Equivalence of Definitions of Real Natural Logarithm
Theorem
The following definitions of the concept of Real Natural Logarithm are equivalent:
Definition 1
The (natural) logarithm of $x$ is the real-valued function defined on $\R_{>0}$ as:
- $\ds \forall x \in \R_{>0}: \ln x := \int_1^x \frac {\d t} t$
Definition 2
Let $x \in \R$ be a real number such that $x > 0$.
The (natural) logarithm of $x$ is defined as:
- $\ln x := y \in \R: e^y = x$
where $e$ is Euler's number.
Definition 3
Let $x \in \R$ be a real number such that $x > 0$.
The (natural) logarithm of $x$ is defined as:
- $\ds \ln x := \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1}$
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Proof 1
Definition 1 implies Definition 2
Let $\map F x$ be $\ds \int_1^x \frac {\d t} t$.
Let $\map f t$ be $\ds \int \frac {\d t} t$.
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Then:
- $\dfrac {\d t} t = \dfrac 1 t$
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Or:
- $\dfrac {\d x} x = \dfrac 1 x$
Also:
- $\map F x = \map f x - \map f 1$
Therefore:
\(\ds \frac {\d \map F x} {\d x}\) | \(=\) | \(\ds \frac {\d \map f x} {\d x} - \frac {\d \map F 1} {\d x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\d \map f x} {\d x}\) | Derivative of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d \map F x}\) | \(=\) | \(\ds x\) | Derivative of Inverse Function |
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Furthermore:
- $\map F 1 = \map f 1 - \map f 1 = 0$
The result follows from the fifth definition of the exponential function:
- $\map F x \equiv e^x$
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$\Box$
Definition 2 implies Definition 1
\(\ds e^{\map F x}\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d \map F x}\) | \(=\) | \(\ds x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d \map F x} {\d x}\) | \(=\) | \(\ds \frac 1 x\) | Derivative of Inverse Function |
Let $\map f t$ be $\ds \int \frac 1 t \rd t$.
Then:
- $\map F x = \map f x + C$
When $\map F x = 0$:
- $x = e^{\map F x} = 1$
- $\map F 1 = \map f 1 + C = 0 \implies \map f 1 = - C$
Therefore:
- $\map F x = \map f x - \map f 1$
Therefore:
- $\ds \map F x = \int_1^x \frac {\d t} t$
$\Box$
Therefore:
\(\ds y = \int_1^x \frac {\d t} t\) | \(\leadsto\) | \(\ds e^y = x\) | Definition 1 implies Definition 2 | |||||||||||
\(\ds e^y = x\) | \(\leadsto\) | \(\ds y = \int_1^x \frac {\d t} t\) | Definition 2 implies Definition 1 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y = \int_1^x \frac {\d t} t\) | \(\iff\) | \(\ds e^y = x\) |
$\Box$
This theorem requires a proof. In particular: Add definition 3 You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
$\blacksquare$
Proof 2
Let $x \in \R$ be such that $x > 0$.
Let $y \in \R$ be the unique number such that:
- $e^y = x$
Definition 1 iff Definition 2
\(\ds \int_1^x \dfrac {\rd t} t\) | \(=\) | \(\ds \int_0^y \dfrac {\map {\phi '} s \rd s} {\map \phi s}\) | by Integration by Substitution with $\map \phi s = e^s$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^y \dfrac {e^s \rd s} {e^s}\) | Derivative of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^y 1 \rd s\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \bigintlimits s 0 y\) | Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds y - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y\) |
$\Box$
Definition 3 iff Definition 2
We shall show:
- $\ds \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1} = y$
If $y = 0$, then $x = e^0 = 1$.
Thus the claim is clear, as:
- $\forall n \in \N : n \paren {\sqrt [n] x - 1} = 0$
If $y \ne 0$, then:
\(\ds \dfrac {n \paren {\sqrt [n] x - 1} } y\) | \(=\) | \(\ds \dfrac {n \paren {\sqrt [n] {e^y} - 1} } y\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n \paren {e^{\frac y n} - 1} } y\) | Definition of Root of Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {e^{\frac y n} - 1} } {\dfrac y n}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 1\) | as $n \to \infty$ by Derivative of Exponential at Zero |
$\blacksquare$