Question

How do you find the Maclaurin series of $f\left( x \right) = \cosh \left( x \right)$?

Answer 1

Hint: Given a trigonometric expression. We have to find the Maclaurin series for the expression. We will first apply the expansion of the Maclaurin series. Then, we will find the derivative of the function at $x = 0$.
Formula used:
The expansion of the Taylor series is given by:
$f\left( a \right) + \dfrac{{f'\left( a \right)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots $
The sigma notation of the series is given by:
\[\sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \]
The series is known as Maclaurin series when $a = 0$

Complete step by step solution:
We are given the function, $f\left( x \right) = \cosh \left( x \right)$
First, we will use the Taylor series to expand the function.
$ \Rightarrow f\left( a \right) + \dfrac{{f'\left( a \right)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots $
Now, we will write the Maclaurin series by substituting $a = 0$
$ \Rightarrow f\left( 0 \right) + \dfrac{{f'\left( 0 \right)}}{{1!}}\left( {x - 0} \right) + \dfrac{{f''\left( 0 \right)}}{{2!}}{\left( {x - 0} \right)^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{\left( {x - 0} \right)^3} + \ldots $
Simplifying the expression, we get:
$ \Rightarrow f\left( 0 \right) + \dfrac{{f'\left( 0 \right)}}{{1!}}\left( x \right) + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{x^3} + \ldots $
Now, we find the value of $f\left( 0 \right)$ by substituting zero for $x$ in the$f\left( x \right)$
$ \Rightarrow f\left( 0 \right) = \cos 0^\circ $
$ \Rightarrow f\left( 0 \right) = 1$
Now, we will find the value of $f'\left( x \right)$by differentiating the function, $f\left( x \right)$
\[ \Rightarrow \cosh x = - \sinh x\]
Now compute the value of $f'\left( 0 \right)$ by substituting zero for $x$ in $f'\left( x \right)$
$ \Rightarrow f'\left( 0 \right) = - \sin 0$
Now, substitute zero for $\sin 0$
$ \Rightarrow f'\left( 0 \right) = 0$
Now, we will find the second derivative of the function.
\[ \Rightarrow f''\left( x \right) = {\left( { - \sinh x} \right)^{\prime \prime }}\]
\[ \Rightarrow f''\left( x \right) = - \cosh x\]
Now compute the value of $f''\left( 0 \right)$ by substituting zero for $x$ in $f''\left( x \right)$
$ \Rightarrow f''\left( 0 \right) = - \cos 0$
Now, substitute one for $\cos 0$
$ \Rightarrow f''\left( 0 \right) = - 1$
Now, we will find the third derivative of the function.
\[ \Rightarrow f'''\left( x \right) = {\left( { - \cosh x} \right)^{\prime \prime \prime }}\]
\[ \Rightarrow f'''\left( x \right) = \sinh x\]
Now compute the value of $f'''\left( 0 \right)$ by substituting zero for $x$ in $f'''\left( x \right)$
$ \Rightarrow f'''\left( 0 \right) = \sin 0$
Now, substitute zero for $\sin 0$
$ \Rightarrow f'''\left( 0 \right) = 0$
Now, we will find the value of ${f^4}\left( x \right)$.
\[ \Rightarrow {f^4}\left( x \right) = {\left( {\sinh x} \right)^\prime }^{\prime \prime \prime }\]
\[ \Rightarrow {f^4}\left( x \right) = \cosh x\]
Now compute the value of ${f^4}\left( 0 \right)$ by substituting zero for $x$ in ${f^4}\left( x \right)$
$ \Rightarrow {f^4}\left( 0 \right) = \cos 0$
$ \Rightarrow {f^4}\left( 0 \right) = 1$
Now, substitute the values in the Maclaurin series, we get:
$ \Rightarrow 1 + \dfrac{0}{{1!}}\left( x \right) - \dfrac{1}{{2!}}{x^2} + \dfrac{0}{{3!}}{x^3} + \dfrac{1}{{4!}}{x^4} \ldots $
$ \Rightarrow 1 + 0 - \dfrac{{{x^2}}}{{2!}} + 0 + \dfrac{{{x^4}}}{{4!}} + \ldots $
$ \Rightarrow 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \ldots $
Hence, the Maclaurin series for $f\left( x \right) = \cosh \left( x \right)$ is $1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \ldots $

Note: Please note that the Taylor series is given by:
$f\left( a \right) + \dfrac{{f'\left( a \right)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots $
The Maclaurin series is a special occurrence of Taylor series in which the series is obtained by substituting $a = 0$. The Taylor series is an expansion of a function about a particular point and a one dimensional series which is an expansion of real function about the point is known as Maclaurin series.