Question

Find the Maclaurin series for the specified function:

$e^x$.

Short answer

The Maclaurin series is,

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$.

Step-by-step solution

Step 1. Given Information.

The function is$e^x$.

Step 2. Finding the Maclaurin series.

Let $f\left(x\right)=e^x$.

Since for any function $f$with derivatives of all orders at the point $x_0=0$, then the Maclaurin series is,

$f\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4+......$

Or we can write the general form Maclaurin series of the function $f$is,$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$.

The table of the Maclaurin series for the function $f\left(x\right)=e^x$

n	$f^n\left(x\right)$	$f^n\left(0\right)$	$\frac{f^n\left(0\right)}{n!}$
0	$e^x$	1	1
1	$e^x$	1	1
2	$e^x$	1	$\frac{1}{2!}$
. . .	. . .	. . .	. . .
k	$e^x$	1	$\frac{1}{k!}$
. . .	. . .	. . .	. . .

The Maclaurin series is,

$f\left(x\right)=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$.