Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\frac{6}{n^2}\cos\frac{n\pi}{2})cos(\frac{nt\pi}{3}) $

Question

I am unsure how to do this, any help would be greatly appreciated.

$G(t) = \sum_\limits{n=1}^{\infty} (\frac{6}{n^2}\cos\frac{n\pi}{2})cos(\frac{nt\pi}{3}) $

Note, I know that for n(even) we see that $\cos\frac{n\pi}{2}$ is $(-1)^n$ but I am unsure how to fit that into the summation.

(This is more a notation and algebraic manipulation question).

Answer 1

The series is absolutly convergent.

\begin{align} G(t)&=\sum_{k=1}^{+\infty}\frac{6}{4k^2}\cos\left(\frac{2k\pi}{2}\right)\cos\left( \frac{4kt\pi}{3} \right)\\&\quad+ \sum_{k=0}^{+\infty}\frac{6}{(2k+1)^2}\cos\left(\frac{(2k+1)\pi}{2}\right)\cos\left(\frac{2(2k+1)t\pi}{3} \right)\\ &=\frac{3}{2}\sum_{k=1}^{+\infty}\frac{(-1)^k}{k^2}\cos\left(\frac{4k\pi t}{3}\right) \end{align} since

$$\cos(k\pi)=(-1)^k\;,\;\cos\left(k\pi+\frac{\pi}{2}\right)=0.$$

Answer 2

Using Octave to plot the first 10,000 terms, I guessed that $G$ is the Fourier series of a function of the form $g(t) = at^2+b$ over the interval $[-{3\over 2}, {3 \over 2}]$.

To compute $a,b$, I do the following: Since $\int_0^{\pi \over 3} G(t)dt = 0$, we get $12a+9b=0$, and since $b=G(0) = 6 \sum_{n=1}^\infty (-1)^n {1 \over n^2} = - {\pi^2 \over 2}$, we get $g(t) = {\pi^2 \over 6} (4x^2-3)$.

Computing the Fourier series shows that $G$ has the form given in the question.

Since $g$ is Lipschitz for all $t$, we see that $G(t) = {\pi^2 \over 6} (4t^2-3)$ for $|t| \le {3 \over2}$.

Hence $\sum_\limits{n=1}^{\infty} (\frac{6}{n^2}\cos\frac{n\pi}{2})cos(\frac{nt\pi}{3}) = {\pi^2 \over 6} (4t^2-3)$ for $|t| \le {3 \over2}$, and is $3$-periodic.