Originally Posted by

**StefanM** I must show that the polynomial$\displaystyle H_{n} $ satisfies a diferential equation.I used some notation(1,2..) so that I don't rewrite the same thing again. By diferentiating $\displaystyle H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x) (1)$and

using induction on n,I'm supposed to show that, for n >= 1,

$\displaystyle H'_n(x) = -nH_{n-1}(x) $(2)

I have to use (2) to express$\displaystyle H_{n-1}$ and$\displaystyle H_{n-2}$ in terms of derivatives of$\displaystyle H_{n}$, and substitute these into (1) to show that

$\displaystyle H''_{n} - xH'_{n} + nH_{n} = 0 $(3)

for n>= 0. Now let$\displaystyle \phi'_n(x) = exp(-(x^2)/4 )H_{n}(x).$ Using (3) I must show that

$\displaystyle \phi'_n(x) +(n+1/2-(x^2)/4)\phi_n(x)=0$

**Also H_n is a polynomial defined as follows:**

$\displaystyle H_{n}(x) for n = 0,1,2.... : $ $\displaystyle H_{0}(x) =1 $ and $\displaystyle H_{1}(x) = -x $; then, for n >=2, H_{n} is defined by the recurrence

$\displaystyle H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x): $ (1)

**Attempt at a solution**

I need help with the entire problem mainly because for example $\displaystyle H'_n(x) = -nH_{n-1}(x) $ I can only prove it for the base case when but fot the general I get $\displaystyle H'_n(x) = -H_{n-1}(x)-xH_{n-1}(x)-(n - 1)H_{n-2}(x)=-H_{n-1}(x)-xH'_{n-1}(x)-nH'_{n-2}(x)-H'_{n-2}(x)$=$\displaystyle -H_{n-1}(x)-xnH'_{n-1}(x)+xH'_{n-2}(x)-n^2H'_{n-3}(x)+2nH'_{n-3}(x)-nH'_{n-3}(x)+2H'_{n-3}(x)$......and I do not know what do do next...if it isn't obvious I just substituted in the relation that I have to prove since I'm assuming that everything from n-1 is true......

Also for the other parts I still need help in showing (3) and (4).Thanks in advance