Equivalence of Definitions of Complex Exponential Function

From ProofWiki
Jump to navigation Jump to search

Theorem

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

As a Sum of a Series

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

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

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:

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

As the Solution of a Differential Equation

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

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

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

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 $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ is absolutely convergent over the entirety of $\C$.

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


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


Sum of Series equivalent to Solution of Differential Equation

Sum of Series implies Solution of Differential Equation

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

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


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

Then:

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


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

$\exp \paren 0 = 1$.


Setting $z = 0$ we find:

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


That is:

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

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

satisfying the initial condition $y \left({0}\right) = 1$.

$\Box$


Solution of Differential Equation implies Sum of Series

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

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

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


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:

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

about any $\xi \in \C$.


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

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

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

$a_n = \dfrac 1 {n!}$

Hence:

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

$\Box$


Sum of Series equivalent to Definition by Real Functions

We have the result:

Sum of Series equivalent to Solution of Differential Equation

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

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

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:

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

$\Box$


Sum of Series equivalent to Limit of Sequence

Let:

$\displaystyle s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$
$a_n = \left({1 + \dfrac z n}\right)^n$

Then we can express $a_n$ as follows:

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


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

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


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 $m \ge M$:

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

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

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

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

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

As an Absolutely Convergent Series is Convergent, $\left\langle{s_n}\right\rangle$ converges.

Then:

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

The result follows.

$\blacksquare$