Showing that the function $F(x)=\sum_{n=1}^\infty e^{-nx}\cos n\pi x$ is differentiable on the interval $[0,\infty)$

Question

I'm trying to solve this problem from an old exam in real analysis. Thus, only such methods may be used. The problem is as follows.

Show that the function $F(x)=\sum_{n=1}^\infty e^{-nx}\cos n\pi x$ is differentiable on the interval $[0,\infty)$. Thereafter calculate the exact numerical value of $F'(1)$.

These are my computations this far. I'm stuck with not being able to show that this limit exists as I can't complete the computations.

That a function is differentiable on an interval means that the limit \begin{align*} f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} \end{align*} exists every point $a$ on that interval. Here, we need to show that this limit exists for all points $x$ in $[0,\infty)$ for $F(x)$ to show that it's differentiable there. Moreover \begin{align*} F'(x)&=\lim_{h\to0}\frac{F(x+h)-F(x)}{h}=\lim_{h\to0}\frac{\sum_{n=1}^\infty e^{-n(x+h)}\cos n\pi(x+h)-\sum_{n=1}^\infty e^{-nx}\cos n\pi x}{h}\\ &=\lim_{h\to0}\frac{\sum_{n=1}^\infty(e^{-n(x+h)}\cos n\pi(x+h)-e^{-nx}\cos n\pi x)}{h}\\ &=\lim_{h\to0}\frac{1}{h}\sum_{n=1}^\infty(\frac{1}{e^{n(x+h)}}\cos n\pi(x+h)-\frac{1}{e^{nx}}\cos n\pi x)\\ &=\lim_{h\to0}\frac{1}{h}\sum_{n=1}^\infty(\frac{1}{e^{n(x+h)}}\cos n\pi(x+h)-\frac{e^{nh}}{e^{n(x+h)}}\cos n\pi x)\\ &=\lim_{h\to0}\frac{1}{h}\sum_{n=1}^\infty\frac{\cos n\pi(x+h)-e^{nh}\cos n\pi x}{e^{n(x+h)}}\\ &=\lim_{h\to0}\frac{1}{he^{n(x+h)}}\sum_{n=1}^\infty(\cos n\pi(x+h)-e^{nh}\cos n\pi x). \end{align*} Both the numerator and the denominator are differentiable and approach zero as $h$ does. Consequently, we can use L'Hôpital's rule. Here \begin{align*} &\lim_{h\to0}\frac{1}{he^{n(x+h)}}\sum_{n=1}^\infty(\cos n\pi(x+h)-e^{nh}\cos n\pi x)=\\ =&\lim_{h\to0}\frac{-n}{he^{n(x+h)}}\sum_{n=1}^\infty(-n\pi\sin n\pi(x+h)+n\pi e^{nh}\sin n\pi x) \end{align*}

I'm not even sure that I'm using the right method. Any help would be greatly appreciated.

Answer 1

If we define \begin{align*} F(x)=\int_a^xf(t)dt \end{align*} then \begin{align*} F'(x)=\frac{d}{dx}\int_a^xf(t)dt=\frac{d}{dx}(F(x)-F(a)). \end{align*} If we differentiate the series termwise and can show that $f_n'$ is uniformly convergent, then \begin{align*} \sum_{n=0}^\infty\int_a^xf_n'(t)dt=\sum_{n=0}^\infty f_n(x)-\sum_{n=0}^\infty f_n(a). \end{align*} Using this together with that uniformly convergent sequences of functions fulfill \begin{align*} \int_a^b\sum_nf_n=\sum_n\int_a^bf_n \end{align*} we obtain \begin{align*} \int_a^x\sum_nf_n'(t)dt=\sum_n\int_a^xf_n'(t)dt=\sum_nf_n(x)-\sum_nf_n(a) \end{align*} which means that \begin{align*} F'(x)=\frac{d}{dx}(\int_a^x\sum_nf_n'(t)dt)=\sum_nf_n'(x)=\frac{d}{dx}(\sum_nf_n(x)-\sum_nf_n(a))=\frac{d}{dx}\sum_nf_n(x) \end{align*} where we set $t=x$ in the second equality.

All of this just means that if we can show that the sequence of functions converges uniformly, then the termwise differentiation equals the derivative of the series, which means that it's differentiable.

Here \begin{align*} f(x)=\lim_{n\to\infty}f_n(x)&=\lim_{n\to\infty}e^{-nx}\cos n\pi x\leq\\ &\leq\lim_{n\to\infty}|e^{-nx}\cos n\pi x|\leq\\ &\leq\lim_{n\to\infty}|e^{-nx}||\cos n\pi x|\leq\\ &\leq\lim_{n\to\infty}|e^{-nx}|=0 \end{align*} as $x\in(0,\infty)$. Now \begin{align*} |f_n(x)-f(x)|=|f_n(x)|&=|e^{-nx}\cos n\pi x|\leq\\ &\leq|e^{-nx}||\cos n\pi x|\leq\\ &\leq|e^{-nx}|. \end{align*} This means that for all $\epsilon>0$ there exists an $N\in\Bbb{N}$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n\geq N$ and all $x\in(0,\infty)$. Thus, $f(x)$ is uniformly convergent, and the above reasoning may be used.

Answer 2

Summing the geometric series $\sum_{n=1}^\infty e^{(-1+i\pi)nx}$ and simplifying shows that $$F(x)={e^x \cos(\pi x)-1\over e^{2x}-2e^x\cos(\pi x)+1}\qquad(x> 0)\ .$$ Since the given series is divergent at $x=0$ it follows that the function $F$ is defined and smooth for $0<x<\infty$.