Equivalence of Definitions of Real Exponential Function/Sum of Series equivalent to Differential Equation

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Theorem

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

As the Sum of a Series

The exponential function can be defined as a power series:

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

As the Solution of a Differential Equation

The exponential function can be defined as the unique solution $y = f \left({x}\right)$ to the first order ODE:

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

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


Proof

Sum of Series implies Solution of Differential Equation

Let $\exp x$ be the real function defined as the sum of the power series:

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


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


Then:

\(\displaystyle \dfrac {\d y} {\d x}\) \(=\) \(\displaystyle \dfrac \d {\d x} \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac \d {\d x} \paren {\frac {x^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!} }\) $\quad$ extracting the zeroth term $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0 + \sum_{n \mathop = 1}^\infty \frac {n x^{n - 1} } {n!}\) $\quad$ Power Rule for Derivatives and Derivative of Constant $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {\paren {n - 1}!}\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) $\quad$ Translation of Index Variable of Summation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle y\) $\quad$ $\quad$


Setting $x = 0$ we find:

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

$\blacksquare$


That is:

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

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

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

$\Box$


Solution of Differential Equation implies Sum of Series

Let $\exp x$ be the real function defined as the solution of the differential equation:

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

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


We have Taylor Series Expansion for Exponential Function:


From Higher Derivatives of Exponential Function, we have:

$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$


Since $\exp 0 = 1$, the Taylor series expansion for $\exp x$ about $0$ is given by:

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


From Radius of Convergence of Power Series over Factorial, we know that this power series expansion converges for all $x \in \R$.

From Taylor's Theorem, we know that

$\displaystyle \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$

where $0 \le \eta \le x$.

Hence:

\(\displaystyle \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }\) \(=\) \(\displaystyle \size {\frac {x^n} {n!} \map \exp \eta}\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \frac {\size {x^n} } {n!} \map \exp {\size x}\) $\quad$ Exponential is Strictly Increasing $\quad$
\(\displaystyle \) \(\to\) \(\displaystyle 0\) \(\displaystyle \text { as } n \to \infty\) $\quad$ Series of Power over Factorial Converges $\quad$


So the partial sums of the power series converge to $\exp x$.

The result follows.

$\blacksquare$