1. ## Polynomial Help

$H_{n}(x)=-xH_{n-1}(x)-(n-1)H_{n-2}(x) ,for\ n\geq2 H_{0}(x)=1\ and H_{1}(x)=-x$.
a)Show that $H_{n}(x)$ is an even function when n is even and an odd function when n is odd.
Also show by induction that:
b) $H_{2k}(x)=(-1)^k(2k-1)(2k-3)...1$.
What is the value of $H_{n}(0)$when n is odd?
a)Now I proved that $H_{n}(x)$ is an even function when n is even and an odd function when n is odd. for the base case but I'm stuck whit the general case as when :
1.n is even=>n+1 odd and by the recurrence relation I'm stuck with the difference between an even function and an odd one.
2.n is odd=>n+1 even and by the recurrence relation again I get the difference between an odd function and an even one.
b)I think it has something to do with a)

2. Originally Posted by StefanM
$H_{n}(x)=-xH_{n-1}(x)-(n-1)H_{n-2}(x) ,for\ n\geq2 H_{0}(x)=1\ and H_{1}(x)=-x$.
a)Show that $H_{n}(x)$ is an even function when n is even and an odd function when n is odd.
Also show by induction that:
b) $H_{2k}(x)=(-1)^k(2k-1)(2k-3)...1$.
What is the value of $H_{n}(0)$when n is odd?
a)Now I proved that $H_{n}(x)$ is an even function when n is even and an odd function when n is odd. for the base case but I'm stuck whit the general case as when :
1.n is even=>n+1 odd and by the recurrence relation I'm stuck with the difference between an even function and an odd one.
2.n is odd=>n+1 even and by the recurrence relation again I get the difference between an odd function and an even one.

The induction here works as follows: for n even, then n+2 even, and for

n odd, n+2 odd. You have to do these two cases separatedly.

b)I think it has something to do with a)

Show the formula in (b) by induction, again. And then remember that for any odd function, if it is defined in zero

then its value there must be zero.

Tonio

.

3. It doesn't matter because even if I do it for n+2 by the recurrence relation I still end up with the difference between an odd function and an even one....

4. Originally Posted by StefanM
It doesn't matter because even if I do it for n+2 by the recurrence relation I still end up with the difference between an odd function and an even one....

So we assume that $H_k(x)$ is odd for odd $k and even for even $k , so:

$H_{n}=-xH_{n-1}(x)-(n-1)H_{n-2}(x)$ , and we're done since:

i) if $n$ is odd then $n-1$ is even, so $H_{n-1}$ is even by the ind. hypothesis, and

thus $-xH_{n-1}$ is even as a product of odd and even functions, and $n-2$ is odd,

so $H_{n-2}$ is odd by the ind. hyp., and thus the whole thing is odd as the sum of two odd functions.

Now do something similar for the other case.

Tonio

5. I need some help with b) because I have to prove by induction that $H_{2k}(0)=(-1)^k(2k-1)(2k-3)...1$ I get the following....for the base case is true since $H_{0}(0)=1$ and for the 2k+1 case we assume that 2k is true=> $H_{2k+1}(0)=-0*H_{2k}(0)-(2k)H_{2k-1}(0)$=> $H_{2k+1}(0)=(2k)H_{2k-1}(0)=>H_{2k+1}(0)=2k(-1)^{k-1}(2k-1-1)(2k-1-3)....(3)(1)$and it is wrong

6. Originally Posted by StefanM
I need some help with b) because I have to prove by induction that $H_{2k}(0)=(-1)^k(2k-1)(2k-3)...1$ I get the following....for the base case is true since $H_{0}(0)=1$ and for the 2k+1 case we assume that 2k is true=> $H_{2k+1}(0)=-0*H_{2k}(0)-(2k)H_{2k-1}(0)$=> $H_{2k+1}(0)=(2k)H_{2k-1}(0)=>H_{2k+1}(0)=2k(-1)^{k-1}(2k-1-1)(2k-1-3)....(3)(1)$and it is wrong

For the inductive step you must go from 2k to 2(k+1)=2k+2 , and then the result is immediate.

Tonio

7. For $n$ odd, $H_n(0)=0$ because $H_n(x)$ is odd.