0
$\begingroup$

In the proof of the Basel problem Euler reaches a step where expansion of $(\sin x)/x$ is $[1-(x^2/\pi^2)][1-(x^2/4\pi^2)][1-(x^2/9\pi^2)]\cdots$ And from that he got to- $1- x^2\{1/\pi^2 + (1/4\pi^2) + (1/9\pi^2) +\cdots\}+ x^4(\cdots)+\cdots$. I don't understand how this happens?

$\endgroup$
  • $\begingroup$ It comes from the second derivative at $x=0$. To make this rigorous you need some complex analysis to show $\pi \cot (\pi z) = \lim_{N \to \infty} \sum_{n=-N}^N \frac{1}{z+n}$. $\endgroup$ – reuns Oct 14 '17 at 8:41
  • $\begingroup$ Well, what do you get as the coefficient of $x^2$ when you multiply everything out? $\endgroup$ – Gerry Myerson Oct 14 '17 at 8:46
  • $\begingroup$ But how do we multiply "everything" when we have to multiply infinite "things"? $\endgroup$ – user167920 Oct 14 '17 at 8:48
  • $\begingroup$ Well x^2 will be preserved only if the term containing x^2 is multiplied by 1. $\endgroup$ – user167920 Oct 14 '17 at 8:50
  • 1
    $\begingroup$ @GerryMyerson Yes I do believe so. Thanks for your help $\endgroup$ – user167920 Oct 15 '17 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.