Question

How do you find the Maclaurin series for \[f\left( x \right)\] using the definition of a Maclaurin series of \[4\sinh \left( 4x \right)\]?

Answer 1

Hint: In this problem, we have to find the Maclaurin series for \[f\left( x \right)\] using the definition of a Maclaurin series, of \[4\sinh \left( 4x \right)\]. We can now derive the Maclaurin series from an infinite series. We can write the infinite series, we can then find the first derivative, second derivative, third derivative, fourth and fifth derivative of the given function. We can substitute the derivatives in the infinite function to get the final answer.

Complete step by step answer:
We can now derive the Maclaurin series from an infinite series. We can write the infinite series.
\[\Rightarrow f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{x}^{3}}+....+\dfrac{{{f}^{n}}\left( 0 \right)}{n!}{{x}^{n}}\] ……. (1)
We know that the given function is,
\[f\left( x \right)=4\sinh \left( 4x \right)\]
We can now find the first term of the infinite series (1) from the above function,
\[\Rightarrow f\left( 0 \right)=4\sinh \left( 0 \right)=0\]…… (2)
We can now find the first derivative of \[f\left( x \right)=4\sinh \left( 4x \right)\] , we get
\[\begin{align}
  & \Rightarrow f'\left( x \right)=4\cosh \left( 4x \right)4 \\
 & \Rightarrow f'\left( x \right)=16\cosh \left( 4x \right) \\
 & \Rightarrow f'\left( 0 \right)=16\cosh \left( 0 \right)=16={{4}^{2}} \\
\end{align}\]
We can now find the second derivative \[f\left( x \right)=4\sinh \left( 4x \right)\], we get
\[\begin{align}
  & \Rightarrow f''\left( x \right)=16\sinh \left( 4x \right)4 \\
 & \Rightarrow f''\left( x \right)=64\sinh \left( 4x \right) \\
 & \Rightarrow f''\left( 0 \right)=64\sinh \left( 0 \right)=0 \\
\end{align}\]
We can now find the third derivative \[f\left( x \right)=4\sinh \left( 4x \right)\], we get
\[\begin{align}
  & \Rightarrow f'''\left( x \right)=64\cosh \left( 4x \right)4 \\
 & \Rightarrow f'''\left( x \right)=256\cosh \left( 4x \right) \\
 & \Rightarrow f'''\left( 0 \right)=256\cosh \left( 0 \right)=256={{4}^{4}} \\
\end{align}\]
We can now find the fourth derivative \[f\left( x \right)=4\sinh \left( 4x \right)\], we get
\[\begin{align}
  & \Rightarrow f''''\left( x \right)=256\sinh \left( 4x \right)4 \\
 & \Rightarrow f''''\left( x \right)=1024\sinh \left( 4x \right) \\
 & \Rightarrow f''''\left( 0 \right)=1024\sinh \left( 0 \right)=0 \\
\end{align}\]
We can now find the fifth derivative \[f\left( x \right)=4\sinh \left( 4x \right)\], we get
\[\begin{align}
  & \Rightarrow f'\left( x \right)=1024\cosh \left( 4x \right)4 \\
 & \Rightarrow f'\left( x \right)=4096\cosh \left( 4x \right) \\
 & \Rightarrow f'\left( 0 \right)=4096\cosh \left( 0 \right)={{4}^{6}} \\
\end{align}\]
We can now observe that, we can see a clear pattern formatting, where
\[{{f}^{\left( n \right)}}\left( 0 \right)=\left\{ \begin{matrix}
   0 & n\text{ even} \\
   4\times {{4}^{n}} & n\text{ odd} \\
\end{matrix} \right.\]
We can now form the Maclaurin series from (1), we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=0+16x+0{{x}^{2}}+\dfrac{256}{6}{{x}^{3}}+0{{x}^{4}}+\dfrac{4096}{120}{{x}^{5}}.... \\
 & \Rightarrow f\left( x \right)=16x+\dfrac{128}{3}{{x}^{3}}+\dfrac{512}{15}{{x}^{5}}....+\dfrac{4\times {{4}^{2n+1}}}{2n+1}{{x}^{2n+1}} \\
 & \Rightarrow f\left( x \right)=16x+\dfrac{128}{3}{{x}^{3}}+\dfrac{512}{15}{{x}^{5}}....+\dfrac{{{4}^{2n+2}}}{2n+1}{{x}^{2n+1}} \\
\end{align}\]
Therefore, the Maclaurin series is \[f\left( x \right)=16x+\dfrac{128}{3}{{x}^{3}}+\dfrac{512}{15}{{x}^{5}}....+\dfrac{{{4}^{2n+2}}}{2n+1}{{x}^{2n+1}}\].

Note: Students make mistakes while finding the derivatives for the given function for which we should know some differentiation formulas. We should also concentrate while substituting each derivative of \[f\left( 0 \right)\] and we should calculate it to get the Maclaurin series.