Hermite Polynomials.

[MathematicalPhysicist]

I need to show that:
$$\sum_{n=0}^{\infty}\frac{H_n(x)}{n!}y^n=e^{-y^2+2xy}$$
where H_n(x) is hermite polynomial.

Now I tried the next expansion:
$$e^{-y^2}e^{2xy}=\sum_{n=0}^{\infty}\frac{(-y)^{2n}}{n!}\cdot \sum_{k=0}^{\infty}\frac{(2xy)^k}{k!}$$
after some simple algebraic rearrangemnets i got:
$$\sum_{n=0}^{\infty}(2x-y)^n\frac{y^n}{n!}$$
which looks similar to what i need to show, the problem is that the polynomial (2x-y)^n satisifes only the condition: H_n'(x)=2nH_n-1(x)
and not the other two conditions, so i guess something is missing, can anyone help me on this?

thanks in advance.

 

[Big-T]

$$e^{-y^2}=\sum_{n=0}^{\infty}\frac{(-y)^{2n}}{n!}$$
Is this correct?

I'm inclined to think that:
$$e^{-y^2}=\sum_{n=0}^{\infty}\frac{(-y^2)^{n}}{n!}$$
$$=\sum_{n=0}^{\infty}\frac{(-1)^n y^{2n}}{n!}$$.?

 

[MathematicalPhysicist]

yes, ofcourse you are correct, i mixed between them, can you please help me on this?

 

[Big-T]

What two other conditions do these polynomials have to satisfy?

 

[MathematicalPhysicist]

H''_n-2xH'_n+2nH_n=0
H_n+1-2xH_n+2nH_n-1=0

according to this exercise.

well I looked at wikipedia, and I guess I only need to use the first definition given at wikipedia.