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