Upper Bound of Natural Logarithm/Proof 2
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Theorem
Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.
Then:
- $\ln x \le x - 1$
Proof
Let $\sequence {f_n}$ denote the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
- $\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Fix $x \in \R_{>0}$.
We first show that $\forall n \in \N : n \paren {\sqrt [n] x - 1} < x - 1 $
Case 1: $0 < x < 1$
Suppose $0 < x < 1$.
Then:
| \(\ds 0\) | \(<\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds 1\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\, \ds \sqrt [n] x^{n - k} \, \) | \(\, \ds < \, \) | \(\ds 1\) | Power Function on Base between Zero and One is Strictly Decreasing/Rational Number | \(\quad\) $\forall k \in \set {0, 1, \ldots, n - 1}$ | |||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\, \ds \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k} \, \) | \(\, \ds < \, \) | \(\ds n\) | Real Number Ordering is Compatible with Addition | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds \frac 1 n \, \) | \(\, \ds < \, \) | \(\ds \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | Ordering of Reciprocals | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds 1 \, \) | \(\, \ds < \, \) | \(\ds \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | multiplying both sides by $n$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds x - 1 \, \) | \(\, \ds > \, \) | \(\ds \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | multiplying both sides by $x - 1 < 0$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds y^n - 1 \, \) | \(\, \ds > \, \) | \(\ds \frac {\paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }\) | substituting $y = \sqrt [n] x$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds y^n - 1 \, \) | \(\, \ds > \, \) | \(\ds n \paren {y - 1}\) | Sum of Geometric Sequence | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds x - 1 \, \) | \(\, \ds > \, \) | \(\ds n \paren {\sqrt [n] x - 1}\) | substituting $\sqrt [n] x = y$ |
$\Box$
Case 2: $x = 1$
Suppose $x = 1$.
Then:
| \(\ds \map \ln x\) | \(=\) | \(\ds \map \ln 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Natural Logarithm of 1 is 0/Proof 3 | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x - 1\) |
$\Box$
Case 3: $x > 1$
Suppose $x > 1$.
Then:
| \(\ds x\) | \(>\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt [n] x^{n - k}\) | \(>\) | \(\ds 1\) | Power Function on Base Greater than One is Strictly Increasing/Rational Number | \(\quad\) $\forall k \in \set { 0, 1, \ldots, n - 1}$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}\) | \(>\) | \(\ds n\) | Real Number Ordering is Compatible with Addition | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 n\) | \(>\) | \(\ds \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | Ordering of Reciprocals | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(>\) | \(\ds \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | Multiply both sides by $n$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - 1\) | \(>\) | \(\ds \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }\) | multiplying both sides by $x - 1 > 1$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y^n - 1\) | \(>\) | \(\ds \frac {n \paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }\) | substituting $y = \sqrt [n] x$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y^n - 1\) | \(>\) | \(\ds n \paren {y - 1}\) | Sum of Geometric Sequence | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - 1\) | \(>\) | \(\ds n \paren {\sqrt [n] x - 1}\) | Substitute $\sqrt [n] x = y$ |
$\Box$
Thus:
- $\forall n \in \N: n \paren {\sqrt [n] x - 1} \le x - 1$
by Proof by Cases.
Thus:
- $\ds \lim_{n \mathop \to \infty} \paren {\sqrt [n] x - 1 } \le x - 1$
from Limit of Bounded Convergent Sequence is Bounded.
Hence the result, from the definition of $\ln$.
$\blacksquare$