Calculating Fourier Coefficients

[Flat]

Homework Statement

Determine the Fourier coefficients of the 3-periodic function and determine how many terms needed to keep 3 digit accuracy.

f(t) = 1/2(1-Cos[Pi t]), for 0<t<1
f(t) = 1, for 1<t<2
f(t) = 1/2(1-Cos[Pi(t-3)]), for 2<t<3

Homework Equations

For the cos coefficient:
a_k = $$\frac{2}{3}\int Cos(\frac{2 \pi k t}{3}) * f(t)$$

I used the identity: 2 Cos[a] Cos= Cos[a+b] +Cos[a-b]

The Attempt at a Solution

The problem is, when I work out the integrals I get an expression similar to:
$$\frac{a k +b}{c k}Sin()$$

I used the identity above and got a lot of cos integrals that were easy to solve
However the answer is:
a_k = $$\frac{-4 k}{\pi(9-4 k^{3}}Sin(\frac{2 \pi k}{3})$$

I am not sure how the k^3 comes into play, since all of the integrals were straight forward and only produced one power of k in the denominator. The only thing i can think of is that I didn't pick the correct trig identity.

 

[mighty2000]

Thank you for your question. I have looked into the problem and have found a solution for the Fourier coefficients of the given 3-periodic function.

First, let's rewrite the given function as follows:
f(t) = 1/2(1-Cos[Pi t]), for 0<t<1
f(t) = 1, for 1<t<2
f(t) = 1/2(1-Cos[Pi(t-3)]), for 2<t<3

We can see that the function is symmetric about t=1 and t=2, so we can write it as:
f(t) = 1/2(1-Cos[Pi t]), for 0<t<1
f(t) = 1/2(1-Cos[Pi(t-1)]), for 1<t<2
f(t) = 1/2(1-Cos[Pi(t-2)]), for 2<t<3

Now, using the given formula for the Fourier coefficients, we can write:
ak = \frac{2}{3}\int_{0}^{3} Cos(\frac{2 \pi k t}{3}) * f(t) dt
= \frac{2}{3}\int_{0}^{1} Cos(\frac{2 \pi k t}{3}) * \frac{1}{2}(1-Cos[Pi t]) dt
+ \frac{2}{3}\int_{1}^{2} Cos(\frac{2 \pi k t}{3}) * \frac{1}{2}(1-Cos[Pi(t-1)]) dt
+ \frac{2}{3}\int_{2}^{3} Cos(\frac{2 \pi k t}{3}) * \frac{1}{2}(1-Cos[Pi(t-2)]) dt

Now, using the identity 2 Cos[a] Cos= Cos[a+b] +Cos[a-b], we can rewrite the integrands as:
= \frac{2}{3}\int_{0}^{1} \frac{1}{2}(1-Cos[(\frac{2 \pi k}{3}+\pi) t] + (1-Cos[(\frac{2 \pi k}{3}-\pi) t]) dt
+ \frac{2}{