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Proving identities of special functions

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Re: Proving identities of special functions

Sat Aug 31, 2013 2:18 pm
Random Variable Integration Guru

Posts: 381
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

Re: Proving identities of special functions

Sat Aug 31, 2013 8:28 pm
Random Variable Integration Guru

Posts: 381
Problem 4

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

Re: Proving identities of special functions

Mon Sep 02, 2013 5:55 am

Posts: 850
Location: Jaipur, India

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*}

Re: Proving identities of special functions

Sat Sep 07, 2013 7:15 am
zaidalyafey
Global Moderator

Posts: 354
Problem 5

Prove that

$\displaystyle \sum_{p=0}^{q-1}\psi(a+p/q)=q(\psi(qa)-\log(q))$
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

Re: Proving identities of special functions

Sat Sep 07, 2013 8:40 am

Posts: 850
Location: Jaipur, India

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 $a$:

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

Re: Proving identities of special functions

Sat Sep 07, 2013 5:23 pm

Posts: 138
Location: North Londinium, UK
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

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

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

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

Re: Proving identities of special functions

Sun Sep 08, 2013 7:53 pm
Random Variable Integration Guru

Posts: 381
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.

Re: Proving identities of special functions

Tue Sep 10, 2013 6:48 pm

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 ( ), 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]

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

Re: Proving identities of special functions

Wed Sep 11, 2013 7:55 pm
Random Variable Integration Guru

Posts: 381
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 $\displaystyle \prod_{k=1}^{n-1} \Gamma \Big(\frac{k}{n} \Big) = \sqrt{\prod_{k=1}^{n-1}\frac{\pi}{\sin\left(\frac{\pi k}{n}\right)}}$

Re: Proving identities of special functions

Wed Sep 11, 2013 11:01 pm
Random Variable Integration Guru

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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