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Proving identities of special functions

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Problem 3 is Gauss' digamma theorem.

But the only proof I know of is from Danish mathematician Johan Jensen. So maybe this should really be called Jensen's digamma theorem.

One place to find the proof is on planetmath.org.

http://planetmath.org/ProofOfGaussDigammaTheorem


It uses a theorem about power series that I don't completely understand, namely that if \(\displaystyle f(x) = \sum_{n=0}^{\infty} a_{n}x^{n}\), then \(\displaystyle \sum_{n=0}^{\infty} a_{p+nq} x^{p+nq} = \frac{1}{q} \sum_{j=0}^{q-1} \omega^{-jp} f(\omega^{j} x)\) where \(\displaystyle \omega\) is a primitive qth root of unity.


EDIT:

http://math.stackexchange.com/questions ... 747#480747

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Problem 4

Show that \(\displaystyle \displaystyle \zeta'(0) = - \frac{1}{2} \ln(2 \pi)\).

Shobhit Site Admin
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Here's a proof for problem 4:

From the functional equational equation of reimann zeta function and the fact that \(\displaystyle \Gamma \left( \frac{1-s}{2}\right)=\frac{2}{1-s}\Gamma \left(\frac{3-s}{2} \right)\), we have

\(\displaystyle -\zeta(1-s) = \pi^{-s+1/2}\frac{\Gamma(s/2)}{2\Gamma((3-s)/2)}(s-1)\zeta(s) \tag{1}\)

Now we know that, \(\displaystyle (s-1)\zeta(s) = 1+(s-1)\gamma-(s-1)^2 \gamma_1 + \cdots \tag{2}\)

where \(\gamma_n\) are the Stieltjes constants.

Taking logarithmic derivative of (1) and using (2), we obtain

\(\displaystyle \frac{\zeta'(1-s)}{\zeta(1-s)}=\log(\pi)-\frac{1}{2}\psi_0 \left(\frac{s}{2} \right)-\frac{1}{2}\psi_0 \left(\frac{3-s}{2} \right)-\frac{\gamma-2\gamma_1 (s-1)+\cdots }{1+\gamma(s-1)+\cdots} \tag{3}\)

Set \(s=1\) in (3):

\[\begin{align*}
\zeta'(0) &= \zeta(0) \left( \log(\pi)-\frac{1}{2}\psi_0 \left(\frac{1}{2} \right)-\frac{1}{2}\psi_0 \left(1\right)-\gamma \right) \\
&= -\frac{1}{2}\left( \log(\pi)+\log(2)\right) \\
&= -\frac{1}{2}\log(2\pi) \tag{4}
\end{align*}\]

zaidalyafey Global Moderator
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Problem 5

Prove that

\(\displaystyle \sum_{p=0}^{q-1}\psi(a+p/q)=q(\psi(qa)-\log(q))\)
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Shobhit Site Admin
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Using the Multiplication theorem of Gamma function, we have

\(\displaystyle \begin{align*}
\sum_{p=0}^{q-1} \log \Gamma \left(a+\frac{p}{q} \right) &= \log \prod_{p=0}^{q-1}\Gamma \left(a+\frac{p}{q} \right) \\
&= \log \left((2\pi)^{\frac{q-1}{2}}q^{\frac{1}{2}-qa}\Gamma(qa) \right)
\end{align*}\)

Now, take derivative with respect to \(a\):

\(\displaystyle \sum_{p=0}^{q-1}\psi_0 \left( a+\frac{p}{q}\right)=-q\log q +q \psi_0(qa)\)


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zaidalyafey wrote:
Problem # 2

\(\displaystyle \sum_{k\geq 0} \frac{(-1)^k}{zk+1}= \frac{1}{2z} \left( \psi_0\left(\frac{z+1}{2z} \right)- \psi_0\left(\frac{1}{2z} \right)\right)\)


There is a much simpler solution to this problem. Start with the series definition of the digamma function:

\(\displaystyle \psi_0(z)=\sum_{n=1}^{\infty}\left( \frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma=\zeta(1)-\gamma-\sum_{n=1}^{\infty}\frac{1}{z+n-1}\)

Then

\(\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k}{(zk+1)}=\frac{1}{z}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+(1/z)}=\)

\(\displaystyle \frac{1}{z}\sum_{k=0}^{\infty}\frac{1}{2k+(1/z)}-\frac{1}{z}\sum_{k=0}^{\infty}\frac{1}{2k+1+(1/z)}=\)

\(\displaystyle \frac{1}{2z}\sum_{k=0}^{\infty}\frac{1}{k+(1/2z)}-\frac{1}{2z}\sum_{k=0}^{\infty}\frac{1}{k+((z+1)/2z)}\)

Now change the summation variable to s=k+1 to get

\(\displaystyle \frac{1}{2z}\sum_{s=1}^{\infty}\frac{1}{s-1+(1/2z)}-\frac{1}{2z}\sum_{s=1}^{\infty}\frac{1}{s-1+((z+1)/2z)}=\)

\(\displaystyle \frac{1}{2z}\left[\zeta(1)-\gamma-\psi_0\left(\tfrac{1}{2z}\right)\right]-\frac{1}{2z}\left[\zeta(1)-\gamma-\psi_0\left(\tfrac{z+1}{2z}\right)\right]=\)

\(\displaystyle \frac{1}{2z}\left[\psi_0\left(\tfrac{z+1}{2z}\right) -\psi_0\left(\tfrac{1}{2z}\right) \right]\)

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Problem 6


EDIT: \(\displaystyle \prod_{k=1}^{n-1} \Gamma \Big(\frac{k}{n} \Big) = \frac{ \sqrt{2 \pi}^{ \ n-1}}{\sqrt{n}}\)
Last edited by Random Variable on Wed Sep 11, 2013 7:55 pm, edited 1 time in total.


Posts: 138
Location: North Londinium, UK
Random Variable wrote:
Problem 6


\(\displaystyle \sum_{k=1}^{n-1} \Gamma \Big(\frac{k}{n} \Big) = \frac{ \sqrt{2 \pi}^{ \ n-1}}{\sqrt{n}}\)


I'm a bit rusty on these types of relations, and have had a little vino ( :roll: ), but I think I'm half-way there...



let \(\displaystyle s=n - k\,\) represent a change of summation index, then

\(\displaystyle \mathcal{S}_n=\sum_{k=1}^{n-1} \Gamma \left(\frac{k}{n} \right)=\sum_{s=(n-1)}^{[n-(n-1)]} \Gamma \left(\frac{n-s}{n} \right)=\sum_{s=(n-1)}^{s=1} \Gamma \left(\frac{n-s}{n} \right)=\sum_{s=1}^{n-1} \Gamma \left(1-\frac{s}{n} \right)\)

Hence

\(\displaystyle 2\sum_{k=1}^{n-1} \Gamma \left(\frac{k}{n} \right)=2\mathcal{S}_n=\sum_{k=1}^{n-1} \Gamma \left(\frac{k}{n} \right)+\sum_{k=1}^{n-1} \Gamma \left(1-\frac{k}{n} \right)\)

\(\displaystyle \text{exp}\left[\log \left(\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)\Gamma\left(1-\frac{k}{n}\right)\right)\right]=2\mathcal{S}\Rightarrow\)

\(\displaystyle \mathcal{S}=\frac{1}{2}\,\prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)\Gamma\left(1-\frac{k}{n}\right)=\frac{1}{2}\,\prod_{k=1}^{n-1}\frac{\pi}{\sin\left(\frac{\pi k}{n}\right)}\)




And that's as far as I'll get for tonight... hic! [Mmmmm... vino] :twisted:


\(\displaystyle 2\sum_{k=1}^{n-1} \Gamma \left(\frac{k}{n} \right)=\prod_{k=1}^{n-1}\frac{\pi}{\sin\left(\frac{\pi k}{n}\right)}\)

Random Variable Integration Guru
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Problem 6 should be \(\displaystyle \prod_{k=1}^{n-1} \Gamma \Big(\frac{k}{n} \Big) = \frac{ \sqrt{2 \pi}^{ \ n-1}}{\sqrt{n}}\) :|

So \(\displaystyle \prod_{k=1}^{n-1} \Gamma \Big(\frac{k}{n} \Big) = \sqrt{\prod_{k=1}^{n-1}\frac{\pi}{\sin\left(\frac{\pi k}{n}\right)}}\)

Random Variable Integration Guru
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Posts: 381
\(\displaystyle \prod_{k=1}^{n-1} \exp \left( \frac{i k \pi}{n} \right) \sin \left(\frac{k \pi}{n} \right) = \left( \frac{i}{2} \right)^{n-1} \ \prod_{k=1}^{n-1} \Bigg( 1- \exp \left( \frac{2 \pi i k}{n} \right) \Bigg) =n \left( \frac{i}{2} \right)^{n-1}\)


And \(\displaystyle \prod_{k=1}^{n-1} \exp \left( \frac{i k \pi}{n} \right) \sin \left(\frac{k \pi}{n} \right) = \prod_{k=1}^{n-1} \exp \left( \frac{i k \pi}{n} \right) \prod_{k=1}^{n-1} \sin \left(\frac{k \pi}{n} \right)= i^{n-1} \prod_{k=1}^{n-1} \sin \left(\frac{k \pi}{n} \right)\)

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