# Fourier Series Problem

1. Nov 6, 2008

### tommyhakinen

1. The problem statement, all variables and given/known data
Given that the fourier series of function f1(t)

= At/π + A , -π < t < 0
= -At/π +A, 0 < t < π

is f(t) = 0.5A +4A/π2(cos t + 1/9 cos 3t + 1/25 cos 5t + ...)

what is the fourier series of function f2(t)
= -At/π, -π < t < 0
= At/π, 0 < t < π

3. The attempt at a solution
for a function f2(t) = f1(t + π) means the function f2(t) is the function of f1(t) shifted by a distance of π to the left. How do I link this with the above problem?

2. Nov 6, 2008

### gabbagabbahey

Hint: what is $$\frac{A(-t+\pi)}{\pi}$$?

3. Nov 6, 2008

### tommyhakinen

Thanks for the reply. The function f1 is a periodic function. The function f2 is a function f1 shifted by a distance π. However, which direction the shifting should be the right answer? I found out that if we shifted f1 either to the right or left by π distance, we still get the same f2. or both answer are acceptable?

my answer is : f2(t) = A/2 + 4A/π2 {cos(t+π) + 1/9 cos(3(t+π)) + 1/25 cos(5(t+π)) + ...}

please correct me if i am wrong. thank you.

4. Nov 6, 2008

### gabbagabbahey

Shouldn't it be $f_2(t)=f_1(-t+\pi)$ i.e. with a negative sign in front of t

5. Nov 6, 2008

### tommyhakinen

why is it negative t? can you elaborate more on this? from my observation, i can see that the function f1 is just shifting to the left or right by π. since f1 is a periodic function, it looks like sin wave on cosine wave which are π distance from each other.

6. Nov 7, 2008

### gabbagabbahey

well, look at the functions....for $0<t<\pi$,

$$f_1(-t+\pi)=\frac{-A(-t+\pi)}{\pi}+A=\frac{At}{\pi}=f_2(t)$$

The same relationship holds for $\pi<t<0$.