Question

How do you find the Taylor series for $f\left( x \right)=\cos x$ centered at $a=\pi $ ?

Answer 1

Hint: To find the Taylor series for $f\left( x \right)=\cos x$ centered at $a=\pi $ , we will use the formula for Taylor series of a function which is given by \[f\left( x \right)=\sum\limits_{n=0}^{N}{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}\] ., where ${{f}^{\left( n \right)}}\left( a \right)$ is the ${{n}^{th}}$ derivative of f(x) at $x=a$ . Then, we will differentiate the given function once, twice, thrice, and so on at $x=a$ and substitute these values in the given formula. Then, we will use the given condition $a=\pi $ and simplify further.

Complete step-by-step solution:
We have to find the Taylor series for $f\left( x \right)=\cos x$ centered at $a=\pi $ . We know that Taylor series of a function is given by
\[f\left( x \right)=\sum\limits_{n=0}^{N}{\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}\]
where ${{f}^{\left( n \right)}}\left( a \right)$ is the ${{n}^{th}}$ derivative of f(x) at $x=a$ .
Let us expand the series.
\[f\left( x \right)=f\left( a \right)+{f}'\left( a \right)\dfrac{x-a}{1!}+{f}''\left( a \right)\dfrac{{{\left( x-a \right)}^{2}}}{2!}+{f}'''\left( a \right)\dfrac{{{\left( x-a \right)}^{3}}}{3!}+...+{{f}^{n}}\left( a \right)\dfrac{{{\left( x-a \right)}^{n}}}{n!}\]
We are given that $f\left( x \right)=\cos x$ . Let us differentiate this function with respect to x at $x=a$ .
${{\left. {f}'\left( x \right) \right|}_{x=a}}=-\sin a$
Now, let us again differentiate the above function.
${{\left. {f}''\left( x \right) \right|}_{x=a}}=-\cos a$
We have to find the third derivative of the given function at $x=a$ .
${{\left. {f}'''\left( x \right) \right|}_{x=a}}=\sin a$
We have to find the fourth derivative of the given function at $x=a$ .
${{\left. {{f}'''}'\left( x \right) \right|}_{x=a}}=\cos a$
Now, let us substitute these values in the Taylor series.
\[\cos x=\cos a+-\sin a\dfrac{x-a}{1!}+-\cos a\dfrac{{{\left( x-a \right)}^{2}}}{2!}+\sin a\dfrac{{{\left( x-a \right)}^{3}}}{3!}+\cos a\dfrac{{{\left( x-a \right)}^{4}}}{4!}+...\]
We are given that $a=\pi $ . Therefore, the above form becomes
\[\Rightarrow \cos x=\cos \pi +-\sin \pi \dfrac{x-\pi }{1!}+-\cos \pi \dfrac{{{\left( x-\pi \right)}^{2}}}{2!}+\sin \pi \dfrac{{{\left( x-\pi \right)}^{3}}}{3!}+\cos \pi \dfrac{{{\left( x-\pi \right)}^{4}}}{4!}+...\]
We know that $\sin \pi =0$ . Therefore, all the off terms will be 0.
\[\Rightarrow \cos x=\cos \pi +-\cos \pi \dfrac{{{\left( x-\pi \right)}^{2}}}{2!}+\cos \pi \dfrac{{{\left( x-\pi \right)}^{4}}}{4!}+...\]
We know that $\cos \pi =-1$ .
\[\Rightarrow \cos x=-1+\dfrac{{{\left( x-\pi \right)}^{2}}}{2!}-\dfrac{{{\left( x-\pi \right)}^{4}}}{4!}+...\]
Therefore, the Taylor series for $f\left( x \right)=\cos x$ centered at $a=\pi $ is \[-1+\dfrac{{{\left( x-\pi \right)}^{2}}}{2!}-\dfrac{{{\left( x-\pi \right)}^{4}}}{4!}+...\].

Note: Students must know the Taylor series and to expand them.They must note that Taylor series is a series which is the reason we did not get a finite result. They must know different functions. Students must be thorough with the values of trigonometric functions at basic angles.