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Proving identities of special functions

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Location: North Londinium, UK
Nicely done, RV! :D

Here's a rather easy but nonetheless important one; Pfaff's hypergeometric tranform:


Problem 7:


For Re(c) > Re(b) > 0


\(\displaystyle _2F_1(a,\,b\,;\,c\,;\,x)=(1-x)^{-a}\,_2F_1\left(a,\,c-b\,;\,c\,;\,\frac{x}{x-1}\right)\)

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\(\displaystyle {}_2F_{1} (a,b;c;x) = \frac{1}{B(b,c-b)} \int_{0}^{1} t^{b-1} (1-t)^{c-b-1} (1-xt)^{-a} \ dt\)

\(\displaystyle = \frac{1}{B(b,c-b)} \int_{0}^{1} (1-u)^{b-1} u^{c-b-1} (1-x+xu) ^{-a} \ du\)

\(\displaystyle = (1-x)^{-a} \frac{1}{\ B(b,c-b)} \int_{0}^{1} u^{c-b-1} (1-u)^{b-1} \Big(1+u \ \frac{x}{1-x} \Big) ^{-a} \ du\)

\(\displaystyle = (1-x)^{-a} \ \frac{B(c-b,b)}{B(b,c-b)} {}_2F_{1} \Big( a,c-b;c; - \frac{x}{1-x} \Big)\)

\(\displaystyle = (1-x)^{-a} \ {}_2F_{1} \Big( a,c-b;c; \frac{x}{x-1} \Big)\)
Last edited by Random Variable on Thu Sep 12, 2013 11:32 am, edited 1 time in total.


Posts: 138
Location: North Londinium, UK
And since the hypergeometric function is symmetric w.r.t. the parameters a and b, we can apply Pfaff's tranform a second time, to the other 'upper' parameter, to obtain Euler's transform:



\(\displaystyle _2F_1(a,\,b\,;\,c\,;\,x)=(1-x)^{c-a-b}\,_2F_1\left(c-a,\,c-b\,;\,c\,;\,x\right)\)

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Nice!

Problem 8

Show that

\(\displaystyle _2F_1 (a,b;2b;z) = (1-z)^{-a/2} {\;}_2F_1 \left( \frac{a}{2}, b-\frac{a}{2}; b+\frac{1}{2}; \frac{z^2}{4z-4} \right)\)

This is a quadratic transformation.

Random Variable Integration Guru
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I've been unable to prove problem 8.

So instead I'm going to share a proof of another quadratic transformation, namely

\(\displaystyle _{2}F_{1} (a,b,a-b+1;z) = (1-z)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, \frac{a}{2} + \frac{1}{2} - b, a-b+1; - \frac{4z}{(1-z)^{2}} \Big)\)


It uses the identity

\(\displaystyle _{3}F_{2} (a,b,-n;c,1+a+b-c-n;1) = \frac{(c-a)_{n} (c-b)_{n}}{(c)_{n} (c-a-b)_{n}}\)

which can be derived from Eulers' transformation.


Here I'm using \(\displaystyle (x)_{n}\) to represent the rising factorial.


\(\displaystyle (1-z)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, \frac{a}{2} + \frac{1}{2} - b, a-b+1; - \frac{4z}{(1-z)^{2}} \Big) = (1-z)^{-a} \sum_{k=0}^{\infty} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} \Big(- \frac{4z}{(1-z)^2} \Big)^{k}\)

\(\displaystyle = \sum_{k=0}^{\infty} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4z)^{k} \frac{1}{(1-z)^{a+2k}} = \sum_{k=0}^{\infty} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} z^{k} \sum_{j=0}^{\infty} \binom{a+2k+j-1}{j} z^{j}\)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \binom{a+k+n-1}{n-k} \ z^{n}\) (using the Cauchy product)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \frac{\Gamma(a+k+n)}{\Gamma(a+2k)(n-k)!} \ z^{n} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \frac{(a+n)_{k} \Gamma(a+n)}{\Gamma(a+2k)(n-k)!} \ z^{n}\)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \frac{(a+n)_{k}}{(n-k)!} \frac{(a)_{n} \Gamma(a)}{\Gamma(a+2k)} \ z^{n}\)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \frac{(a+n)_{k} (a)_{n}}{(n-k)!} \frac{2^{a-1} \Gamma (\frac{a}{2} ) \Gamma (\frac{a}{2} + \frac{1}{2})}{2^{a+2k-1} \Gamma(\frac{a}{2}+k) \Gamma(\frac{a}{2}+k + \frac{1}{2})} \ z^{n}\) (using the duplication formula twice)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{ (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2} - b)_{k}}{k! (a-b+1)_{k}} (-4)^{k} \frac{(a+n)_{k} (a)_{n}}{(n-k)!} \frac{1}{4^{k} (\frac{a}{2})_{k} (\frac{a}{2} + \frac{1}{2})_{k} } \ z^{n}\)

\(\displaystyle = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(a)_{n} (\frac{a}{2} + \frac{1}{2} - b)_{k} (a+n)_{k}}{k! (a-b+1)_{k} (\frac{a}{2} + \frac{1}{2} )_{k}} \frac{(-1)^{k}}{(n-k)!} z^{n}\)


\(\displaystyle (-n)_{k} = (-n)(-n+1) \cdots (-n+k-1) = (-1)^{k} n(n+1) \cdots (n-k+1)= \frac{(-1)^{k} n!}{(n-k)!}\)


So \(\displaystyle (1-z)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, \frac{a}{2} + \frac{1}{2} - b, a-b+1; - \frac{4z}{(1-z)^{2}} \Big) = \sum_{n=0}^{\infty} \frac{(a)_{n}}{n!} \sum_{k=0}^{n} \frac{(\frac{a}{2} + \frac{1}{2} - b)_{k} (a+n)_{k} (-n)_{k}}{k! (a-b+1)_{k} (\frac{a}{2} + \frac{1}{2} )_{k}} z^{n}\)


Now let \(\displaystyle c= a-b+1\) and use that identity.


\(\displaystyle = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \frac{1}{(a-b+1)_{n}} \frac{(\frac{a}{2}- \frac{1}{2})_{n}(1-b-n)_{n}}{(\frac{1}{2}-\frac{a}{2}-n)_{n}} z^{n} = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \frac{1}{(a-b+1)_{n}} \frac{\Gamma(\frac{a}{2}-\frac{1}{2}+n)}{\Gamma(\frac{a}{2}-\frac{1}{2})} \frac{\Gamma(\frac{1}{2} - \frac{a}{2}-n)}{\Gamma(\frac{1}{2}-\frac{a}{2})} \frac{\Gamma(1-b)}{\Gamma(1-b-n)} z^{n}\)

\(\displaystyle = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \frac{1}{(a-b+1)_{n}} \frac{\csc[\pi(\frac{a}{2}-\frac{1}{2}+n)]}{\csc[\pi(\frac{a}{2}-\frac{1}{2}]} \frac{\Gamma(1-b)}{\Gamma(1-b-n)} z^{n}\) (using the reflection formula)

\(\displaystyle = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} \frac{1}{(a-b+1)_{n}} (-1)^{n} \frac{\Gamma(1-b)}{\Gamma(1-b-n)} z^{n}\)


\(\displaystyle \frac{\Gamma(1-b)}{\Gamma(1-b-n)} \frac{\Gamma(b)}{\Gamma(b+n)} = (-1)^{n}\) (again using the reflection formula)


At last we have

\(\displaystyle (1-z)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, \frac{a}{2} + \frac{1}{2} - b, a-b+1; - \frac{4z}{(1-z)^{2}} \Big) = \sum_{n=0}^{\infty} \frac{(a)_{n} (b)_{n}}{n! (a-b+1)_{n}} z^{n} = \ _{2}F_{1}(a,b;a-b+1;z)\)

Random Variable Integration Guru
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If you apply Pfaff's transformation to another quadratic transformation, namely


\(\displaystyle _{2}F_{1} (a,b;2a;x) = \left( 1 - \frac{x}{2} \right)^{-b} \ _{2}F_{1} \Big( \frac{b}{2}, \frac{b+1}{2} ; a + \frac{1}{2} ; \frac{x^{2}}{(2-x)^{2}} \Big)\),

you get an transformation that has a different term out front that the one in problem 8.



\(\displaystyle _{2}F_{1} (a,b;2b;z) = \left( 1 - \frac{x}{2} \right)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, \frac{a+1}{2} ; b + \frac{1}{2} ; \frac{x^{2}}{(2-x)^{2}} \Big)\)

\(\displaystyle = \left( 1 - \frac{x}{2} \right)^{-a} \left( 1- \frac{x^{2}}{(2-x)^{2}} \right)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, b + \frac{1}{2} - \frac{a+1}{2}; b + \frac{1}{2}; \frac{\frac{x^{2}}{(2-x)^{2}}}{\frac{x^{2}}{(2-x)^{2}}-1} \Big)\)

\(\displaystyle = \left(\frac{2-x}{2} \right)^{-a} \left( \frac{4(1-x)}{(2-x)^{2}} \right)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, b - \frac{a}{2}; b + \frac{1}{2}; \frac{x^{2}}{x^{2}-(2-x)^{2}} \Big)\)

\(\displaystyle = \left( \frac{2(1-x)}{(2-x)} \right)^{-a} \ _{2}F_{1} \Big( \frac{a}{2}, b - \frac{a}{2}; b + \frac{1}{2}; \frac{x^{2}}{4x-4} \Big)\)

\(\displaystyle = \left( \frac{4(1-x)^{2}}{(2-x)^{2}} \right)^{-\frac{a}{2}} \ _{2}F_{1} \Big( \frac{a}{2}, b - \frac{a}{2}; b + \frac{1}{2}; \frac{x^{2}}{4x-4} \Big)\) :?

Random Variable Integration Guru
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Problem 9


Show that for \(\displaystyle a \ne b\) and \(\displaystyle a\) and \(\displaystyle b\) not negative integers, \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{ \psi(a) - \psi(b)}{a-b}\).


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Location: North Londinium, UK
Random Variable wrote:
Problem 9


Show that for \(\displaystyle a \ne b\) and \(\displaystyle a\) and \(\displaystyle b\) not negative integers, \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{ \psi(a) - \psi(b)}{a-b}\).



\(\displaystyle \sum_{k=0}^{\infty} \frac{1}{(k+a)(k+b)}=\sum_{k=0}^{\infty} \frac{1}{(k+a)(k+b)}\frac{(k+a)-(k+b)}{(a-b)}=\)


\(\displaystyle \frac{1}{(a-b)}\Bigg\{\sum_{k=0}^{\infty}\frac{1}{k+b}-\sum_{k=0}^{\infty}\frac{1}{k+a}\Bigg\}\)


On the other hand,


\(\displaystyle \sum_{k=0}^{\infty} \frac{1}{(k+a)(k+b)}=\sum_{k=0}^{\infty} \frac{1}{(k+a)(k+b)}\frac{(k+a)-(k+b)}{(a-b)}=\)


\(\displaystyle \psi_0(x)=\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{x+k-1}\right)-\gamma=\zeta(1)-\gamma-\sum_{k=0}^{\infty}\frac{1}{k+x}\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \frac{1}{(a-b)}\Bigg\{\sum_{k=0}^{\infty}\frac{1}{k+b}-\sum_{k=0}^{\infty}\frac{1}{k+a}\Bigg\}\)


\(\displaystyle \frac{1}{(a-b)}\Bigg\{-\psi_0(b)+\psi_0(a)\Bigg\}=\frac{\psi_0(a)-\psi_0(b)}{(a-b)}\, . \, \Box\)


Posts: 138
Location: North Londinium, UK
Problem 10




Assuming the existence of the limit


\(\displaystyle \gamma=\lim_{n\to\infty}\, \left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}-\log n \right)\)


And - possibly - the Euler integral form for the Gamma function \(\displaystyle \Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt\), prove Weierstrauss' canonical product for the Gamma Function:


\(\displaystyle \frac{1}{\Gamma(x)}=xe^{\gamma x}\, \prod_{k=1}^{\infty}\Bigg\{e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\)

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I'm starting with Euler's limit definition of the Gamma function.

\(\displaystyle \Gamma(x) = \lim_{n \to \infty} \frac{n! n^{x}}{x(x+1)(x+2) \cdots (x+n)} = \lim_{n \to \infty} n^{x} \frac{1}{x} \frac{1}{x+1} \frac{2}{x+2} \cdots \frac{n}{x+n}\)

\(\displaystyle = \lim_{n \to \infty} e^{x \ln n} \frac{1}{x} \frac{1}{1+x} \frac{1}{1+\frac{x}{2}} \cdots \frac{1}{1+\frac{x}{n}}= \lim_{n \to \infty} e^{x \ln x} \frac{1}{x} \prod_{k=1}^{n} \frac{1}{1+\frac{x}{k}}\)

\(\displaystyle = \lim_{n \to \infty} \frac{e^{x \ln n -x - \frac{x}{2} - \ldots - \frac{x}{n}}}{x} \prod_{k=1}^{n} \frac{e^{\frac{x}{k}}}{1+\frac{x}{k}} = \lim_{n \to \infty} \frac{e^{-x (H_{n}-\ln n)}}{x} \prod_{k=1}^{n} \frac{e^{\frac{x}{k}}}{1+\frac{x}{k}}\)

\(\displaystyle = \frac{e^{-\gamma x}}{x} \prod_{k=1}^{\infty} \frac{e^{\frac{x}{k}}}{1+\frac{x}{k}}\)

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