Legendre polynomial of zero

  • #1

Homework Statement


Using the Generating function for Legendre polynomials, show that:
##P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}##

Homework Equations


Generating function: ##(1-2xt+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(x) t^n##

The Attempt at a Solution


I put ##x=0## in the generating function and got ##(1+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(0) t^n## Now, I think I should expand ##(1+t^2)^{-1/2}## using the binomial theorem, then I get an expression that's only valid for even values of ##n##, but it's not the expression stated in the question. So, how can I solve this?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Using the Generating function for Legendre polynomials, show that:
##P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}##

Homework Equations


Generating function: ##(1-2xt+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(x) t^n##

The Attempt at a Solution


I put ##x=0## in the generating function and got ##(1+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(0) t^n## Now, I think I should expand ##(1+t^2)^{-1/2}## using the binomial theorem, then I get an expression that's only valid for even values of ##n##, but it's not the expression stated in the question. So, how can I solve this?

So, what does that tell you about ##P_n(0)## for odd ##n##?
 
  • #3
vela
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The Attempt at a Solution


I put ##x=0## in the generating function and got ##(1+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty P_n(0) t^n## Now, I think I should expand ##(1+t^2)^{-1/2}## using the binomial theorem, then I get an expression that's only valid for even values of ##n##, but it's not the expression stated in the question. So, how can I solve this?
"Valid" isn't the right word here. The expansion only contains terms with even exponents, so what does that tell you about the coefficients of the odd terms?
 

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