Proof that $f$ is differentiable

Let $$f(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n} \cos{nx}$$ Proof that $$f$$ is differentiable on $$(-2,2)$$

my approach

let $$m := \frac{x}{2}$$ so $$m<1$$ $$\left| \frac{x^n}{2^n} \cos{nx} \right| \le m^n \rightarrow 0$$ So by comparison we have point convergence. But I don't know how to deal with diffentiable..

• It's not enough that $m^n \to 0$, you need convergence of $\sum\limits_{n=1}^\infty m^n$ to use comparison. – mihaild May 21 at 15:19

Series of derivatives, $$\sum\limits_{n=1}^\infty \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$$ converges uniformly on any subsegment of $$(-2, 2)$$ (per comparison again) and series itself converges. Then the series converges to differentiable function (and it's derivative is sum of series of derivatives) - see, for example, theorem 7.17 of Rudin, "Principles of Mathematical Analysis".
To prove the series is uniformly convergent on $$(-2 + \alpha, 2 - \alpha)$$, we can note that $$|n x^{n - 1} \cos nx - nx^n \sin nx| \leqslant 2 \cdot n \cdot (2 - \alpha)^n$$, and it is less then $$(2 - \frac{\alpha}{2})^n$$ for large enough $$n$$. And the series $$\sum_{n=1}^\infty \frac{(2 - \frac{\alpha}{2})^n}{2^n} = \sum\limits_{n=1}^\infty \left(1 - \frac{\alpha}{4}\right)^n$$ converges.
• how you got $\sum\limits_{n=1}^\infty -\frac{n^2 x^{n - 1}}{2^n}\sin nx$? – trolley May 21 at 15:34
• By differentiating each term we get $2^{-n} n x^{n-1} \cos (n x)-2^{-n} n x^n \sin (n x) \neq \frac{n^2 x^{n - 1}}{2^n}\sin nx$ – trolley May 21 at 15:37
• how can I prove uniform convergence of that $\sum \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$ ? $\frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n} = \frac{nx^{n-1}}{2^n}(\cos{nx} - x \sin {nx}) < \frac{3n}{2^n}\cdot 2^{n-1} = \frac{3}{2}n$ – trolley May 21 at 15:57