How Can Fourier Series be Used to Expand a Continuous Function of Period 2L?

[blueyellow]

please help on this question
Any continuous function of period 2L can be expanded as a Fourier series

f(x)=a0/2+∑(from n=1 to∞) (ancos(n pi x/L)+bnsin(n pi x/L))


Using ∫(from -L to +L) sin(m pi x/L)sin(n pi x/L)dx=L kronecker delta m n

Show that
Bn=1/L∫(from -L to+L) sin(n pi x/L)f(x) dx



i am seriously stuck on this - kinda can't stand proof questions
thanks in advance

 

[accatagliato]

you must multiply f(x) by (1/L)sin(m pi x/L) and then integrate from -L to L in order to obtein:
<br /> \frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})f(x)dx=\frac{a_{0}}{2L}\int_{-L}^{L}sin(\frac{m\pi x}{L})dx+\sum a_{n} \frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})cos(\frac{n\pi x}{L})dx+\sum b_{n} \frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})sin(\frac{n\pi x}{L})dx<br />

then just apply the previous property that you mentioned and the fact that sin and cos are ortogonal function:

<br /> \int_{-L}^{L}sin(\frac{m\pi x}{L})cos(\frac{n\pi x}{L})dx=0<br />

 

[blueyellow]

but how do i get rid of th a0 and summation signs?
i tried what accatagliato said but 1)i couldn't get rid of the a0, and 2) i ended up with sin (n pi x/L) on a denominator

 

[accatagliato]

<br /> \int_{-L}^{L}sin(\frac{m\pi x}{L})dx=0<br />
because the sin function is an odd function and the interval of integration is symmetric. In fact:

<br /> \int_{-L}^{L}sin(\frac{m\pi x}{L})dx=\int_{-L}^{0}sin(\frac{m\pi x}{L})dx+\int_{0}^{L}sin(\frac{m\pi x}{L})dx<br />
changing x --> -x in the second integral

<br /> \int_{-L}^{0}sin(\frac{m\pi x}{L})dx+\int_{0}^{-L}sin(\frac{m\pi x}{L})dx=0<br />

For the summation:

<br /> \sum_{n} b_{n}\delta_{nm}=b_{m}<br />

 

[blueyellow]

thanks
im so sorry but what am i supposed to now do with b m (b subscript m)?
if i substitute the last equation in, the b n disappears, so then i can't actually do what they ask me to do in the question cos in the question they ask me to show that b n equals something

 

[accatagliato]

after substitution you obteined:

<br /> b_{m}=\frac{1}{L}\int_{-L}^{L}sin(\frac{m\pi x}{L})f(x)d<br />

nothing changes if there is n or m in the final result, you can call the index however you want, so this result transforms into:


<br /> b_{n}=\frac{1}{L}\int_{-L}^{L}sin(\frac{n\pi x}{L})f(x)d<br />