# Thread: legendre polynomial problem (pls help!)

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

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

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.

2. ## 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?

3. ## 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.