Hermite Polynomials

  • #1
Im stuck on this question :(


The Hermite polynomials can be defined through

[tex]\displaystyle{F(x,h) = \sum^{\infty}_{n = 0} \frac{h^n}{n!}H_n(x)}[/tex]

Prove that the [tex]H_n[/tex] satisfy the hermite equation

[tex]\displaystyle{H''_n(x) - 2xH'_n(x) + 2nH_n(x) = 0}[/tex]

Using

[tex]\displaystyle{\sum^{\infty}_{n = 0} \frac{h^n}{n!}nH_n(x) = h\frac{\partial}{\partial h}F(x,h)}[/tex]

Can someone give me a bit of a push in the right direction?
 

Answers and Replies

  • #2
dextercioby
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Who's [itex] F(x,h) [/itex] ...?

IIRC, it has to have some exponential, right...?

Daniel.
 
  • #3
HallsofIvy
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No Name Required said:
Im stuck on this question :(


The Hermite polynomials can be defined through

[tex]\displaystyle{F(x,h) = \sum^{\infty}_{n = 0} \frac{h^n}{n!}H_n(x)}[/tex]

Prove that the [tex]H_n[/tex] satisfy the hermite equation

[tex]\displaystyle{H''_n(x) - 2xH'_n(x) + 2nH_n(x) = 0}[/tex]

Using

[tex]\displaystyle{\sum^{\infty}_{n = 0} \frac{h^n}{n!}nH_n(x) = h\frac{\partial}{\partial h}F(x,h)}[/tex]

Can someone give me a bit of a push in the right direction?

What's wrong with "just do it!"? Just go ahead and differentiate that sum, term by term, (your hint essentially says that you can do that), plug into the differential equation and see what happens.
 

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