On Legendre's Polynomial












1












$begingroup$


I want to show that the coefficient of $x^n$ in $P_n(x)$ is $(2n)!/ (2^n(n!)^2)$



my problem is that I cannot find the the $n$-th derivative of $(x^2-1)^n$ to be able to simplify Rodrigues' formula of pn! can someone give me a hint?






legendre-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:35












  • $begingroup$
    Maybe look here
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41










  • $begingroup$
    Possible duplicate of Proving a property of Legendre Polynomials
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41
















1












$begingroup$


I want to show that the coefficient of $x^n$ in $P_n(x)$ is $(2n)!/ (2^n(n!)^2)$



my problem is that I cannot find the the $n$-th derivative of $(x^2-1)^n$ to be able to simplify Rodrigues' formula of pn! can someone give me a hint?






legendre-polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:35












  • $begingroup$
    Maybe look here
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41










  • $begingroup$
    Possible duplicate of Proving a property of Legendre Polynomials
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41














1












1








1





$begingroup$


I want to show that the coefficient of $x^n$ in $P_n(x)$ is $(2n)!/ (2^n(n!)^2)$



my problem is that I cannot find the the $n$-th derivative of $(x^2-1)^n$ to be able to simplify Rodrigues' formula of pn! can someone give me a hint?






legendre-polynomials







share|cite|improve this question











$endgroup$




I want to show that the coefficient of $x^n$ in $P_n(x)$ is $(2n)!/ (2^n(n!)^2)$



my problem is that I cannot find the the $n$-th derivative of $(x^2-1)^n$ to be able to simplify Rodrigues' formula of pn! can someone give me a hint?






legendre-polynomials




legendre-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 12:50









TonyK

42.6k355134




42.6k355134










asked Jan 18 at 12:19









user635977user635977

63




63












  • $begingroup$
    WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:35












  • $begingroup$
    Maybe look here
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41










  • $begingroup$
    Possible duplicate of Proving a property of Legendre Polynomials
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41


















  • $begingroup$
    WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:35












  • $begingroup$
    Maybe look here
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41










  • $begingroup$
    Possible duplicate of Proving a property of Legendre Polynomials
    $endgroup$
    – Viktor Glombik
    Jan 18 at 12:41
















$begingroup$
WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
$endgroup$
– Viktor Glombik
Jan 18 at 12:35






$begingroup$
WolframAlpha says $$sum_{k=0}^{n} sum_{j=0}^{k} frac{(-1)^j x^{2 k - n} (-1 + x^2)^{n-k} (1 - 2 j + 2 k - n)_n (1 - k + n)_k}{j! (k - j)!},$$where $(a_n) = frac{Gamma( a + n)}{Gamma(a)}$ is the pochhammer symbol and $Gamma(x) = (x - 1)!$ is the gamma function. In summary, there should be a different, easier way.
$endgroup$
– Viktor Glombik
Jan 18 at 12:35














$begingroup$
Maybe look here
$endgroup$
– Viktor Glombik
Jan 18 at 12:41




$begingroup$
Maybe look here
$endgroup$
– Viktor Glombik
Jan 18 at 12:41












$begingroup$
Possible duplicate of Proving a property of Legendre Polynomials
$endgroup$
– Viktor Glombik
Jan 18 at 12:41




$begingroup$
Possible duplicate of Proving a property of Legendre Polynomials
$endgroup$
– Viktor Glombik
Jan 18 at 12:41










1 Answer
1






active

oldest

votes


















2












$begingroup$

According to Rodrigue's formula,



$$P_n(x)=left(frac{1}{2^n n!}right) frac{d^n}{dx^n}left(x^2-1right)^n$$



The degree of the polynomial $q_n(x)=left(x^2-1right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) dots (n+1) = frac{(2n)!}{n!}$.



Multiplying by the coefficient $left(frac{1}{2^n n!}right)$ you get the desired result $frac{(2n)!}{2^n (n!)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
    $endgroup$
    – user635977
    Jan 18 at 12:51











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

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2












$begingroup$

According to Rodrigue's formula,



$$P_n(x)=left(frac{1}{2^n n!}right) frac{d^n}{dx^n}left(x^2-1right)^n$$



The degree of the polynomial $q_n(x)=left(x^2-1right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) dots (n+1) = frac{(2n)!}{n!}$.



Multiplying by the coefficient $left(frac{1}{2^n n!}right)$ you get the desired result $frac{(2n)!}{2^n (n!)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
    $endgroup$
    – user635977
    Jan 18 at 12:51
















2












$begingroup$

According to Rodrigue's formula,



$$P_n(x)=left(frac{1}{2^n n!}right) frac{d^n}{dx^n}left(x^2-1right)^n$$



The degree of the polynomial $q_n(x)=left(x^2-1right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) dots (n+1) = frac{(2n)!}{n!}$.



Multiplying by the coefficient $left(frac{1}{2^n n!}right)$ you get the desired result $frac{(2n)!}{2^n (n!)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
    $endgroup$
    – user635977
    Jan 18 at 12:51














2












2








2





$begingroup$

According to Rodrigue's formula,



$$P_n(x)=left(frac{1}{2^n n!}right) frac{d^n}{dx^n}left(x^2-1right)^n$$



The degree of the polynomial $q_n(x)=left(x^2-1right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) dots (n+1) = frac{(2n)!}{n!}$.



Multiplying by the coefficient $left(frac{1}{2^n n!}right)$ you get the desired result $frac{(2n)!}{2^n (n!)^2}$






share|cite|improve this answer









$endgroup$



According to Rodrigue's formula,



$$P_n(x)=left(frac{1}{2^n n!}right) frac{d^n}{dx^n}left(x^2-1right)^n$$



The degree of the polynomial $q_n(x)=left(x^2-1right)^n$ is equal to $2n$. Its $n$-th derivative is a polynomial of degree $n$. You just have to find the coefficient of the term of degree $n$. And that is $(2n)(2n-1) dots (n+1) = frac{(2n)!}{n!}$.



Multiplying by the coefficient $left(frac{1}{2^n n!}right)$ you get the desired result $frac{(2n)!}{2^n (n!)^2}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 at 12:46









mathcounterexamples.netmathcounterexamples.net

26.8k22157




26.8k22157












  • $begingroup$
    Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
    $endgroup$
    – user635977
    Jan 18 at 12:51


















  • $begingroup$
    Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
    $endgroup$
    – user635977
    Jan 18 at 12:51
















$begingroup$
Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
$endgroup$
– user635977
Jan 18 at 12:51




$begingroup$
Thank you! I had some sort of the same analysis but I discarded it as I forgot to add the n! term in the denominator!
$endgroup$
– user635977
Jan 18 at 12:51


















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