a) Find the half range Fourier sine series of cos(x) on 0 < x < $\frac{\pi}{2}$.
b) Use this extension to show that $\sum_{m=0}^{\infty}\frac{(2m+1)}{4(2m+1)^{2}-1}(-1)^{m}$ = $\frac{\pi}{8\sqrt2}$
For a) I have solved it by using:
$$b_n = \frac{2}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} cos(x)sin(2nx) dx $$
So, $$b_n = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}} cos(x)sin(2nx) dx $$
I solved $b_n$ by using the formula for $sinAcosB = \frac{1}{2}(sin(A+B)+sin(A-B)$.
So I got, $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} sin(2nx+x)sin(2nx-x) dx $$
This gives: $$\frac{sin(\pi n)+1}{2n+1}+\frac{1-sin(\pi n)}{2n-1}$$
After simplifying, I got:
$$b_n = \frac{4}{\pi}\frac{-sin(\pi n)+2n}{4n^{2}-1} $$
This is where my problem is. I'm thinking that what I get from part a) should be similar to part b), but with my n=2m+1. From part b), it seems the answer should be:
$$\frac{nsin(\frac{\pi n}{2})}{4n^{2}-1}$$.
Can someone tell me where I went wrong with part a)? Thanks.
Best Answer
$b_n = \dfrac{8n}{\pi(4n^{2}-1)} $ and $$\cos x=\sum_{n=1}^\infty \dfrac{8n}{\pi(4n^{2}-1)}\sin 2nx$$ set $x=\dfrac{\pi}{4}$: $$\cos \dfrac{\pi}{4}=\sum_{n=1}^\infty \dfrac{8n}{\pi(4n^{2}-1)}\sin n\dfrac{\pi}{2}$$ if $n=2m$ then $\sin n\dfrac{\pi}{2}=0$. if $n=2m+1$ then $\sin n\dfrac{\pi}{2}=(-1)^m$ thus $$\dfrac{1}{\sqrt{2}}=\sum_{m=0}^\infty \dfrac{8(2m+1)}{\pi(4(2m+1)^{2}-1)}(-1)^m$$