Question

How do you find the Maclaurin series expansion of $f\left( x \right) = \cos \left( {5{x^2}} \right)$?

Answer 1

Maclaurin series expansion of a cosine function is $\cos \left( y \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{{\left( y \right)}^{2k}}}}{{\left( {2k} \right)!}}}$, where the interval of convergence is the whole real number. This expansion has infinite terms but still converges depending on the argument $y$.

Complete step by step solution:
Write the given function whose Maclaurin series has to be found.
$\Rightarrow f\left( x \right) = \cos \left( {5{x^2}} \right)$

The given function $f\left( x \right) = \cos \left( {5{x^2}} \right)$ can be written as a composite function of two functions $g\left( y \right) = \cos y$ and $h\left( x \right) = 5{x^2}$ as shown below.

$\begin{array}{c}g \circ h\left( x \right) = g\left( {h\left( x \right)} \right)\\\ = g\left( {5{x^2}} \right)\\\ = \cos \left( {5{x^2}} \right)\\\ = f\left( x \right)\end{array}$

Now use the fact that the Maclaurin series for $g\left( y \right) = \cos y$ function can be written as shown below.
$\Rightarrow g\left( y \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{{\left( y \right)}^{2k}}}}{{\left( {2k} \right)!}}}$

Substitute $y$ as $h\left( x \right) = 5{x^2}$ and obtain the Maclaurin for $f\left( x \right) = g \circ h\left( x \right)$ as shown below.
$\Rightarrow f\left( x \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{{\left( {5{x^2}} \right)}^{2k}}}}{{\left( {2k} \right)!}}}$

Therefore, the Maclaurin series for $f\left( x \right) = \cos \left( {5{x^2}} \right)$ can be written as shown below.
$\Rightarrow \cos \left( {5{x^2}} \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{{\left( {5{x^2}} \right)}^{2k}}}}{{\left( {2k} \right)!}}}$

Simplify the series as shown below.
$\begin{array}{c}\cos \left( {5{x^2}} \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{{\left( 5 \right)}^{2k}}{{\left( {{x^2}} \right)}^{2k}}}}{{\left( {2k} \right)!}}} \\\ = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{5^{2k}}{x^{4k}}}}{{\left( {2k} \right)!}}} \end{array}$
Now, expand the series with few terms as shown below.
$\cos \left( {5{x^2}} \right) = 1 - \dfrac{{25{x^4}}}{2} + \dfrac{{625{x^8}}}{{24}} - \cdots$

Thus, the Maclaurin series for $f\left( x \right) = \cos \left( {5{x^2}} \right)$ is $f\left( x \right) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{5^{2k}}{x^{4k}}}}{{\left( {2k} \right)!}}}$ in compact form and $f\left( x \right) = 1 - \dfrac{{25{x^4}}}{2} + \dfrac{{625{x^8}}}{{24}} - \cdots$ in expanded form.

Note: Most of the Maclaurin series are the composition and combination of elementary or basic functions. So always memorise the Maclaurin series of elementary functions. Maclaurin series are differentiable and integrable.