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This is from a book im reading.

$ 1-\frac{x^2}{3!}+ \frac{x^2}{5!} -\frac{x^2}{7!}+ \frac{x^2}{9!}-...$

$= [1-\frac{x^2}{\pi^2}]\ [1-\frac{x^2}{4\pi^2}]\ [1-\frac{x^2}{9\pi^2}]\ [1-\frac{x^2}{16\pi^2}]... $

$= 1-(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \frac{1}{16\pi^2}+...)x^2+(...)x^4-\ ... $

"once Euler had multiplied out the infinite product to get two infinite sums equaling each other, nothing would be more natural than to equate the like powers of x. Note that both series begin with 1 . Next comes the $x^2$ term in each series, and so their coefficients must be equal . That is,"

$$-\frac{1}{3!}= -(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \frac{1}{16\pi^2}\,+\,...) $$

What I dont understand is that where did this $-\frac{1}{3!}$ came from?? please help.

Thanks.

Asim
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1 Answers1

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On one hand you have the series $1-\frac1{3!}x^2 + \cdots$. On the other you have the series $1 - \left(\frac1{\pi^2} + \frac1{4\pi^2} + \cdots\right)x^2 + \cdots$ These series are thought to be equal. The coefficient of $x^2$ on one side is $-\frac1{3!}$, while on the other side it's $\frac1{\pi^2} + \frac1{4\pi^2} + \cdots$. If the series as a whole are equal, then these two coefficients ought to be equal. So that's where the final equality comes from.

Arthur
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  • but what about other coefficients, and how does whole series is equal to only single coefficient, sorry but i haven't any know-how about series rules. Thanks. – Asim Nov 11 '19 at 11:24
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    If $$a_0 + a_1x + a_2x^2 + \cdots = b_0 + b_1x + b_2x^2 + \cdots$$and we assume some basic convergence properties, then $a_0 = b_0, a_1 = b_1, a_2 = b_2$ and so on. That's all that's happening here. We are specifically looking at $a_2 = b_2$, and forget about the rest. Now, in this specific case, $b_2$ happens to be an infinite sum, but that's not really an issue. – Arthur Nov 11 '19 at 11:28