Show power series converges for every $x$.

Question

Let $$f(x) = 1 + a_{1}x^{1}+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+...$$ be a solution of the differential equation $f'(x)=xf(x).$ Now I need to explain that the power series that define $f(x)$ converges for every $x$.

This problem is introduced by two smaller proofs which I will give (or my attempts for them) so the track of thought is clear.

First I found a simple expression for $\frac{a_{2n}}{a{2n-2}}$ for every $n \in \mathbb{N}$. Since this fraction is $\frac{1}{2}$ when $n=1$, it is $\frac{1}{4}$ when $n=2$, it is $\frac{1}{6}$ when $n=3$ and so on, I let

$$\frac{a_{2n}}{a_{2n-2}}=\frac{1}{2n}.$$

Next I had to show that for every fixed $x \in \mathbb{R}$ there exists a $N \in \mathbb{B}$ such that $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ whenever $n \geq N$. This is my attempt:

Let $x \in \mathbb{R}$ be given and let $N > x^{2}$, then for all $n \geq N : \frac{1}{n} < \frac{1}{x^{2}}.$ Then

$$\frac{|a_{2n}|}{|a_{2n-2}|}=\frac{1}{2} \frac{1}{|n|} < \frac{1}{2} \frac{1}{|x^{2}|}=\frac{1}{2} \left| \frac{x^{2n-2}}{x^{2n}} \right|\\ \Longrightarrow\\ |a_{2n}||x^{2n}| < \frac{1}{2} |x^{2n-2}||a_{2n-2}|. $$

I expect to have to use $|a_{2n}x^{2n}| \leq \frac{1}{2} |a_{2n-2}x^{2n-2}|$ to show the convergence, but I have no idea how. Any suggestions?

Answer 1

We can write the differential equation as $$ \sum_{n\ge1}na_{n}x^{n-1}=\sum_{n\ge0}a_nx^{n+1} $$ Changing indices, we obtain $$ a_1+\sum_{n\ge1}(n+1)a_{n+1}x^{n}=\sum_{n\ge1}a_{n-1}x^n $$ so that $a_1=0$ and $$ (n+1)a_{n+1}=a_{n-1} $$ Therefore all odd terms are zero, and $(n+2)a_{n+2}=a_n$ for even $n$, starting with $a_0=1$. The first terms are $$ a_2=\frac{1}{2},\qquad a_4=\frac{1}{4}\frac{1}{2},\qquad a_6=\frac{1}{6}\frac{1}{4}\frac{1}{2} $$ and we can conjecture that $$ a_{2n}=\frac{1}{2^nn!} $$ Indeed, $(2n+2)a_{2n+2}=a_{2n}$ and so $$ a_{2n+2}=\frac{1}{2n+2}\frac{1}{2^nn!}=\frac{1}{2^{n+1}(n+1)!} $$ Using the ratio test, we have $$ \left|\frac{a_{2n+2}x^{2n+2}}{a_{2n}x^{2n}}\right|=\frac{2^nn!}{2^{n+1}(n+1)!}|x^2|= \frac{x^2}{2(n+1)} $$ whose limit is $0$. Thus the series converges everywhere.

By the way, $f(x)=e^{x^2/2}$.

Answer 2

With $a_{2n}= \frac{1}{2}a_{2n-2}$ you can write your series as the sum of two geometric series. First take n even. Define $b_n= a_{2n}$. Then $b_{n}= \frac{1}{2}b_{n-1}$ so this is a geometric series: $b_0= a_0$, $b_1= a_2= \frac{1}{2}a_0$, $b_2= a_4= \frac{1}{2}a_2= \left(\frac{1}{2}\right)^2a_0$, etc.

And similarly for n odd. Let $c_n= a_{2n+1}$ so that $c_0= a_1$. $c_1= a_3= \frac{1}{2}a_1$, $c_2= a_5= \frac{1}{2}a_3= \left(\frac{1}{2}\right)^2a_1$ etc.