Context/background:
I am self-studying series, first in the context of generating functions and now in the context of functional/differential equations. As such, I like to set myself practise problems, which is what I bring you today. Of course, being self-studying, it can be hard to find anyone to give feedback on my problems and solutions, which is why I bring this one to you today. Thanks for your time.
Problem statement:
Given the differential equation $xf(x)-f'(x)=0$ and $f(0)=1$, find $f(x)$ in power series form. Bonus questions: find the closed form of $f(x)$ (if it exists) and the radius of convergence of the power series.
Solution:
Let $f(x)=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$
Then: $xf(x)=\sum_{n=0}^{\infty}a_nx^{n+1}=a_0x+a_1x^2+a_2x^3+a_3x^4+...$
And: $f'(x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+...$
Furthermore: $xf(x)-f'(x)=-a_1+(a_0-2a_2)x+(a_1-3a_3)x^2+(a_2-4a_4)x^3+(a_3-5a_5)x^4+...$
Given that $f(0)=1$, we must have that $a_0=1$. Moreover, since the coefficients of $xf(x)-f'(x)$ must all be zero, we have the following (infinite) system of equations:
$$-a_1=0\Longrightarrow a_1=0$$ $$a_0-2a_2=0\Longrightarrow a_2=\frac{a_0}{2}$$ $$a_1-3a_3=0\Longrightarrow a_3=0$$ $$a_2-4a_4=0\Longrightarrow a_4=\frac{a_2}{4}=\frac{a_0}{8}$$ $$a_3-5a_5=0\Longrightarrow a_5=0$$ $$a_4-6a_6=0\Longrightarrow a_6=\frac{a_4}{6}=\frac{a_0}{48}$$
The pattern continues, with every odd coefficient equal to $0$ and every odd coefficient equal to $a_0$ divided by the product of every even number up to and including $n$. As such, it makes sense to define a new sequence of coefficients $c_n$ such that $c_n=a_{2n}$; then $c_n=\frac{a_0}{(2n)!!}=\frac{1}{(2n)!!}$ and the power series solution is $f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!!}=1+\frac{x^2}{2}+\frac{x^4}{8}+\frac{x^6}{48}+...$
Radius of convergence
Consider $\lim_{n\to\infty}|\frac{c_{n+1}}{c_n}|$. This is equal to $\lim_{n\to\infty}|\frac{\frac{1}{(2n+2)!!}}{\frac{1}{(2n)!!}}|=\lim_{n\to\infty}|\frac{(2n)!!}{(2n+2)!!}|=\lim_{n\to\infty}|\frac{2n(2n-2)(2n-4)...2}{(2n+2)(2n)(2n-2)(2n-4)...2}|=\lim_{n\to\infty}|\frac{1}{2n+2}|=0$
Since the limit of the ratio of consecutive coefficients is $0$, the radius of convergence for this series is infinite; i.e. it converges for all $x$.
Closed form solution
I'll admit, this one stumped me for a bit. Then I realised that $(2n)!!=2^nn!$, so we have $\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!!}=\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}$. Setting $t=\frac{x^2}{2}$, we have $\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}=\sum_{n=0}^{\infty}\frac{t^n}{n!}$. This last series has the very well-known closed form of $e^t$, so the closed form for our series solution (in $x$) is $f(x)=e^{\frac{x^2}{2}}$. To check said solution: $xf(x)=xe^{\frac{x^2}{2}}$, and $f'(x)=xe^{\frac{x^2}{2}}$, so $xf(x)-f'(x)=xe^{\frac{x^2}{2}}-xe^{\frac{x^2}{2}}=0$.
My question:
As I mentioned earlier, I'm a self-studier, so it can be hard to find (qualified) feedback. So if there's any you could offer me in regards to my solution — in terms of rigour, flow, leaps of logic, whatever you can think of — I'd really appreciate it. I want to improve!