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

Theorem

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

As a Power Series Expansion

The exponential function can be defined as a power series:

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

As the Solution of a Differential Equation

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

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

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

Proof

Power Series Expansion implies Solution of Differential Equation

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

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

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

Then:

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

Setting $x = 0$ we find:

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

$\blacksquare$

That is:

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

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

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

$\Box$

Solution of Differential Equation implies Power Series Expansion

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

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

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

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:

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

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

The result follows.

$\blacksquare$