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I know that $$\frac{\sin x}x=\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right).$$ Why exactly can I take the product and factor $x^2$? $$\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)=1-x^2\left(\frac1{\pi^2}+\frac1{4\pi^2}+\frac1{\pi^2 9}+\ldots \right)-\ldots.$$ This question rose from Euler's Basel problem proof.

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I'd say that is an Eulerian way to work: just just "open" parentheses in that infinite product, take first the product of all the first terms in each parentheses, which is only $\;1\cdot1\cdot\ldots=1\;$ , and then oberve that in any other product the factor $\;x^2\;$ appears so you factor it out, leaving the expression within the parentheses in the right side.

I'd say there are quite a few things to justify formally before one can produce such an equality, yet Euler used to do lots of things in this way.

Edit added : take logarithms in your first infinite product:

$$\log\left[\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)\right]=\sum_{n=1}^\infty\log\left(1-\frac{x^2}{n^2\pi^2}\right)$$

and observe that

$$\lim_{n\to\infty}\frac{\left|\log\left(1-\frac{x^2}{n^2\pi^2}\right)\right|}{\frac1{n^2}}=\frac{2x^2}{\pi^2}$$

so the series converges absolutely and so does the product, and thus you can rearrange the series (and thus the product) any way you can and it still will converge and to the same value. Perhaps there's a little more to say about justification, but I think this is already pretty close.

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  • $\begingroup$ I'm looking for a formal justification. $\endgroup$ – user221028 Mar 29 '16 at 7:45
  • $\begingroup$ @user221028 I added some info to my answer that can probably be helpful to you. $\endgroup$ – DonAntonio Mar 29 '16 at 8:03
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Suppose that you have $$P_k=\prod_{n=1}^k \left(1-a_n{x^2}\right)$$ Expand the product for a few terms.

Now, build a Taylor expansion around $x=0$ to get $$P_k=1-x^2\sum_{i=1}^k a_i+x^4 \sum_{i=1}^{k-1}\sum_{j=i+1}^k a_ia_j+O\left(x^6\right)$$ Replace $a_n$ by $\frac 1{n^2\pi^2}$

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$$\prod_{k=1}^n(1-a_k)= 1-(a_1+a_2+\cdots a_n)+(a_1a_2+a_1a_3+\cdots a_{n-1}a_n)-\cdots+(-1)^na_1a_2\cdots a_n$$

so that

$$\prod_{k=1}^n(1-x^2a_k)=\\ 1-x^2(a_1+a_2+\cdots a_n)+x^4(a_1a_2+a_1a_3+\cdots a_{n-1}a_n)-\cdots+(-1)^nx^{2n}a_1a_2\cdots a_n.$$

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