Q. 23 - Calculus

Question

Show that when you take the derivative of the Maclaurin series for the exponential function term by term, you obtain the same series you started with. Why does that make sense? 

Step-by-Step Solution

Answer

The Maclaurin series derivate for the exponential function term by term is the same series as before.

Step 1. Given information

The function is $f\left(x\right)=e^x$

Step 2. Calculation

Let us consider the function $f\left(x\right)=e^x$

For function, the Maclaurin series is as follows:

$e^x=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^{2k}$

The derivative of the function$f\left(x\right)$ is

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

It can also be written as:

$f^{\text{'}}\left(x\right)=0+1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$

Therefore, the Maclaurin series derivate for the exponential function term by term is the same series as before.