Hermite Polynomials.

  • #1
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Main Question or Discussion Point

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.
 
Last edited:

Answers and Replies

  • #2
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[tex]e^{-y^2}=\sum_{n=0}^{\infty}\frac{(-y)^{2n}}{n!}[/tex]
Is this correct?

I'm inclined to think that:
[tex]e^{-y^2}=\sum_{n=0}^{\infty}\frac{(-y^2)^{n}}{n!}[/tex]
[tex]=\sum_{n=0}^{\infty}\frac{(-1)^n y^{2n}}{n!}[/tex].?
 
  • #3
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yes, ofcourse you are correct, i mixed between them, can you please help me on this?
 
  • #4
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What two other conditions do these polynomials have to satisfy?
 
  • #5
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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.
 

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