Upper Bound of Natural Logarithm/Proof 1
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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
From Logarithm is Strictly Concave:
- $\ln$ is (strictly) concave.
From Mean Value of Concave Real Function:
- $\ln x - \ln 1 \le \paren {\dfrac \d {\d x} \ln 1} \paren {x - 1}$
From Derivative of Natural Logarithm:
- $\dfrac \d {\d x} \ln 1 = \dfrac 1 1 = 1$
So:
- $\ln x - \ln 1 \le \paren {x - 1}$
But from Logarithm of 1 is 0:
- $\ln 1 = 0$
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.3 \ (2)$