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Constant area

June 2, 2021 / Leave a comment

Let a be a positive real number. The parabolas defined by y_1=ax^2 and y_2^2=ax intersect at the points \mathrm{O} and \mathrm{A}.

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Prove that the area enclosed by the two curves is constant. Explain why.

Solution

First of all we note that

    \begin{align*} y_1 = y_2 &\Leftrightarrow \left ( ax^2 \right )^2 = ax \\ &\Leftrightarrow a^2 x^4 = ax \\ &\!\!\!\!\!\overset{a>0}{\Leftarrow \! =\! =\! \Rightarrow } a x^4 - x =0 \\ &\Leftrightarrow x \left ( ax^3 -1 \right ) =0 \\ &\Leftrightarrow \left\{\begin{matrix} x & = & 0\\ x &= & \sqrt[3]{\frac{1}{a}} \end{matrix}\right. \end{align*}

Hence,

    \begin{align*} \mathrm{E}\left ( \Omega \right ) &= \int_{0}^{\sqrt[3]{1/a}} \left | ax^2 - \sqrt{ax} \right |\, \mathrm{d}x\\ &=\int_{0}^{\sqrt[3]{1/a}} \left ( \sqrt{ax} - ax^2 \right )\, \mathrm{d}x \\ &=\frac{2}{3} - \frac{1}{3} \\ &= \frac{1}{3} \end{align*}

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Binomial sum

May 26, 2021 / Leave a comment

Let \mathbb{N} \ni k >1. Evaluate the sum

    \[\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{\binom{n+k}{k}}\]

Solution

We have successively

\begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{\binom{n+k}{k}} &= \sum_{n=1}^{\infty} \frac{k!H_n}{(n+1) \cdots (n+k)} \\ &= \sum_{n=1}^{\infty} \frac{k!H_n}{k-1} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1}\sum_{n=1}^{\infty} \sum_{i=1}^n \frac{1}{i} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i} \sum_{n=i}^{\infty} \left(\frac{1}{(n+1)\cdots (n+k-1)} - \frac{1}{(n+2) \cdots (n+k)} \right) \\ &= \frac{k!}{k-1} \sum_{i=1}^{\infty} \frac{1}{i(i+1) \cdots (i+k-1)} \\ &= \frac{k!}{(k-1)^2} \sum_{i=1}^{\infty} \left(\frac{1}{i(i+1) \cdots (i+k-2)} - \frac{1}{(i+1) \cdots (i+k-1)}\right) \\ &= \frac{k!}{(k-1)^2} \frac{1}{(k-1)!} \\ &= \frac{k}{(k-1)^2} \end{aligned}

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Sophomore’s dream constant

May 14, 2021 / Leave a comment

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \int_0^1 (xy)^{xy} \,\mathrm{d}(x, y)\]

Solution

Let t=xy and s=y. The Jacobian is

    \[\frac{\partial (s, t)}{\partial (x, y)} = y \Rightarrow \left (\frac{\partial (s, t)}{\partial (x, y)} \right )^{-1} = \frac{1}{y} = \frac{1}{s}\]

Hence,

    \begin{align*} \mathcal{J} &= \iint \limits_{[0, 1]^2} \left ( xy \right )^{xy} \, \mathrm{d}(x, y) \\ &=\int_{0}^{1} \int_{0}^{s} \frac{t^t}{s} \, \mathrm{d}(t, s) \\ &= \int_{0}^{1} \int_{t}^{1} \frac{t^t}{s} \, \mathrm{d} ( s, t)\\ &= -\int_{0}^{1} t^t \log t \, \mathrm{d}t \end{align*}

However , since \displaystyle \int_{0}^{1} t^t \left ( 1 + \log t \right ) \, \mathrm{d}t =0 we conclude that

    \[\mathcal{J} = \int_0^1 t^t \, \mathrm{d}t = \mathcal{S}\]

where \mathcal{S} is Sophomore’s dream constant.

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Differential equation (II)

May 4, 2021 / Leave a comment

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)

May 4, 2021 / Leave a comment

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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