2
$\begingroup$

I got this expression for my $b_n$ to a Fourier series: $$b_n=\frac{(2- \pi^2 n^2)\cos(\pi n) -2}{4( \pi n)^3}$$ Now I want to write it in a closed form without the use of $\text{when } n \text{ is even}$ and $\text{when } n \text{ is odd}$. How could this be done?

$\endgroup$
3
  • $\begingroup$ You mean you want to do more than replace $\cos\pi n$ by $(-1)^n$? $\endgroup$
    – almagest
    Sep 3, 2014 at 15:34
  • $\begingroup$ @almagest Yes, I know this one already. $\endgroup$
    – E Be
    Sep 3, 2014 at 15:35
  • $\begingroup$ Put that is unlikely to be possible. Try looking at the first dozen values. We have $b_n>0$ for $n$ odd and $b_n<0$ for $n$ even. $\endgroup$
    – almagest
    Sep 3, 2014 at 15:41

1 Answer 1

1
$\begingroup$

If n is even then $b_n$ will have the form

$$ b_{2m}=\frac{(2 - 4\pi^2 m^2)\cos(2\pi m) -2}{4( 2\pi m)^3}= -\frac{ 1 }{8\pi m}.$$

I eave it for you to do the odd case. Note that you need $\cos( (2m+1)\pi ) =(-1)^{m+1}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .