# Power Series Expansion for Exponential Function

## Theorem

Let $\exp x$ be the exponential function.

Then:

 $\ds \forall x \in \R: \,$ $\ds \exp x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ $\ds$ $=$ $\ds 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots$

## Proof

From Higher Derivatives of Exponential Function, we have:

$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$

Since $\exp 0 = 1$, the Taylor series expansion for $\exp x$ about $0$ is given by:

$\ds \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

From Radius of Convergence of Power Series over Factorial, we know that this power series expansion converges for all $x \in \R$.

From Taylor's Theorem, we know that

$\ds \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$

where $0 \le \eta \le x$.

Hence:

 $\ds \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }$ $=$ $\ds \size {\frac {x^n} {n!} \map \exp \eta}$ $\ds$ $\le$ $\ds \frac {\size {x^n} } {n!} \map \exp {\size x}$ Exponential is Strictly Increasing $\ds$ $\to$ $\ds 0$ $\ds \text { as } n \to \infty$ Series of Power over Factorial Converges

So the partial sums of the power series converge to $\exp x$.

The result follows.

$\blacksquare$

## Historical Note

The power series expansion for $e^x$ was first established by Isaac Newton in $1665$.