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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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Fibonacci series

Let \{F_n\}_{n \in \mathbb{N}} denote the Fibonacci sequence such that F_1=F_2=1 and F_3=2. Prove that

    \[\sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{1}{F_{2n}}  =\frac{\log 3}{2}\]

Solution

It holds that

    \[{\rm arctanh}\; x - {\rm arctanh} \; y= {\rm arctanh} \left ( \frac{x-y}{1-xy} \right )\]

It follows from Cassini’s identity that

(1)   \begin{equation*} {F_{n-1} F_{n+1} - F_n^2 = (-1)^n \end{equation*}

Setting n \mapsto 2n back at (1) we get

(2)   \begin{equation*} F^2_{2n}=F_{2n-1} F_{2n+1} -1 \end{equation*}

Since F_n =F_{n-1} + F_{n-2} we get

(3)   \begin{equation*} F_{2n+1} =F_{2n} + F_{2n-1} \Leftrightarrow F_{2n} = F_{2n+1} - F_{2n-1} \end{equation*}

Hence,

    \begin{align*} \sum_{n=2}^{\infty} {\rm arctanh}\;   \frac{1}{F_{2n}}  &= \sum_{n=2}^{\infty} {\rm arctanh} \; \frac{F_{2n}}{F_{2n}^2}  \\ &=\sum_{n=2}^{\infty} {\rm arctanh} \frac{F_{2n-1} - F_{2n+1}}{1-F_{2n-1}F_{2n+1}} \\ &= \sum_{n=2}^{\infty} {\rm arctanh} \;  \frac{\frac{1}{F_{2n-1}} - \frac{1}{F_{2n+1}}}{1- \frac{1}{F_{2n-1} F_{2n+1}}} \\ &=\lim_{m \rightarrow +\infty} \sum_{n=2}^{m} \bigg [ {\rm arctanh} \left ( \frac{1}{F_{2n-1}} \right ) - \\ &\quad \quad \quad - {\rm arctanh} \left ( \frac{1}{F_{2n+1}} \right ) \bigg ] \\ &= \lim_{m \rightarrow +\infty} \left [ {\rm arctanh} \left ( \frac{1}{F_3} \right ) - {\rm arctanh} \left ( \frac{1}{F_{2m+1}} \right )\right ]\\ &= {\rm arctanh} \left ( \frac{1}{2} \right ) \\ &= \frac{\log 3}{2} \end{align*}

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Logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_{0}^{1} \ln^2 \left | \sqrt{x} - \sqrt{1-x} \right |\, \mathrm{d}x\]

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