Question

How do you write the \[{{5}^{th}}\] degree Taylor polynomial for \[\sin (x)\]?

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 \[x=a\] . We are asked to write the Taylor polynomial of \[{{5}^{th}}\] degree for the function \[\sin (x)\].

Complete step by step answer:
The given function is \[\sin (x)\], we are asked to write \[{{5}^{th}}\] degree Taylor polynomial 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, \[f(x)=\sin (x)\] and \[n=5\], so the series will be \[{{T}_{5}}(x)=\sum\limits_{i=0}^{5}{\dfrac{{{\sin }^{(i)}}(a)}{i!}}{{\left( x-a \right)}^{i}}\]. We have to find the required 6 derivatives of \[\sin (x)\] to compute the polynomial,
\[\begin{align}
  & {{\sin }^{(0)}}(a)=\sin (a) \\
 & {{\sin }^{(1)}}(a)=\cos (a) \\
 & {{\sin }^{(2)}}(a)=-\sin (a) \\
 & {{\sin }^{(3)}}(a)=-\cos (a) \\
 & {{\sin }^{(4)}}(a)=\sin (a) \\
 & {{\sin }^{(5)}}(a)=\cos (a) \\
\end{align}\]
Substitute these in the polynomial series we get,
\[{{T}_{5}}(x)=\sum\limits_{i=0}^{5}{\dfrac{{{\sin }^{(i)}}(a)}{i!}}{{\left( x-a \right)}^{i}}\]\[\Rightarrow \dfrac{{{\sin }^{(0)}}(a)}{0!}+\dfrac{{{\sin }^{(1)}}(a)}{1!}\left( x-a \right)+\dfrac{{{\sin }^{(2)}}(a)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{{{\sin }^{(3)}}(a)}{3!}{{\left( x-a \right)}^{3}}+\dfrac{{{\sin }^{(4)}}(a)}{4!}{{\left( x-a \right)}^{4}}+\dfrac{{{\sin }^{(5)}}(a)}{5!}{{\left( x-a \right)}^{5}}\]
\[\Rightarrow \dfrac{{{\sin }^{(0)}}(a)}{0!}+\dfrac{\cos (a)}{1!}\left( x-a \right)+\dfrac{-\sin (a)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{-\cos (a)}{3!}{{\left( x-a \right)}^{3}}+\dfrac{\sin (a)}{4!}{{\left( x-a \right)}^{4}}+\dfrac{\cos (a)}{5!}{{\left( x-a \right)}^{5}}\]
This is the series of \[{{5}^{th}}\] degree Taylor polynomials of function \[\sin (x)\].
The general case of this will be when a is zero, in that case, the \[{{T}_{5}}(x)\] series will become, we know that \[\sin 0=0\] and \[\cos 0=1\].
\[\Rightarrow {{T}_{5}}(x)=0+x+0-\dfrac{{{x}^{3}}}{3!}+0+\dfrac{{{x}^{5}}}{5!}\]
\[\Rightarrow {{T}_{5}}(x)=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}\]

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.