# Complex power series: radius of convergence and derivative

Let $z\in\mathbb{C}$ and define the power series by: $$\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+2}}{(2n+1)!}$$

Notice that the sum starts at $n=0$ so the first term is $z^2$

a) show that the series has a radius of convergence $R = \infty$.

Now let $$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)!} \ , \ x\in\mathbb{R}$$

b) Find the power series of the functions $f'(x)$ and $f''(x)$ and show that $f''(x)+f(x) = 2\cos(x) \ \forall x\in\mathbb{R}$

a) I know that I can find the radius of convergence by using $R = \lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}$. However, when the series on the form $\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+2}}{(2n+1)!}$ do I have to perform a index shift of $z^{2n+2}$ or is there another way to calculate the radius of convergence?

b) \begin{align}f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^nx^{2n+2}}{(2n+1)!}\right) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\cdot(2n+2)x^{2n+1}\\=\sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)x^{2n+1}}{(2n+1)!}\right)\end{align}

\begin{align}f''(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)x^{2n+1}}{(2n+1)!}\right) &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\cdot(2n+2)\cdot (2n+1)x^{2n}\\&= \sum_{n=0}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n(2n+2)(2n+1)x^{2n}}{(2n+1)!}\right) \end{align}

How am I suppose to conclude that $f''(x) + f(x) = 2\cos(x)$?

• Perhaps it may help to note that the numerator can also be written as $z^2(-z^2)^n$ – Rhys Hughes Jun 5 '18 at 11:21
• I am not entirely sure how it helps to rewrite the numerator when it is on the form $z^{2n+2}$ wouldn't it be possible to do a index shift of so that we get it on the form $w^n$, which makes it possible to equation given for R above? – Simbörg Jun 5 '18 at 14:35
• Tbh I was rushed so just wrote out what I spotted regardless of whether it was any use – Rhys Hughes Jun 5 '18 at 16:26

$$\sum_{n=0}^\infty\frac{(-1)^nw^{n+1}}{(2n+1)!}$$ Then, $a_n=\dfrac{(-1)^n}{(2n+1)!}$, and $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=0,$ and hence the about power series, for all $w$, and so does is we replace $w$ by $z^2$.
• I understand the index shift, but I am unsure about how you got your limit, since I get: $\lim_{n\to\infty}\left|\frac{a_{n}}{a_{n+1}} \right| = \lim_{n\to\infty}\left| \frac{\frac{(-1)^n}{(2n+1)!}}{\frac{(-1)^{n+1}}{(2n+2)!}} \right| = \lim_{n\to\infty}\left|\frac{(-1)^n\cdot (2n+2)!}{(-1)^{n+1} \cdot (2n+1)!} \right| = \lim_{n\to\infty} \left| -2n-2 \right| = \infty$ – Simbörg Jun 5 '18 at 15:01
• And because the $\lim_{n\to\infty}$ for $w$ we have that $R_w = \infty$ which implies that $\sqrt(R_w)=R_z = \infty$? – Simbörg Jun 5 '18 at 19:24
• In a sense what you say is correct. My argument is the following: If the series converges for some $x_0\ne 0$, then its radius of convergence $R$ satisfies $R\ge |x_0|$. Hence, is the p.s. converges for all $x_0$, then $R=\infty$. – Yiorgos S. Smyrlis Jun 5 '18 at 19:34