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Contents Index Dark Mode Prev Up Next \(\def\versionnumber{1.54}
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Section 10.2 Binomial Coefficients
The binomial coefficient \(\binom n k = \frac{ 1 }{ k! } n(n-1)(n-2) \cdots (n-k+1)\) is a natural candidate for being explored analytically, as the binomial theorem
\begin{equation*}
(x+y)^n \ = \ \sum_{k=0}^n \binom n k x^k \, y^{n-k}
\end{equation*}
(for \(x,y \in \C\) and \(n \in \Z_{ \ge 0 }\) ) tells us that \(\binom n k\) is the coefficient of \(z^k\) in \((z+1)^n\text{.}\) You will derive two sample identities in the course of the exercises below.
Convince yourself that
\begin{equation*}
\binom{n}{k} \ = \ \frac 1 {2 \pi i} \int_\gamma \frac{
(z+1)^{n} }{ z^{k+1} } \, \diff{z}
\end{equation*}
where \(\gamma\) is any simple closed piecewise smooth path such that 0 is inside \(\gamma\text{.}\)
Derive a recurrence relation for binomial coefficients from the fact that
\(\frac 1 z + 1 = \frac{ z+1 }{ z }\text{.}\) (
Hint : Multiply both sides by
\(\frac{ (z+1)^n }{ z^k }\text{.}\) )
Now suppose \(x \in \R\) with \(|x|\lt 1/4\text{.}\) Find a simple closed path \(\gamma\) surrounding the origin such that
\begin{equation*}
\s \left( \frac{ (z+1)^2 }{ z } \, x \right)^k
\end{equation*}
converges uniformly on \(\gamma\) as a function of \(z\text{.}\) Evaluate this sum.
Keeping
\(x\) and
\(\gg\) from
3 , convince yourself that
\begin{equation*}
\s \binom{2k}{k} x^k \ = \ \frac 1 {2 \pi i} \s \int_\gamma
\frac{ (z+1)^{2k} }{ z^{k+1} } \, x^k \, \diff{z} \, \text{,}
\end{equation*}
Use (
3 ) to interchange summation and integral, and use the Residue
Theorem 9.2.2 to evaluate the integral, giving an identity for
\(\s \binom{2k}{k} x^k\text{.}\)
