Taylor Series Expansion for Exponential Function

Theorem

Real Numbers

Let $\exp x$ be the exponential function.

Then:

$\displaystyle \forall x \in \R: \exp x = \sum_{n=0}^\infty \frac {x^n} {n!}$

Proof

Proof for Real Numbers

From Higher Derivatives of Exponential Function, we have:

$\forall n \in \N: f^{\left({n}\right)} \left({\exp x}\right) = \exp x$

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

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

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

From Taylor's Theorem, we know that

$\displaystyle \exp x = 1 + \frac {x} {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1}} {\left({n-1}\right)!} + \frac {x^n} {n!} \exp \left({\eta}\right)$

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

Hence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left \vert {\exp x - \left({1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} }\right)} \right \vert\)	\(=\)	\(\displaystyle \left \vert {\frac {x^n} {n!} \exp \left({\eta}\right)} \right \vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \frac {\left \vert {x^n} \right \vert} {n!} \exp \left({\left \vert {x} \right \vert}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\to\)	\(\displaystyle 0 \text { as } n \to \infty\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by Series of Power over Factorial Converges

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

The result follows.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 15.5$