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I recently read and immediately loved Euler's solution to the Basel problem. In essence, he took the polynomial on the LHS of the below equation and equated it to the infinite product of its factors by noting that the polynomial is just $\frac{sin(x)}{x}$, and that $sin(x)$ has roots at integer multiples of $\pi$.

$$1 - \frac{x^2}{3!} + \frac{x^4}{5!} -... = (1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4 \pi^2})...$$

You can then expand and equate the $x^2$ term to find the solution to the zeta function of $2$. This part makes sense to me, but the article then stated that he further expanded the product to find (in theory) all even values of the zeta function.

I was trying to do this myself to find the solution to the zeta function of $4$, but have thus far been unsuccessful. I tried expanding further on the RHS, but didn't end up with inverse fourth power terms like I wanted. I also tried multiplying both sides by the infinite product $(1+\frac{x^2}{\pi^2})(1+\frac{x^2}{4 \pi^2})...$ to get the desired RHS, but the LHS seemed to get quite messy in the process.

I would greatly appreciate some help!

2 Answers 2

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Hint:

Well, you know this much:

$$\frac{\sin(x)}x=\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\dots$$

From that, you should be able to deduce the following:

$$\frac{\sin(ix)}{ix}=\left(1+\frac{x^2}{1^2\pi^2}\right)\left(1+\frac{x^2}{2^2\pi^2}\right)\dots$$

And multiply them together, you have

$$\frac{\sin(x)}x\frac{\sin(ix)}{ix}=\left(1-\frac{x^4}{1^4\pi^4}\right)\left(1-\frac{x^4}{2^4\pi^4}\right)\dots$$

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    And of course, if you want more explanation and less hint, feel free to ask. – 2017-01-05
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    Perfect, thank you! I was able to get the correct result with that hint. I had the right idea about what to do on the RHS, but failed to see the connection between the infinite product of factors and $\frac{sin(ix)}{ix}$ – 2017-01-05
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    @infinitylord :D No problem. And you will probably have noticed how this works for $\zeta(6,8,\dots)$? – 2017-01-05
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    I'm not entirely sure how to use it for all even powers, but I believe I see how to use it for all $\zeta(2^{n})$. For example, $\zeta (8)$ can be found by expanding the product $\frac{sin(x)}{x} \frac{sin(ix)}{ix} \frac{sin(\sqrt{i}x)}{\sqrt{i}x} \frac{sin(i\sqrt{i}x)}{i \sqrt{i}x}$, correct? – 2017-01-06
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    @infinitylord Yes, basically. :D – 2017-01-06
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One way to do that would be to partially multiply out the product up to the term $x^{4}$ and compare it to the $x^{4}$ in $\frac{\sin\left(\pi x\right)}{\pi x}$

If you begin to multiply and collect the terms in the fashion of $\sum_{j>i>0}^{} \frac{1}{i^{2}j^{2}}$, Then you have the following equality; \begin{equation} 2\sum_{j>i>0}^{} \frac{1}{i^{2}j^{2}}=\sum_{n>0}^{} \frac{1}{n^{2}}^{2}-\sum_{n>0}^{} \frac{1}{n^{4}} \end{equation}

From there, it is possible to find $\zeta(4)$ if you know $\zeta(2)$ and the coefficient of $x^{4}$ in $\frac{\sin\left(\pi x\right)}{\pi x}$

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    Take a look at [this approach](https://en.wikipedia.org/wiki/Basel_problem#Generalizations_of_Euler's_method_using_elementary_symmetric_polynomials) which is more in line with Euler's original proof in the Basel problem. – 2018-05-24