Prove $f(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n} \cos{nx}$ is differentiable at $(-2, 2)$.

Question

Prove $$f(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n} \cos{nx}$$ is differentiable at $(-2, 2)$.

I can't use formula for radius of convergence, because it's not a power series ($x$ is present also in $\cos$).

Answer 1

I don't want to spoil all of the fun, so here are the general steps that you typically want to follow for a problem such as this.

First, as stated in the other answer, restrict to $[-2+\epsilon, 2-\epsilon]$ for arbitrary $\epsilon>0$ small. Define $f_n(x)=\frac{x^n}{2^n}\cos nx.$ Show that the sum of the sequence $(f_n')$ converges uniformly, using e.g. the Weierstrass M-test (you'll see when doing this why we cannot go up to the boundary of $(-2,2)$). Then, if you can show the original series converges at a single point, you'll have that your sum converges uniformly on $[-2+\epsilon, 2-\epsilon]$ to a differentiable function, and the derivative commutes with the summation.

Answer 2

HINT: Try to prove uniform convergence on $[-2+\alpha,2-\alpha]$ for any $0<\alpha<2$

Answer 3

Hint

$f(x)$ is the real part of

$$-1+\sum_{r=0}^\infty\left(\dfrac{xe^{ix}}2\right)^r$$