Binomial Expansion

Table of Contents

Introduction

The Binomial Expansion tells us how to expand a power of two summed terms such as (a+b)n(a + b)^n.

(a+b)1=a+b(a+b)2=a2+2ab+b2(a+b)3=a3+3a2b+3ab2+b3(a+b)4=a4+4a3b+6a2b2+4ab3+b4\begin{aligned} (a + b)^1 &= a + b \\ (a + b)^2 &= a^2 + 2ab + b^2 \\ (a + b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\ (a + b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{aligned}

The coefficients (1,2,1  −  1,3,3,1  −  1,4,6,4,11, 2, 1 \;-\; 1, 3, 3, 1 \;-\; 1, 4, 6, 4, 1) form a pattern known as Pascal's Triangle.

Pascal's Triangle

Pascal's Triangle is a simple recursive structure where each number inside the triangle is the sum of the two numbers above it.

nnCoefficients of (a+b)n(a + b)^n
01
11 1
21 2 1
31 3 3 1
41 4 6 4 1
51 5 10 10 5 1

This triangle encodes the number of ways each term can appear in the product (a+b)n(a + b)^n. For example, in (a+b)3(a + b)^3, there are 3 ways to choose which of the three factors contributes bb (and the rest contribute aa), producing the term 3a2b3a^2b.

Counting the Terms

Let's look deeper the combinatorial logic more.

When expanding (a+b)n(a + b)^n, each term in the result corresponds to a unique choice of where the bb appear among the nn factors.

For a given term with bkb^k, we are choosing kk positions (out of nn) where bb comes from, and the remaining n−kn - k positions where aa appears. The number of such combinations is:

Number of ways=n!k!(n−k)!\text{Number of ways} = \frac{n!}{k!(n - k)!}

This is the binomial coefficient, often written as:

(nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n - k)!}

So the general Binomial Expansion becomes:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^{n - k} b^k

Approximating exe^x

We can now use this foundation to approach the exponential function.

The value of ee itself can be understood as a special case of exe^x where x=1x = 1.

First, we can define exe^x as:

lim⁡n→∞(1+1n)nx\lim\limits_{n \to \infty} (1 + \frac{1}{n})^{nx}

Then we can apply the binomial theorem:

ex=lim⁡n→∞∑k=0n(nxk)1nx−k(1n)k=lim⁡n→∞∑k=0n(nx)!k!(nx−k)!(1n)k=∑k=0nlim⁡n→∞(nx)!(nx−k)!⋅1k!nk\begin{split} e^x &= \lim\limits_{n \to \infty} \sum_{k = 0}^{n} \binom{nx}{k} 1^{nx - k} \left(\frac{1}{n}\right)^k \\ &= \lim\limits_{n \to \infty} \sum_{k = 0}^{n} \frac{(nx)!}{k!(nx - k)!} \left(\frac{1}{n}\right)^k \\ &= \sum_{k = 0}^{n} \lim\limits_{n \to \infty} \frac{(nx)!}{(nx - k)!} \cdot \frac{1}{k!n^k} \end{split}

Just like with n!(n−k)!\frac{n!}{(n - k)!}, we can expand (nx)!(nx−k)!\frac{(nx)!}{(nx - k)!}. As nn approaches infinity, nx−knx - k becomes closer to nxnx:

lim⁡n→∞(nx)!(nx−k)!⋅1nk=lim⁡n→∞nxn⋅nx−1n⋯nx−k+1n=x⋅x⋯x=xk\begin{split} \lim\limits_{n \to \infty} \frac{(nx)!}{(nx - k)!} \cdot \frac{1}{n^k} &= \lim\limits_{n \to \infty} \frac{nx}{n} \cdot \frac{nx - 1}{n} \cdots \frac{nx - k + 1}{n} \\ &= x \cdot x \cdots x = x^k \end{split}

Plugging this back into the sum:

ex=∑k=0nlim⁡n→∞(nx)!(nx−k)!⋅1k!nk=∑k=0∞xkk!\begin{split} e^x &= \sum_{k = 0}^{n} \lim\limits_{n \to \infty} \frac{(nx)!}{(nx - k)!} \cdot \frac{1}{k!n^k} \\ &= \sum_{k = 0}^{\infty} \frac{x^k}{k!} \end{split}

This series gives us the infinite expansion of the exponential function.

The Self-Derivative Property

A beautiful property of exe^x is that its derivative equals itself.

We can verify this from the series:

ex=∑k=0∞xkk!dexdx=∑k=1∞kxk−1k!=∑k=1∞xk−1(k−1)!=∑k=0∞xkk!=ex\begin{split} e^x &= \sum_{k = 0}^{\infty} \frac{x^k}{k!} \\ \frac{de^x}{dx} &= \sum_{k = 1}^{\infty} \frac{kx^{k - 1}}{k!} \\ &= \sum_{k = 1}^{\infty} \frac{x^{k - 1}}{(k - 1)!} \\ &= \sum_{k = 0}^{\infty} \frac{x^k}{k!} = e^x \end{split}