Zero determinant

May 24, 2020 / Leave a comment

Consider the real numbers x_i, x_j for 1 \leq i , j \leq 3 . Prove that

    \[\mathcal{D} = \begin{vmatrix} \sin (x_1+y_1) & \sin (x_1+y_2) & \sin (x_1+y_3)\\ \sin (x_2+y_1) & \sin (x_2+y_2) & \sin (x_2+y_3)\\ \sin (x_3+y_1) & \sin (x_3+y_2) & \sin (x_3+y_3) \end{vmatrix}=0\]

Solution

Using the identity \sin (a+b)=\sin a\cos b+\sin b\cos a in combination with \det AB = \det A \det B we have:

    \[\mathcal{D}=\begin{vmatrix} \sin x_1 & \cos x_1 & 0\\ \sin x_2 & \cos x_2 & 0\\ \sin x_3 & \cos x_3 & 0 \end{vmatrix} \cdot \begin{vmatrix} \cos y_1 & \cos y_2 & \cos y_3 \\ \sin y_1 & \sin y_2 & \sin y_3 \\ 0 & 0 & 0 \end{vmatrix} =0\]

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Arctan squared integral

May 23, 2020 / Leave a comment

Evaluate the integral

    \[\mathcal{J} =\int_0^\pi \arctan^2 \left( \frac{\sin x}{2+\cos x} \right) \, \mathrm{d}x\]

Solution

First of all we note that for x \in (0, \pi)

    \begin{align*} \arctan \left ( \frac{\sin x}{2 + \cos x} \right ) &= \mathfrak{Im} \log \left ( 2 + e^{ix} \right ) \\ &= \mathfrak{Im} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n 2^n} e^{inx} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} }{n 2^n} \sin nx \end{align*}

Hence by Parseval we get that

    \[\int_{0}^{\pi} \arctan^2 \left(\frac{\sin x}{2+\cos x}\right)\,\mathrm{d} x=\frac{\pi}{2} \cdot \sum_{n=1}^{\infty} \frac{1}{n^2 4^n}=\frac{\pi}{2} \cdot \mathrm{Li}_2 \left ( \frac{1}{4} \right )\]

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Values of parameter

May 23, 2020 / Leave a comment

Find all values of \lambda \in \mathbb{R} such that

    \[e^x - \lambda x \geq 0 \quad \text{forall} \; x \in \mathbb{R}\]

Solution

We’re invoking the same technique as in the problem here. Successively we have:

    \begin{align*} e^x - \lambda x \geq 0 &\Leftrightarrow e^x \geq \lambda x \\ &\Leftrightarrow \lambda x e^{-x} \leq 1 \\ &\Leftrightarrow f(x) \leq 1 \end{align*}

Obviously f is differentiable in \mathbb{R} and its derivative is

    \[f'(x) = \lambda e^{-x} \left ( 1-x \right )\]

It is f'(x) =0 \Leftrightarrow x=1. We distinguish cases:

  • If \lambda>0 then f attains global maximum at x=1 equal to f(1)=\frac{\lambda}{e}. Since f(x) \leq 1 it follows that \lambda \leq e.
  • If \lambda<0 then f attains global minimum at x=1 equal to f(1)=\frac{\lambda}{e}. In this case , however , the inequality cannot hold for all \lambda <0; since f is continuous its range is:

        \begin{align*} f(\mathbb{R}) &= \left [ f(1), \lim_{x \rightarrow -\infty} f(x) \right ) \cup \left [ f(1), \lim_{x \rightarrow +\infty} f(x) \right ) \\ &= \left [ \frac{\lambda}{e}, +\infty \right ) \cup \left [ \frac{\lambda}{e} , 0 \right ) \\ &= \left [ \frac{\lambda}{e}, +\infty \right ) \end{align*}

    Hence this case is rejected.

  • For \lambda=0 the inequality obviously holds for all x \in \mathbb{R}.

Summing up , \lambda \in [0, e].

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Convergence of a series

May 12, 2020 / 1 Comment on Convergence of a series

Examine the convergence of a series

    \[\sum_{n=1}^{\infty} \left ( 1 - 2\exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) \right )\]

Solution

From Taylor’s theorem with integral remainder we have that

    \[-\ln 2 = \sum_{k=1}^{n} \frac{(-1)^k}{k} + (-1)^{n+1} \int_0^1 \frac{x^n}{1+x} \, \mathrm{d}x\]

However it is known that \displaystyle \int_{0}^{1} \frac{x^n}{1+x} \, \mathrm{d}x = \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right ). Hence,

    \[\sum_{k=1}^{n} \frac{(-1)^k}{k} = -\ln 2 + \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Exponentiating we get

    \[1- 2 \exp \left ( \sum_{k=1}^{n} \frac{(-1)^k}{k} \right ) = \frac{(-1)^n}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Thus the series converges.

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Rational function and polynomial

May 9, 2020 / Leave a comment

Prove that there does not exist a rational function f with real coefficients such that

    \[f \left ( \frac{x^2}{x+1} \right ) = \mathrm{P}(x)\]

where \mathrm{P}(x) \in \mathbb{R}[x] is a non constant polynomial.

Solution

Since polynomials are defined on x=-1 we have that

    \begin{align*} \mathbb{R} \ni \mathrm{P}(-1) & =\lim_{x \rightarrow -1^+} \mathrm{P}(x) \\ &=\lim_{x \rightarrow -1^+}f \left(\frac{x^2}{x+1} \right) \\ &=\lim_{x \rightarrow \infty} f\left(\frac{x^2}{x+1}\right) \\ &=\lim_{x \rightarrow \infty} \mathrm{P}(x) \end{align*}

Since \mathrm{P} tends to a finite value as x \rightarrow \infty it must be a constant polynomial. In particular, f must be constant in the range of \frac{x^2}{x+1} which is an infinite set, implying that f must also be constant. This proves what we wanted.

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