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Math Help - legendre polynomial problem (pls help!)

  1. #1
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    legendre polynomial problem (pls help!)

    show that P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^n}

    effort made as follows;
    given P_{n}(x)=\frac{1}{2^n}\sum_{k=0}^{\frac{1}{2}n}(-1)^{k} {{^n}}C_k{^{2n-2k}C_n}x^{n-2k}

    let n = 2n+1
    we have P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{k=0}^{{n}+\fra  c{1}{2}}(-1)^{k} {{^{2n+1}}}C_k{^{2(2n+1)-2k}C_{2n+1}}x^{2n+1-2k}

    let k=n

    P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\fra  c{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2(2n+1)-2n}C_{2n+1}}x^{2n+1-2n}

    P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\fra  c{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}x^{1}

    P_{2n+1}^{'}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+  \frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}

    simplifying

    P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^{2n}}

    this result is at variance with the one given, missed out something obviously, need help pls.
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  2. #2
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    Re: legendre polynomial problem (pls help!)

    Hey lawochekel.

    Why did you just let k = n? Shouldn't you sum out the series and then differentiate the result?
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  3. #3
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    Re: legendre polynomial problem (pls help!)

    I don't understand some of the notation, what is that C_{n} ?

    Since the polynomial is of odd order, it will be of the form a_{0}x + a_{1}x^{3} + \dots, so after differentiating we get a_{0} + 3a_{1}x^{2}+ \dots, in which case P_{2n+1}(0) = a_{0}.

    That is, you simply need to pick out the coefficient of the x term from the original summation.
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