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June 4

Minor confusion on "if G has faithful complex irreducible representation then Z(G) is cyclic"

Hello all,

I am confronted with the following proof:

Claim: If G has faithful complex irreducible representation then the centre Z(G) is cyclic.

Proof: Let $\rho :G\to GL(V)$ be a faithful irreducible complex representation, and $z\in Z(G)$: so $zg=gz\,\forall \,v\in V$. This is a G-endomorphism on V and so is multiplication by some scalar $\tau _{z}$ say, by Schur's lemma. Then the map $Z(G)\to \mathbb {C} ^{\times },\,z\to \tau _{z}$ is a representation of Z and is faithful as $\rho$ is faithful. Thus, Z(G) is isomorphic to a finite subgroup of $\mathbb {C} ^{\times }$ and so is cyclic.

Now my question is this: where do we actually use the fact that we're working with the centre? I'm sure it's in there somewhere, but I can't spot why this only works with elements commuting with everything in G, and why we can't just apply the argument to the whole of G. Could anyone explain? Thanks very much. Mathmos6 (talk) 12:31, 4 June 2011 (UTC)

Schur's lemma only applies for z in the center. More specifically, you're using the fact that the image of z is in the center of End(V).--RDBury (talk) 21:30, 4 June 2011 (UTC)

Rotation of conic sections in polar form

Hi Reference Desk,

I have been given a conic section in polar form and I was asked to find the angle of rotation and then sketch the graph.

r = 1 / (1 - 2 cos(t) + 2 sin(t))

I rewrote r into

r = 1 / (1 - e cos(t + t'))

where e is the eccentricity

I solved for e and t' and this gives me

r = 1 / (1 - 2 cos(t + pi/4))

Does this mean the conic section is rotated anticlockwise by pi/4 or clockwise by pi/4? Does the sign in front of the cos or sin matter?

Are there any general rules for rotation of conic sections in the polar form? I have read a lot articles on rotation of conic sections but they were all written in terms of the standard form and not the polar form.

Thanks a lot! — Preceding unsigned comment added by 169.232.101.13 (talk) 22:39, 4 June 2011 (UTC)

You probably meant r = 1 / (1 - 2 √2 cos(t + pi/4)) so the eccentricity is 2 √2. In general, replacing t by t+α in a polar equation rotates the graph clockwise by α, just as replacing x by x+a in a Cartesian equation shifts left by a.--RDBury (talk) 05:49, 5 June 2011 (UTC)

Problem with Lagrange multipliers

Question:

Find the maximum and minimum values of f(x,y) = x^2 + y^2 subject to the constraint g(x,y) = x^6 + y^6 = 1

This is how I went about solving it:

grad f = t grad g (where t is a constant) 2x = 6t x^5 2y = 6t y^5

1 = 3 t x^4 --> (1/3t)^(1/4)=x 1 = 3 t y^4 --> (1/3t)^(1/4)=y

I substituted x and y into g(x,y)to determine t

(1/3t)^(3/2) + (1/3t)^(3/2) = 1 (1/3t)^(3/2) = 1/2 1/3t = (1/4)^(1/3)

Substituting 1/(3t) back into x and y

x=(1/4)^(1/12) y=(1/4)^(1/12)

To determine whether this is a maximum or a minimum, I need to use the second derivative test

fx=2x fxx=2 fxy=0

fy=2y fyy=2 fyx=0

The Hessian matrix D = 4 > 0, fxx = 2 > 0. This usually means that the point is a minimum, but the problem is that we did not even substitute x = (1/4)^(1/12) into the determinant of the Hessian matrix.

How should this problem be done? — Preceding unsigned comment added by 169.232.101.13 (talk) 22:52, 4 June 2011 (UTC)

First, you're missing a few solutions. What Lagrange really says is that the vectors (2x,2y) and (6x^5,6y^5) are linearly dependent. So solve by setting the determinant (2x)(6y^5)-(2y)(6x^5) equal to 0. This gives x=0, y=0 and x=±y, and putting these back into the original equation gives 8 solutions (0,±1), (±1,0), (±2^−1/6,±2^−1/6). Second, while there probably is a second derivative test for Lagrange multipliers somewhere, I can't find it the books I have. The Hessian only applies for unconstrained problems. In any case, the second derivative test only tells you if a point is a local maximum or minimum and the problem is asking for global. So just evaluate the function at the critical points; the ones where the values is highest are the maxima and the ones where the values is lowest are the maxima. The constraint curve is bounded so you don't have to worry about what happens at infinity.

Another way to solve this is by setting x=cos^1/3t, y=sin^1/3t. This reduces the problem is one dimension and you can use simpler methods on it. (The algebra might get more complicated though.)--RDBury (talk) 06:25, 5 June 2011 (UTC)

Question about 'a convenient way to type symbols?'

I searched on google but can't find good result, so I came here for help. The only way I type symbols like 'sets relation symbol' 'predicate logic symbol' 'very strange operators' is opening the unicode list then copy it to my sheet. Is there some trick to make this operation faster? (Also I found this 'uncommon symbols' on Wikipedia are showed as PNG pictures.)Nilman (talk) 23:00, 4 June 2011 (UTC)

The Edit tools, which appear below the edit window, have various sets including 'Math and logic'. Select that from the popup menu then click on a symbol to insert it as you type. If a symbol you use a lot doesn't appear there you can request that it's added.--JohnBlackburne^words_deeds 23:14, 4 June 2011 (UTC)

More generally there are two ways of getting symbols that don't appear on your keyboard: (1) Use something like <math> mode that allows you to use special descriptive names such as \alpha for the Greek letter α, or (2) find an input method that does what you need. (The Edit tools are an input method supplied by the Wikipedia editor.) Looie496 (talk) 23:47, 4 June 2011 (UTC)

Thank you.Nilman (talk) 11:35, 5 June 2011 (UTC)

You can also use a program like AutoHotkey to set up keyboard shortcuts to specific symbols. This has the advantage of working in (almost) any program. -Elmer Clark (talk) 15:35, 7 June 2011 (UTC)

June 5

a simple ODE

I have a question about solving a simple linear ODE

${\dot {Y}}=\alpha (v(Y)-s(Y))$

where $\alpha >0$ and v(Y) and s(Y) are linear functions of Y that intersect at one point.

Taking the simplest example

${\dot {Y}}=\alpha (v-s)Y$

gives a solution of $Y(t)=e^{\alpha (v-s)t}$ which will go to infinity if $v>s$ and zero (which is the point of intersection) if $v<s$.

What if zero is not the point of intersection, so that $v(Y)=v_{0}+vY$ and $s(Y)=s_{0}+sY$?

Then the point of intersection is at $Y={\frac {s_{0}-v_{0}}{v-s}}$ and the ODE is

${\dot {Y}}=\alpha (v_{0}-s_{0}+(v-s)Y)$

How do I show the solution to the ODE in this case? I want to be able to show that depending on the parameters of the linear functions, the solution will tend to a stable equilibrium at that point of intersection. — Preceding unsigned comment added by 130.102.78.164 (talk) 00:13, 5 June 2011 (UTC)

This is a case of Linear differential equation#Nonhomogeneous equation with constant coefficients but in this case you can find a particular solution by assuming Y is constant. The general solution is then found by adding solutions to the homogeneous case and you've already done that.--RDBury (talk) 06:34, 5 June 2011 (UTC)

what if any statement implied its opposite?

what if there were a proof that didn't require any axioms (was true for all systems), that from any statement, its opposite followed? --86.8.139.65 (talk) 18:56, 5 June 2011 (UTC)

That would mean that the rules of inference you are using are not a very good choice. --Tango (talk) 19:27, 5 June 2011 (UTC)

What do you mean by "opposite"? What's the "opposite" of 2 + 3 = 5? Michael Hardy (talk) 20:17, 5 June 2011 (UTC)

In this case I would say the opposite is "two plus three very much does not equal five". I don't know how to mark that up. 86.8.139.65 (talk) 20:36, 5 June 2011 (UTC)

Is this like the Liar paradox? Grandiose (me, talk, contribs) 20:40, 5 June 2011 (UTC)

Then that would also apply to the statement that says that: "from any statement, its opposite follows". Count Iblis (talk) 22:34, 5 June 2011 (UTC)

Ex falso quodlibet 83.134.160.239 (talk) 09:42, 7 June 2011 (UTC)

I think that this may be different. The OP was asking about the possibility of there existing a $\phi$ such that $\phi \to \neg \phi$. The link you mention is about something different. Ex falso quodlibet is about being able to imply anything from an assumed contradiction, e.g. $(\phi \land \neg \phi )\to \psi$. Although if there exists a φ such that $\phi \to \neg \phi$ then we would have $\phi \land \neg \phi$ which would then imply any statement you could imagine by ex falso quodlibet. — Fly by Night (talk) 15:40, 7 June 2011 (UTC)

For many statements there's no problem with them implying their opposite. If we then showed the opposite implied the first thing again you start having problems. For instance if we showed that ants are three meters tall implied ants are not three meters tall then there's no problem. Dmcq (talk) 20:50, 7 June 2011 (UTC)

Could you give some examples? If (all) ants are three meters tall then then "(all) ants are not three meters tall" is clearly false, so you have $\phi \not \to \neg \phi$. They all cause problems: "if A is true then A is false" is what the OP was asking about. — Fly by Night (talk) 21:13, 7 June 2011 (UTC)

For instance, let $\phi$ be the proposition $(\psi \land \neg \psi )$. By ex falso quodlibet, $\phi \to \neg \phi$.

Conversely, if φ were any proposition such that $\phi \to \neg \phi$ then by conditional exchange, we can conclude $\neg \phi$. Sławomir Biały (talk) 22:10, 7 June 2011 (UTC)

Well actually in that last bit you can't conclude $\neg \phi$ as ψ may be false and false implies false is true. Dmcq (talk) 22:20, 7 June 2011 (UTC)

Sorry, the psi should have been a phi. I've corrected it. Sławomir Biały (talk) 22:33, 7 June 2011 (UTC)

Niels Bohr is quoted for saying: "The negation of a truth is a falsehood, but the negation of a deep truth is another deep truth". I don't know whether this is a truth, a falsehood, or a deep truth. Bo Jacoby (talk) 09:01, 8 June 2011 (UTC).

Tango gave the correct answer as the first post. If you can prove that no consistent system exists then you should restrict the rules of inference. Taemyr (talk) 11:41, 8 June 2011 (UTC)

As I showed above, there is nothing inconsistent about the existence of a proposition $\phi$ for which $\phi \to \neg \phi$. In fact, it's easy to construct such a proposition in standard propositional logic. You might take issue with the rules of inference of propositional calculus (many logicians have), but it's not for want of consistency. Sławomir Biały (talk) 11:54, 8 June 2011 (UTC)

Indeed. We have ${\text{False}}\to P$ for any proposition P because $\neg ({\text{False}}\land \neg P)$. In particular, we therefore have ${\text{False}}\to \neg {\text{False}}$.Gandalf61 (talk) 12:55, 8 June 2011 (UTC)

That's what I meant when I asked for an example. I can't think of a statement that implies its negation. The ant example wasn't quite right. I think that a correct "if A then not A" is inconsistent. If A implies not A then both A and not A are true. That's a contradiction. If we could find an A such that A implies not A then we could imply any statement (by ex falso quodlibet). — Fly by Night (talk) 21:00, 8 June 2011 (UTC)

"If 2 is odd, then 2 is even" is a true statement, using the standard mathematical meaning of "if…then". Since the premise is false, the statement as a whole is true, regardless of whether the conclusion is true or false. See Material conditional. —Bkell (talk) 21:26, 8 June 2011 (UTC)

Heh, whaddya know, "if 2 is odd then 2 is even" is used verbatim in that article as an example. :-) —Bkell (talk)

I think we've got our wires crossed somewhere. "if 2 is odd then 2 is even" is vacuous because 2 is not odd. (Notice that it's equivalent to its own contrapositive.) The point is when you have "if A then not A" holding while at the same time A is true. That's when the problems start. Otherwise, like I said, it's vacuous. — Fly by Night (talk) 21:36, 8 June 2011 (UTC)

Certainly, a true statement of the form "if A then not A" can cause a contradiction, but only if A is true. Above you said you couldn't think of a statement that implies its negation, so I gave you an example of one that does, namely, the statement "2 is odd". You also said, "If A implies not A then both A and not A are true"—that holds only if you additionally assume that A is true. Alone, the statement "A implies not A", even if true, does not inherently cause a contradiction, because it can be the case that A is false. —Bkell (talk) 21:54, 8 June 2011 (UTC)

But that's my hang-up. The statement "2 is odd" does not imply that "2 is even". The statement "2 is odd" is false, so a statement of the form "if 2 is odd then B" does not imply anything. It tells us nothing about the validity of B. Think about it: "if 2 is odd"… well it's not… so I stop. I agree with the last point, but that just says that a vacuous statement does not imply a contradiction. So yes: there are many statements of the form "if A then not A" that do not imply a contradiction; but that's because they are either false or they do not imply anything (i.e. are vacuous). — Fly by Night (talk) 22:48, 8 June 2011 (UTC)

From a strictly logical point of view, vacuous truth is still truth, and "2 is odd" does imply "2 is even", because the statement "if 2 is odd, then 2 is even" is true (vacuously). Your difficulties seem to be based on what you think the English construction "if…then" should mean (in ordinary language, it usually indicates cause and effect) and the idea that a logical implication of the form "A implies B" should tell you something about B (it doesn't, if A is false). Perhaps you should read Material conditional#Philosophical problems with material conditional, which addresses some of these issues. —Bkell (talk) 23:31, 8 June 2011 (UTC)

If you don't like this treatment of implication and vacuous truth, there are different logical systems that may appeal better to your intuition, like relevance logic. These are not the standard logical system used in most mathematics, however. —Bkell (talk) 23:45, 8 June 2011 (UTC)

It might happen that a universally quantified statement is a true statement about the mathematical world, but that when it is actually instantiated, it produces a vacuous implication at certain values of the quantified variable. For instance,

$\forall x((x{\text{is even}})\to \neg (2x^{2}-1{\text{is even}})).$

is clearly a true sentence. But instantiating at $x=1$ gives

$(1{\text{is even}})\to \neg (1{\text{is even}}).$

-- Sławomir Biały (talk) 01:26, 9 June 2011 (UTC)

I don't have a "difficulty". Just because I don't agree with your point of view does not mean that I have a difficulty understanding. — Fly by Night (talk) 11:11, 9 June 2011 (UTC)

I'm sorry, I didn't mean to imply that you had difficulty understanding. My use of the word "difficulty" was in response to your use of the word "hang-up", which, according to wikt:hang-up, means "an emotional difficulty or a psychological inhibition". I interpreted that to mean that you were having difficulty reconciling mathematical ideas with your intuition, which is common for everyone from time to time. —Bkell (talk) 11:29, 9 June 2011 (UTC)

Ha ha ha :-) You're not far off there. I do have many emotional difficulties, and psychological inhibitions; but I try to carry on regardless. The medication helps a lot too. — Fly by Night (talk) 14:18, 9 June 2011 (UTC)

To Slavomir. OP did not ask for the existence of a statement that implies it's opposite. He asked for the implications of a proof that *any* statement implies it's opposite. So not only do you have ${\text{False}}\to \neg {\text{False}}$, which is not a problem, you also have $\neg {\text{False}}\to \neg \neg {\text{False}}$, which is a bit more of a problem.Taemyr (talk) 08:45, 10 June 2011 (UTC)

How to average ranks?

As an example, there are several listings of the best 100 (or whatever) novels of the 20th. century. A particular novel is not going to have the same rank in every list. To get the "average" rank of each novel, is there any better method than merely taking the mean average of its ranks? Thanks 92.28.240.238 (talk) 20:24, 5 June 2011 (UTC)

Average rank is not a very good way of doing it unfortunately. The main problems are that firstly it is very bad at dealing with incomplete orders and secondly in general the rank can vary more in the middle rather than the beginning and end so it is not combining things with similar ranges. I tried doing a quick search for merging rank order and combining rank order but didn't find the one I was looking for which is iterative and assigns a real number value rather than an integer order. Other ways of doing it involve setting up a matrix of wins/losses and getting the overall order from that. Dmcq (talk) 23:59, 5 June 2011 (UTC)

Instead of attempting to order them, you can group them. For example, assume you have 200 books that appear in 10 different top 100 listings. You can place that data in a matrix and use singular value decomposition on the matrix. Then (assuming the books are on the left vertical column of your matrix) the first table will have 5 columns - one per book. Replace all negative values with 0 and all positive values with 1. Each book will have an identifier like 01101. All books with the same identifier are in the same group. With little work, you can sort the groups into the most popular, somewhat popular, and less popular groups. Further, SVD allows you to fill in a rather good estimate for missing data. So, if a book doesn't appear on a particular top 100 list, you can estimate where it is on the list: 101, 110, 250... -- kainaw™ 00:06, 6 June 2011 (UTC)

Just looking again at the type of stuff you want to order I think you can probably get a quick and fairly good order by averaging the logarithm of the order including something like 1+log(100) for all the missing ones. Rather empirical but far better than just averaging the order without getting the log. Dmcq (talk) 00:27, 6 June 2011 (UTC)

If I were doing this, I would probably just use the median of the reported ranks. The main problem with using the mean is that it can be badly affected by outliers -- a single ranking that widely differs from the others can significantly change the result. The median does not suffer from that problem, and is simple to compute. Looie496 (talk) 00:42, 6 June 2011 (UTC)

Isnt there any maths/stats theory which would indicate the best method to use? 2.97.212.124 (talk) 12:17, 6 June 2011 (UTC)

Well there's lots of studies of different algorithms which are based on different assumptions! I found something mentioning what I was looking for above in http://deepblue.lib.umich.edu/bitstream/2027.42/66929/2/10.1177_001316447303300104.pdf , I'm sure there's much better now with people using Monte-Carlo methods since they've oodles of performance to waste on computers. Dmcq (talk) 13:41, 6 June 2011 (UTC)

There is a hell of a lot of information on this topic. You are dealing with three main issues: synonymy, polysemy, and sparsity. In your specific case, synonymy and polysemy are not too bad, but there is the issue of how books are listed. For example, what if one list has both the hardcover and softcover version of a book listed? What if one list places an entire series as a single book while another list breaks each independent book out? Your main issue is sparsity. Many books will not be on all lists. So, assume you use median. A book is, for some reason, on one and only one list at rank 25. It didn't even make the top 100 list for any of the lists but one. You end up ranking it 25 because that is the book's median rank. You need to figure out how to handle sparsity such that it doesn't inhibit the result you want to get. -- kainaw™ 14:16, 6 June 2011 (UTC)

Before you get hat far you've of course got to ask yourself how the various lists of novels or whatever were compiled. Were they done by a wide poll or seeing library returns or by a journalist asking a few friends? Was it done by some teenagers or some mothers group? So you have the problems that some might have much more weight than others and some might be much more compatible with your purposes than others. Dmcq (talk) 15:56, 6 June 2011 (UTC)

This is pretty similar to the questions investigated in social choice theory—you're trying to take several rankings and produce one overall ranking. This is difficult to do well. For example, you would probably like the method you use to have nice properties, such as these:

• If every list ranks The Grapes of Wrath higher than A Clockwork Orange, then the overall ranking should also rank The Grapes of Wrath higher than A Clockwork Orange.
• If the various lists were different, but the relative rankings of The Grapes of Wrath and A Clockwork Orange do not change (in other words, every list that previously ranked The Grapes of Wrath higher than A Clockwork Orange still does so, and vice versa), then the relative rankings of The Grapes of Wrath and A Clockwork Orange should remain unchanged in the overall ranking.
• No one list determines the overall ranking; every list is taken into account.

Unfortunately, it is impossible to satisfy all of these criteria simultaneously; see Arrow's impossibility theorem. —Bkell (talk) 16:49, 6 June 2011 (UTC)

Yep there's no sound mathematical method with even very simple assumptions. It's up to people to make choices on what they find works for them. You just have to make an admission like Google who now say their page ranks reflect their judgement of what will be most relevant for a user's query rather than just that it is determined by an algorithm. Dmcq (talk) 10:04, 7 June 2011 (UTC)

Converting between the forms of equations for an ellipse

I know that the standard form for an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1 or (x-h)^2/b^2+(y-k)^2/a^2=1 depending on the major axis and the general form is ax^2+cy^2+dx+ey=f. I want to be able to convert between these two forms with a simple set of equations. Can anyone help me? --Melab±1 ☎ 21:01, 5 June 2011 (UTC)

It doesn't help that you've used the constant "a" in both forms, making it impossible to get an exact correspondence. If you take the general form as cx^2+dy^2+ex+fy=g, just expand the first form and match coefficients of x^2, y^2, x, y and the constant term.→86.132.165.117 (talk) 21:40, 5 June 2011 (UTC)

To go from the former to latter, just expand as 86.132 suggests. To go the other way, you complete the square for each variable. SemanticMantis (talk) 20:35, 6 June 2011 (UTC)

• An ellipse in the plane is given by F = ax² + by² + 2hxy + 2gx + 2fy + c = 0 where a, b, c, f, g and h are real numbers and h² – ac < 0. (If h² – ac = 0 you have a parabola and if h² – ac > 0 you have a hyperbola.) To calculate the centre of the ellipse (or more generally any conic) you have to solve ∂F/∂x = ∂F/∂y = 0 with respect to x and y. Once you know the centre, say x = p and y = q, you translate the centre to the origin by a substitution x = x + p and y = y + q. After that, you'll see that F(x+p,y+q) = Ax² + 2Bxy + Cy² + D, where A, B, C and D are real numbers (in fact they are functions of the old a, b, c, f, g and h). To calculate the axes of the conic you need to consider the matrix

$\left[{\begin{array}{cc}A&B\\B&C\end{array}}\right],$

which is the matrix of the quadratic form Ax² + 2Bxy + Cy². The eigenvectors of that matrix give the axes of the conic. Finally, you need to calculate an orthogonal change of basis matrix that takes the axes of the conic onto the coordinate axes. It needs to be orthogonal so that you don't stretch the conic in anyway. This is just a rotation about the origin. Then your conic will be in a normal form, e.g. (x/α)² + (y/β)² = 1. I know it might sound complicated, but all that we have done is slide the conic and then rotate it to put it into normal form. — Fly by Night (talk) 18:13, 9 June 2011 (UTC)

June 6

Fermat's Principle

Can Fermat's principle be used as a definition of the straight line? 128.232.240.223 (talk) 22:58, 6 June 2011 (UTC)

Plato's definition of a line was basically that which blocks the view between two points, or in a more modern interpretation, the path followed by a ray of light. (See Heath's commentary on Euclid Book I Definition 4.) Since Euclid however this kind of definition is frowned upon because it appeals to physical theory rather than mathematical truth. This is especially true in more recent times, where space itself is considered curved. The modern viewpoint is that a line is an undefined concept whose behavior is determined by the axioms of geometry.--RDBury (talk) 00:22, 7 June 2011 (UTC)

Fermat's principle is just a physics-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line". That holds in Euclidean geometry, due to the triangle inequality. (In some respects, the triangle inequality is the math-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line".) For non-Euclidean geometry, at least for metric spaces, the triangle inequality is a convenient property to have (although not essential). For those metric spaces with the triangle inequality, it's perfectly feasible to use it to define a line-like object, although that might not match the conventional definition of a straight line that others use for that space. -- 174.31.219.218 (talk) 15:57, 7 June 2011 (UTC)

What is the definition of the straight line that does not require discussion of linear equations then? Your article is somewhat unclear. 128.232.240.223 (talk) 19:46, 7 June 2011 (UTC)

A straight line in Euclidean space is an example of a geodesic. — Fly by Night (talk) 20:01, 7 June 2011 (UTC)

I appreciate that yours is a correct answer but I would ask, is it a helpful one? If pressed for the definition of a real number, one might define it as an example of a tensor. I don't think we have learnt anything from this definition though. Is there a definition, which mentions linear equations neither explicitly nor implicitly (unless you regard any two, apparently, distinct definitions of a mathematical object as implicitly mentioning the other by virtue of the fact that they are equivalent definitions), of a more basic nature that is still unambiguous? Or are we in rather murky murky waters here, since the straight line underpins Euclidean geometry, and, as such, is challenging to rigorously define? 128.232.240.223 (talk) 21:25, 7 June 2011 (UTC)

If you're willing to accept the "distance between two points" as a primitive concept, then the definition of a line as a geodesic is a natural one. This is, by definition, the shortest continuous curve between two points. I don't see what your issue is. Sławomir Biały (talk) 22:16, 7 June 2011 (UTC)

"A straight line is the shortest distance between two points" is the colloquial way of putting it. "128.232.240.223", what do you find so unhelpful about that? Michael Hardy (talk) 01:25, 9 June 2011 (UTC)

Surely we are relying on concepts here that themselves depend upon the straight line. I would argue that 'shortest' implies a minimisation, which implies calculus. Elementary calculus necessitates an understanding of the straight line, ie when you let consider the gradient of the straight line passing through two points on a curve and then let one point tend towards the other. Similarly, surely 'distance' can only properly be discussed once we have an understanding of the straight line. I like the OP's point about scalars and tensors. Generally, we proceed from the special case to the general case, eg we start with scalars, move onto vectors, onto matrices, onto tensors, and so it strikes me unhelpful to define a simpler concept in terms of a more complicated one. Or is the question itself just unhelpful; can a concept as fundamental as the straight line ever have an entirely rigorous definition in terms of simpler concepts, since no such concepts exist? meromorphic [talk to me] 16:50, 12 June 2011 (UTC)

As the geodesic article says: "In the presence of an affine connection, geodesics are defined to be curves whose tangent vectors remain parallel if they are transported along it." Don't worry too much about the technical language if it's too much. In Euclidean space (parallel) transport is just translation. So a geodesic in Euclidean space is a smooth curve whose tangent vectors are all parallel (in the ordinary sense of the word). Physically, a tangent vector represents the velocity of an object. So a geodesic in Euclidean space, i.e. a straight line, is given by the path of a particle whose velocity is parallel at each and every moment. — Fly by Night (talk) 19:19, 9 June 2011 (UTC)

You can use an infinitesimal argument to show that this curve is a straight line. We want the velocity vectors to be parallel at each point. Consider a parametrisation: $\scriptstyle \gamma (t):=(x(t),y(t)).$ Consider two nearby velocity vectors, say $\scriptstyle {\dot {\gamma }}(t)$ and $\scriptstyle {\dot {\gamma }}(t+\delta t).$ Notice that

${\dot {\gamma }}(t+\delta t)={\dot {\gamma }}(t)+{\ddot {\gamma }}(t)\,\delta t+O(\delta t^{2})\,.$

If you want $\scriptstyle {\dot {\gamma }}(t)$ and $\scriptstyle {\dot {\gamma }}(t+\delta t)$ to be parallel you need

$\det \left[{\begin{array}{cc}{\dot {x}}(t)&{\dot {y}}(t)\\{\dot {x}}(t)+{\ddot {x}}(t)\,\delta t+O(\delta t^{2})&{\dot {y}}(t)+{\ddot {y}}(t)\,\delta t+O(\delta t^{2})\end{array}}\right]=0\,.$

Expanding we see that we need $\scriptstyle ({\dot {x}}(t){\ddot {y}}(t)\,-\,{\ddot {x}}(t){\dot {y}}(t))\,\delta t\,+\,O(\delta t^{2})\,=\,0.$ For the limiting process we divide by $\delta t$ and then let $\delta t$ tend to zero. This gives $\scriptstyle {\dot {x}}(t){\ddot {y}}(t)\,-\,{\ddot {x}}(t){\dot {y}}(t)=0,$ i.e. the velocity and acceleration are parallel for each t, i.e. the plane curve curvature of γ is zero for all t, i.e. we have a straight line. — Fly by Night (talk) 19:45, 9 June 2011 (UTC)

June 8

Example of a graph

Can someone please give me an example of a k-regular connected graph G, (k>2) which has order 4 or more and contains a triangle but no square?-Shahab (talk) 06:14, 8 June 2011 (UTC)

Markström graph

I have an example of order 10, but I'm going to have to draw it. I did a quick search on Commons to see if there was already an image of it there, and I found the Markström graph (see right), which is another example. —Bkell (talk) 07:20, 8 June 2011 (UTC)

The Dürer graph is also an example. —Bkell (talk) 07:22, 8 June 2011 (UTC)

And Tietze's graph. (I'm running through the list at commons:Graphs in graph theory, by the way.) —Bkell (talk) 07:25, 8 June 2011 (UTC)

Three of the truncated Platonic solids, too. The truncated tetrahedron is pretty close to the example I came up with; here's a poor attempt to draw my example (the O's are the vertices):

   O---------O
  / \       / \
 /   \     /   \
O-----O   O-----O
 \     \ /     /
  \     O     /
   \    |    /
    \   O   /
     \ / \ /
      O---O

—Bkell (talk) 07:40, 8 June 2011 (UTC)

June 9

Total ordering

Can ZF (without choice) prove every set can be totally ordered? Not well ordered, just total. Money is tight (talk) 04:28, 9 June 2011 (UTC)

No. If every set can be totally ordered, then you can always find a choice function for any set of unordered pairs, which is not provable in ZF. (Suppose you have a set of unordered pairs, then linearly order the corresponding set of ordered pairs, and from a pair {a,b}, choose a if (a,b) appears before (b,a) in the linear order, otherwise b.) --Trovatore (talk) 04:33, 9 June 2011 (UTC)

Actually I guess it's a bit simpler than that — just linearly order the union of all the pairs, and then from each pair, take the lesser one in the linear order. Of course that just pushes the question back to why you can't prove in ZF that there's always a choice function on sets of pairs. I don't know the answer to that for sure. I want to say that a choice function on pairs of equivalence classes for the Vitali relation will somehow give you something that can't exist in a model of ZF+AD, say a non-measurable set of reals or one without the property of Baire, but the details escape me. --Trovatore (talk) 04:51, 9 June 2011 (UTC)

Ah, I think I can do it. Suppose there's a choice function on all pairs of equivalence classes of the Vitali equivalence relation. Graph that on the plane, meaning that we put (x,y) into the graph just in case, when you apply the choice function to the pair ([x],[y]), where [x] means the equivalence class of x, the choice function gives you back the equivalence class of x, not the equivalence class of y.

OK, what can we say about this graph? It has to be invariant under rational shifts: If (x,y) is in the graph, then so is (x+r,y+s) for any rational r and s; that's just by the definition of the Vitali relation. Moreover, it's complemented when you reflect it across the main diagonal, in the sense that if (x,y) is in the graph, then (y,x) is not in the graph (except for the case that x is a rational shift of y, which we should easily be able to argue is a meager set of exceptions).

So now suppose this set in the plane has the property of Baire. That implies that either (i) it's meager, which it can't be because then the plane would be the union of three meager sets, namely the set, its reflection across the main diagonal, and the set such that x is a rational shift of y. Or (ii), there's some little neighborhood on which the set is comeager.

But in case (ii), cut that little neighborhood down to something nice, say a circular disk or a square. Then when you reflect across the main diagonal, you have to get a neighborhood on which the set is meager, by the complementing property we discussed.

But surely there's some rational shift that takes the neighborhood mostly onto the reflected neighborhood. Now you have a contradiction.

Anyone see any mistakes? --Trovatore (talk) 11:01, 9 June 2011 (UTC)

Thanks. Money is tight (talk) 00:15, 10 June 2011 (UTC)

How many observations are needed to reach a certain accuracy?

Say you wanted to see what cloud cover was for each month of the year and put up a machine that took 1 reading an hour for 5 years.

So you might get something like Jan 0/10 38%, 1/10 5% ... 9/10 4%, 10/10 25%;  Feb 0/10 34% etc. or whatever.

You'd be 50% sure of being how close to the real value? What about 95%? What formula do I use? Is it 1/√(trials)? Then again, I've seen months when a weather system or high pressure stays stuck over the area for 10, 14 days in a row when days are supposed to be close to a random bag here. Get one of those in your month and your accuracy would be pretty much thrown off, right? Maybe you really need 30 years like the climatologists use?

The observations are not independent, that is if you have clouds on day one this changes the probability of clouds on day two, compared to no data for day one or no clouds on day one.  So before you can say anything about the accuracy of your observations you would need a model for that dependency.  Taemyr (talk) 19:36, 9 June 2011 (UTC)

I don't know what the dependency is but from about 1 decade of noticing I'd say streaks are usually 1, 2, sometimes 3 days, within the same day gets more correlated, 5 days in a row on the weather forcast and I start noticing, last month there were 10 days in a row when clouds rolled in maybe twice, and only for a few hours (no rain that long makes allergies really bad), it was pretty much overcast 10/11 days in a row in June of '09, and I saw almost no sun for a half month in 2005. Dependency is like this year round, except in early summer much of the rain comes from quick thunderstorms so an otherwise sunny day will be dark+pouring and back again in 3 hours. Sagittarian Milky Way (talk) 22:40, 9 June 2011 (UTC)

I don't completely understand your question. What do you mean by "what cloud cover was for each month of the year"? What does "Jan 0/10" mean? --Tango (talk) 19:36, 9 June 2011 (UTC)

They list cloud coverage for all 12 months in "percent of the time that the sky is 0/10ths covered", "percent of the time it's 1/10ths", "percent of the time that it's 2/10ths", 3/10ths, etc. up to 10/10ths., presumably because the instrument that asseses area is not that accurate. Sagittarian Milky Way (talk) 22:40, 9 June 2011 (UTC)

Ok, so where does the error come in? If you are measuring it, then surely you are getting it right... What is it you are actually trying to work out? --Tango (talk) 23:09, 9 June 2011 (UTC)

Ignoring the details, our articles on standard error (statistics) and confidence intervals probably address your question. Looie496 (talk) 22:52, 9 June 2011 (UTC)

...if you can make sense of them, that is. The quality of writing in those articles is pretty pathetic. Looie496 (talk) 22:55, 9 June 2011 (UTC)

Triangular calculation equation

Probably due to lack of sleep, I just can't wrap my brain around what I am sure is a simple problem. I have an array indexed 1, 2, 3, 4... Each index in the array has an integer such that the value in the index is less than the index before it. An example: {5,4,3,2,1}. That is optimal. Each index is exactly one less than the one before it. However, this program sometimes has one (and only one) occurrence where one index will be two less than the one before it: {6,5,4,2,1}. In this case, I want to remove the far right value and distribute it among the other values to get the optimal distribute: {6,5,4,2+1} = {6,5,4,3}. Sometimes, it won't work easily: {6,5,3,2} = {6,5,3+2} = {6,5,4} with a remainder of 1. I am doing this with a loop, removing one column on the right at a time. Isn't there a way, given the number of columns, value of the left column, and the column in which the value is 2 less than the previous column, that I can perform a single calculation to see how many columns I need to remove and distribute? -- kainaw™ 12:41, 9 June 2011 (UTC)

I think it depends on the factorization of the sum, call it p say. If p is a prime then there are only going to be one or two possible arrays. For example with p=23 there are only {23} and {12,11}. If you have a factorization of p then there is a calculation, but without it it's probably not possible.--RDBury (talk) 15:55, 9 June 2011 (UTC)

2ⁿ has only a single representation as a sum of consecutive integers: {2ⁿ}. Gandalf61 (talk) 16:11, 9 June 2011 (UTC)

Thanks, but the problem I was working on doesn't require a perfect distribution. It looks for an optimal distribution in which there is a remainder left over that, compared to the size of the columns, is negligible. As I showed with {6,5,3,2}, the optimal solution is {6.5.4} with one left over. So, given the number of columns (C=4), the size of the left-most column (S=6), and the column in which the error occurs (E=3 if we count them 1, 2, 3, 4), how many columns will be left in the optimal solution (O=3) - and the remainder can be included (R=1). This is an easy problem to solve by iterating over "remove the far-right column and place them on E, E+1, E+2, E+3... until all columns have been filled. If E!=O, do it again." I am trying to get O=f(C,S,E) without the iteration because the size I'm working with is around C=1,000,000, S=1,000,000, and E=250,000. After calculating O and R, I have to go to the next step of grouping the R's together for another task. -- kainaw™ 17:18, 9 June 2011 (UTC)

Now that I'm awake, I can explain why this is hard for me to think about... Assume you look at it and you realize that you need 10 to fill the top row (and also assume the far right column has a value of 1). So, it is easy to calculate that 1+2+3+4 = 10. Therefore, removing the right 4 columns will give you the necessary amount to fill the top row - but that won't work. You just removed 4 columns. You only need 6 to fill the top row now. The correct answer was that removing 1+2+3 = 6. But, you only removed 3 columns this time. You need 7 to fill the top row. The original answer was the correct one. Remove 4 columns and have a remainder of 4. It is the removal of columns that makes it harder than simply calculating 1+2+3+...+n = X for some value X. -- kainaw™ 19:08, 9 June 2011 (UTC)

Take your case {6,5,3,2}. Why can you not go {6,5,3,2}->{6+1,5,3+1}->{7+2,5+2}->{9+7}->{16} with no remainder? — Preceding unsigned comment added by Taemyr (talk • contribs) 19:29, 9 June 2011 (UTC)

The goal is not to trick the problem into a trivial solution. The goal is to find out how many columns are required to fill the incomplete top row. -- kainaw™ 19:36, 9 June 2011 (UTC)

OK. So you can only add points below the gap? (Or rather what you are trying to optimize is maximizing the number of colums, the remainder is irrelevant). Take the case where you need 10 points. This means that your set goes {...12,10,9,8,7,6,5,4,3,2,1}? you remove the one so you are left with needing 8(the value of the one+ one colum less to fill). then the 2, at which point you need 5. After the 3 you need 1, and when you remove the 4 you are done and have a remainder of 4.

If this is correct then:

(n+1)+(n+2)+...+(n+m)=n*m+(1+m)m/2 which will be the value you redistribute if you remove m colums where the least has value (n+1).

To this we must add the number of colums removed so (n+1)*m+(1+m)m/2.

We want to find the lowest m such that (n+1)*m+(1+m)m/2>=target.

0.5m^2+(1.5+n)m-target>=0.

Switching n such that we look at the least value beeing n, rather than n+1, we get

0.5m^2+(.5+n)m-target>=0

So roof(.5+n+sqrt((.5+n)^2+2*target)) columns needs to be removed.

Baring misunderstandings or outright mistakes on my part. Taemyr (talk) 20:19, 9 June 2011 (UTC)

Thanks. I was having extreme difficulty thinking yesterday. I just couldn't make the jump to reducing the columns required to fill as the columns were being used. -- kainaw™ 12:38, 10 June 2011 (UTC)

Tetrahedrons and triangular graphs

I've used triangular graphs before, and the requirements for one (as I know the term, seems there's another) to be valid could be said to be {x,y,z} where x+y+z=100%. I've been trying to picture, without success, a hollow tetrahedron (OK so far!). Then there's some point within the tetrahedron with lines coming from it perpendicular to each of its sides - four of them. At the point they touch the sides, that point is read like a triangular graph. How many unique values does it create, I can't think how many of the 12 values are duplicates? If so, how does that change the requirement for it to be valid? Thanks, Grandiose (me, talk, contribs) 19:23, 9 June 2011 (UTC)

I don't understand this question. Could you explain what you mean by "triangular graph"? —Bkell (talk) 18:12, 10 June 2011 (UTC)

Something like this. So each point in the triangle has three components that sum to 100%. Grandiose (me, talk, contribs) 09:22, 11 June 2011 (UTC)

You reference is an awful graph. The shadows on the font makes things very hard to read. Anyway your triangular graph could just as easily be draw as a standard x,y graph (at right angles) and you can come along afterwards and mark in the derived z coordinate lines. Similarly a hollow tetrahedron could be drawn as a standard x,y,z 3-D graph. Everything else is derived from that - any u v or w coordinates you may choose to invent. My "tetrahedron" will not have the "right" sold angle (0.55 steradians), but will have 0.5pi (1.57)steradians and again you can come along afterwards and mark angled planes of constant u, v, or w. Does this help at all? -- SGBailey (talk) 20:34, 11 June 2011 (UTC)

Other than explaining why no concept occurs in mathematics, not really (it's a bit over me :)). It's the "Similarly a hollow tetrahedron could be drawn as a standard x,y,z 3-D graph." I'm struggling, mostly as to the consquences, as with the original tetrahedron. If our original triangle was actually two values, and a third derived one, and you think the tetrahedron can be shown to be a 3D one, does that mean that the tetrahedron is actually based on three values, and therefore of the 12, it is possible to identify a point with only three? That would be a very useful answer to my question. Grandiose (me, talk, contribs) 20:51, 11 June 2011 (UTC)

A hollow tetrahedron is surface. There are simple functions that will map this surface onto a flat plane - just unfold it. Once on the flat plane, then x,y is sufficient to define a point. Thus two coordinates is still sufficient to define all coordinates on the tetrahedron. However there will now be a mixup of unfolded x and y against whatever the parameters you have on the tetrahedrons edges are - you have 6 edges, so you have 3 triangular graphs abc on the base, ade on one side, bef on another side and cfd on the third side. I've no idea what such a shape could meaningfully plot and I don't know where you get "12" from. -- SGBailey (talk) 23:15, 11 June 2011 (UTC)

The triangular graph is an example of barycentric coordinates. This can be extended to three dimensions. Gandalf61 (talk) 21:27, 11 June 2011 (UTC)

Uncountable Godel Theorem

As I understand it, Godel's theorems only apply to logical systems in which it is possible to enumerate all the theorems. Suppose we had a system of uncountably many mutually independent axioms for arithmetic. Then, it would be impossible to enumerate all theorems, or even all axioms, simply because there would be too many of them. Would Godel's theorem, or some modern variant, still apply to this fattened system? (I think that such a system could be constructed as follows: Let 0 stand for PA, and N stand for "for M<N, all the axioms of M plus 'M is consistent.'" Then by induction up the countable ordinals, there must be an uncountable such system, in which all axioms are independent and in some sense self-evident. Is this correct?) Black Carrot (talk) 23:20, 9 June 2011 (UTC)

To clarify, you're asking about Godel's Incompleteness Theorem. Godel has another famous theorem confusingly known as the Completeness Theorem.

To answer your question, yes, the incompleteness theorem requires in an essential way that the theorems can be effectively enumerated. You don't need to go uncountable to see this, though; just let T be the theory consisting of all sentences which are true in $(\mathbb {N} ,+,*,0,1)$. Then it's trivially complete and consistent.

Your method of building an uncountable system doesn't work, though; just by counting, there are only countably many sentences expressible in the language of PA, so it can't go uncountable. The reason your recursive construction breaks is that at some countable ordinal, there's no way to code "M is consistent", since M isn't a computable set. I suspect this happens at $\epsilon _{0}$, but I'm not sure.--121.74.111.168 (talk) 04:53, 10 June 2011 (UTC)

You can go way past ε₀. It's pretty clear what to do up to $\omega _{1}^{CK}$, the first non-recursive ordinal. My understanding is that if you iterate to $\omega _{1}^{CK}$, you get a (non-computable) theory that proves all true $\Pi _{1}^{0}$ statements.

I imagine someone has tried to push it further, but I don't know exactly in what way. Some new idea would be required. --Trovatore (talk) 20:23, 10 June 2011 (UTC)

Interesting. Do you know where I could look to find out more about that? Is that a new result, or has it been known for a long time? Black Carrot (talk) 23:08, 10 June 2011 (UTC)

It seems like I saw a paper, or at least a citation of a paper, that mentioned it, sometime in the last year, but I can't remember the details and I'm not sure even what search terms to use. If I come across it I'll try to let you know. I think it might be a reasonably standard result in the theory of proof-theoretic ordinals, which is something I don't know too much about. --Trovatore (talk) 10:20, 11 June 2011 (UTC)

Try Beklemishev, L. D. (2003). "Proof-theoretic analysis by iterated reflection". Arch. Math. Logic. 42: 515–552. doi:10.1007/s00153-002-0158-7. --Trovatore (talk) 01:21, 12 June 2011 (UTC)

If you feel like trying to prove it yourself (almost always more enlightening than reading someone else's proofs), you might first learn about ordinal notations, if you don't already know about them, and think a bit about how you might do your iteration relative to an ordinal notation. Then think about whether two notations for the same ordinal give you the same theory if you iterate along them. I'm not sure whether they do or not. If they don't, maybe they give the same $\Pi _{1}^{0}$ theory; try to prove that. --Trovatore (talk) 10:24, 11 June 2011 (UTC)

June 10

Functional powers

Beginning with the usual definition of f⁻¹ for a function f and

$f^{n}:=\underbrace {f\circ f\circ \dots \circ f} _{n}$

led me to the idea of functional roots (i.e. if g² = f, then g = f^1/2) and then to rational powers of functions (i.e. f^a/b = (f^1/b)^a). From here, I had several questions:

1. How could you extend the definition to real or complex powers of functions? Is it even possible, given the duplicity of things like f^1/2 (i.e. if f(x) = g(x) = x and h(x) = 1/x, then g(g(x)) = h(h(x)) = f(x), so g and h are both "f^1/2" for the right domain)? If it is possible, you could consider taking functions to the power of other functions, which is an interesting concept (to me, anyway).
2. How do you go about graphing or finding formulaic approximations for functions like rin = sin^1/2 (see here)? How was it determined that as n goes to infinity, sin^1/n goes to the sawtooth function?
3. Do these notions have any particular application?

Thanks in advance for taking the time with these loaded and naïve questions. —Anonymous Dissident^Talk 12:44, 10 June 2011 (UTC)

Your questions will take you into the area of functional equations. One complication you will encounter is that the functional square root of a function is usually far from unique - for example, the function

$y(x)=x$

has functional square roots

$y_{(1/2)}={\frac {x+a}{ax-1}}$

for any real (or complex) number a. Gandalf61 (talk) 14:13, 10 June 2011 (UTC)

The set of all "square roots" of the identity function on an arbitrary set $X$ is in natural one-to-one correspondence with the set of all partitions of $X$ into sets of size 1 or 2. --COVIZAPIBETEFOKY (talk) 15:00, 10 June 2011 (UTC)

You can define a generator of a function as follows. If:

$f(x)=\lim _{n\to \infty }\left[x+{\frac {g(x)}{n}}\right]^{\circ n}$

then we can consider g(x) to be a generator of f(x). Count Iblis (talk) 15:44, 10 June 2011 (UTC)

I get a one-to-one correspondence between the complement of a projective algebraic variety in the complex projective plane and the Möbius transformations that are functional square roots of g(z) = z. Let [a : b : c] be in CP², with a² − bc ≠ 0, and define

$f(z):={\frac {az+b}{cz-a}}\,.$

we see that ƒ has the property that (ƒ ∘ ƒ)(z) = z. — Fly by Night (talk) 11:32, 11 June 2011 (UTC)

If you're just interested in constructing the functional square root, then the method of Kneser (referenced in our article) seems to be worth studying. This paper constructs a functional square root of the exponential function, and the method seems like one could work it out without a great deal of specialized knowledge. If you're interested in looking at the whole semigroup (if there is one—which seems to me a little unlikely) of functional roots $\sin ^{\alpha }$, then some basic papers on this subject appear to be Erdos and Jabotinsky (1960) [1] and Szekeres (1958) "Regular iteration of real and complex functions", Volume 100, Numbers 3-4, 203-258, DOI: 10.1007/BF02559539. It seems to be a theorem that there is no way to define non-integer iterates of the exponential function so that the semigroup property holds (maybe along with analyticity in the iteration parameter, it's not clear to me what the rules are). The following paper also seems to be worth looking at: Levy, (1928) "Fonctions à croissance régulière et itération d'ordre fractionnaire", Annali di Matematica Pura ed Applicata Volume 5, Number 1, 269-298, DOI: 10.1007/BF02415428. I haven't found any (reliable) papers that specifically address the sine function. There's this, which I find a little dubious. Sławomir Biały (talk) 13:55, 12 June 2011 (UTC)

June 11

Proof of Inequality

How to prove the following inequality?

If $p>0$ is a real number, then there exists a constant $C_{p}$ dependent only on $p$ such that for all real numbers $x,y\geq 0$, we have the inequality:

$x^{p}+y^{p}\leq C_{p}(x+y)^{p}$

This isn't homework. I was reading a proof in a paper and I think this inequality is used and I'm wondering how to prove it. Can anyone enlighten me? — Preceding unsigned comment added by 49.2.4.186 (talk) 08:47, 11 June 2011 (UTC)

As stated, it isn't true. For $p=1/2$, $y=1$, there's no $C_{p}$ that works for all $x$.--203.97.79.114 (talk) 12:32, 11 June 2011 (UTC)

Sorry, ignore that. I just had a really dumb moment.--203.97.79.114 (talk) 12:43, 11 June 2011 (UTC)

Okay, trying again. First, if $p\geq 1$, then $C_{p}=1$ will suffice: the inequality clearly holds for $x=0$, and the derivative of the right (with respect to $x$) is always at least as much as the derivative of the left.

If $p<1$, then $C_{p}=2$ will suffice. Let's assume by symmetry that $x\geq y$. If $x=y$, then the inequality becomes $2x^{p}\leq 2\cdot 2^{p}x^{p}$, and since $p>0$, this holds. Now we again take the derivatives with respect to $x$. The left gets us $px^{p-1}$, while the right gives $2p(x+y)^{p-1}\geq 2p(2x)^{p-1}>px^{p-1}$.--203.97.79.114 (talk) 12:55, 11 June 2011 (UTC)

Excellent. Thanks! It didn't occur to me to differentiate. But do you think that this can be proven without differentiation. This is just out of curiosity since I understand your proof and am content with it but would like to know whether there is a nice trick that solves it. — Preceding unsigned comment added by 49.2.4.186 (talk) 02:04, 12 June 2011 (UTC)

I think perhaps an easier approach is to divide both sides by $y^{p}$, which converts the expressions to functions of a single variable equal to $x/y$. Looie496 (talk) 16:44, 11 June 2011 (UTC)

Alternative Proof

Does anyone know of a way to evaluate the following:

$\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k^{2}}}\right)$

that does not involve Fourier series? — Fly by Night (talk) 21:48, 11 June 2011 (UTC)

By evaluating the double integral $\int _{0}^{1}\int _{0}^{1}{\frac {dx\,dy}{1-xy}}$. Sławomir Biały (talk) 22:33, 11 June 2011 (UTC)

Sławomir, how does that double integral relate to the sum? I don't see how the dilog function relates to the sum in question in a simple, non-convoluted way. Could you show the steps for getting from the sum I gave to that double integral please? — Fly by Night (talk) 00:56, 12 June 2011 (UTC)

Expand the integrand as a geometric series and integrate it term-by-term. That gives your series. The integral itself can be evaluated by rotating the coordinate system through an angle of $\pi /4$, which transforms the integrand into something of the form $(1-x^{2}/2+y^{2}/2)^{-1}$. The resulting integral can be dealt with by elementary means. Sławomir Biały (talk) 02:00, 12 June 2011 (UTC)

Here is a link. Sławomir Biały (talk) 02:28, 12 June 2011 (UTC)

There is another elementary proof at our article Basel problem (along with Euler's proof, mentioned below, and a proof using Fourier series). Sławomir Biały (talk) 02:47, 12 June 2011 (UTC)

Excellent. How very cunning. Thanks for that. — Fly by Night (talk) 14:27, 12 June 2011 (UTC)

There's a description of Euler's proof here. AndrewWTaylor (talk) 23:25, 11 June 2011 (UTC)

Thanks. I'll take a look in the morning. The non-LaTeX font makes it hard to read at this time of night. — Fly by Night (talk) 00:56, 12 June 2011 (UTC)

See question 8 here: http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-mathematics-extension-2.pdf . (I don't really know anything about Fourier series, but the proof which the student is led through doesn't explicitly mention them; judge for yourself whether they are "involved" implicitly.) —Anonymous Dissident^Talk 23:27, 11 June 2011 (UTC)

Thanks a lot for taking the time to find that link. It looks to me that Question 8 is basically asking people to compute the integrals involved in calculating the Fourier series. — Fly by Night (talk) 00:56, 12 June 2011 (UTC)

What the hell? Have you heard of Parseval's identity you guys? — Preceding unsigned comment added by 49.2.4.186 (talk) 02:08, 12 June 2011 (UTC)

The OP wants a proof that does not involve Fourier series. Sławomir Biały (talk) 02:23, 12 June 2011 (UTC)

Quite! — Fly by Night (talk) 14:27, 12 June 2011 (UTC)

HAHA 49.2.4.186 IS AN IDIOT! LOL! — Preceding unsigned comment added by 72.179.51.84 (talk) 02:21, 12 June 2011 (UTC)