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Polynomial that satisfies a differential equation

  • Thread starter TiberiusK
  • Start date
  • #1
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Homework Statement


I must show that t H_n satisfies a diferential equation. By diferentiating H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x) (1) and
using induction on n, show that, for n >= 1,
H'_n(x) = -nH_(n-1)(x) (2)
I have to use (2) to express H_(n-1) and H_(n-2) in terms of derivatives of H_n, and substitute these into (1) to show that
H''_n - xH'_n + nH_n = 0 (3)
for n>= 0. Now let O_n(x) = exp(-(x^2)/4 )H_n(x). Using (3) I must show that
O''_n +(n+1/2-(x^2)/3)O_n=0

Homework Equations


H_n(x) for n = 0,1,2.... : H_0(x) =1 and H_1(x) = -x; then, for n >=2, H_n is defined by the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x): (1)


The Attempt at a Solution


I need some help on getting started ,my first problem is how to differentiate the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x) ....because I get
H_n'(x) =-H_(n-1)'(x) -xH_(n-1)(x) - (n - 1)H_(n-2)'(x) =>-H_(n-1)'(x)[1+x]- (n - 1)H_(n-2)'(x) and I do not know how to continue ....
 

Answers and Replies

  • #2
24
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any ideas?
 
  • #3
HallsofIvy
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Homework Statement


I must show that t H_n satisfies a diferential equation. By diferentiating H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x) (1) and
using induction on n, show that, for n >= 1,
H'_n(x) = -nH_(n-1)(x) (2)

I have to use (2) to express H_(n-1) and H_(n-2) in terms of derivatives of H_n, and substitute these into (1) to show that
H''_n - xH'_n + nH_n = 0 (3)
for n>= 0. Now let O_n(x) = exp(-(x^2)/4 )H_n(x). Using (3) I must show that
O''_n +(n+1/2-(x^2)/3)O_n=0

Homework Equations


H_n(x) for n = 0,1,2.... : H_0(x) =1 and H_1(x) = -x; then, for n >=2, H_n is defined by the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x): (1)
So that H_0'= 0 and H_1'= -1.
The formula you want to prove says that H'_1(x)= -(1)H_0(x)= -(1) so it is correct for n= 1.

The Attempt at a Solution

[/B]
I need some help on getting started ,my first problem is how to differentiate the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x) ....because I get
H_n'(x) =-H_(n-1)'(x) -xH_(n-1)(x) - (n - 1)H_(n-2)'(x) =>-H_(n-1)'(x)[1+x]- (n - 1)H_(n-2)'(x) and I do not know how to continue ....
Well, you are using the product rule incorrectly.
H_n'(x)= -H_(n-1)(x)- xH_(n-1)'(x)- (n-1)H_(n-2)'(x)
 

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