Question

How do you find the Maclaurin Series for $f(x) = \cos (5{x^2})?$

Answer 1

The Maclaurin series for the function about the point $a = 0$ is expressed as $f(x) = f\left( a \right) + f'\left( a \right)x + \dfrac{{f''\left( a \right)}}{{2!}}{x^2} + \dfrac{{f'''\left( a \right)}}{{3!}}{x^3} + \dfrac{{f''''\left( a \right)}}{{4!}}{x^4} + .....$ This series contains the derivatives and the factorial. For writing this series, we have to calculate the higher-order derivatives of the function at the point using the basic differentiation methods.

Complete step by step answer:
First step is to find the Maclaurin Series of$cosx$. Once we find this, we will substitute $5x$ in place of $x$ , and find the series.
Let us find first find the Maclaurin Series of $\cos (x)$
$\cos (x) = \dfrac{d}{{dx}}\sin (x)$
We know that, expansion of $\sin (x)$ is $\sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right)!}}} {x^{2k + 1}}$ . Therefore,
$\Rightarrow \dfrac{d}{{dx}}\sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right)!}}} {x^{2k + 1}}$
We just open the summation,
$\Rightarrow \dfrac{d}{{dx}}\left( {x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .....} \right)$
Finding the derivatives of all these terms, we will get,
$\Rightarrow 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - .....$
This whole expression can b used like this or can be written in the form of summation as
$\Rightarrow \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}}}$
Now, we just have to substitute $5x$ in place of $x$ in the expression we have obtained above.
Therefore, $\cos (x)$ becomes $\cos \left( {5{x^2}} \right)$
$\Rightarrow \cos \left( {5{x^2}} \right) = 1 - \dfrac{{{{\left( {5{x^2}} \right)}^2}}}{2} + \dfrac{{{{\left( {5{x^2}} \right)}^4}}}{{24}} - \dfrac{{{{\left( {5{x^2}} \right)}^6}}}{{720}} + .....$
Simplify and making the terms in their standard form by cancelling the common factors we get
$\Rightarrow 1 - \dfrac{{25}}{2}{x^4} + \dfrac{{625}}{{24}}{x^8} - \dfrac{{3125}}{{144}}{x^{12}} + .....$

Hence, the Maclaurin series for $\cos \left( {5{x^2}} \right)$ is $1 - \dfrac{{25}}{2}{x^4} + \dfrac{{625}}{{24}}{x^8} - \dfrac{{3125}}{{144}}{x^{12}} + .....$

Note: The Maclaurin series expansion for $\cos (x)$ is given by
$\Rightarrow \cos (x) = \sum\limits_{k = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^k}{x^{2k}}}}{{\left( {2k} \right)!}}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}}....$
This formula is valid for all real values of $x$ .
The Taylor Series, or Taylor Polynomial, is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point.
A McLaurin series is a special occurrence of the Taylor Series where the series is constructed around x=0. Taylor series and Taylor polynomials allow us to approximate functions that are otherwise difficult to calculate.
The Taylor series for sine may not seem very useful to us, since we are used to hitting the sine function on our calculator which then spits out an answer. But our calculators actually make use of similar series to approximate the trigonometric functions, as well as other functions, to provide us with a decimal approximation. Likewise, physicists often take measurements and produce curves that do not clearly resemble a known function. However, they can use Taylor series to come up with a working model, even if it is not exact.