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Bessel series

February 22, 2020 / Leave a comment

Let J_n denote the Bessel function of the first kind. Prove that:

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]

Solution

The Jacobi – Anger expansion tells us that

    \[e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}\]

Hence by Parseval’s Theorem it follows that:

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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Pell – Lucas series

February 22, 2020 / Leave a comment

The Pell – Lucas numbers \mathcal{Q}_n are defined as follows \mathcal{Q}_0 = \mathcal{Q}_1 =2 and for every n \geq 2 it holds that

    \[\mathcal{Q}_n = 2\mathcal{Q}_{n-1} +\mathcal{Q}_{n-2}\]

Prove that

    \[\sum_{n=1}^{\infty} \arctan \frac{2}{\mathcal{Q}_n} \arctan \frac{2}{\mathcal{Q}_{n+1}} = \frac{\pi^2}{32}\]

 

Gamma infinite product

January 8, 2020 / Leave a comment

Prove that

    \[\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}} = \frac{8\sqrt{\pi}}{e^2}\]

Solution

Converting the product to a sum and using duplication formula for the gamma function and telescoping,

\displaystyle \sum_{n=1}^{N}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-N}\right)\ln\left(2\sqrt{\pi}\right)-2N\ln2+\frac{2}{2^{N+1}}\ln\Gamma(2^{N+1})

Using Stirling formula

    \[\frac{1}{N}\ln\Gamma(N)=\ln N-1+\mathcal{O}\left(\frac{\ln N}{N}\right)\quad \text{as}\;\; N\rightarrow\infty\]

we get that

    \[\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}\]

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Digamma and Trigamma functions

December 16, 2019 / Leave a comment

Let \psi^{(0)} and \psi^{(1)} denote the digamma and trigamma functions respectively. Prove that:

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

We begin with the recently discovered identity:

    \begin{align*} \log \Gamma(1+x) &= \frac{\ln 2 \pi -1}{2} -\gamma \left ( x+\frac{1}{2} \right ) \\ &\quad + \frac{2x-1}{2} + \sum_{n=1}^{\infty} \left ( \psi^{(0)}(n+1) -\ln(x+n) +\frac{2x-1}{2(1+n)} \right ) \end{align*}

Letting x=1 we get that

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n +\frac{1}{2n} \right ) = \frac{1+\gamma-\ln 2\pi}{2}\]

Now combining this result here we conclude the exercise.

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Trigamma series

December 16, 2019 / Leave a comment

Let \psi^{(1)} denote the trigamma function. Prove that

    \[\sum_{n=1}^{\infty} \left (\psi^{(1)} (n) - \frac{1}{n} \right )=1\]

Solution

    \begin{align*} \sum_{n=1}^{\infty} \left ( \psi^{(1)}(n) - \frac{1}{n} \right ) &= \lim_{N \rightarrow +\infty}\sum_{n=1}^{N} \left ( \psi^{(1)}(n) - \frac{1}{n} \right ) \\ &=\lim_{N \rightarrow +\infty} \left ( \log N + 1 + \gamma + \mathcal{O}\left( \frac{1}{N} \right) - \mathcal{H}_n \right ) \\ &=\lim_{N \rightarrow +\infty} \left ( \log N - \mathcal{H}_n + \mathcal{O}\left(\frac{1}{N}\right) \right ) + 1 + \gamma\\ &=-\gamma + 1 + \gamma \\ &= 1 \end{align*}

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