Equivalence of Exponential Definitions
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Contents |
Theorem
All the definitions of the exponential function are equivalent.
- $(1): \quad y = \exp x \iff \ln y = x$:
- $\quad \quad \ln y = \displaystyle \int_{t=1}^{t=y} \frac 1 t \ \mathrm dt$
- $(2): \quad \exp x = e^x$
- $(3): \quad \exp x = \displaystyle \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$
- $(4): \quad \exp x = \displaystyle \sum_{n = 0}^\infty \frac {x^n} {n!}$
- $(5): \quad y = f\left({x}\right)$: $y = \dfrac {\mathrm dy}{\mathrm dx}$, $f\left({0}\right) = 1$
Proof
1 implies 5
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \ln y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \int_{t=1}^{t=y} \frac 1 t \ \mathrm dt\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle D_y \left({x}\right)\) | \(=\) | \(\displaystyle D_y \int_{t=1}^{t=y} \frac 1 t \ \mathrm dt\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | differentiate WRT $y$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm dx}{\mathrm dy}\) | \(=\) | \(\displaystyle \frac 1 y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Fundamental Theorem of Calculus/First Part | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm dy}{\mathrm dx}\) | \(=\) | \(\displaystyle y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Derivative of an Inverse Function |
This proves that $y$ is a solution.
Now, let's check if it fulfills the initial condition.
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle f\left({0}\right)\) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle f^{-1}\left({1}\right)\) | \(=\) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Inverse Mapping Image | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \ln y \vert_{y=1}\) | \(=\) | \(\displaystyle \int_{t=1}^{t=1} \frac 1 t \ \mathrm dt\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Integral on Zero Interval |
... indeed it does.
$\blacksquare$
5 implies 1
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm dy}{\mathrm dx}\) | \(=\) | \(\displaystyle y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {\mathrm dx}{\mathrm dy}\) | \(=\) | \(\displaystyle \frac 1 y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Derivative of an Inverse Function | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \int \ \mathrm dx\) | \(=\) | \(\displaystyle \int \frac 1 y \ \mathrm dy\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Separation of Variables | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle x + C\) | \(=\) | \(\displaystyle \int_{t=1}^{t=y}\frac 1 t \ \mathrm dt\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Fundamental Theorem of Calculus/Second Part |
Now to solve for $C$, put $\left(x_0,y_0\right) = \left(0,1\right)$:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle 0 + C\) | \(=\) | \(\displaystyle \int_{t=1}^{t=1}\frac 1 t \ \mathrm dt\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle C\) | \(=\) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Integral on Zero Interval |
$\blacksquare$
1 implies 3
We have:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \ln \left({\left({1 + \frac x n}\right)^n}\right)\) | \(=\) | \(\displaystyle n \ln \left({1 + xn^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Logarithms of Powers | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle x \frac {\ln \left({1 + x n^{-1} }\right)} {x n^{-1} }\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | multiply by $1 = \dfrac { xn^{-1} }{ xn^{-1} }$ |
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:
- $\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} x = 1$
But $x n^{-1} \to 0$ as $n \to \infty$ from Power of Reciprocal.
Thus:
- $\displaystyle x \frac {\ln \left({1 + x n^{-1}}\right)} {x n^{-1}} \to x$
as $n \to \infty$.
Since the exponential function is continuous at every point, it follows that:
- $\displaystyle \left({1 + \frac x n}\right)^n = \exp \left({n \ln \left({1 + \frac x n}\right)}\right) \to \exp x = e^x$
as $n \to \infty$.
$\blacksquare$
5 implies 4
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:
- $\displaystyle \exp x = \sum_{n=0}^\infty \frac {x^n} {n!}$
From 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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left \vert {\exp x - \left({1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} }\right)} \right \vert\) | \(=\) | \(\displaystyle \left \vert {\frac {x^n} {n!} \exp \left({\eta}\right)} \right \vert\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \frac {\left \vert {x^n} \right \vert} {n!} \exp \left({\left \vert {x} \right \vert}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\to\) | \(\displaystyle 0 \text { as } n \to \infty\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by Series of Power over Factorial Converges |
So the partial sums of the power series converge to $\exp x$.
The result follows.
$\blacksquare$
3 implies 4
From the Binomial Theorem:
- $\displaystyle \left({1 + \frac x n}\right)^n = 1 + x + \frac{n\left({n-1}\right)x^2}{2! \ n^2} + \frac{n\left({n-1}\right)\left({n-2}\right)x^3}{3! \ n^3} + \cdots$
- $ = \displaystyle \frac {x^0}{0!} + \frac {x^1}{1!} + \left({\frac {n-1} {n} }\right) \frac {x^2}{2!} + \left({{\frac {\left({n-1}\right)\left({n-2}\right)} {n^2} }}\right) \frac {x^3}{3!} + \cdots$
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \left({1 + \frac x n}\right)^n - \frac {x^0}{0!} + \frac {x^1}{1!} + \left({\frac {n-1} {n} }\right) \frac {x^2}{2!} + \left({{\frac {\left({n-1}\right)\left({n-2}\right)} {n^2} }}\right) \frac {x^3}{3!} + \cdots\) | \(=\) | \(\displaystyle 0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
From Power of a Number Less Than One, this converges to:
- $\displaystyle \exp x - \frac {x^0}{0!} + \frac {x^1}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots = 0$
as $n \to +\infty$
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \exp x\) | \(=\) | \(\displaystyle \sum_{n = 0}^\infty \frac {x^n} {n!}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
Compare Series of Power over Factorial Converges.
4 implies 5
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \dfrac {\mathrm d}{\mathrm dx}\sum_{n = 0}^\infty \frac {x^n} {n!}\) | \(=\) | \(\displaystyle \dfrac {\mathrm d}{\mathrm dx} \left({\frac {x^0}{0!} + \frac {x^1}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots } \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 0 + 1 + \frac {2x} {2 \cdot 1} + \frac {3x}{3 \cdot 2 \cdot 1} + \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {x^0}{0!} + \frac {x^1}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n = 0}^\infty \frac {x^n} {n!}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{n = 0}^\infty \frac {0^n} {n!}\) | \(=\) | \(\displaystyle \frac {0^0} {0!} + 0 + 0 + \cdots\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | definition of $0^0$ |
$\blacksquare$
3 implies 2
Let the restriction of the exponential function to the rationals be defined as:
- $\exp \restriction_{\Q}: x \mapsto \displaystyle \lim_{n \to +\infty}\left (1 + \frac x n\right)^n$
Let $e$ be Euler's Number defined as:
- $e = \displaystyle \lim_{n \to +\infty}\left (1 + \frac 1 n\right)^n$
For $x=0$:
| \(\displaystyle \) | \(\displaystyle \exp \restriction_{\Q} \ 0\) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \to +\infty}\left (1 + \frac 0 n\right)^n\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle e^0\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\Box$
For $x \ne 0$:
| \(\displaystyle \) | \(\displaystyle \exp \restriction_{\Q} \ x\) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \to +\infty}\left (1 + \frac x n\right)^n\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{\left({n/x}\right) \to +\infty}\left (\left (1 + \frac 1 { \left({n/x}\right) } \right)^{\left({n/x}\right)}\right)^x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Exponent Combination Laws | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle e^x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
3 implies 5
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle D_x \left({\exp x}\right)\) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {\exp \left({x+h}\right) - \exp x} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by definition of derivative | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Exponent of Sum | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} h\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \exp x \left({\lim_{h \to 0} \frac {\exp h - 1} h}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Multiple Rule for Limits of Functions, as $\exp x$ is constant | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \exp x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Derivative of Exponential at Zero |
The application of Derivative of Exponential at Zero isn't circular as the referenced proof does not depend on $D_x \exp x = \exp x$.
$\Box$
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \exp 0\) | \(=\) | \(\displaystyle \lim_{n \to +\infty} \left({1 + \frac 0 n}\right)^n\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
