Question

How do you find the maclaurin series expansion of \[\cos {{\left( x \right)}^{2}}\]?

Answer 1

Hint: Given a function \[f(x)\], a specific point \[x=a\] (called the center), and a positive integer n, the Taylor polynomial of \[f(x)\]at \[a\], of degree n, is the polynomial T of degree n that best fits the curve \[y=f(x)\] near the point a, in the sense that T and all its first n derivatives have the same value at\[x=a\] as \[f\] does. The general formula for finding the Taylor polynomial is as follows, \[{{T}_{n}}(x)=\sum\limits_{i=0}^{n}{\dfrac{{{f}^{(i)}}(a)}{i!}}{{\left( x-a \right)}^{i}}\] , here \[{{f}^{(i)}}(a)\]represents \[{{i}^{th}}\]derivative of \[f(x)\]with respect to \[x\]at\[x=a\]. If \[a=0\] then the expansion is called a maclaurin series.

Complete step-by-step answer:
The given function is \[\cos {{\left( x \right)}^{2}}\], we are asked to find maclaurin for this. As we know that the formula for finding the Taylor polynomial is \[{{T}_{n}}(x)=\sum\limits_{i=0}^{n}{\dfrac{{{f}^{(i)}}(a)}{i!}}{{\left( x-a \right)}^{i}}\], here \[{{f}^{(i)}}(a)\]represents \[{{i}^{th}}\]derivative of \[f\]with respect to \[x\]at\[x=a\]
For this question we have, \[f(x)=\cos {{\left( x \right)}^{2}}\] and \[n=n\] because, we are not told to find any specific number of terms, so the series will be \[{{T}_{n}}(x)=\sum\limits_{i=0}^{n}{\dfrac{{{\cos }^{(i)}}\left( {{a}^{2}} \right)}{i!}}{{\left( x-a \right)}^{i}}\]. We will find some of the initial terms of the series,
\[\begin{align}
  & {{\cos }^{\left( 0 \right)}}\left( {{a}^{2}} \right)=\cos \left( {{a}^{2}} \right)=1,\because a=0 \\
 & {{\cos }^{\left( 1 \right)}}\left( {{a}^{2}} \right)=-2a\sin \left( {{a}^{2}} \right)=0,\because a=0 \\
 & {{\cos }^{\left( 2 \right)}}\left( {{a}^{2}} \right)=-4{{a}^{2}}\cos \left( {{a}^{2}} \right)-2\sin \left( {{a}^{2}} \right)=0,\because a=0 \\
 & {{\cos }^{\left( 3 \right)}}\left( {{a}^{2}} \right)=8{{a}^{3}}\sin \left( {{a}^{2}} \right)-12a\cos \left( {{a}^{2}} \right)=0,\because a=0 \\
 & {{\cos }^{\left( 4 \right)}}\left( {{a}^{2}} \right)=16{{a}^{4}}\cos \left( {{a}^{2}} \right)+48{{a}^{2}}\sin \left( {{a}^{2}} \right)-12\cos \left( {{a}^{2}} \right)=-12,\because a=0 \\
\end{align}\]
Thus, as we can see that all the derivatives other than the order of \[4k\] are becoming zero.
Substituting the values in the expression for maclaurin series, we get
\[{{T}_{n}}(x)=\sum\limits_{i=0}^{n}{\dfrac{{{\cos }^{(i)}}\left( {{a}^{2}} \right)}{i!}}{{\left( x-a \right)}^{i}}\]
\[\Rightarrow {{T}_{n}}(x)\approx \dfrac{1}{0!}{{x}^{0}}+\dfrac{0}{1!}{{x}^{1}}+\dfrac{0}{2!}{{x}^{2}}+\dfrac{0}{3!}{{x}^{3}}+\dfrac{-12}{4!}{{x}^{4}}\]
\[\Rightarrow {{T}_{n}}(x)\approx 1-\dfrac{1}{2}{{x}^{4}}\]
We can also find the terms ahead of this using the formula \[{{T}_{n}}(x)=\sum\limits_{k=0}^{n}{\dfrac{{{\cos }^{(4k)}}\left( {{a}^{2}} \right)}{\left( 4k \right)!}}{{\left( x-a \right)}^{4k}}\].

Note: The general formula for Taylor polynomial is\[{{T}_{n}}(x)=\sum\limits_{i=0}^{n}{\dfrac{{{f}^{(i)}}(a)}{i!}}{{\left( x-a \right)}^{i}}\]
At first glance, we know it may look difficult at first, but there's a step-by-step procedure for creating a Taylor polynomial. As long as you've had plenty of experience with derivatives, and if you know your way around factorials, then it shouldn't be too hard.