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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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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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Binomial sum

Let m, n be positive numbers with n > m . Prove that

    \[\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{m+n-2k}{n-1} = \binom{n}{m+1}\]

Solution

Using the exercise here we have that

    \[\binom{n}{m} = \frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^n}{z^{m+1}}\, \mathrm{d}z\]

Hence,

\begin{aligned} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{m+n-2k}{n-1} &= \frac{1}{2\pi i } \sum_{k=0}^{n} (-1)^k \binom{n}{k} \oint \limits_{\left | z \right |=1} \frac{\left ( 1+z \right )^{m+n-2k}}{z^n}\, \mathrm{d}z \\ &=\frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^{m+n}}{z^n} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{\mathrm{d}z}{\left ( z+1 \right )^{2k}} \\ &= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=1} \frac{\left ( z+1 \right )^{m+n}}{z^n} \left ( 1 - \frac{1}{\left ( z+1 \right )^2} \right )^n \, \mathrm{d}z \\ &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\left ( z+2 \right )^n}{\left ( z+1 \right )^{n-m}}\, \mathrm{d}z \\ &=\mathfrak{Res}_{z =-1} \frac{\left ( z+2 \right )^n}{\left ( z+1 \right )^{n-m}} \\ &= \lim_{z \rightarrow -1} \frac{1}{\left ( n-m-1 \right )!} \frac{\mathrm{d}^{n-m-1} }{\mathrm{d} z^{n-m-1}} \left (\left ( z+2 \right )^n \right ) \\ &= \binom{n}{m+1} \end{aligned}

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Logarithmic inequality

Let 0< p \leq q. Prove that

    \[\ln \frac{p}{q} \leq \frac{p-q}{\sqrt{pq}}\]

Solution

Let f(x)=\frac{1}{x} and g(x)=1. Thus,

    \[\left(\int_{q}^{p} \frac{1}{x} \, \mathrm{d}x \right)^2 \leq \int_{q}^{p} \frac{1}{x^2}\, \mathrm{d}x \int_{q}^{p} 1 \, \mathrm{d}x\]

Thus,

    \begin{align*} (\ln p-\ln q)^2 &\leq \left(-\frac{1}{p}+\frac{1}{q}\right)(p-q)\\ &=\frac{(p-q)^2}{pq} \end{align*}

The result follows.

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Nested binomial sum

Prove that

    \[\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\ldots+b_n}{a_1+a_2+\ldots+a_n}}=b_1+b_2+\cdots+b_n+1\]

Solution

We may begin with the beta function identity for non negative integer values of a, b.

    \[\int_0^1 x^{b-a}(1-x)^a \, \mathrm{d}x = \frac{1}{(b+1)\binom{b}{a}}\]

Hence, for non-negative integers a', b'

    \begin{align*} \sum_{a = 0}^{b}\frac{\binom{b}{a}}{\binom{b+b'}{a+a'}} &= (b+b'+1)\sum_{a = 0}^{b}\binom{b}{a}\int_0^1 x^{b+b'-a-a'}(1-x)^{a+a'}\,\mathrm{d}x\\ &= (b+b'+1)\int_0^1 x^{b'-a'}(1-x)^{a'}\, \mathrm{d}x\\ &= \frac{b+b'+1}{(b'+1)\binom{b'}{a'}} \end{align*}

As a result we may compute the nested summation as,

\begin{aligned} \sum_{a_1 = 0}^{b_1}\cdots \sum_{a_n = 0}^{b_n}\frac{\binom{b_1}{a_1}\cdots \binom{b_n}{a_n}}{\binom{b_1+\cdots+b_n}{a_1+\cdots+a_n}} &= \frac{b_1+\cdots + b_n +1}{b_1+\cdots+b_{n-1}+1}\sum_{a_1 = 0}^{b_1}\cdots \sum_{a_{n-1} = 0}^{b_{n-1}}\frac{\binom{b_1}{a_1}\cdots \binom{b_{n-1}}{a_{n-1}}}{\binom{b_1+\cdots+b_{n-1}}{a_1+\cdots+a_{n-1}}} \\ &= \cdots \\ &= \prod_{j=0}^{n-2}\frac{b_1+\cdots+b_{n-j}+1}{b_1+\cdots+b_{n-j-1}+1}\sum_{a_1=0}^{b_1}\frac{\binom{b_1}{a_1}}{\binom{b_1}{a_1}}\\ &= b_1+\cdots+b_n+1 \end{aligned}

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