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The result $$\zeta(2) = \frac{\pi^2}{6},$$ tends to amaze young students because of its beauty.

However, although in literature there are many proofs of this result, I find that none is suitable for someone who didn't take some advanced classes in calculus. Moreover, as far as I know, they do not offer an intuitive explanation of why this result should be true.

So my question is:

What is the key intuition -- that is, the picture -- behind the result? Are there any visual of anyway intuitive proofs (*) of the statement that can be used to convey a better (or, at least, different) understanding of it (to people who haven't taken advanced calculus yet)?

(*) Note that full rigour is not compulsory for the scope of this question (it is very appreciated by me personally, though).

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  • $\begingroup$ This question was put on hold at MO: mathoverflow.net/questions/185020/… $\endgroup$ – Steven Gubkin Oct 23 '14 at 15:42
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    $\begingroup$ One of the things about intuition is that you can build it. Part of what makes mathematics so beautiful is that you can take a result like this which you currently have no intuition for and custom build intuition to include it. It does not have to fit into your old limited version of intuition. Any of the proofs of this result could, with enough effort, become intuitive. It seems like you are asking for an answer which would be intuitive to you currently. But no one knows what you find intuitive. $\endgroup$ – Steven Gubkin Oct 23 '14 at 15:46
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    $\begingroup$ @Alizter I have a good explanation for that one. If you define $e^x = \lim_{n \to \infty} (1+x/n)^n$ (which I have a lot of good intuitive reasons to do), then $e^{i\theta} \approx (1+\frac{i\theta}{n})^n$. But $1+\frac{i\theta}{n} \approx cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n})$ by geometry. Then by de Moivre, $e^{i\theta} \approx \cos(\theta)+i\sin(\theta)$. This approximation should get better and better as $n \to \infty$, so it is reasonable to conclude that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$. $\endgroup$ – Steven Gubkin Oct 23 '14 at 15:57
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    $\begingroup$ @Alizter see also en.wikipedia.org/wiki/Euler%27s_formula#mediaviewer/… $\endgroup$ – Steven Gubkin Oct 23 '14 at 15:58
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    $\begingroup$ @Alizter BTW, if you define $e^x$ as I like to, as the solution to the diff EQ $y'=y$ with initial condition $y(0)=1$, then $(1+\frac{x}{n})^n$ is simply what you get by applying Euler's method to the interval $[0,x]$ $\endgroup$ – Steven Gubkin Oct 23 '14 at 16:01
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You probably already know that $1/\zeta(2)$ is the probability of two "random" integers being relatively prime. Since this constant is related to number theory, you should expect any proof of the Basel formula to be hard work.

Probably the best proof from the point of view of transparency is obtained by evaluating $$\int_0^1\int_0^1 \frac{1}{1-xy}{\rm d}x{\rm d}y$$ in two different ways; one gives $\zeta(2)$, and the other $\pi^2/6$. A version appears in

Aigner, Martin; Ziegler, Günter (2009). Proofs from THE BOOK (4th ed.). Berlin, New York: Springer-Verlag. ISBN 978-3-642-00855-9.

But a shorter version appears in

Apostol, Tom M., A proof that Euler missed: evaluating $\zeta (2)$ the easy way, The Mathematical Intelligencer, vol 5, #3, 1983, 59--60.

It has the advantage of pointing out that the second evaluation of the double integral is achieved via a very appealing (read tricky) change of variables; namely rotating the coordinate axes by $\pi/4$.

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Euler's original heuristic may "fly" with people who don't know about calculus: just as polynomials (e.g., with all real roots) are essentially products of linear factors $x-\alpha$ where $\alpha$ runs through the roots, one might imagine that $\sin \pi z$ (using radian measure) is a products of something like $1-{z\over n}$ for $n$ non-zero integer, and also a factor of $z$ for the zero at $0$. We can get the constant right by looking at the limit as $z\to 0$. Then it is plausible that the next-order term is the sum of inverses of roots... which would give the desired formula.

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Interpreting

What is the ... the picture...?

far too literally, an image of $1/\zeta(2)$ was included in the question, "What fraction of the integer lattice can be seen from the origin?" The answer is $\frac{6}{\pi^2} \approx 0.6079$, i.e., about 61%:


         


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There isn't any "intuition" behind it, it was mind-blowing when Euler came up with this ("What?! The reciprocals of squares give something belonging to the circle??", remember in Euler's time calculus was only really starting to untangle itself from geometry). And precisely because of that is that I consider this result of a strange beauty.

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