User:Anghel/Sandbox

Theorem
All the definitions of the complex exponential function $\exp: \C \to \C \setminus \left\{ {0}\right\}$ are equivalent.


 * $(1): \quad \displaystyle \exp z = \sum_{n \mathop = 0}^\infty \dfrac{z^n}{n!}$


 * $(2): \quad \exp z = e^x \left({ \cos \left({y}\right) + i \sin \left({y}\right) }\right)$


 * with $x, y \in \R, z = x+iy$.


 * $(3): \quad \displaystyle \exp z = \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^n$


 * $(4): \quad y = \exp z: y = \dfrac {\mathrm dy}{\mathrm dz}, \exp \left({0}\right) = 1$

Proof
From Power Series over Factorial/Complex Case, it follows that the power series $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{z^n}{n!}$ is absolutely convergent.

Hence, definition $(1)$ is valid.

$(1)$ is equivalent to $(2)$
Let $e$ denote the real exponential function, and let $\sin$ and $\cos$ denote the real sine and cosine functions.

If we put $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{z^n}{n!}$, we have:

$(1)$ is equivalent to $(3)$
Put $\displaystyle s_n = \sum_{k \mathop = 0}^n \dfrac{z^k}{k!}$, and $a_n = \left({1 + \dfrac z n}\right)^n$.

Then, we can express $a_n$ as follows:

The limit of the difference between the $k$th terms of $a_n$ and $s_n$ is:

To show that $s_n$ and $a_n$ has the same limit, let $\epsilon \in \R_{>0}$.

From Tail of Convergent Series, it follows that we can find $M \in \N$ such that for all $m \ge M$:


 * $\displaystyle \sum_{k \mathop = m}^n \left\vert{\dfrac{z^k}{k!} }\right\vert < \dfrac \epsilon 2$

For all $k \in \left\{ {0, 1, \ldots, M-1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$:


 * $\displaystyle \left\vert{\dfrac{z^k}{k!} - \dfrac{z^k}{k!} \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }\right\vert < \dfrac {\epsilon}{2M}$

Then for all $n \ge \max \left({M, N_0, N_1, \ldots, N_{M-1} }\right)$, we have:

As an Absolutely Convergent Series is Convergent, we have:

The result follows.

$(1)$ implies $(4)$
We show that $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{z^n}{n!}$ is a solution to the differential equation:

We show that $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{z^n}{n!}$ satisfies the initial condition:


 * $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{0^n}{n!} = \dfrac{0^0}{0!} = 1$

$(4)$ implies $(1)$
Suppose that $f: \C \to \C$ is a solution to the differential equation $\dfrac{\mathrm df}{\mathrm dz} = f$ with $f \left({0}\right) = 1$.

Then Complex-Differentiable Function is Analytic shows that $f$ can be expressed as a power series $\displaystyle \sum_{n \mathop = 0}^\infty a_n \left({z - \xi}\right)^n$ about any $\xi \in \C$.

When $\xi = 0$, we have for all $n \in \N_{\ge 1}$:

As $a_0 = \dfrac{f^{\left({0}\right) } \left({0}\right) }{ 0! } = 1$ by the initial condition, it follows inductively that:


 * $a_n = \dfrac{1}{n!}$

Hence, $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty \dfrac{1}{n!} z^n$.