LUCAS WILLEMS

A 25 year-old student passionate about maths and programming

Exponential serie

Article

In the 18th century, Leonard Euler found this formula to write explicitly exponential :

$$x \in \mathbb{C} \qquad e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$

But how to find it ? It is what I'm going to show you.

Summary

Proof

This proof is the one that I found on my own (but isn't most probably the one that Euler found) and only uses exponential properties and the assumption that $$e^x$$ can be written like a polynomial of infinite degree, i.e. these 3 equalities :

$$\begin{gather*} \exp' = exp \\ e^0 = 1 \\ e^x = \sum_{k = 0}^{\infty} a_k x^k \end{gather*}$$

From these equalities, let's use a proof by induction with the following statement :

$$\forall n \in \mathbb{N} \quad H_n : a_n = \frac{1}{n!}$$

that leads us to the following reasoning :

Bases :
For $$n = 0$$, $$e^0 = a^0 = 1 = \frac{1}{0!}$$. So, $$H_0$$ holds.

Inductive step :
As $$\exp = \exp'$$, we can write the following :$$\sum_{k = 0}^{\infty} a_k x^k = (\sum_{k = 0}^{\infty} a_k x^k)' = \sum_{k = 0}^{\infty} (k+1) a_{k+1} x^k$$which means that for $$m \in \mathbb{N}$$,

$$a_m = (m+1)a_{m+1} \Leftrightarrow a_{m+1} = \frac{a_m}{m+1}$$

And, as we assume $$H_m$$ holds,

$$a_{m+1} = \frac{a_m}{m+1} = \frac{\frac{1}{m!}}{m+1} = \frac{1}{(m+1)!}$$

Consequently, $$H_m \Rightarrow H_{m+1}$$.

Conclusion :
$$H_n$$ holds $$\forall n \in \mathbb{N}$$. Thus :

$$e^x = \sum_{k = 0}^{\infty} \frac{1}{k!} x^k = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$