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

## 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:

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

Setting $x = 0$ 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 $0^0$

$\blacksquare$

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 $f \left({0}\right) = 1$.

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:

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

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

The result follows.

$\blacksquare$