Question

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

Answer 1

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,

& {{\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{aligned}$$ Thus, as we can see that all the derivatives other than the order of $$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.