Finding the Fourier Series of f(x)

[bon]

Homework Statement

Trying to find the Fourier series for the function

f(x) = 0 for -pi<x<0 and f(x) = sinx for 0<x<pi

The Attempt at a Solution

im having a little trouble working it out..

are any of the sets of coefficients = 0?

Im getting two non-zero integrals for the coefficients an and bn which are difficult to work out :(

 

[micromass]

You need to calculate

$$a_0= \frac{1}{2\pi} \int_0^\pi{sin(x) dx}$$
$$a_n= \frac{1}{\pi} \int_0^\pi{sin(x)cos(nx) dx}$$
$$b_n= \frac{1}{\pi} \int_0^\pi{sin(x)sin(nx) dx}$$

These integrals are not that diffucult are they? I mean $$a_0$$ is easy. And for $$a_n,b_n$$ just apply the product-to-sum formulas (or Simpson's formula, whatever you call it).

 

[bon]

ahh that comes out all horrible though :(

the an integral is like cos pin + 1 / pi - pi n^2

 

[micromass]

Yeah, the result is not very beautiful, I know ):

 

[bon]

is bn = 0 since n is integer?

 

[micromass]

I don't think b1 is 0. The other bn probably are 0.

 

[bon]

but bn = sinpi n / pi - pi n^2 so when n=1 we have sin pi on the top which = 0

 

[bon]

sorry - ignore..

 

[bon]

ok so i think i have it now..but when i graph it, it doesn't seem to fit that well...

trying the first few terms, f(x) = 1/pi -2/3pi cos2x -2/15pi cos4x + ... + 1/2sinx

For one thing, it doesn't go through (0,0)! why?

 

[micromass]

That it doesn't go through (0,0) is expect. The value at 0 is a series and as such is not expected to go through (0,0).

As for the convergence problems. All I can say is that the convergence is probably slow... I do think you have the right solution however...

 

[bon]

thanks ok

my next question asks me to expand f(x) = xsinx 0<x<pi as a Fourier sine series

the hint is that i should extend the interval to -pi<x<pi and then require that f(-x) = -f(x) i.e. f(x) is odd.. but i don't get it...f(x) isn't odd.. :S

thanks

 

[bon]

oh i think i get it..

i say that f(x) = -xsinx for -pi to 0 and xsinx for 0 to pi?

Then can i just say all an = 0 as it is odd? Do i need to be careful about case n=1? so in general is it always true that if it is odd, an = 0 ? including n=1?

 

[micromass]

They just mean that u define f(x)=xsin(x) for 0<x<pi. And then extend this function to [-pi,pi] so that the function is odd. So define f(x)=-xsin(x) for -pi<x<0.

 

[micromass]

If a function is odd, then all an are indeed 0. So only the bn matter now.
If a function is even, then all bn are 0.

 

[bon]

Then can i just say all an = 0 as it is odd? Do i need to be careful about case n=1? so in general is it always true that if it is odd, an = 0 ? including n=1?

 

[bon]

Yay okay thanks

 

[bon]

am i right in thinking bn = 2/pi times the integral from 0 to pi of xsinx sin nx

that is a horrible integral..cant be right..?

 

[micromass]

Yes, it IS a horrible integral

Try first Simpsons formula to split up sin(x)sin(nx) in sums.
Then do partial integration.

It's even less beautiful then last time