How Do Polynomial Recurrence Relations Determine Function Parity?

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Homework Statement


[H_{n}(x)=-xH_{n-1}(x)-(n-1)H_{n-2}(x) ,for,n>=2 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
hat is the value o H_{n}(0) when n is odd

Homework Equations


The Attempt at a Solution


a)Show that is an even function when n is even and an odd function when n is odd.
Also show by induction that:
b).
What is the value of 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)
 
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TiberiusK said:

Homework Statement


[H_{n}(x)=-xH_{n-1}(x)-(n-1)H_{n-2}(x) ,for,n>=2 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
hat is the value o H_{n}(0) when n is odd

Homework Equations


The Attempt at a Solution


a)Show that is an even function when n is even and an odd function when n is odd.
Also show by induction that:
b).
What is the value of 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)

I think that you are mixing up the two parts. Part a doesn't have anything to do with induction, but what you're saying about base cases implies that you think it does.

Part b definitely should be done using induction. For the base case you can use k = 1, and evaluate H2(x).

BTW, your partial LaTeX is difficult to read. Since all you're using are exponents and subscripts, you can write your stuff using the X2 and X2 buttons. Click Go Advanced to see the expanded menu across the top of the entry area.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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