Question

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

Answer 1

The Maclaurin series is the Taylor series expansion of the function about 0. So, use Taylor’s series expansion about 0 for the function $\cos x$ and expand it by using the formula of Taylor’s series expansion.

Complete Step by Step Solution:
We have to find the Maclaurin series for the function $\cos x$. So, first, we should know about the Maclaurin series and find its formula to expand.
Maclaurin series can be defined as Taylor’s series expansion about 0. Therefore, we must know about the Taylor series. So, the Taylor series of the function is an infinite sum of terms that are expressed in terms of the derivatives of the function at a single point.
The Taylor series of the real or complex – valued function $f\left( x \right)$ that is infinitely differentiable at a real or complex number $a$ is the power series
$f\left( a \right) + \dfrac{{f'\left( a \right)}}{{1!}}\left( {x - a} \right) + \dfrac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots$.
In the more compact sigma edition, we can write the above series as –
$\sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}}$
where, ${f^n}\left( a \right)$ denotes the $nth$ derivative of $f$ evaluated at point $a$.
When $a = 0$, then the above-written series is termed as Maclaurin series.
So, putting $a = 0$ in the above Taylor’s series, we get –
$f\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \dfrac{{f''\left( 0 \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{x^3} + \ldots + \dfrac{{{f^n}\left( 0 \right)}}{{n!}}{x^n} + \ldots$.
The function given in the question is $f\left( x \right) = \cos \left( x \right)$. So, using the Maclaurin’s series for the expansion of the given function.
Finding $f\left( 0 \right),f'\left( 0 \right),f''\left( 0 \right),f'''\left( 0 \right)$ and ${f^4}\left( 0 \right)$ for the given function, we get –
$\Rightarrow f\left( 0 \right) = \cos 0 = 1 \\\ \Rightarrow f'\left( x \right) = - \sin 0 = 0 \\\ \Rightarrow f''\left( x \right) = - \cos 0 = - 1 \\\ \Rightarrow f'''\left( x \right) = \sin 0 = 0 \\\ \Rightarrow {f^4}\left( x \right) = \cos 0 = 1 \\\$
Since, $f\left( x \right) = {f^4}\left( x \right)$ , the cycle of {1, 0, -1, 0} repeats itself.
Putting all these values in the Maclaurin series, we get –
$\Rightarrow f\left( x \right) = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \ldots \\\ \Rightarrow f\left( x \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\dfrac{{{x^{2n}}}}{{\left( {2n} \right)!}}} \\\$

Hence, it is the expansion for the function $\cos x$.

Note: Many students continue expanding the Maclaurin series for the function but the expansion of the series should be stopped when the cycle starts repeating itself. This series is the type of series expansion in which all terms are nonnegative integer powers of the variable.