Question

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

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 $x=a$ . 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,

& {{\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{aligned}$$ Substitute these in the polynomial series we get, $${{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.