In this post, we will define binomial coefficients using integration of complex valued functions and prove high school identities using integration methods. 12th standard students who are learning calculus should be able to follow this post. You can consider this as a practice in integration theory. I usually call such fun extras as JEE side quests.
The following is the school definition of binomial coefficients.
Definition:
Let $k$ be a whole number and $n$ be any complex number, then define $$\binom{n}{k} := \dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}.$$
In this post, we will define binomial coefficients using integration (Fourier theory) and proving identities will reduce to fun integration problems.
Fourier Definition
If $n$ is a whole number, then the binomial coefficient is the number of different ways of choosing $k$ identical objects out of $n$ distinct objects. In a separate post I will introduce different ways of thinking about this notion. But here I want to talk about a crazy formula using complex numbers for whole numbers $n,k$:
Blog definition:
Let $k$ be a whole number and $n$ is any natural number, then define $$\binom{n}{k}=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^n e^{-ik\theta}\,\, d\theta$$
The proof simply follows from binomial theorem. If you know Fourier transform theory, then we are simply saying that the Fourier coefficients $(1+e^{i\theta})^n$ are the binomial coefficients.
OK, so what?
Binomial identities
We know and love many binomial identities, but how do we prove them using the Fourier definition?
Identity 1: $$\binom{n}{k} = \binom{n}{n-k}$$
Proof: We know that $$\binom{n}{k}=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^n e^{-ik\theta}\,\, d\theta$$
Substitute $\theta=-\alpha$, we get $$\begin{aligned} \binom{n}{k}=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^n e^{-ik\theta}\,\, d\theta &= \dfrac1{2\pi}\displaystyle \int_0^{-2\pi} (1+e^{-i\alpha})^n e^{ik\alpha}\,\, (-d\alpha)\\ &\stackrel{a}{=} -\dfrac1{2\pi}\displaystyle \int_{2\pi}^{0} (1+e^{-i\alpha’+i2\pi })^n e^{ik(\alpha’-2\pi)}\,\, d\alpha’ \\ &\stackrel{b}{=} -\dfrac1{2\pi}\displaystyle \int_{2\pi}^{0} (1+e^{-i\alpha’})^n e^{ik\alpha’}\,\, d\alpha’\\ &\stackrel{c}{=} +\dfrac1{2\pi}\displaystyle \int_{0}^{2\pi} (1+e^{-i\alpha’})^n e^{ik\alpha’}\,\, d\alpha’ \\ &= \dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\alpha’})^n e^{-i(n-k)\alpha’}\,\, d\alpha’\\ &= \binom{n}{n-k} \hspace{4cm} \blacksquare \end{aligned} $$
Quick explanations of some steps:
(a) Substitution $\alpha’ = 2\pi +\alpha$.
(b)$e^{\pm i 2 \pi}=1$
(c) $\displaystyle \int_b^a f \, dx = -\int_a^b f \, dx$
Identity 2: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
Proof: We know that $$\binom{n}{k}=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^n e^{-ik\theta}\,\, d\theta$$
We will start with the right hand side and simplify it, $$\begin{aligned} \binom{n-1}{k} + \binom{n-1}{k-1}&=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^{n-1} e^{-ik\theta}\,\, d\theta + \dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^{n-1} e^{-i(k-1)\theta}\,\, d\theta \\ &= \dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^{n-1}(e^{-ik\theta}+e^{-i(k-1)\theta})\,\, d\theta\\ &=\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^{n-1} e^{-ik\theta}(1+e^{i\theta})\,\, d\theta\\ &= \dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{i\theta})^n e^{-ik\theta}\,\, d\theta\\ &= \binom{n}{k} \hspace{4cm} \blacksquare \end{aligned} $$
I leave you with a puzzle:
Compute the integral:
$$\dfrac1{2\pi}\displaystyle \int_0^{2\pi} (1+e^{-i\theta})^n \left(\dfrac{1}{1-e^{i\theta}}\right) \,\, d\theta$$