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Differential equation

Let f:(0, +\infty) \rightarrow \mathbb{R} be a twice differentiable function such that

    \[x^2 f''(x) + x f'(x) = x - 2 \quad \text{for all} \;\; x>0\]

If f(1) =0 , f'(1)=1 find an explicit formula of f.

Solution

We have successively

    \begin{align*} x^2 f''(x) + x f'(x) = x - 2 &\Rightarrow x f''(x) + f'(x) = 1 - \frac{2}{x} \\ &\Rightarrow \left ( x f'(x) \right )' = \left ( x - 2 \ln x \right )' \\ &\Rightarrow x f'(x) = x - 2 \ln x + c_1 \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c_1=0}{=\! =\! =\!=\! =\! =\!\Rightarrow } x f'(x) = x - 2 \ln x \\ &\Rightarrow f'(x) = 1 - \frac{2 \ln x}{x} \\ &\Rightarrow \left ( f(x) \right ) ' = \left ( x - \ln^2 x \right )' \\ &\Rightarrow f(x) = x - \ln^2 x + c \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c=-1}{=\! =\! =\!=\! =\! =\!\Rightarrow } f(x) = x - \ln^2 x - 1 \end{align*}

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Polylogarithm series

Let \mathrm{Li}_n denote the polylogarithm. Prove that

    \[\sum_{n=2}^{\infty} \left( \mathrm{Li}_n(1) - 1 \right) =1\]

Solution

Since \displaystyle \zeta(n) = \sum_{k=1}^{\infty} \frac{1}{k^n} we have successively

    \begin{align*} \sum_{n=2}^{\infty} \left ( \mathrm{Li}_n(1) - 1 \right ) &= \sum_{n=2}^{\infty} \left ( \zeta(n) - 1 \right ) \\ &= \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{k^n} \\ &=\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^n} \\ &= \sum_{k=2}^{\infty} \left ( \frac{1}{k-1} - \frac{1}{k} \right ) \\ &= 1 \end{align*}

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Convex function

Let f be a convex function on a convex domain \Omega and g a convex non-decreasing function on \mathbb{R}. Prove that the composition of g \circ f is convex on \Omega.

Solution

We want to prove that for x, y \in \Omega it holds that

    \[(g \circ f)\left(\lambda x + (1 - \lambda) y\right) \le \lambda (g \circ f)(x) + (1 - \lambda)(g \circ f)(y)\]

We have:

    \begin{align*} (g \circ f)\left(\lambda x + (1 - \lambda) y\right) &= g\left(f\left(\lambda x + (1 - \lambda) y\right)\right) \\ &\le g\left(\lambda f(x) + (1 - \lambda) f(y)\right) & \text{(} f \text{ convex and } g \text{ nondecreasing)} \\ &\le \lambda g(f(x)) + (1 - \lambda)g(f(y)) & \text{(} g \text{ convex)} \\ &= \lambda (g \circ f)(x) + (1 - \lambda)(g \circ f)(y) \end{align*}

 

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Polynomial equation

Let \Phi denote the golden ratio. Solve the equation

    \[x^3 +x^2-\Phi^5 x +\Phi^5=0\]

Solution

First of all we note that

    \[\left\{ \begin{array}{l} \Phi ^2 = \Phi + 1\\ \Phi ^3 = \Phi ^2 + \Phi = 2\Phi + 1 \end{array} \right. \Rightarrow \Phi ^5 = 2 \Phi ^2 + 3\Phi + 1 \Rightarrow \bold{\Phi ^5 = 5\Phi + 3}\]

We easily note that \Phi is one root of the equation, hence using Horner we get that

    \begin{align*} x^2 + \Phi ^2x - (2\Phi ^2 + \Phi ) = 0 &\Leftrightarrow x = \frac{ - \Phi ^2 \pm \sqrt {9\Phi ^2 + 6\Phi + 1}}{2} \\ &\Leftrightarrow x = \frac{ - (\Phi + 1) \pm (3\Phi + 1)}{2} \end{align*}

Hence \Phi is a double root and the other root is x=-2-\sqrt{5}.

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Similarity implies equality?

Let A\in \mathbb{C}^{n\times n} be similar to A^2. Does A=A^2 hold?

Solution

No! Take A=\begin{pmatrix}1 &0 \\1&1 \end{pmatrix} then A^2=\begin{pmatrix}1 &0\\ 2&1 \end{pmatrix}. The matrices are similar but not equal.

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