Question

How do you find the Maclaurin series for $e^x$?

Answer 1

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....+\frac{x^n}{n!}$
The Maclaurin series is obtained by the Power Series:
$f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)\frac{x}{0!}+f^{\prime \prime }\left(0\right)\frac{x^2}{2!}+f^{\left(3\right)}\left(0\right)\frac{x^3}{3!}+...$
so with $f\left(x\right)=e^x$ we have
$f\left(x\right)=e^x\Rightarrow f\left(0\right)=1$
$f^{\prime }\left(x\right)=e^x\Rightarrow f^{\prime }\left(0\right)=1$
$f^{\prime \prime }\left(x\right)=e^x\Rightarrow f^{\prime \prime }\left(0\right)=1$
$f^{\left(3\right)}\left(x\right)=e^x\Rightarrow f^{\left(3\right)}\left(0\right)=1$
$f^{\left(n\right)}\left(x\right)=e^x\Rightarrow f^{\left(n\right)}\left(0\right)=1$
so the maclaurin series is:
$e^x=1+1\frac{x}{0!}+1\frac{x^2}{2!}+1\frac{x^3}{3!}+1\frac{x^4}{4!}+....+1\frac{x^n}{n!}+...$
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....+\frac{x^n}{n!}+...$