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Hermite Polynomials.

  1. May 17, 2008 #1
    I need to show that:
    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:
    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: May 17, 2008
  2. jcsd
  3. May 17, 2008 #2
    Is this correct?

    I'm inclined to think that:
    [tex]=\sum_{n=0}^{\infty}\frac{(-1)^n y^{2n}}{n!}[/tex].?
  4. May 17, 2008 #3
    yes, ofcourse you are correct, i mixed between them, can you please help me on this?
  5. May 17, 2008 #4
    What two other conditions do these polynomials have to satisfy?
  6. May 17, 2008 #5

    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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