A problem with Hermite Polynomials We define $ H_n(x)=(-1)^n e^{x^2} \frac {d^n}{dx^n} (e^{-x^2})$ ,
and $\phi _n(x)=e^{{-x^2}/2}H_n(x)$
Now we have to prove $\phi _n^"-x^2 \phi _n+(2n+1)\phi _n=0$.
Anyway after two pages of calculation I have got $\phi _n^"-x^2 \phi _n+(2n+1)\phi _n=2(1+n)\phi _n - 2x \phi _{n+1} + \phi _{n+2}$
And if I take an other expression then it will be same to show $H_n^{"} -2x H_n^{'}+ 2n H_n=0$.
So how to prove that? What I am thinking that I have to use some recurrence relation in induction. Will anyone please prove this?
 A: Well, let's see.
$\varphi_n(x)=e^{-x^2/2}H_n(x)=(-1)^n e^{x^2/2} \frac {d^n}{dx^n} (e^{-x^2})$, hence 
$\frac{d^n}{dx^n} (e^{-x^2})=(-1)^n e^{-x^2/2}\varphi_n(x)$.
Now, $$\varphi_{n+1}(x)=(-1)^{n+1} e^{x^2/2}\frac {d^{n+1}}{dx^{n+1}} (e^{-x^2})=
(-1)^{n+1} e^{x^2/2}\frac d{dx}\left(\frac {d^n}{dx^n} (e^{-x^2})\right)=
(-1)^{n+1} e^{x^2/2}\frac d{dx}\left((-1)^n e^{-x^2/2}\varphi_n(x)\right)=
-e^{x^2/2}\frac d{dx}\left(e^{-x^2/2}\varphi_n(x)\right)=
-e^{x^2/2}\left(-xe^{-x^2/2}\varphi_n(x)+e^{-x^2/2}\varphi'_n(x)\right)=
x\varphi_n(x)-\varphi'_n(x)$$
Now plug this into the expression that we want to be 0:
$$
\underbrace{\varphi_{n+1}''-x^2\varphi_{n+1}+(2(n+1)+1)\phi_{n+1}}_\text{is it 0, really?}=\\
=(x\varphi_n(x)-\varphi'_n(x))''-x^2(x\varphi_n(x)-\varphi'_n(x))+(2n+3)(x\varphi_n(x)-\varphi'_n(x))=\\
=2\varphi'_n(x)+x\varphi''_n(x)-\varphi'''_n(x)-x^3\varphi_n(x)+x^2\varphi'_n(x)+(2n+3)x\varphi_n(x)-(2n+3)\varphi'_n(x)=\\
=x(\underbrace{\varphi''_n(x)-x^2\varphi_n(x)+(2n+1)\varphi_n(x)}_\text{hey, I've seen this one before})+2\varphi'_n(x)-\varphi'''_n(x)+x^2\varphi'_n(x)+2x\varphi_n(x)-(2n+3)\varphi'_n(x)=\\
=2\varphi'_n(x)-{d\over dx}\Big( x^2\varphi_n(x)-(2n+1)\varphi_n(x)\Big)+x^2\varphi'_n(x)+2x\varphi_n(x)-(2n+3)\varphi'_n(x)=\\
=2\varphi'_n(x)-2x\varphi_n(x)-x^2\varphi'_n(x)+(2n+1)\varphi'_n(x)+x^2\varphi'_n(x)+2x\varphi_n(x)-(2n+3)\varphi'_n(x)=\dots
$$
well, I think I've done enough.
A: Think of a polynomial as living in a vector space. Differentiating on a vector of it's coefficients is a linear operation. Differentiating on exponentials of a polynomial is a multiplication with the derivative of the exponent, in this case multiplying with $x$, which is also a linear operation. Denoting the matrice performing differentiation : $\bf D$ , and multiplication by x : $\bf X$.
$\phi_n$ can be acquired by a sequence of linear operations ${\bf B}^n = ({\bf X-D})^n$, where the recursion step $\bf X - D$ is easy to show. Now the proposition $$H_n'' + 2xH_n' + 2nH_n = 0$$ is the same as to say that the vector multiplied by $$ {\bf O} = {\bf D}^2 + 2{\bf XD} + 2n{\bf I}$$ equals the zero vector. Now how can the concatenated operation (${\bf OB}^n$) be simplified?
