Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Differential Equation

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Theorem

The following definitions of the concept of Complex Exponential Function are equivalent:

As a Power Series Expansion

The exponential function can be defined as a (complex) power series:

\(\ds \forall z \in \C: \, \) \(\ds \exp z\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}\)
\(\ds \) \(=\) \(\ds 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots\)

As the Solution of a Differential Equation

The exponential function can be defined as the unique particular solution $y = \map f z$ to the first order ODE:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map f 0 = 1$.

That is, the defining property of $\exp$ is that it is its own derivative.


Proof

Power Series Expansion implies Solution of Differential Equation

Let $\exp z$ be the complex function defined as the power series:

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


Let $y = \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$.

Then:

\(\ds \dfrac {\d y} {\d z}\) \(=\) \(\ds \dfrac \d {\d z} \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {z^{n - 1} } {\paren {n - 1}!}\) Derivative of Complex Power Series
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds y\)


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

$\exp \paren 0 = 1$.


Setting $z = 0$ we find:

\(\ds y \paren 0\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {0^n} {n!}\)
\(\ds \) \(=\) \(\ds \frac {0^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {0^n} {n!}\)
\(\ds \) \(=\) \(\ds \frac {0^0} {0!}\) as $0^n = 0$ for all $n > 0$
\(\ds \) \(=\) \(\ds 1\) Definition of $0^0$


That is:

$\exp z$ is the particular solution of the differential equation:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map y 0 = 1$.

$\Box$


Solution of Differential Equation implies Power Series Expansion

Let $\exp z$ be the complex function defined as the particular solution of the differential equation:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map y 0 = 1$.


Let $f: \C \to \C$ be a solution to the differential equation $\dfrac {\d f} {\d z} = f$ with $f \paren 0 = 1$.

Then Holomorphic Function is Analytic shows that $f$ can be expressed as a power series:

$\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$

about any $\xi \in \C$.


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

\(\ds a_n\) \(=\) \(\ds \dfrac {\map {f^{\paren n} } 0} {n!}\) Power Series is Taylor Series
\(\ds \) \(=\) \(\ds \dfrac {\map {f^{\paren {n - 1} } } 0} {n!}\) as $\map {f^{\paren n} } 0 = \map f 0 = \map {f^{\paren {n - 1} } } 0$
\(\ds \) \(=\) \(\ds \dfrac 1 n a_{n - 1}\) Power Series is Taylor Series

As $a_0 = \dfrac {\map {f^{\paren 0} } 0} {0!} = 1$ by the initial condition, it follows inductively that:

$a_n = \dfrac 1 {n!}$

Hence:

$\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$

$\blacksquare$

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