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Prove that the Legendre polynomial holds true using the generating function and a binomial expansion.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding the number of integral solutions to an equation given constraints.Binomial theorem / Generating functionHow to prove the Legendre Polynomials are unique up to a scalar multiple?Proving binomial summation identity using generating functionsAsymptotic expansion of Legendre polynomialUsing Rodriguez to prove LegendreProve Fibonacci Identity using generating functionsObtain the probability generating function from a binomial looking functionExpansion of fibonacci generating functionGenerating function and binomial coefficient
$begingroup$
I tried using that
$$P_2n(z)=frac12^2n(2n)!fracd^2ndz^2n(z^2-1)^2n $$
but I am having trouble taking the derivative when $n$ is unknown and $z$ is $0$.
Any help would be greatly appreciated!
generating-functions legendre-polynomials
$endgroup$
add a comment |
$begingroup$
I tried using that
$$P_2n(z)=frac12^2n(2n)!fracd^2ndz^2n(z^2-1)^2n $$
but I am having trouble taking the derivative when $n$ is unknown and $z$ is $0$.
Any help would be greatly appreciated!
generating-functions legendre-polynomials
$endgroup$
add a comment |
$begingroup$
I tried using that
$$P_2n(z)=frac12^2n(2n)!fracd^2ndz^2n(z^2-1)^2n $$
but I am having trouble taking the derivative when $n$ is unknown and $z$ is $0$.
Any help would be greatly appreciated!
generating-functions legendre-polynomials
$endgroup$
I tried using that
$$P_2n(z)=frac12^2n(2n)!fracd^2ndz^2n(z^2-1)^2n $$
but I am having trouble taking the derivative when $n$ is unknown and $z$ is $0$.
Any help would be greatly appreciated!
generating-functions legendre-polynomials
generating-functions legendre-polynomials
edited Apr 3 at 16:50
Paul Enta
5,49611435
5,49611435
asked Apr 2 at 0:49
JetJet
82
82
add a comment |
add a comment |
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