- #1

- 4,699

- 369

I need to show that:

[tex]\sum_{n=0}^{\infty}\frac{H_n(x)}{n!}y^n=e^{-y^2+2xy}[/tex]

where H_n(x) is hermite polynomial.

Now I tried the next expansion:

[tex]e^{-y^2}e^{2xy}=\sum_{n=0}^{\infty}\frac{(-y)^{2n}}{n!}\cdot \sum_{k=0}^{\infty}\frac{(2xy)^k}{k!}[/tex]

after some simple algebraic rearrangemnets i got:

[tex]\sum_{n=0}^{\infty}(2x-y)^n\frac{y^n}{n!}[/tex]

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.

[tex]\sum_{n=0}^{\infty}\frac{H_n(x)}{n!}y^n=e^{-y^2+2xy}[/tex]

where H_n(x) is hermite polynomial.

Now I tried the next expansion:

[tex]e^{-y^2}e^{2xy}=\sum_{n=0}^{\infty}\frac{(-y)^{2n}}{n!}\cdot \sum_{k=0}^{\infty}\frac{(2xy)^k}{k!}[/tex]

after some simple algebraic rearrangemnets i got:

[tex]\sum_{n=0}^{\infty}(2x-y)^n\frac{y^n}{n!}[/tex]

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.

Last edited: