Differentiation of Fourier series for $f(x)$

Question

Let $$f(x)=\sum_{n=1}^{\infty}\left( \dfrac{\cos nx}{n^3} + (-1)^n \dfrac{\sin nx}{n+1} \right)$$ Show that $f(x)$ is differentiable on $(-\pi,\pi)$ and find the Fourier series for $f^{'}(x)$

Answer 1

All the Fourier series below are for $-\pi <x<\pi$ as required.

Write $f(x)=f_1(x)+f_2(x), f_1(x)=\sum_{n \ge 1} \frac {\cos nx}{n^3}, f_2(x)=\sum_{n \ge 1}(-1)^n \frac{\sin nx}{n+1}$ which is legitimate since the series in $f_1$ is absolutely convergent.

Noting that $\sum_{n \ge 1} \frac {\cos nx}{n^3}$ can be differentiated term by term since the derivative series is absolutely convergent,

we get $f_1'(x)=-\sum_{n \ge 1} \frac {\sin nx}{n^2}$

Writing $\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n^2}+\frac{1}{n^2(n+1)}$, we get (again because of the absolute convergence of the last $2$ series):

$f_2(x)=\sum_{n \ge 1}(-1)^n \frac{\sin nx}{n}+\sum_{n \ge 1}(-1)^{n+1} \frac{\sin nx}{n^2}+\sum_{n \ge 1}(-1)^n \frac{\sin nx}{n^2(n+1)}=$

$=-\frac{x}{2}+\int_0^x\log (2\cos(y/2))dy+f_3(x)$, where $f_3(x)=\sum_{n \ge 1}(-1)^n \frac{\sin nx}{n^2(n+1)}$

As in step one above $f_3$ is differentiable term by term while the fundamental theorem of calculus applies to $\int_0^x\log (2\cos(y/2))dy$ since $\log (2\cos{y/2})$ is continuos and integrable on $(-\pi,\pi)$, so $f_2$ is differentiable and:

$f_2'(x)=-1/2+\log (2\cos(x/2))+\sum_{n \ge 1}(-1)^n \frac{\cos nx}{n(n+1)}=$

$=-1/2+\sum_{n \ge 1}(-1)^{n+1} \frac{\cos nx}{n}+\sum_{n \ge 1}(-1)^n \frac{\cos nx}{n(n+1)}=1/2-\sum_{n \ge 1}(-1)^{n} \frac{\cos nx}{n+1}$

Putting all together we get that $f$ is differentiable and

$f'(x)=-1/2-\sum_{n \ge 1}(-1)^{n} \frac{\cos nx}{n+1}-\sum_{n \ge 1} \frac {\sin nx}{n^2}$