Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Differential Equation
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:
- $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$
$\blacksquare$