Equivalence of Definitions of Complex Exponential Function

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Theorem

The following definitions of the concept of the complex exponential function are equivalent:

As a Power Series Expansion

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

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

By Real Functions

The exponential function can be defined by the real exponential, sine and cosine functions:

$\exp z := e^x \paren {\cos y + i \sin y}$

where $z = x + i y$ with $x, y \in \R$.

Here, $e^x$ denotes the real exponential function, which must be defined first.

As a Limit of a Sequence

The exponential function can be defined as a limit of a sequence:

$\ds \exp z := \lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^n$

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

From Radius of Convergence of Power Series over Factorial: Complex Case, it follows that the power series $\ds \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ is absolutely convergent over the entirety of $\C$.

Hence, the definition of $\exp z$ as a power series is valid.


It remains to demonstrate the logical equivalence of all the definitions.


Power Series Expansion equivalent to Solution of Differential Equation

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:

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

$\Box$


Power Series Expansion equivalent to Definition by Real Functions

We have the result:

Power Series Expansion equivalent to Solution of Differential Equation

which gives that the definition of $\exp z$ as the power series:

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

is equivalent to the definition of $\exp z$ as the solution of the differential equation:

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

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


Let:

$e: \R \to \R$ denote the real exponential function
$\sin: \R \to \R$ denote the real sine function
$\cos: \R \to \R$ denote the real cosine function.


Then:

\(\ds \exp z\) \(=\) \(\ds \map \exp {x + i y}\) where $x, y \in \R$
\(\ds \) \(=\) \(\ds \map \exp x \map \exp {i y}\) Exponential of Sum: Complex Numbers
\(\ds \) \(=\) \(\ds e^x \map \exp {i y}\) Definition (as Power Series Expansion) agrees with Definition of Real Exponential for all $x \in \R$
\(\ds \) \(=\) \(\ds e^x \paren {\cos y + i \sin y}\) Euler's Formula, which can be proven using Definition as Power Series Expansion

$\Box$


Power Series Expansion equivalent to Limit of Sequence

Let:

$\ds s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$
$a_n = \paren {1 + \dfrac z n}^n$

Then we can express $a_n$ as follows:

\(\ds a_n\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {\dfrac z n}^k 1^{n - k}\) Binomial Theorem: Integral Index
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \dfrac {z^k} {k!} \dfrac {n \paren {n - 1} \cdots \paren {n - k + 1} }{n^k}\)
\(\ds \) \(=\) \(\ds 1 + \sum_{k \mathop = 1}^n \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n}\) by algebraic manipulations


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

\(\ds \lim_{n \mathop \to +\infty} \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }\) \(=\) \(\ds \cmod {\lim_{n \mathop \to +\infty} \paren {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }\) Modulus of Limit
\(\ds \) \(=\) \(\ds \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \lim_{n \mathop \to +\infty} \dfrac j n} } }\) Combination Theorem for Limits of Complex Functions
\(\ds \) \(=\) \(\ds \cmod {\dfrac {z^k} {k!} \paren {1 - 1} }\) Sequence of Powers of Reciprocals is Null Sequence
\(\ds \) \(=\) \(\ds 0\)


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

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

$\ds \sum_{k \mathop = m}^n \cmod {\dfrac {z^k} {k!} } < \dfrac \epsilon 2$

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

$\ds \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } < \dfrac \epsilon {2 M}$

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

\(\ds \cmod {s_n - a_n}\) \(=\) \(\ds \cmod {\sum_{k \mathop = 1}^{M - 1} \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } + \sum_{k \mathop = M}^n \dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k- 1 } \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} } \cmod {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }\) Triangle Inequality for Complex Numbers
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = 1}^{M - 1} \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } } + \sum_{k \mathop = M}^n \cmod {\dfrac {z^k} {k!} }\)
\(\ds \) \(<\) \(\ds \sum_{k \mathop = 1}^{M - 1} \dfrac \epsilon {2 M} + \dfrac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

As an Absolutely Convergent Series is Convergent, $\sequence {s_n}$ converges.

Then:

\(\ds 0\) \(=\) \(\ds \lim_{n \mathop \to +\infty} \cmod {s_n - a_n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to +\infty} s_n - \lim_{n \mathop \to +\infty} a_n\) Combination Theorem for Limits of Complex Functions
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {z^k} {k!} - \lim_{n \mathop \to +\infty} \paren {1 + \dfrac z n}^n\)

The result follows.

$\blacksquare$