1. Fourier Series

Can someone help me with these two Fourier Series problems please? Also, can you explain how you did the problems because I dont understand how to do Fourier Series ...Thanks in advance!

#1 Use the Fourier Series
f(x) = {0, -pi < x < 0; x^2, 0 < x < pi

to show that
1 - 1/4 + 1/9 - 1/16 +.... = (pi^2)/12

#2 Establish the result in the problem below, where m and n are positive integers. (Hint: sinA sinB = (1/2) [cos(A-B) - cos(A+B)].)
L
∫ sin((n pi x)/L) sin((m pi x)/L) dx = {0, m≠n; L, m=n
-L

2. Originally Posted by cherry106
Can someone help me with these two Fourier Series problems please? Also, can you explain how you did the problems because I dont understand how to do Fourier Series ...Thanks in advance!

#1 Use the Fourier Series
f(x) = {0, -pi < x < 0; x^2, 0 < x < pi

to show that
1 - 1/4 + 1/9 - 1/16 +.... = (pi^2)/12
The Fourier series representation of $\displaystyle f(x)$ is:

$\displaystyle f(x)=a_0+\sum_1^{\infty}a_n \cos(nx) +b_n \sin(nx)$,

where:

$\displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) dx =\frac{1}{2\pi}\int_0^{\pi} x^2 dx=\frac{\pi^2}{6}$

$\displaystyle a_n=\frac{1}{\pi}\int_{-pi}^{\pi}f(x) \cos(nx) dx=\frac{1}{\pi}\int_{0}^{\pi}x^2 \cos(nx) dx=\frac{(-1)^n 2}{n^2}$

and

$\displaystyle b_n=\frac{1}{\pi}\int_{-pi}^{\pi}f(x) \sin(nx) dx=\frac{1}{\pi}\int_{0}^{\pi}x^2 \sin(nx) dx$

So:

$\displaystyle f(0)=a_0+2\sum_1^{\infty} b_n=\frac{\pi^2}{6}+2\sum_1^{\infty} \frac{(-1)^n}{n^2}=0$.

Rearranging gives:

$\displaystyle 1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+...=\frac{\pi^2}{12}$

RonL

3. Originally Posted by cherry106
#2 Establish the result in the problem below, where m and n are positive integers. (Hint: sinA sinB = (1/2) [cos(A-B) - cos(A+B)].)
L
∫ sin((n pi x)/L) sin((m pi x)/L) dx = {0, m≠n; L, m=n
-L
$\displaystyle \int_L^L \sin(n \pi x/L) \sin(m \pi x/L) dx=$$\displaystyle \frac{1}{2} \int_L^L [\cos((n-m) \pi x/L) -\cos((n+m) \pi x/L)]dx$

Now if $\displaystyle n \ne m$ the integral on the RHS is over an integer number of cycles of each of the cosines
which sum to the integrand and so is zero.

When $\displaystyle n=m$ the integand on the RHS is one plus a cosine whose period
is a sub-multiple of the interval of integration and so the integral is $\displaystyle L/2$ and when
this is multipled by two gives the intregral on the LHS is $\displaystyle L$

RonL

4. Show that,
$\displaystyle 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+...=\frac{ \pi^2}{8}$