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Math Help - Polynomial Help

  1. #1
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    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)
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  2. #2
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    Quote Originally Posted by StefanM View Post
     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




    .
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  3. #3
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    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....
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  4. #4
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    Quote Originally Posted by StefanM View Post
    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<n and even for even k<n , 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
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  5. #5
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    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
    Last edited by StefanM; February 18th 2011 at 09:20 AM.
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  6. #6
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    Quote Originally Posted by StefanM View Post
    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
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  7. #7
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    For n odd, H_n(0)=0 because H_n(x) is odd.
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