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PostPosted: Thu, 18 Mar 2010 05:02:17 UTC 
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Joined: Fri, 27 Jan 2006 22:01:08 UTC
Posts: 116
I don't know if I am posting this is the right place, but I am trying to prove the validity of the following formula and was wondering if any of you might
be able to offer me some advice:

\displaystyle
\frac{(-\nu)_k}{k!}=\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{ \bigl(-\frac{\nu}{2}\bigr)_{k-2n}
}{
\bigl(\frac{d}{2}-1\bigr)_{k-2n}
}
\frac{\bigl(d-2\bigr)_{k-2n}}
{(k-2n)!}
\frac{\bigl(k-2n-\frac{\nu}{2}\bigr)_n}{\bigl(k-2n+\frac{d}{2}\bigr)_n}
\frac{\bigl(1-\frac{\nu+d}{2}\bigr)_n}
{n!},

where (z)_k is the Pochhammer symbol for rising factorials used in hypergeometric series, and
d\in\{2,3,\ldots\}, k\in\{0,1,\ldots\} and \nu\in{\mathbf C}.
As is indicated by the formula, the sum seems to be independent of d and
I have verified this formula using Mathematica for 0\le k\le 7. I have been able to show that this result can be re-written as

\displaystyle 
\frac{(-\nu)_k}{k!}=
\frac{1}{1}
\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{\sqrt{\pi}2^{2-d}(d-2+2k-4n)\Gamma(d-2+k-2n)\Gamma(k-n-\nu/2)\Gamma(1-d/2+n-\nu/2)}
{\Gamma((d-1)/2)\Gamma(1+k-2n)\Gamma(d/2+k-n)\Gamma(1-(d+\nu)/2)\Gamma(-\nu/2)n!},


where \Gamma is the Gamma function. I was looking for a proof and I was wondering if you had ever encountered series such as this.
I need this result in order to complete a proof of a Gegenbauer polynomial expansion for powers of the distance between two points in {\mathbf R}^d.


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PostPosted: Thu, 18 Mar 2010 17:14:49 UTC 
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Moderator
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6896
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
howi wrote:
I don't know if I am posting this is the right place, but I am trying to prove the validity of the following formula and was wondering if any of you might
be able to offer me some advice:

\displaystyle
\frac{(-\nu)_k}{k!}=\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{ \bigl(-\frac{\nu}{2}\bigr)_{k-2n}
}{
\bigl(\frac{d}{2}-1\bigr)_{k-2n}
}
\frac{\bigl(d-2\bigr)_{k-2n}}
{(k-2n)!}
\frac{\bigl(k-2n-\frac{\nu}{2}\bigr)_n}{\bigl(k-2n+\frac{d}{2}\bigr)_n}
\frac{\bigl(1-\frac{\nu+d}{2}\bigr)_n}
{n!},

where (z)_k is the Pochhammer symbol for rising factorials used in hypergeometric series, and
d\in\{2,3,\ldots\}, k\in\{0,1,\ldots\} and \nu\in{\mathbf C}.
As is indicated by the formula, the sum seems to be independent of d and
I have verified this formula using Mathematica for 0\le k\le 7. I have been able to show that this result can be re-written as

\displaystyle 
\frac{(-\nu)_k}{k!}=
\frac{1}{1}
\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{\sqrt{\pi}2^{2-d}(d-2+2k-4n)\Gamma(d-2+k-2n)\Gamma(k-n-\nu/2)\Gamma(1-d/2+n-\nu/2)}
{\Gamma((d-1)/2)\Gamma(1+k-2n)\Gamma(d/2+k-n)\Gamma(1-(d+\nu)/2)\Gamma(-\nu/2)n!},


where \Gamma is the Gamma function. I was looking for a proof and I was wondering if you had ever encountered series such as this.
I need this result in order to complete a proof of a Gegenbauer polynomial expansion for powers of the distance between two points in {\mathbf R}^d.


Hypergeometric identities? Looks like the methods in book A=B would help you.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Wed, 24 Mar 2010 03:11:29 UTC 
Offline
Senior Member

Joined: Fri, 27 Jan 2006 22:01:08 UTC
Posts: 116
outermeasure wrote:
howi wrote:
I don't know if I am posting this is the right place, but I am trying to prove the validity of the following formula and was wondering if any of you might
be able to offer me some advice:

\displaystyle
\frac{(-\nu)_k}{k!}=\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{ \bigl(-\frac{\nu}{2}\bigr)_{k-2n}
}{
\bigl(\frac{d}{2}-1\bigr)_{k-2n}
}
\frac{\bigl(d-2\bigr)_{k-2n}}
{(k-2n)!}
\frac{\bigl(k-2n-\frac{\nu}{2}\bigr)_n}{\bigl(k-2n+\frac{d}{2}\bigr)_n}
\frac{\bigl(1-\frac{\nu+d}{2}\bigr)_n}
{n!},

where (z)_k is the Pochhammer symbol for rising factorials used in hypergeometric series, and
d\in\{2,3,\ldots\}, k\in\{0,1,\ldots\} and \nu\in{\mathbf C}.
As is indicated by the formula, the sum seems to be independent of d and
I have verified this formula using Mathematica for 0\le k\le 7. I have been able to show that this result can be re-written as

\displaystyle 
\frac{(-\nu)_k}{k!}=
\frac{1}{1}
\sum_{n=0}^{\displaystyle \left\lfloor k/2\right\rfloor}
\frac{\sqrt{\pi}2^{2-d}(d-2+2k-4n)\Gamma(d-2+k-2n)\Gamma(k-n-\nu/2)\Gamma(1-d/2+n-\nu/2)}
{\Gamma((d-1)/2)\Gamma(1+k-2n)\Gamma(d/2+k-n)\Gamma(1-(d+\nu)/2)\Gamma(-\nu/2)n!},


where \Gamma is the Gamma function. I was looking for a proof and I was wondering if you had ever encountered series such as this.
I need this result in order to complete a proof of a Gegenbauer polynomial expansion for powers of the distance between two points in {\mathbf R}^d.


Hypergeometric identities? Looks like the methods in book A=B would help you.


Hi outermeasure. You were absolutely correct. Zeilberger's algorithm directly produces a proof of the identity. Thanks so much for pointing me to these methods.

Its incredible what a computer is capable of. Apparently, the method produces sister identities as well...I am still trying to obtain these, although they are not necessary for my main interest.


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