Q. 21 - Calculus

Question

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

Step-by-Step Solution

Answer

The Maclaurin series for the cosine function is the derivate of the Maclaurin series for the sine function term by term.

Step 1 : Given information

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

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

Consider the function $f\left(x\right)=\sin \left(x\right)$

For function, the Maclaurin series is as follows:

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

The function's$f\left(x\right)$ derivation is

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

Since then, the Maclaurin series for cosine has been as follows:

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

Thus, the Maclaurin series for the cosine function is the derivate of the Maclaurin series for the sine function term by term.