Constant area

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

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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Independent of n

Prove that the sum

    \[\mathcal{S}_n = \sum_{j=1}^{n} \sum_{k=1}^{n} (-1)^{j+k} \frac{1}{j! k!} \binom{n-1}{j-1} \binom{n-1}{k-1} \left ( j+k \right )!\]

is independent of n.

Sophomore’s dream constant

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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Area of triangle

Evaluate the area of the given triangle:

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