Question

How do you find the coefficient of \[{{x}^{6}}\] in the Taylor series expansion?

Answer 1

Hint: These types of problems are pretty straight forward and are very easy to solve. For such problems, the most primary thing that we need to keep in mind is the Taylor’s formula. We need to have an in-depth knowledge of the Taylor’s theorem and have a clear understanding of its application in different problems. The Taylor’s theorem is expressed as,
\[\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}\]. Here \[{{f}^{n}}\left( a \right)\] is the nth derivative of \[f\left( x \right)\] at \[x=a\], where ‘a’ is any constant. Since in this problem we need to find only the coefficient, the value of ‘a’ doesn’t matter. Thus we take the value as ‘a’ only and proceed for the problem.

Complete step by step solution:
We start off with the solution to the problem by writing it as, we first of all find all the values of the possible derivatives of the function \[{{x}^{6}}\] , we do,
\[\begin{align}
  & f\left( x \right)={{x}^{6}} \\
 & {{f}^{'}}\left( x \right)=6{{x}^{5}} \\
 & {{f}^{''}}\left( x \right)=30{{x}^{4}} \\
 & {{f}^{'''}}\left( x \right)=120{{x}^{3}} \\
 & {{f}^{''''}}\left( x \right)=360{{x}^{2}} \\
 & {{f}^{'''''}}\left( x \right)=720x \\
 & {{f}^{''''''}}\left( x \right)=720 \\
 & {{f}^{'''''''}}\left( x \right)=0 \\
\end{align}\]
Now, putting all these values in the Taylor’s equation, we get,
The Taylor series expansion of the function \[{{x}^{6}}\] is given by,
\[\begin{align}
  & =\dfrac{f\left( a \right)}{0!}{{\left( x-a \right)}^{0}}+\dfrac{{{f}^{'}}\left( a \right)}{1!}{{\left( x-a \right)}^{1}}+\dfrac{{{f}^{''}}\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{{{f}^{'''}}\left( a \right)}{3!}{{\left( x-a \right)}^{3}} \\
 & +\dfrac{{{f}^{''''}}\left( a \right)}{4!}{{\left( x-a \right)}^{4}}+\dfrac{{{f}^{'''''}}\left( a \right)}{5!}{{\left( x-a \right)}^{5}}+\dfrac{{{f}^{''''''}}\left( a \right)}{6!}{{\left( x-a \right)}^{6}}+\dfrac{{{f}^{'''''''}}\left( a \right)}{7!}{{\left( x-a \right)}^{7}}+0 \\
\end{align}\]
Putting the respective values we get,
\[\begin{align}
  & =\dfrac{{{a}^{6}}}{0!}{{\left( x-a \right)}^{0}}+\dfrac{6{{a}^{5}}}{1!}{{\left( x-a \right)}^{1}}+\dfrac{30{{a}^{4}}}{2!}{{\left( x-a \right)}^{2}}+\dfrac{120{{a}^{3}}}{3!}{{\left( x-a \right)}^{3}} \\
 & +\dfrac{360{{a}^{2}}}{4!}{{\left( x-a \right)}^{4}}+\dfrac{720a}{5!}{{\left( x-a \right)}^{5}}+\dfrac{720}{6!}{{\left( x-a \right)}^{6}}+\dfrac{0}{7!}{{\left( x-a \right)}^{7}}+0 \\
\end{align}\]
Thus, from this above equation we can very easily find out the value of the coefficients. We just need to evaluate the value of each of the equations that precedes the power term. Thus calculating those we get the coefficients as,
\[\begin{align}
  & =\dfrac{{{a}^{6}}}{0!}={{a}^{6}},\dfrac{6{{a}^{5}}}{1!}=6{{a}^{5}},\dfrac{30{{a}^{4}}}{2!}=15{{a}^{4}},\dfrac{120{{a}^{3}}}{3!}=20{{a}^{3}},\dfrac{360{{a}^{2}}}{4!}=15{{a}^{2}}, \\
 & \dfrac{720a}{5!}=6a,\dfrac{720}{6!}=1,\dfrac{0}{7!}=0 \\
\end{align}\]

Note: For such types of problems we first of all need to have a clear cut idea of the Taylor theorem and we need to have a look at some of the applications of it. We also need to have an understanding of differentiation which is a part of this problem. Since in this problem we need to find out the coefficients only, since no value of a particular point is given, so with the change in the value of ‘a’, the value of the coefficients also changes.