Binomial sum

February 1, 2021 / Leave a comment

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

January 26, 2021 / Leave a comment

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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Contour integral

January 20, 2021 / Leave a comment

Evaluate the integral

    \[\mathcal{J}_{n, k} = \oint \limits_{\left | z \right |=1} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z\]

Solution

The function \displaystyle \frac{\left ( 1+z \right )^n}{z^{k+1}} is meromorphic on \mathbb{C}. Its only pole is 0 of order k+1. Hence,

    \begin{align*} \mathfrak{Res} \left ( f; z=0 \right ) &=\frac{1}{k!} \lim_{z \rightarrow 0} \left ( z^{k+1} f(z) \right )^{(k)} \\ &=\frac{1}{k!} \lim_{z\rightarrow 0} \left ( \left ( 1+z \right )^n \right )^{(k)} \\ &= \frac{n \left ( n-1 \right ) \cdots \left ( n- k+1 \right )}{k!} \\ &= \binom{n}{k} \end{align*}

Therefore,

    \[\mathcal{J}_{n, k} = 2 \pi i \binom{n}{k}\]

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Nested binomial sum

January 20, 2021 / Leave a comment

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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An arcosine integral

January 20, 2021 / Leave a comment

Evaluate the integral

    \[\mathcal{J} = \int_{-1}^{1} \frac{\arccos x}{\sqrt{3x^4+2x^2+3}} \, \mathrm{d}x\]

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