# Maclaurin Series for The Natural Exponential Function

Recall that a Maclaurin series of a function $f$ is Taylor series centered about $x = 0$ and that has the form $\sum_{n=0}^{\infty} f^{(n)}(0) \frac{x^n}{n!} = f(0) + f'(0)\frac{x}{1!} + f''(0)\frac{x^2}{2!} + ...$. We will now verify the Maclaurin series for the natural exponential function $f(x) = e^x$.

Property: If $f(x) = e^x$, then $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ is the Maclaurin series for $f(x)$ and is valid for all $x \in \mathbb{R}$. |

First consider the following table of derivatives of $f$ and the derivatives of $f$ evaluated at $x = 0$ (since we're dealing with a Maclaurin series):

$f(x) = e^x$ | $f(0) = e^0 = 1$ |

$f'(x) = e^x$ | $f'(0) = e^0 = 1$ |

$f''(x) = e^x$ | $f''(0) = e^0 = 1$ |

$f'''(x) = e^x$ | $f'''(0) = e^0 = 1$ |

$f^{(4)}(x) = e^x$ | $f^{(4)}(0) = e^0 = 1$ |

We see that a general formula for the $n^{\mathrm{th}}$ derivative of $f$ is $f^{(n)}(x) = e^x$ and that a general formula for the $n^{\mathrm{th}}$ derivative of $f$ evaluated at $x = 0$ is $f^{(n)}(0) = e^0 = 1$. Substituting this into the formula for a Maclaurin series and we obtain that:

(1)Now let's determine the interval of convergence of this Maclaurin series using the ratio test as follows:

(2)Therefore the radius of convergence is $R = \infty$, and so this power series converges for all $x \in \mathbb{R}$. The last thing that we need to show is that $e^x$ is analytic and hence show that the sum of this series is precisely $e^x$. Suppose that the sum of our power series is some function $g(x)$, that is:

(3)Notice that the derivative of $g(x)$ is:

(4)Therefore we have the differential equation $g(x) = g'(x)$, and so $g(x) = Ce^x$ for some constant $C$. Notice that if we plug $x = 0$ we have that $g(0) = 1 = Ce^0 = C$ and so $C = 1$, therefore $g(x) = e^x$ and so we have shown that:

(5)