Q. 22 - Calculus

Question

Show that when you take the derivative of the Maclaurin series for the cosine function term by term you obtain the negative of the Maclaurin series for the sine. 

Step-by-Step Solution

Answer

The  Maclaurin series term derivative term for the cosine function  is a negative value for the Maclaurin series for the sine function. 

Step 1. Given information

Given function : $f\left(x\right)=\cos x$

Step 2 : Showing that the derivative of the Maclaurin series for f ( x ) = cos x is the negative of the Maclaurin series for f ( x ) = sin x

Let us consider the function.

$f\left(x\right)=\cos x$

For function, the Maclaurin series is as follows:

$\cos x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$

Now, the derivate of the function $f\left(x\right)$ is

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left(\sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}\right)\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}\frac{d}{dx}(x{)}^{2k}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}\left(2k\right)x^{2k-1}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k-1)!}{x}^{2k-1}\end{array}$

Or, it can be written as:

$f^{\text{'}}\left(x\right)=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+\dots$

Hence, the Maclaurin series for sine is:

$\sin x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$

This implies that:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$

Hence, the  Maclaurin series term derivative term for the cosine function  is a negative value for the Maclaurin series for the sine function.