Deduce Fourier Series of $t\cos(t)$ from the $f(t)=t$ Fourier series.

Question

I'm stuck with an exercise of Fourier series where I'm asked to deduce a Fourier series from a given one.

I have to deduce that:

$$t\cos(t)= -\frac{1}{2}\sin(t)+2\sum_{n\geq 2} \frac{(-1)^n}{n^2-1} \sin(nt), \qquad -\pi<t<\pi $$

from the Fourier series of $f(t)=t$ that I had calculated in an exercise before.

$$\sum_{n>=1} \frac{(-1)^{n-1}}{n} \sin(nt)$$

The approach I've been trying was to calculate a Fourier series for $\cos(t)$ but I'm getting coefficients $ a_n=0 $ $ b_n=0 $. My idea was to calculate that Fourier series and then multiply it with the $f(t)=t$ Fourier series and simplify to try to get the result I need.

How should I calculate it?

Answer 1

Taking the derivarive of $t\cos(t)$ gives you $\cos(t)-t\sin(t)$, and the second derivative gives you $-2\sin(t)-t\cos(t)$ so calling $g(t)=t\cos(t)$ you have that: $$g''(t)=-2\sin(t)-g(t)$$ I think that if you substitute $g(t)$ by its generic fourier series in that differential equation, you can obtain the fourier coefficients.

Answer 2

Hint: It is useful to recall/derive the following product-to-sum identity:

\begin{align} 2\sin A\cos B &=\frac{1}{2i}(e^{i A}-e^{-i A})(e^{i B}+e^{-i B})\\ &=\frac{1}{2i}(e^{i (A+B)}-e^{-i (A+B)}+e^{i (A-B)}-e^{-i (A-B)})\\ &=\sin(A+B)+\sin(A-B). \end{align} What does this tell you about $\sin(nt)\cos(t)$?