Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ of a power series defined as $$ f(x) := \sum_{n=0}^\infty \frac{(-1)^n2^{-n}x^{2n+1}}{n!}, \ x \in \mathbb{R} $$
Question:
Find the Power series for $f'(x)$ and $f''(x)$ and prove that $$ -f''(x) + x^2f(x) = 3f(x) \ \text{for all} \ x \in \mathbb{R} $$
In an earlier question I have argued for that $f \in C^\infty(\mathbb{R})$ as we have a Power Series of the form $\sum_{n=0}^\infty a_n(z-a)^n$ with $a = 0 \in \mathbb{R}$ and convergence radius $R = \infty > 0$ which means that $f \in C^\infty(\mathbb{R})$ in the interval $]-\infty, \infty[$. Thus we have that $$ f'(x) = \sum_{n=1}^\infty \frac{(2n+1)(-1)^n2^{-n}x^{2n}}{n!} \ \text{for} \ -\infty < x < \infty $$ and $$ f''(x) = \sum_{n=1}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} \ \text{for} \ -\infty < x < \infty $$ but then to show that $$ -f''(x) + x^2f(x) = 3f(x) \ \text{for all} \ x \in \mathbb{R} $$ I really have no idea how to. My TA told us that this question was pretty difficult and that we had to manipulate the sum and so on, but we have not really learned how to manipulate sums in general as this is not taught in my university (I think we are expected learn this on our own).
If I just plug in what we have I get \begin{align} -\sum_{n=1}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} + x^2 \sum_{n=0}^\infty \frac{(-1)^n2^{-n}x^{2n+1}}{n!} & = \\ -\sum_{n=1}^\infty \frac{(4n^2+2n)(-1)^n2^{-n}x^{2n-1}}{n!} +\sum_{n=0}^\infty \frac{(-1)^n2^{-n}x^{2n+3}}{n!} \end{align} but I really have no idea how to proceed from here. Can you help me?
From $\displaystyle f(x)=\sum_{n=0}^\infty\,\frac{(-1)^n}{2^nn!}\,x^{2n+1}$, we have $$f'(x)=\sum_{n=0}^\infty\,\frac{(-1)^n(2n+1)}{2^nn!}\,x^{2n}$$ and $$\begin{align}f''(x)&=\sum_{n=0}^\infty\,\frac{(-1)^n(2n+1)(2n)}{2^nn!}\,x^{2n-1}\\&=\sum_{n=1}^\infty\,\frac{(-1)^n(2n+1)}{2^{n-1}(n-1)!}\,x^{2n-1}\\&=\sum_{n=0}^\infty\,\frac{(-1)^{n-1}(2n+3)}{2^nn!}\,x^{2n+1}\,.\end{align}$$ Consequently, $$\begin{align}(x^2-3)\,f(x)&=x^2\,\sum_{n=0}^\infty\,\frac{(-1)^n}{2^nn!}\,x^{2n+1}-3\,\sum_{n=0}^\infty\,\frac{(-1)^n}{2^nn!}\,x^{2n+1} \\&=\sum_{n=1}^\infty\,\frac{(-1)^{n-1}}{2^{n-1}(n-1)!}\,x^{2n+1}-\sum_{n=0}^\infty\,\frac{3(-1)^n}{2^nn!}\,x^{2n+1}\\&=-3x+\sum_{n=1}^\infty\,\left(\frac{(-1)^{n-1}}{2^{n-1}(n-1)!}-\frac{3(-1)^n}{2^nn!\,}\right)\,x^{2n+1}\\ &=-3x+\sum_{n=1}^\infty\,\frac{(-1)^{n-1}(2n+3)}{2^nn!}\,x^{2n+1}\\&=\sum_{n=0}^\infty\,\frac{(-1)^{n-1}(2n+3)}{2^nn!}\,x^{2n+1}=f''(x)\,.\end{align}$$
Note that if the power series $\displaystyle g(x)=\sum_{n=0}^\infty\,b_n\,x^n$ satisfies $$g''(x)=(x^2-3)\,g(x)\,,\tag{#}$$ then $$\sum_{n=0}^\infty\,(n+1)(n+2)\,b_{n+2}\,x^n=-3b_0-3b_1x+\sum_{n=2}^\infty\,\big(b_{n-2}-3\,b_n\big)\,x^n\,.$$ Therefore, $b_2=-\dfrac{3}{2}\,b_0$, $b_3=-\dfrac12\,b_1$, and $$(n+1)(n+2)\,b_{n+2}+3\,b_n-b_{n-2}=0\tag{*}$$ for all integers $n\geq 2$.
Write $e_n:=n!\,b_{2n}$ and $o_n:=n!\,b_{2n+1}$ for $n=0,1,2,\ldots$. From (*), we get $$(2n+2)(2n+3)\,\frac{o_{n+1}}{(n+1)!}+3\,\frac{o_n}{n!}-\frac{o_{n-1}}{(n-1)!}=0\,,$$ or equivalently, $$2\,(2n+3)\,o_{n+1}+3\,o_n-n\,o_{n-1}=0\,,$$ for all $n=0,1,2,\ldots$, with $o_0=b_1$ and $o_1=-\dfrac12\,b_1$. By induction, $o_n=\left(-\dfrac12\right)^n\,b_1$, whence $$b_{2n+1}=n!\,o_n=\frac{(-1)^n}{2^nn!}\,b_1$$ for all $n=0,1,2,\ldots$.
On the other hand, we have $$(2n+1)(2n+2)\,\frac{e_{n+1}}{(n+1)!}+3\,\frac{e_n}{n!}-\frac{e_{n-1}}{(n-1)!}=0\,,$$ or $$2\,(2n+1)\,e_{n+1}+3\,e_n-n\,e_{n-1}=0$$ for $n=0,1,2,\ldots$, with $e_0=b_0$ and $e_1=-\dfrac{3}{2}\,b_0$. Unfortunately, there is no simple expression for $e_n$ as far as I know.
However, as Lutz Lehmann remarked, $f(x)=x\,\exp\left(-\dfrac{x^2}{2}\right)$. We can use this to find all solutions $g(x)$ to (#) by using the technique known as Reduction of Order.
Consider the differential equation $$u'(x)+\frac{2\,f'(x)}{f(x)}\,u(x)=0\,.$$ We see that $$u'(x)-\frac{2(x^2-1)}{x}\,u(x)=0\,.$$ Hence, $$\frac{\text{d}}{\text{d}x}\,\big(x^2\,\exp(-x^2)\,u(x)\big)=0\,.$$ Therefore, $$u(x)=-\dfrac{A}{x^2}\,\exp\left(x^2\right)$$ for some constant $A$. That is, $$\int\,u(x)\,\text{d}x=-A\,\left(\sqrt{\pi}\,\text{erfi}(x)-\frac{\exp(x^2)}{x}\right)+B$$ for some constant $B$, where $\text{erfi}$ is the imaginary error function: $$\text{erfi}(x):=\dfrac{2}{\sqrt{\pi}}\,\int_{-x}^{+x}\,\exp(t^2)\,\text{d}t\,.$$ Since $$g(x)=f(x)\,\int\,u(x)\,\text{d}x\,,$$ we get $$g(x)=A\,\tilde{f}(x)+B\,{f}(x)\,,$$ where $$\tilde{f}(x):=\exp\left(\frac{x^2}{2}\right)-\sqrt{\pi}\,x\,\text{erfi}(x)\,\exp\left(-\frac{x^2}{2}\right)\,.$$ Indeed, $A=g(0)$ and $B=g'(0)$, so $$\begin{align}g(x)&=g(0)\,\tilde{f}(x)+g'(0)\,f(x)\\ &=g(0)\,\Biggl(\exp\left(\frac{x^2}{2}\right)-\sqrt{\pi}\,x\,\text{erfi}(x)\,\exp\left(-\frac{x^2}{2}\right)\Biggr)+g'(0)\,x\,\exp\left(-\frac{x^2}{2}\right)\,.\end{align}$$ By the way, (#) is a parabolic cylinder differential equation.