Differential equation (II)

Let f:(0, +\infty) \rightarrow \mathbb{R} be a differentiable function such that f(1) = \frac{1}{2} and

    \[f'(x) + e^{f(x)} = x + \frac{1}{x} \quad \text{for all} \;\; x >0\]

Find an explicit formula for f.

Solution

Let us consider the function g(x) = f(x) - \ln x - \frac{x^2}{2} which is clearly differentiable. Hence,

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

Thus,

    \[f(x) = \ln x + \frac{x^2}{2} - \ln \left ( e^{x^2/2} + 1 - \sqrt{e} \right )\]

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Differential equation (I)

Let f:\mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f(0)=1 and

    \[f'(x) \sqrt{x^2+1} = f(x) \quad \text{for all} \;\; x \in \mathbb{R}\]

Find an explicit formula for f.

Solution

We have successively

    \begin{align*} f'(x)\sqrt{x^2+1}=f(x) &\Leftrightarrow f'(x)\sqrt{x^2+1}-f(x)=0 \\ &\Leftrightarrow f'(x)\sqrt{x^2+1}-xf'(x)+\frac{xf(x)}{\sqrt{x^2+1}}-f(x)=0 \\ &\Leftrightarrow f'(x)\left ( \sqrt{x^2+1}-x \right )+f(x)\left ( \frac{x}{\sqrt{x^2+1}}-1 \right )=0 \\ &\Rightarrow \left [ f(x)\left ( \sqrt{x^2+1}-x \right ) \right ]'=(c)'\\ &\!\!\!\!\!\!\!\!\!\!\!\!\overset{f(0)=1\Rightarrow c=1}{=\!\!=\!=\!=\!=\!=\!=\!=\! \Rightarrow}f(x)=x+\sqrt{x^2+1}, \; x \in \mathbb{R} \end{align*}

which satisfies the given conditions.

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An infimum

Let \mathcal{F} be the vector space of all continuous functions f:[0, 1] \rightarrow \mathbb{R} such that f(0)=0 . Evaluate

    \[\mathrm{I} = \inf_{f \in \mathcal{F}} \left ( \sup_{0 \leq t \leq 1} \left | 1 - f(t) \right | + \int_{0}^{1} \left | 1 - f(t) \right | \, \mathrm{d}t \right )\]

Maximum value of function

Let p, q \in \mathbb{N}. Find the maximum vale of the function

    \[f(x) = \sin^p x \cos^q x\]

Solution

Using the AM – GM inequality we have

    \begin{align*} \frac{1}{p+q} &= \frac{\overbrace{\frac{\sin^2 x}{p}+ \frac{\sin^2 p}{p} + \cdots + \frac{\sin^2 p}{p}}^{p} + \overbrace{\frac{\cos^2 x}{q} + \frac{\cos^2 x}{q} + \cdots + \frac{\cos^2 x}{q}}^{q}}{p+q} \\ &\geq \sqrt[p+q]{\left ( \frac{\sin^2 x}{p} \right )^p \left ( \frac{\cos^2 x}{q} \right )^q} \\ &=\sqrt[p+q]{\frac{f^2(x)}{p^p q^q}} \end{align*}

Hence,

    \[f(x) = \sqrt{\frac{p^p q^q}{\left ( p+q \right )^{p+q}}}\]

Equality holds when

    \[\frac{\sin^2 x}{p} = \frac{\cos^2 x}{q} \Leftrightarrow \cos^2 x = \frac{p}{p+q}\]

Since \frac{p}{p+q} \in (0, 1) there exists an x \in \mathbb{R} such that the equation has at least one root. The minimum follows.

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Infinite product

Prove that

    \[\prod_{n=0}^{\infty} \frac{\cosh \left ( n^2 + n + \frac{1}{2} \right ) + i \sinh \left ( n + \frac{1}{2} \right )}{\cosh \left ( n^2 + n + \frac{1}{2} \right ) - i \sinh \left ( n + \frac{1}{2} \right )} = i\]

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