Taylor and Maclaurin Series
Solved Problems
Example 3.
Find the Maclaurin series for \({e^{kx}},\) \(k\) is a real number.
Solution.
Calculate the derivatives:
\[f'\left( x \right) = {\left( {{e^{kx}}} \right)^\prime } = k{e^{kx}},\;\; f^{\prime\prime}\left( x \right) = {\left( {k{e^{kx}}} \right)^\prime } = {k^2}{e^{kx}}, \ldots \;\; {f^{\left( n \right)}}\left( x \right) = {k^n}{e^{kx}}.\]
Then, at \(x = 0\) we have
\[f\left( 0 \right) = {e^0} = 1,\;\; f'\left( 0 \right) = k{e^0} = k,\;\; f^{\prime\prime}\left( 0 \right) = {k^2}{e^0} = {k^2}, \ldots \;\; {f^{\left( n \right)}}\left( 0 \right) = {k^n}{e^0} = {k^n}.\]
Hence, the Maclaurin expansion for the given function is
\[{e^{kx}} = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 0 \right)\frac{{{x^n}}}{{n!}}} = 1 + kx + \frac{{{k^2}{x^2}}}{{2!}} + \frac{{{k^3}{x^3}}}{{3!}} + \ldots = \sum\limits_{n = 0}^\infty {\frac{{{k^n}{x^n}}}{{n!}}} .\]
Example 4.
Find the Taylor series of the cubic function \({x^3}\) about \(x = 2.\)
Solution.
We denote \(f\left( x \right) = {x^3}.\) Then
\[f'\left( x \right) = {\left( {{x^3}} \right)^\prime } = 3{x^2},\;\; f^{\prime\prime}\left( x \right) = {\left( {3{x^2}} \right)^\prime } = 6x,\;\; f^{\prime\prime\prime}\left( x \right) = {\left( {6x} \right)^\prime } = 6,\;\; {f^{IV}}\left( x \right) = 0,\]
and further \({f^{\left( n \right)}}\left( x \right) = 0\) for all \(x \ge 4.\)
Respectively, at the point \(x = 2,\) we have
\[f\left( 2 \right) = 8,\;\; f'\left( 2 \right) = 12,\;\; f^{\prime\prime}\left( 2 \right) = 12,\;\; f^{\prime\prime\prime}\left( 2 \right) = 6.\]
Hence, the Taylor series expansion for the cubic function is given by the expression
\[{x^3} = \sum\limits_{n = 0}^\infty {{f^{\left( n \right)}}\left( 2 \right)\frac{{{{\left( {x - 2} \right)}^n}}}{{n!}}} = 8 + 12\left( {x - 2} \right) + \frac{{12{{\left( {x - 2} \right)}^2}}}{{2!}} + \frac{{6{{\left( {x - 2} \right)}^3}}}{{3!}} = 8 + 12\left( {x - 2} \right) + 6{\left( {x - 2} \right)^2} + {\left( {x - 2} \right)^3}.\]
Example 5.
Find the Maclaurin series for \[{\left( {1 + x} \right)^\mu }.\]
Solution.
Let \(f\left( x \right) = {\left( {1 + x} \right)^\mu },\) where \(\mu\) is a real number and \(x \ne -1.\) Then we can write the derivatives as follows
\[f'\left( x \right) = \mu {\left( {1 + x} \right)^{\mu - 1}},\]
\[f^{\prime\prime}\left( x \right) = \mu \left( {\mu - 1} \right){\left( {1 + x} \right)^{\mu - 2}},\]
\[f^{\prime\prime\prime}\left( x \right) = \mu \left( {\mu - 1} \right)\left( {\mu - 2} \right) \cdot {\left( {1 + x} \right)^{\mu - 3}},\]
\[{f^{\left( n \right)}}\left( x \right) = \mu \left( {\mu - 1} \right)\left( {\mu - 2} \right) \cdots \left( {\mu - n + 1} \right){\left( {1 + x} \right)^{\mu - n}}.\]
For \(x = 0,\) we obtain
\[f\left( 0 \right) = 1,\;\; f'\left( 0 \right) = \mu ,\;\; f^{\prime\prime}\left( 0 \right) = \mu \left( {\mu - 1} \right), \ldots \;\; {f^{\left( n \right)}}\left( 0 \right) = \mu \left( {\mu - 1} \right) \cdots \left( {\mu - n + 1} \right).\]
Hence, the series expansion can be written in the form
\[\left( {1 + x} \right)^\mu = 1 + \mu x + \frac{{\mu \left( {\mu - 1} \right)}}{{2!}}{x^2} + \frac{{\mu \left( {\mu - 1} \right)\left( {\mu - 2} \right)}}{{3!}}{x^2} + \ldots + \frac{{\mu \left( {\mu - 1} \right) \cdots \left( {\mu - n + 1} \right)}}{{n!}}{x^n} + \ldots \]
This series is called the binomial series.
Example 6.
Determine the Maclaurin series for \[f\left( x \right) = \sqrt {1 + x}.\]
Solution.
Using the binomial series found in the previous example and substituting \(\mu = {\frac{1}{2}},\) we get
\[\sqrt {1 + x} = {\left( {1 + x} \right)^{\frac{1}{2}}} = 1 + \frac{x}{2} + \frac{{\frac{1}{2}\left( {\frac{1}{2} - 1} \right)}}{{2!}}{x^2} + \frac{{\frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right)}}{{3!}}{x^3} + \ldots = 1 + \frac{x}{2} - \frac{{1 \cdot {x^2}}}{{{2^2}2!}} + \frac{{1 \cdot 3 \cdot {x^3}}}{{{2^3}3!}} - \frac{{1 \cdot 3 \cdot 5 \cdot {x^3}}}{{{2^4}4!}} + \ldots + {\left( { - 1} \right)^{n + 1}}\cdot\frac{{1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right){x^n}}}{{{2^n}n!}}.\]
Keeping only the first three terms, we can write this series as
\[\sqrt {1 + x} \approx 1 + \frac{x}{2} - \frac{{{x^2}}}{8}.\]
