Math Help - Fourier series help (odd/even)

1. Fourier series help (odd/even)

Q1, 2: Options are Odd or Even extensions.

My attempt.
1. Odd extension
2. Even extension (guessing, just looking at graph)

2. Re: Fourier series help (odd/even)

I think you have it wrong. With (A), f(-t)=f(t), so it is an even extension, and with (B), f(-t)=-f(t), so it is an odd extension.

- Hollywood

3. Re: Fourier series help (odd/even)

Originally Posted by hollywood
I think you have it wrong. With (A), f(-t)=f(t), so it is an even extension, and with (B), f(-t)=-f(t), so it is an odd extension.

- Hollywood
1. Even
2. odd
3. 2/n
4. pi
5. 4/((pi)n^2) or 0

4. Re: Fourier series help (odd/even)

I'm pretty sure #4 should be the average value of the function, which is $\frac{pi}{2}$. How did you calculate #3 and #5?

- Hollywood

5. Re: Fourier series help (odd/even)

Originally Posted by hollywood
I'm pretty sure #4 should be the average value of the function, which is $\frac{pi}{2}$. How did you calculate #3 and #5?

- Hollywood
3) b(n) = (2/π) ∫(t = 0 to π) (π - t) sin(nπt/π) dt
............= (2/π) [(π - t) * -cos(nt)/n - (-1)(-sin(nt)/n^2)] {for t = 0 to π}
............= 2/n.

4) a(0) = (2/π) ∫(t = 0 to π) (π - t) dt
............= (2/π) [(πt - (1/2)t^2) {for t = 0 to π}
............= π.

5 )For n > 0:
a(n) = (2/π) ∫(t = 0 to π) (π - t) cos(nπt/π) dt
.......= (2/π) [(π - t) * sin(nt)/n - (-1)(-cos(nt)/n^2)] {for t = 0 to π}
.......= (2/π) * -((-1)^n - 1)/n^2
.......= 0 if n is even and 4/(πn^2) if n is odd.