Title: Default for 16.3 The Binomial Series
Course: HELM Workbooks 2023-2024

The binomial series

The binomial series is a more general form of the binomial theorem, that is, the expansion of \((x+1)^n\). When \(n\) is a natural number then we can expand \((x+1)^n\) using the usual laws of algebra and get a finite sum of terms. If \(n\) is not a natural number then the binomial theorem does not apply but we can get a series expansion. It is important to note that the series typically won't converge for all \(x\).

A very important infinite series which occurs often in applications and in algebra has the form:
\[
1+p x+\dfrac{p(p-1)}{2!}x^2+\dfrac{p(p-1)(p-2)}{3!}x^3+\cdots
\]
in which \(p\) is a given number and \(x\) is a variable. By using the ratio test it can be shown that this series converges, irrespective of the value of \(p\), as long as \(|x|<1\). In fact, as we shall see in Maclaurin and Taylor series that the given series converges to the value \((1+x)^p\) as long as \(|x|<1\).

Key Point

The Binomial Series

\[ (1+x)^p = 1+px+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\cdots \qquad |x|<1 \]

The binomial theorem can be obtained directly from the binomial series if \(p\) is chosen to be a positive integer (here we need not demand that \(|x|<1\) as the series is now finite and so is always convergent irrespective of the value of \(x\)). For example, with \(p=2\) we obtain
\[
\begin{aligned}
(1+x)^2&=& 1+2x+\dfrac{2(1)}{2}x^2+0+0+\cdots\\ &=&1+2x+x^2\ \ \ \text{as is well known.}
\end{aligned}
\]

With \(p=3\) we get
\[
\begin{aligned} (1+x)^3&=& 1+3x+\dfrac{3(2)}{2}x^2+\dfrac{3(2)(1)}{3!}x^3+0+0+\cdots &=&1+3x+3x^2+x^3
\end{aligned}
\]
Generally if \(p=n\) (a positive integer) then

\[ (1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots +x^n \]

which is a form of the binomial theorem. If \(x\) is replaced by \(\dfrac{b}{a}\) then
\[
\left(1+\dfrac{b}{a}\right)^n=1+n\left(\dfrac{b}{a}\right)+\dfrac{n(n-1)}{2!}\left(\dfrac{b}{a}\right)^2+\cdots +\left(\dfrac{b}{a}\right)^n
\]
Now multiplying both sides by \(a^n\) we have the following Key Point:

A generalisation of the third law of indices states:

Key Point

If \(n\) is a positive integer then the expansion of \( (a+b) \) raised to the power \(n\) is given by:

\[ (a+b)^n=a^n+na^{n−1}b+\frac{n(n−1)}{2!}a^{n−2}b^2+ \cdots +b^n.\]

This is known as the binomial theorem.

Exercise

Use the binomial theorem to obtain

1. \(([value]+x)^{[value]}=\)
2. \( (a+b)^{[value]}=\)
Feedback: In each case we apply the binomial theorem, which can be stated as \[(a+b)^n =\sum_{i=0}^n \binom{n}{i}a^ib^{n-i}\] where \(\displaystyle{\binom{n}{i}=\dfrac{n!}{(n-i)!i!}}\). 1. \(([value]+x)^{[value]}=[value]\) 2. \( (a+b)^{[value]}=[value]\)

Exercise

As a real number \(x\) approaches zero we note hiigher powers of  \(x\) become even closer to zero and may even fall within a margin of error we can tolerate.

Assume, for this exercise, that \(x^3\) and higher powers may be ignored in comparison to lower order terms.

Find a quadratic polynomial approximating \( (1−x)^{[value]} \). Hint: First expand \( (1+x)^{[value]} \) using the binomial series with \(p=\displaystyle[value] \)and with \( x \) replaced by \( (−x) \):

\( (1-x)^[value] \approx \)
Feedback: The binomial series gives \[ (x-1)^[value] \approx [value].\] What we have done in this exercise is to approximate the function \((1-x)^{[value]}\) by the simpler (polynomial) function \(\displaystyle [value]\) which is reasonable provided \(x\) is very small. In the case of \((1-x)^{\frac{1}{2}}\)this approximation is well illustrated geometrically by drawing the curves \(y=(1-x)^{\frac{1}{2}}\) and \(y=1-\dfrac{1}{2}x-\dfrac{1}{8}x^2\). The two curves coincide when \(x\) is ‘small’. [Image: plot of x minus 1 over 8 x squared with dotted line 1 minus x all to the power of a half]

Exercise

In this exercise we will obtain a cubic approximation of \(\displaystyle[value]\).

1. First we write the expression \(\displaystyle[value]\) in a form suitable for the binomial series, \(a(1+y)^p\) for some constant \(a\) and some \(y\) which depends on \(x\).
\(\displaystyle[value]=\)
2. Now use the binomial series to approximate \(\displaystyle[value]\) by a cubic polynomial.
\(\displaystyle[value]\approx\)
Feedback: 1. \({\displaystyle[value]=\frac{1}{[value]}\left(1+\frac{x}{[value]}\right)^{-1}}\) 2. \(\displaystyle[value]\approx[value]\) Below illustrates the close correspondence (when \(x\) is ‘small’) between the curves \(y=\frac{1}{2}(1 +x^2)^{−1}\) and \(y=\frac{1}{2}−\frac{x} {4}+\frac{x^2}{8}−\frac{x^3}{16}\). [Image: graph of two plus x to the power minus one with dotted line of binomial expansion]

Exercise

Determine the binomial expansion of each of the following. If there is finite expansion then give the full expansion, if there is an infinite expansion then give the first four terms.

1. \([value]=\)
2. \([value]=\)
3. \([value]\approx\)
Feedback: 1. \([value]=[value]\) 2. \([value]=[value]\) 3. \([value]=[value]\)

Exercise

Obtain a cubic approximation (valid if \(x\) is small) of the function [value]. Give the first four terms.

\( [value] \approx \)
Feedback: We need to obtain the cubic approximation of the function [value]. We expand using he binomial series expansion \[ (1+x)^p = 1+px+\frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\cdots \] with \( p =[value]\) and [value] instead of \(x\), to include terms up to \( x^3.\) Hence, \[ [value] \approx\] \[1+ [value]([value])+ \frac{[value]([value]-1)}{2!}([value])^2+ \frac{[value]([value]-1)([value]-2)}{3!}([value])^3 \approx \] \[[value]\]