The equation with angular velocity is

2 dq = ω q dt,    ω is the angular velocity vector

Perhaps that will suffice. --KSmrq^T 20:15, 3 April 2006 (UTC)

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

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

Boundary conditions:

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

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

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

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

Deprit’s variables

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

I think this (click on either of the Full Refereed Article links depending on whether you want PS/PDF or scanned GIFs) might be the original paper in question. Way over my head, though. --Bth 09:11, 3 April 2006 (UTC)

Or maybe this series of papers (with a mysteriously missing number III)? --Bth 09:21, 3 April 2006 (UTC)

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

Formula for compound mitres

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

Can you help?

---70.50.197.126

I'm not sure I understood your description. Can you clarify a bit? -- Meni Rosenfeld (talk) 15:59, 3 April 2006 (UTC)

The questioner needs to know the angle between faces of a pyramid as specified (and then, presumably, divide by two for the mitring). It's a bit unclear how regular the pyramid is. Are the three triangular boards isosceles and the same shape? If so, we should be able to give a simple enough formula. —Blotwell 01:47, 4 April 2006 (UTC)

If your interpretation is right (and I suspect it is), I'm confused by the "based on number of sides" bit in the question. Maybe he wants to generalise after making his three-sided one? --Bth 12:38, 4 April 2006 (UTC)

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

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

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

(I am fully aware the following formula could be simplified. I've just lost any will to do it. It is offered with no warranties, etc., though it does reproduce StuRat's results for equilateral triangle side pieces and $n$ of 3 and 4. [I also lost the will to continue testing.])

Here's what I've assumed: a regular polygon of $n$ sides of length $a$ for the base, isosceles triangles for the sides of base $a$ and whatever side length is necessary to get the required height of the pyramid, $h$.

Make a definition of $\theta ={180(n-2) \over n}$, which is simply the internal angle of the polygon in question (we could write this out in full in the equation, but it would drive us mad). This is for degrees, of course; for radians, we'd replace the 180 with a $\pi$ -- NB that Excel works natively in radians.

The mitre angle, which I'm going to call $\psi$ is then (assuming I haven't made any hideous mistakes, which I probably have):

$\psi ={1 \over 2}\left(180-\arccos \left({a^{2}h^{2}\cos \theta -{a^{4} \over 4}\tan {\theta \over 2}(\sin \theta -\cos \theta \tan {\theta \over 2}) \over {\sqrt {(a^{2}h^{2}+{a^{4} \over 4}\tan ^{2}{\theta \over 2})(a^{2}h^{2}+({a^{2} \over 2}\sin \theta -{a^{2} \over 2}\cos \theta \tan {\theta \over 2})^{2})}}}\right)\right)$

Some other things which might be useful: the formula for the side length of the triangles is ${\sqrt {h^{2}+{a^{2} \over 2}\left(1+\tan ^{2}{\theta \over 2}\right)}}$ and the mitring angle for the join between the faces and the base if you had a base piece as well would be ${1 \over 2}\arctan \left({2h \over a\tan {\theta \over 2}}\right)$ (if it's free standing, you need double this, obviously).

There is almost certainly at least one mistake in the foregoing, but I hope it helps. --Bth 11:49, 8 April 2006 (UTC)

Thanks, it helps alot. ---70.50.197.146

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

Approximating a parallel bezier

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

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

April 4

product a functor?

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

Well, at least you can say this: if a category has binary products, then it has a product functor: a real, honest functor. We can't define it uniquely, but such functors exist... and they're all adjoint to the diagonal functor on whichever side.

I'm sure this isn't news to you, but I think that's what the sources mean. Melchoir 06:06, 4 April 2006 (UTC)

Read the fine print at categorical product#Discussion; the fact that a product object is unique only up to isomorphism is not an obstacle to defining a functor. --KSmrq^T 06:15, 4 April 2006 (UTC)

So I assume you're referring to the sentence "it is possible to choose the products in a compatible fashion so that the product turns into a functor CI → C". I guess whoever wrote that sentence had in mind the issue that's bothering me, and knows the solution to it. Somehow it's possible to choose particular representatives of the isomorphism classes of product objects so that the pseudofunctorial natural isomorphisms all get straightened out into honest equations. Well, I'm pleased that someone knows my woes and has an answer, but I'd be even happier if someone told me what it is. For example, do we require the axiom of choice to hold in the index category in order to be able to make this choice? -lethe ^{talk +} 13:41, 4 April 2006 (UTC)

Oh, but whatever this construction is, even if it does require the axiom of choice for general limits, I guess we wouldn't need the axiom of choice for binary products, since the index category here is finite. Right? -lethe ^{talk +} 13:53, 4 April 2006 (UTC)

I found a paper online which defines functors whose values are only defined up to isomorphism to be anafunctors, not pseudofunctors. I'm not positive that the definition is equivalent to the one I know for pseudofunctors, so now I have to think about that some more. The paper also mentions that the existence of honest functors adjoint to the diagonal functor relies on the Freyd Adjoint Functor Theorem, which does rely on the axiom of choice. I might be satisfied with that. -lethe ^{talk +} 14:30, 4 April 2006 (UTC)

That paper looks like it was written with MS Word. Hard to take such a paper seriously. -lethe ^{talk +} 17:41, 4 April 2006 (UTC)

Pardon the ignorance, but now that you mention it, which software should be used for serious papers? And for that matter, what's wrong with Word? -- Meni Rosenfeld (talk) 05:58, 5 April 2006 (UTC)

LaTeX should be used for math papers. The arXiv and many (most?) reputable journals do not accept word documents. Many journals use house formatting styles which are latex macros. In addition to not looking very nice, Word doesn't support such formatting macros. -lethe ^{talk +} 06:19, 5 April 2006 (UTC)

palindrome

4 digit palindrome number two more than a perfect square

7227 = 85^2+2

(by manual brute force search with Windows calculator). I hope that wasn't your homework, or a particularly easy Enigma or something. --Bth 09:24, 4 April 2006 (UTC)

Or, you could have taken a calculator like casio fx-82TL or similar, start with 1026 and repeatedly calculate

$({\sqrt {Ans-2}}+1)^{2}+2$

That is, of course, unless you have any programming language handy. -- Meni Rosenfeld (talk) 09:51, 5 April 2006 (UTC)

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

Well, you could narrow down the search with this table:

x	(1st digit = 4th digit) in x^2+2	4th digit in x^2	2nd digit in x	candidates
32-44	1	9	3,7	33,37,43
45-54	2	0	0	50
55-63	3	1	1,9	59,61
64-70	4	2	\	\
71-77	5	3	\	\
78-83	6	4	2,8	78,82
84-89	7	5	5	85
90-94	8	6	4,6	94
95-99	9	7	\	\

Checking the candidates will reveal that only 85 produces a palindrome. -- Meni Rosenfeld (talk) 16:02, 5 April 2006 (UTC)

what's up

what's up

A direction. <rimshot /> —Keenan Pepper 13:06, 4 April 2006 (UTC)

There is a complexity class UP, which I only mention in a vague attempt to get this section on-topic for the Maths RD. -- AJR | Talk 17:00, 4 April 2006 (UTC)

Nice. So there's no proof that UP != P or that UP != NP? —Keenan Pepper 23:58, 4 April 2006 (UTC)

The sky. —OneofThem^(talk)_(contribs) 17:34, 4 April 2006 (UTC)

There's UPS, trousers, scissors, but there's no single up, trouser, or scissor. How strange! – b_jonas 12:15, 5 April 2006 (UTC)

I wonder if any of the people in Seven Up ever drank 7-Up. :--) JackofOz

too late to see answer

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

Check the Archive. Isopropyl 23:59, 4 April 2006 (UTC)

Note that questions and answers are normally archived after a week. However, you really should check back each day to see if there are any clarifications needed. StuRat 02:29, 5 April 2006 (UTC)

Also, any answers will be in the page history. The same IP address that asked this question also created a section headed "Subject: naming points on graphs (house numbers on twisty roads)", this is the last version before it was archived. -- AJR | Talk 19:23, 5 April 2006 (UTC)

WHAT IS 123098+1322

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

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

I suspect there's more to the question than meets the I. It is the 5th of April, after all. JackofOz 08:58, 5 April 2006 (UTC)

Go see what google says. – b_jonas 12:12, 5 April 2006 (UTC)

The Golden Ratio

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

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

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

Explain to me why you would waste time writing that here instead of just looking at the article. —Keenan Pepper 17:23, 5 April 2006 (UTC)

Some research has suggested that it may not really be the most aesthetic, and we may just be obsessed with the ancient Greeks. Superm401 - Talk 20:27, 5 April 2006 (UTC)

bound on number points symmetric block design

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

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

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

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

Any suggestions in avoiding this problem would be greatly appreciated.

Sum of individual numbers

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

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

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

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

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

April 6

Matrices

What is expansion by minors?

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

Thank you very much for the speedy response! :)

STATISTICS

WHAT IS MEASURE OF DISPERSION?

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

Melting snowball

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

Fast enough that it'll be gone before you get your homework done. --Trovatore 19:43, 6 April 2006 (UTC)

So is this the way snowballs actually melt? Would not the surface area play a role, if so I would expect a quadratic term rather than just a cubic term in the homework. As to solving the homework you need to the volume of a Sphere, first, then apply some elementary algebra to find the solution. --Salix alba (talk) 20:39, 6 April 2006 (UTC)

If you start with the volume formula you're going to need calculus. Why not just evaluate the surface area of a sphere of diameter 10cm and divide? —Blotwell 03:42, 7 April 2006 (UTC)

This idealized situation is a related rates-type equation. You should relate the equations for the rates of change for volume and diameter with respect to time. As for real life, there's a billion other factors, like the ambient temperature, heat capacity of the surrounding air and the surface it's in contact with, and so forth. Isopropyl 20:51, 6 April 2006 (UTC)

April 7

Zero Radicals and the Transitive Property

To solve the equation of

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

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

$\ x=y^{0}$

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

So,

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

where y is any number.

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

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

I'm not sure where transitivity comes into it. If you're suggesting that we make exceptions to transitivity of = in order to avoid paradox here, that's not how we do it (I think transitivity of = is too important to give up, don't you?). As with division by zero, we simply don't allow taking 0th roots. (Evaluating ${\sqrt[{0}]{x}}$ is a lot like evaluating $x^{\infty }$, so you could also look at indeterminate form.) —Blotwell 03:37, 7 April 2006 (UTC)

To emphasize, the problem is not with transitivity of = (which always holds), but with working with operations which aren't defined. In math, you must only deal with things which are defined and follow the definition. You can in some cases use your intuition to find a plausible extension for a definition (see for example Real projective line, a structure where 1/0 is defined), but you still have to follow the definition rigorously. So your problem is dealing with $1^{\infty }$ which isn't defined (and can't be plausibly defined, for that matter). If you'll show us the exact square root argument you have in mind, we can show you what's wrong with it too. -- Meni Rosenfeld (talk) 09:42, 7 April 2006 (UTC)

sqrt(1)=1, sqrt(1)=1-, but we obviously can't say that since 1 and -1 both solve the equation sqrt(1)=x, that they're equal.

65.31.80.100 12:01, 7 April 2006 (UTC)

Depends how you define sqrt(). Square root is usually defined to return the positive square root. See Root of unity for techniques to handle multiple solutions. --Salix alba (talk) 12:28, 7 April 2006 (UTC)

To put it differently: ${\sqrt {1}}=+1$, period. This is what follows from the definition. It's true that also $(-1)^{2}=1$ but that doesn't mean that ${\sqrt {1}}=-1$. Both +1 and -1 solve the equation $x^{2}=1$ but not the equation $x={\sqrt {1}}$. -- Meni Rosenfeld (talk) 12:40, 7 April 2006 (UTC)

Yes, taking a zeroeth root is like exponentiating by the reciprocal of zero, can't be done. I don't see where transitivity comes in though. Ozone 19:09, 7 April 2006 (UTC)

What the questioner meant was: Assuming that, as argued, :${\sqrt[{0}]{1}}=y$ for any y, then for example ${\sqrt[{0}]{1}}=1$ and ${\sqrt[{0}]{1}}=2$. By transitivity we get 1=2, a contradiction. The suggestion was that perhaps transitivity doesn't hold in such cases. What we have shown is that it is not transitivity that fails, but the equalities themselves. -- Meni Rosenfeld (talk) 17:44, 8 April 2006 (UTC)

Concatenating two space curves

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

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

Thanks, that's what I was doing, but didn't get the right curve. Maybe I'm making a mistake elsewhere. deeptrivia (talk) 04:35, 8 April 2006 (UTC)

Minuscule orthodromic equals loxodromic (expressed as limit?)

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

$\lim _{H\to 0}{\overline {Az}}=\lim _{\Delta {\hat {O}}\to 0}{\widehat {Az}}\,\!$ ?  ~Kaimbridge~16:39, 7 April 2006 (UTC)

I'm confused as to why the symbol for the orthodrome has a delta and the symbol for the loxodrome doesn't. —Keenan Pepper 17:04, 7 April 2006 (UTC)

Because "ΔÔ" is the orthodromic arc segment ("angular distance") endpoint difference (ΔÔ = Ô₂ - Ô₁), whereas H is the loxodromic hypotenuse (here, an approximation, as cos(Lat_m) is the approximation of the reciprocal of the inverse Gudermannian function's divided difference): $H={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!$ AFAIK, there are no explicitly denotable endpoints to the the loxodromic arc segment ("hypotenuse")——it is a "side".  ~Kaimbridge~18:25, 7 April 2006 (UTC)

Now I'm even more confused. I thought hypotenuse meant a straight line segment, the longest side of a right triangle. A loxodrome is a curve on the surface of a sphere. Even in spherical geometry, great circles (a.k.a. orthodromes) play the role of straight lines, and loxodromes aren't straight. I have no idea what you mean by "there are no explicitly denotable endpoints to the loxodromic arc segment".

Right, the formal expression for the loxodrome is $\Delta Lat\sec({\overline {Az}})\,\!$, and since $\tan({\overline {Az}})={\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\,\!$ and sec = [1 + tan²]^.5, then $\Delta Lat\sec({\overline {Az}})=\Delta Lat{\sqrt {1+\left({\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\right)^{2}}}={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!$ When you look at a Mercator map, you are drawing a right triangle (The FCC expresses it this way, too: [1]).

Why can't you just say $\lim _{d\to 0}$ where d is the distance between the two points, measured any way you want? —Keenan Pepper 21:39, 7 April 2006 (UTC)

Because, since the orthodromic distance is different than the loxodromic distance (except along meridians and the equator), shouldn't the limits be different, too (perhaps $\lim _{Dist_{lox}\to 0}{\overline {Az}}=\lim _{Dist_{orth}\to 0}{\widehat {Az}}\,\!$)?  ~Kaimbridge~14:53, 8 April 2006 (UTC)

Fourier transform of a Hilbert space

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

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

Evaluating e^2πi

When you evaluate e^2πi, you get 1. Therefore

ln e^2πi = ln 1

2πi = 0

2 = 0/πi = 0

π = 0/2i = 0

i = 0/2π = 0

How can this be true? M@$+@ Ju ~ ♠ 20:59, 7 April 2006 (UTC)

Wrong. ln 1 = 0 + 2πi*k, where k is integer. ln e^2πi=2πi + 2πi*k, where k is integer. The resulting sets are the same.  Grue  21:08, 7 April 2006 (UTC)

Unfortunately, you've made a mistake in going from the first line to the second. In general, when you have an equation of the form $f(x)=f(y),$ you cannot deduce that $x=y$. For example, if you take the equation $x^{2}=y^{2}$, it does not follow that $x=y$, for example choosing $x=3$ and $y=-3$ solves $x^{2}=y^{2}.$ The problem here is that the exponential function is not one to one, in fact it is periodic in the imaginary direction. --Deville (Talk) 21:23, 7 April 2006 (UTC)

Thanks for the explanation guys. It's still funny to confuse my teachers! M@$+@ Ju ~ ♠ 22:46, 7 April 2006 (UTC)

Yeah, you'll be laughing all the way til detention...:-) --Deville (Talk) 04:28, 8 April 2006 (UTC)

Your teachers got confused on this! What's happening to our world's academic standards!! deeptrivia (talk) 04:34, 8 April 2006 (UTC)

Today's teachers were yesterday's students. If you were never taught it, you can't teach it. Sad but true. JackofOz 04:04, 9 April 2006 (UTC)

long division

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

1) 3265324 / 4124124124

2) 2312 / 312545

3) 757 / 14244

4) 141565 / 234

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

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

The answer to

3) 757 / 14244

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

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

infinitely many solutions

Hi,

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

Thanks in advance, lynton

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

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

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

Sorry, but this is not a mathematically meaningful question. As already pointed out, we have available myriad different kinds of "number", so a choice must be made. Typically cryptography would use natural numbers or positive integers, in which case the number of solutions is finite. Sadly, I must agree with your self-assessment of not being a competent mathematician, and serious cryptography research requires such compentence. As an amateur you can still have fun, but it is unrealistic to expect your work to be of much interest to professionals. --KSmrq^T 02:46, 8 April 2006 (UTC)

As my examples show, there can be no proof of infinitely many solutions in a general additive number system, because it's not true in a general additive number system; there are number systems for which the equation has a finite number of solutions. Generally things that are not true do not have proofs. -lethe ^{talk +} 03:14, 8 April 2006 (UTC)

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

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

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

I merely agreed (in part) with your own description of yourself. You, however, have violated Wikipedia's core rule of no personal attacks in your remarks towards me. Don't do it again; you may be blocked for such behavior.

Since the reals include the natural numbers, a proof by induction would trivally show a solution for every natural number a. Namely, let b equal c and let a equal 0. Now increment a and decrement b. Or do you also want a proof that the natural numbers have infinite cardinality?

Ask smart questions if you expect good answers. --KSmrq^T 12:20, 8 April 2006 (UTC)

Wait, what do you mean, "failed to produce even a rudimentary attempt"? I gave you a proof that there were infinitely many solutions over the integers, and showed by counterexample that there is no proof in general? What more do you want? -lethe ^{talk +} 04:47, 8 April 2006 (UTC)

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

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

Thanks a lot, greatly appreciated.

April 8

improvement

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

Practice, practice, and more practice.

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

Form an equation

3 and -3

Thankyou

Um... what equation? What do 3 and -3 have to do with anything? —Keenan Pepper 17:52, 8 April 2006 (UTC)

I guess the request was to find an equation the solutions of which are 3 and -3. And in that case, the equation

$x^{10}-45x^{8}+810x^{6}-7290x^{4}+32805x^{2}-59049=0$

Will satisfy the homework assignment's requirements. -- Meni Rosenfeld (talk) 17:58, 8 April 2006 (UTC)

So will

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

w00t--Deville (Talk) 19:18, 8 April 2006 (UTC)

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

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

How about the lowly quadratic (x-3)(x+3)=0? Am I missing something here?

Mathematics/measurement

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

5.391 × 10−44 seconds (see Planck time for more information). —Ruud 20:16, 8 April 2006 (UTC)

Not that this has anything to do with mathematics, but whether or not Planck time is the smallest time you can measure, or what it even means to measure time, is a rather subtle, and perhaps model-depentent, question. To measure something smaller than the Planck time would require a clock with higher frequency, thus greater than Planck mass. String theory probably predicts such particles (it predicts an infinite tower of particles starting at nearly Planck mass), but there are probably a lot of subtleties to address before you could use such a particle as a clock. Anyway, most "time" measurements of nuclear decays are actually measurements of the dispersion of energy. I guess with modern clocks, you can measure things as small as one electron decay of Cesium, since that's how we build our atomic clocks. -lethe ^{talk +} 20:39, 8 April 2006 (UTC)

What? Planck mass is a quite large amount. From our article on planck mass: "Unlike most of the other Planck units, the Planck mass is on a scale more or less conceivable to humans, as it is roughly the mass of some fleas." ☢ Ҡi∊ff⌇↯ 08:53, 9 April 2006 (UTC)

symmetric power and alternating power commute?

If

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

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

$\bigwedge ^{q}V$

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

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

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

No, if my arithmetic is right. I have

$dim\bigwedge ^{2}\sum ^{2}F^{2}=3,\quad dim\sum ^{2}\bigwedge ^{2}F^{2}=1.$

Melchoir 22:56, 8 April 2006 (UTC)

Whoa, well I guess that puts a nail in that coffin. Thank you. -lethe ^{talk +} 17:57, 9 April 2006 (UTC)

April 9

Existence of characteristic functions for decision problems

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

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

Do I understand the proof correctly?