Binomial Theorem

Can you expand on $(x+y)^{2}$? I guess you would find that is quite easy to do. You can easily find that $(x+y)^{2} = x^{2}+ 2xy +y^{2}$.

How about the expansion of $(x+y)^{10}$. It is no longer easy.

It is no longer easy isn’t it. However, if we use Binomial Theorem, this expansion becomes an easy problem.

Binomial Theorem is a very intriguing topic in mathematics and it has wide range of applications.

Theorem

Let $x$$y$ be real numbers (or complex, or polynomial). For any positive integer $n$, we have:

$$\begin{align*} (x+y)^{n} &= \binom{n}{0}x^{n} + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^{n}\\ &= \sum_{k=0}^{n} x^{n-k}y^{k} \end{align*}$$

where,

$$\begin{align*} \binom{n}{k} = \frac{n!}{k!(n-k)!} \end{align*}$$

Proof:

We will use prove by induction. The base case $n=1$ is obvious. Now suppose that the theorem is true for the case $n-1$, that is assume that:

$$\begin{align*} (x+y)^{n-1} = \binom{n-1}{0}x^{n-1} + \binom{n-1}{1}x^{n-2}y + \dots + \binom{n-1}{n-2}xy^{n-2} + \binom{n-1}{n-1}y^{n-1} \end{align*}$$

 

we will need to  show that, this is true for

$$\begin{align*} (x+y)^{n} &= \binom{n}{0}x^{n} + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^{n} \label{eq1} \end{align*}$$

Let us consider the left hand side of equation above

$$\begin{align*} (x+y)^{n} &= (x+y) (x+y)^{n-1} \\ &= (x+y) \bigg( \binom{n-1}{0}x^{n-1} + \binom{n-1}{1}x^{n-2}y + \dots \\ &+ \binom{n-1}{n-2}xy^{n-2} + \binom{n-1}{n-1}y^{n-1}\bigg) \\ &= x^{n} + \bigg( \binom{n-1}{0} + \binom{n-1}{1}\bigg) x^{n-1}y + \bigg( \binom{n-1}{1} + \binom{n-1}{2}\bigg) x^{n-2}y^{1} + \dots \\ &+\bigg( \binom{n-1}{n-2} + \binom{n-1}{n-1}\bigg) xy^{n-1} + y^{n} \end{align*}$$

We can now apply Pascal’s identity:

 

$$\begin{align*} \binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k} \end{align*}$$

The equation above can be simplified to:

$$\begin{align*} (x+y)^{n} &= x^{n} + \binom{n}{1}x^{n-1}y + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n-1}xy^{n-1} +y^{n}\\ & = \binom{n}{0}x^{n} + \binom{n}{1}x^{n-1}y + \dots + \binom{n}{n-1}xy^{n-1} + \binom{n}{n}y^{n} \end{align*}$$

as we desired.

Example 1:  Power rule in Calculus

 

In calculus, we always use the power rule that $\frac{d}{dx} x^{n} = n x^{n-1}$

 

We can prove this rule using Binomial Theorem.

Proof:

Recall that derivative for any continuous function f(x) is defined as:

 

$$\begin{align*} \frac{d}{dx} f(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}. \end{align*}$$

Let $n$ be a positive integer and let $f(x) = x^{n}$

 

The derivative of f(x) is:

 

$$\begin{align*} &\frac{d}{dx} x^{n} = \lim_{h \rightarrow 0} \frac{(x+h)^{n} - x^{n}}{h}\\ &= \lim_{h \rightarrow 0} \frac{\bigg( \binom{n}{0} x^{n} + \binom{n}{1}x^{n-1}h + \dots + \binom{n}{n} h^{n} \bigg) - x^{n}}{h}\\ & = \lim_{h \rightarrow 0} \frac{ \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^{2}+ \dots + \binom{n}{n} h^{n}}{h}\\ & = \lim_{h \rightarrow 0} \bigg( \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2}h+ \dots + \binom{n}{n} h^{n-1} \bigg)\\ &= n \lim_{h \rightarrow 0} \bigg( \binom{n-1}{0}x^{n-1} + \binom{n-1}{1}x^{n-2}h+ \dots + \binom{n-1}{n-1}h^{n-1} \bigg)\\ &= n \lim_{h \rightarrow 0} (x+h)^{n-1}\\ &= n x^{n-1}. \end{align*}$$

Example 2:  Binomial Distribution 

Let X be the number of Head a sequence of n independent coin tossing. X is usually model by binomial distribution in probability model. Let $ p \in [0,1]$ be the probability that a head show up in a toss, and let $k = 0,1,\dots,n$. The probability that there is $k$ head in the sequence of $n$ toss is:

$$\begin{align*} P(X = k) = \binom{n}{k} p^{k}(1-p)^{n-k} \end{align*}$$

We know that sum of all the probability must equal to 1. In order to show this, we can use Binomial Theorem. We have:

 

$$\begin{align*} \sum_{k=0}^{n} P(X = k) &= \sum_{k=0}^{n} \binom{n}{k} p^{k}(1-p)^{n-k}\\ & = (p + 1 - p)^{n}\\ &= 1. \end{align*}$$

Please also check another article Gaussian Samples and N-gram language models ,Bayesian model , Monte Carlo for statistics knowledges.