# Equivalence of Definitions of Real Exponential Function

## Contents

- 1 Theorem
- 2 Proof
- 2.1 Inverse of Natural Logarithm implies Limit of Sequence
- 2.2 Limit of Sequence implies Sum of Series
- 2.3 Limit of Sequence implies Extension of Rational Exponential
- 2.4 Extension of Rational Exponential implies Solution of Differential Equation
- 2.5 Sum of Series equivalent to Solution of Differential Equation
- 2.6 Inverse of Natural Logarithm equivalent to Solution of Differential Equation

- 3 Also see

## Theorem

The following definitions of the concept of **Real Exponential Function** are equivalent:

### As a 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 a Limit of a Sequence

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

- $\exp x := \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$

### As an Extension of the Rational Exponential

Let $e$ denote Euler's number.

Let $f: \Q \to \R$ denote the real-valued function defined as:

- $f \left({ x }\right) = e^x$

That is, let $f \left({ x }\right)$ denote $e$ to the power of $x$, for rational $x$.

Then $\exp : \R \to \R$ is defined to be the unique continuous extension of $f$ to $\R$.

$\exp \left({ x }\right)$ is called the **exponential of $x$**.

### As the Inverse of the Natural Logarithm

Consider the natural logarithm $\ln x$, which is defined on the open interval $\left({0 \,.\,.\, +\infty}\right)$.

From Logarithm is Strictly Increasing:

- $\ln x$ is strictly increasing.

From Inverse of Strictly Monotone Function:

- the inverse of $\ln x$ always exists.

The inverse of the natural logarithm function is called the **exponential function**, which is denoted as $\exp$.

Thus for $x \in \R$, we have:

- $y = \exp x \iff x = \ln y$

### 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

### Inverse of Natural Logarithm implies Limit of Sequence

Let $\exp x$ be the real function defined as the inverse of the natural logarithm:

- $y = \exp x \iff x = \ln y$

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as:

- $x_n = \paren {1 + \dfrac x n}^n$

First it needs to be noted that $\left \langle {x_n} \right \rangle$ does indeed converge to a limit.

From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Sum of Series, we have:

- $\displaystyle \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

From Series of Power over Factorial Converges, the right hand side is indeed shown to converge to a limit.

It will next be shown that:

- $\displaystyle \ln \paren {\lim_{n \mathop \to \infty} \left \langle {x_n} \right \rangle} = x$

We have:

\(\displaystyle \ln \paren {\paren {1 + \frac x n}^n}\) | \(=\) | \(\displaystyle n \ln \paren {1 + x n^{-1} }\) | $\quad$ Logarithms of Powers | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \frac {\ln \paren {1 + x n^{-1} } } {x n^{-1} }\) | $\quad$ multiplying by $1 = \dfrac {x n^{-1} } {x n^{-1} }$ | $\quad$ |

From Limit of Sequence is Limit of Real Function, we can consider the differentiable analogue of the sequence.

From Derivative of Logarithm at One we have:

- $\displaystyle \lim_{x \mathop \to 0} \frac {\ln \paren {1 + x} } x = 1$

But $x n^{-1} \to 0$ as $n \to \infty$ from Power of Reciprocal.

Thus:

- $\displaystyle x \frac {\ln \paren {1 + x n^{-1} } } {x n^{-1} } \to x$

as $n \to \infty$.

From Exponential Function is Continuous:

- $\paren {1 + \dfrac x n}^n = \exp \paren {n \ln \paren {1 + \dfrac x n} } \to \exp x = e^x$

as $n \to \infty$.

$\blacksquare$

### Limit of Sequence implies Sum of Series

Let $\exp x$ be the real function defined as the limit of the sequence:

- $\exp x := \displaystyle \lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n$

From the General Binomial Theorem:

\(\displaystyle \paren {1 + \frac x n}^n\) | \(=\) | \(\displaystyle 1 + x + \frac {n \paren {n - 1} x^2} {2! \ n^2} + \frac {n \paren {n - 1} \paren {n - 2} x^3} {3! \ n^3} + \cdots\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \paren {1 + \frac x n}^n - \frac {x^0} {0!} + \frac {x^1} {1!} + \paren {\frac {n - 1} n} \frac {x^2} {2!} + \paren {\frac {\paren {n - 1} \paren {n - 2} } {n^2} } \frac {x^3} {3!} + \cdots\) | $\quad$ | $\quad$ |

From Power over Factorial, this converges to:

- $\exp x - \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \dfrac {x^3} {3!} + \cdots = 0$

as $n \to +\infty$:

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exp x\) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) | $\quad$ | $\quad$ |

$\blacksquare$

### Limit of Sequence implies Extension of Rational Exponential

Let the restriction of the exponential function to the rationals be defined as:

- $\displaystyle \exp \restriction_\Q: x \mapsto \lim_{n \mathop \to +\infty}\left ({1 + \frac x n}\right)^n$

Thus, let $e$ be Euler's Number defined as:

- $e = \displaystyle \lim_{n \mathop \to +\infty}\left ({1 + \frac 1 n}\right)^n$

For $x = 0$:

\(\displaystyle \exp \restriction_\Q \paren 0 \ \ \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e^0\) | $\quad$ | $\quad$ |

For $x \ne 0$:

\(\displaystyle \exp \restriction_\Q \paren x \ \ \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to +\infty}\paren {1 + \frac x n}^n\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{\paren {n/x} \mathop \to +\infty}\paren {\paren {1 + \frac 1 {\paren {n/x} } }^{\paren {n/x} } }^x\) | $\quad$ Exponent Combination Laws | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e^x\) | $\quad$ | $\quad$ |

For $x \in \R \setminus \Q$, we invoke Power Function to Rational Power permits Unique Continuous Extension.

$\blacksquare$

### Extension of Rational Exponential implies Solution of Differential Equation

Let $\exp x$ be the real function defined as the extension of rational exponential.

Then we have:

\(\displaystyle D_x \left({\exp x}\right)\) | \(=\) | \(\displaystyle \lim_{h \mathop \to 0} \frac {\exp \left({x + h}\right) - \exp x} h\) | $\quad$ Definition of Derivative | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) | $\quad$ Exponent of Sum | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \left({\exp h - 1}\right)} h\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \exp x \left({\lim_{h \mathop \to 0} \frac {\exp h - 1} h}\right)\) | $\quad$ Multiple Rule for Limits of Functions, as $\exp x$ is constant | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \exp x\) | $\quad$ Derivative of Exponential at Zero: Proof 2 | $\quad$ |

The application of Derivative of Exponential at Zero is not circular as the referenced proof does not depend on $D_x \exp x = \exp x$.

$\Box$

\(\displaystyle \exp 0\) | \(=\) | \(\displaystyle \lim_{n \mathop \to +\infty} \left({1 + \frac 0 n}\right)^n\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

$\blacksquare$

### Sum of Series equivalent to Solution of Differential Equation

#### 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:

- $\forall n \in \N: f^{\left({n}\right)} \left({\exp x}\right) = \exp x$

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

- $\displaystyle \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1}} {\left({n - 1}\right)!} + \frac {x^n} {n!} \exp \left({\eta}\right)$

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

Hence:

\(\displaystyle \left \vert {\exp x - \left({1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\left({n - 1}\right)!} }\right)} \right \vert\) | \(=\) | \(\displaystyle \left \vert {\frac {x^n} {n!} \exp \left({\eta}\right)} \right \vert\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \frac {\left \vert {x^n} \right \vert} {n!} \exp \left({\left \vert {x} \right \vert}\right)\) | $\quad$ Exponential is Strictly Increasing | $\quad$ | |||||||||

\(\displaystyle \) | \(\to\) | \(\displaystyle 0 \text { as } n \to \infty\) | $\quad$ Series of Power over Factorial Converges | $\quad$ |

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

The result follows.

$\blacksquare$

### Inverse of Natural Logarithm equivalent to Solution of Differential Equation

#### Inverse of Natural Logarithm implies Solution of Differential Equation

Let $\exp x$ be the real function defined as the inverse of the natural logarithm:

- $y = \exp x \iff x = \ln y$

Then:

\(\displaystyle x\) | \(=\) | \(\displaystyle \ln y\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t\) | $\quad$ Definition 1 of Natural Logarithm | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle D_y \left({x}\right)\) | \(=\) | \(\displaystyle D_y \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t\) | $\quad$ Differentiation with respect to $y$ | $\quad$ | ||||||||

\(\displaystyle \frac {\d x} {\d y}\) | \(=\) | \(\displaystyle \frac 1 y\) | $\quad$ Fundamental Theorem of Calculus | $\quad$ | |||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle \frac {\d y} {\d x}\) | \(=\) | \(\displaystyle y\) | $\quad$ Derivative of Inverse Function | $\quad$ |

This proves that $y$ is a solution of the differential equation.

It remains to be proven that $y$ fulfils the initial condition:

\(\displaystyle f \left({0}\right)\) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle f^{-1} \left({1}\right)\) | \(=\) | \(\displaystyle 0\) | $\quad$ Image of Element under Inverse Mapping | $\quad$ | ||||||||

\(\displaystyle \ln y \Big \vert_{y \mathop = 1}\) | \(=\) | \(\displaystyle \int_{t \mathop = 1}^{t \mathop = 1} \frac 1 t \rd t\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | $\quad$ Integral on Zero Interval | $\quad$ |

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 Inverse of Natural Logarithm

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

Thus:

\(\displaystyle \frac {\d y} {\d x}\) | \(=\) | \(\displaystyle y\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d x} {\d y}\) | \(=\) | \(\displaystyle \frac 1 y\) | $\quad$ Derivative of Inverse Function | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \int \rd x\) | \(=\) | \(\displaystyle \int \frac 1 y \rd y\) | $\quad$ Separation of Variables | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x + C\) | \(=\) | \(\displaystyle \int_{t \mathop = 1}^{t \mathop = y} \frac 1 t \rd t\) | $\quad$ Fundamental Theorem of Calculus | $\quad$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \ln y\) | $\quad$ Definition 1 of Natural Logarithm | $\quad$ |

To solve for $C$, put $\left({x_0, y_0}\right) = \left({0, 1}\right)$:

\(\displaystyle 0 + C\) | \(=\) | \(\displaystyle \int_{t \mathop = 1}^{t \mathop = 1}\frac 1 t \rd t\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle C\) | \(=\) | \(\displaystyle 0\) | $\quad$ Integral on Zero Interval | $\quad$ |

That is:

- $y = \exp x \iff x = \ln y$

$\blacksquare$