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On Euler’s totient function series

Let \phi denote Euler’s totient function. Prove that for s>2 it holds that:

    \[\sum_{n=1}^{\infty} \frac{(-1)^n \phi(n)}{n^s} =-\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-3}{2^s-1}\]

where \zeta stands for the Riemann zeta function.

Solution

Well by Euler’s product we have,

    \[\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_{p}\left(1+\frac{\phi(p)}{p^s}+\frac{\phi(p^2)}{p^{2s}}+\frac{\phi(p^3)}{p^{3s}}+\cdots\right)= \prod_{p}\frac{p^s-1}{p^s-p}\]

thus,

(1)   \begin{equation*} \sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_p \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}  \end{equation*}

and

(2)   \begin{equation*} \sum_{\substack{n\geq 1\\n\text{ odd}}}\frac{\phi(n)}{n^s} = \prod_{p>2} \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-2}{2^s-1}  \end{equation*}

Combining we get the result.

Note: It also holds that

    \[\sum_{n=1}^\infty \frac{\phi(2n-1)}{(2n-1)^s} = \frac{(1-2^{1-s})\zeta(s-1)}{1-2^{-s} \zeta(s)}\]

 

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Logarithmic mean inequality

Let x, y>0 such that x \neq y. Prove that

    \[\sqrt{xy} \leq \frac{x-y}{\ln x- \ln y} \leq \frac{x+y}{2}\]

Solution

We are invoking the Hermite – Hadamard Inequality for the convex function f(x)=e^x \; , \; x \in \mathbb{R}. Thus,

    \begin{align*} f\left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{1}{\beta-\alpha} \int_{\alpha}^{\beta} f(x) \; \mathrm{d}x \leq \frac{f(\alpha)+f(\beta)}{2} &\Rightarrow \\ \exp \left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{1}{\beta-\alpha} \int_{\alpha}^{\beta} e^x \; \mathrm{d}x \leq \frac{e^\alpha+e^\beta}{2} &\Rightarrow \\ \exp \left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{e^{\beta}-e^{\alpha}}{\beta-\alpha} \leq \frac{e^{\alpha}+e^{\beta}}{2}&\overset{\alpha=\ln x \;, \; \beta=\ln y}{=\! =\! =\! =\! =\! =\! =\!\Rightarrow } \\ \exp \left ( \frac{\ln x + \ln y}{2} \right ) \leq \frac{e^{\ln y} - e^{\ln x}}{\ln y- \ln x} \leq \frac{e^{\ln x} + e^{\ln y}}{2} \\ \exp \left ( \frac{\ln xy}{2} \right )\leq \frac{y-x}{\ln y- \ln x} \leq \frac{x+y}{2}&\Rightarrow \\ \sqrt{xy} \leq \frac{x-y}{\ln x - \ln y} \leq \frac{x+y}{2} \end{align*}

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On an integral

Evaluate the integral

    \[\mathcal{J} = \int_{0}^{\pi/4} \left ( \frac{x}{x \sin x + \cos x} \right )^2 \, \mathrm{d}x\]

Solution

Successively we have:

    \begin{align*} \mathcal{J} &=\int_{0}^{\pi/4} \left ( \frac{x}{x \sin x + \cos x} \right )^2 \, \mathrm{d}x \\ &=-\int_{0}^{\pi/4} \frac{-x \cos x}{\left ( x \sin x + \cos x \right )^2} \cdot \frac{x}{\cos x} \, \mathrm{d}x \\ &=-\int_{0}^{\pi/4} \left ( \frac{1}{x \sin x+ \cos x} \right )' \frac{x}{\cos x} \, \mathrm{d}x \\ &=-\left [ \frac{1}{x\sin x + \cos x } \cdot \frac{x}{\cos x} \right ]_{0}^{\pi/4} + \int_{0}^{\pi/4} \frac{1}{x \sin x + \cos x} \cdot \left ( \frac{x}{\cos x} \right )' \, \mathrm{d}x \\ &= -\frac{2\pi}{\pi+4} + \int_{0}^{\pi/4} \frac{1}{x \sin x + \cos x} \cdot \frac{x \sin x+ \cos x}{\cos^2 x} \, \mathrm{d}x \\ &= -\frac{2\pi}{\pi+4} + \int_{0}^{\pi/4} \frac{\mathrm{d}x}{\cos^2 x}\\ &= -\frac{2\pi}{\pi+4} + \left [ \tan x \right ]_{0}^{\pi/4} \\ &= 1-\frac{2\pi}{\pi+4} \\ &=\frac{4-\pi}{4+\pi} \end{align*}

We used the simple observation that

    \[\left ( \frac{1}{x \sin x + \cos x} \right ) ' = - \frac{x \cos x}{\left ( x \sin x + \cos x \right )^2}\]

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Limit of a sequence

Let f:\mathbb{R}\rightarrow \mathbb{R} be a function such that f(0)=0 and f is differentiable at 0. Let us set

    \[a_n =\sum_{k=1}^{n} f \left( \frac{k}{n^2} \right)\]

Evaluate the limit \lim \limits_{n \rightarrow +\infty} a_n.

Solution

Since f is differentiable at 0 , there is some \varepsilon: x \mapsto \varepsilon(x) such that

    \[f(x) = f(0) + x f'(0) + x \varepsilon(x) \quad , \quad \varepsilon(0) =0\]

and \varepsilon is of course continuous.

Thus,

    \[\sum_{k=1}^{n} f\left ( \frac{k}{n^2} \right ) = \frac{f'(0)}{n} \sum_{k=1}^{n} \frac{k}{n} + \sum_{k=1}^{n} \frac{k}{n^2} \varepsilon \left ( \frac{k}{n^2} \right )\]

Let \epsilon>0. There exists \delta>0 such that |x| \leq \delta which in return means that |\varepsilon(x)| \leq \epsilon. Hence , for n larger than \frac{1}{\delta}+1 it holds that

    \[\left| \sum_{k=1}^n \frac{k}{n^2}\varepsilon \left( \frac{k}{n^2} \right) \right|\leq \epsilon \frac {n}{n}  = \epsilon\]

On the other hand , the sum \displaystyle \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} is a Riemann sum and converges to \frac{1}{2}.

In conclusion,

    \[\lim_{n \rightarrow +\infty} a_n = \frac{1}{2}\]

 

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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]

Solution

We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}

Hence,

    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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