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January 27, 2020

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### Section 4-18 : Binomial Series

In this final section of this chapter we are going to look at another series representation for a function. Before we do this let’s first recall the following theorem.

#### Binomial Theorem

If $$n$$ is any positive integer then,

\begin{align*}{\left( {a + b} \right)^n} & = \sum\limits_{i = 0}^n {n \choose i} {a^{n - i}}\,{b^i} \,\\ & = {a^n} + n{a^{n - 1}}b + \frac{{n\left( {n - 1} \right)}}{{2!}}{a^{n - 2}}{b^2} + \cdots + na{b^{n - 1}} + {b^n}\end{align*}

where,

\begin{align*}{n \choose i} & = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - i + 1} \right)}}{{i!}}\hspace{0.25in}i = 1,2,3, \ldots n\\ {n \choose 0} & = 1\end{align*}

This is useful for expanding $${\left( {a + b} \right)^n}$$ for large $$n$$ when straight forward multiplication wouldn’t be easy to do. Let’s take a quick look at an example.

Example 1 Use the Binomial Theorem to expand $${\left( {2x - 3} \right)^4}$$
Show Solution

There really isn’t much to do other than plugging into the theorem.

\begin{align*}{\left( {2x - 3} \right)^4} & = \sum\limits_{i = 0}^4 { {4 \choose i} \,{{\left( {2x} \right)}^{4 - i}}\,{{\left( { - 3} \right)}^i}} \\ & = {4 \choose 0}{\left( {2x} \right)^4} + {4 \choose 1}{\left( {2x} \right)^3}\left( { - 3} \right) + {4 \choose 2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + {4 \choose 3}\left( {2x} \right){\left( { - 3} \right)^3} + {4 \choose 4}{\left( { - 3} \right)^4}\\ & = {\left( {2x} \right)^4} + 4{\left( {2x} \right)^3}\left( { - 3} \right) + \frac{{4\left( 3 \right)}}{2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + 4\left( {2x} \right){\left( { - 3} \right)^3} + {\left( { - 3} \right)^4}\\ & = 16{x^4} - 96{x^3} + 216{x^2} - 216x + 81\end{align*}

Now, the Binomial Theorem required that $$n$$ be a positive integer. There is an extension to this however that allows for any number at all.

#### Binomial Series

If $$k$$ is any number and $$\left| x \right| < 1$$ then,

\begin{align*}{\left( {1 + x} \right)^k} & = \sum\limits_{n = 0}^\infty { {k \choose n} {x^n}} \,\\ & = 1 + kx + \frac{{k\left( {k - 1} \right)}}{{2!}}{x^2} + \frac{{k\left( {k - 1} \right)\left( {k - 2} \right)}}{{3!}}{x^3} + \cdots \end{align*}

where,

\begin{align*}{k \choose n} & = \frac{{k\left( {k - 1} \right)\left( {k - 2} \right) \cdots \left( {k - n + 1} \right)}}{{n!}}\hspace{0.25in}n = 1,2,3, \ldots \\ {k \choose 0} & = 1\end{align*}

So, similar to the binomial theorem except that it’s an infinite series and we must have $$\left| x \right| < 1$$ in order to get convergence.

Let’s check out an example of this.

Example 2 Write down the first four terms in the binomial series for $$\sqrt {9 - x}$$
Show Solution

So, in this case $$k = \frac{1}{2}$$ and we’ll need to rewrite the term a little to put it into the form required.

$\sqrt {9 - x} = 3{\left( {1 - \frac{x}{9}} \right)^{\frac{1}{2}}} = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}$

The first four terms in the binomial series is then,

\begin{align*}\sqrt {9 - x} & = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}\\ & = 3\sum\limits_{n = 0}^\infty { {\frac{1}{2} \choose n} {{\left( { - \frac{x}{9}} \right)}^n}} \,\\ & = 3\left[ {1 + \left( {\frac{1}{2}} \right)\left( { - \frac{x}{9}} \right) + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)}}{2}{{\left( { - \frac{x}{9}} \right)}^2} + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)}}{6}{{\left( { - \frac{x}{9}} \right)}^3} + \cdots } \right]\\ & = 3 - \frac{x}{6} - \frac{{{x^2}}}{{216}} - \frac{{{x^3}}}{{3888}} - \cdots \end{align*}