Power Series Expansion for Exponential Function

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Theorem

Let $\exp x$ be the exponential function.


Then:

\(\displaystyle \forall x \in \R: \ \ \) \(\displaystyle \exp x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots\)


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:

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


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

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

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

Hence:

\(\displaystyle \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }\) \(=\) \(\displaystyle \size {\frac {x^n} {n!} \map \exp \eta}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac {\size {x^n} } {n!} \map \exp {\size x}\) Exponential is Strictly Increasing
\(\displaystyle \) \(\to\) \(\displaystyle 0\) \(\displaystyle \text { as } n \to \infty\) 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 $1665$.


Sources