- #1

TiberiusK

- 24

- 0

## Homework Statement

The polynomials are defined as follows H_n(x) for n = 0; 1; : : : as follows: fi rst, set 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)

I have to use (1) to verify that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3, and calculate H_4(x)and H_5(x).

I also have to show that that H_n is an even function when n is even, and that it is an odd function when n is odd (maybe using (1) and induction on n).

Also that also (maybe I should still use induction and (1)) that

H_2k(0) = (-1)^k(2k -1)(2k -3) ...1:

What is the value of H_n(0) when n is odd?

## Homework Equations

All above...I'm having trouble proving that the functions is odd for n odd and even for n even.Also I need to know if my calculations are correct

## The Attempt at a Solution

I proved that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3 by substituting in the formula.

Also H_4(x)==6x^2 +x^4-3 and H_5(x)=-6x^3-x^5-3x

For n=odd=2k+1

Base case H_1(-x)=-x=>it is true.We pressume that n is true and prove for n+1.

H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)

=we know that H_n(-x) is odd so xH_(n)(-x)...and therefore (n)H_(n-1)(x) is also odd...here I need help?

For n=even=2k

ase case H_0(-x)=1=>it is true.We pressume that n is true and prove for n+1.

H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)...again same thing

And of course 'la piece de resistance'

What is the value of H_n(0) when n is odd?