Galois theory … of the Euler’s totient function

Let n>2 and let \omega \in \mathbb{C} be an n-th primitive root of unity. Prove that

    \[[\mathbb{Q}(\omega + \omega^{-1}) :\mathbb{Q}]=\frac{\varphi (n)}{2}\]

where \phi denotes the Euler’s totient function.

Solution

We have

    \[\phi(n)=[\mathbb{Q}(\omega):\mathbb{Q}]=[\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})][\mathbb{Q}(\omega +\omega^{-1}):\mathbb{Q}]\]

where [\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})]=2 since t^2-(\omega+\omega^{-1})t+1 is the minimal polynomial of \omega over \mathbb{Q}(\omega+\omega^{-1}).

The result follows.

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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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Trigonometric sum

Prove that

    \[\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}=\frac{1}{2}\]

Solution

Consider the tridiagonal matrix A =\begin{pmatrix} 1&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix} . Its eigenvalues are \displaystyle 2 \cos \frac{\pi}{7} \; , \; 2 \cos \frac{3\pi}{7} \; , \; 2 \cos \frac{5 \pi}{7}. Hence,

    \[\Tr\left ( A \right ) = 1 = 2 \left ( \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} \right )\]

and the result follows.

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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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