Intuition for an upper bound on $\ln(x)$ for $x \geq 1$

Question

Today I came across the following bound. For $x \geq 1$,

$$\ln(x) \leq \frac{x^2 - 1}{2x}.$$

(it provides a nice way of showing that $\ln(2) \leq 3/4$ when you don't have a calculator and you don't want to check the first 8 or so terms of the Taylor expansion for $\ln(x + 1)$) It is fairly easy to prove using some pre-calculus tools, but can anyone give an idea of how someone might come up with a bound of this form?

Please show the proof. This might help to understand how we can arrive at this inequality. — Peter, Nov 09 '17 at 22:51
BTW . $10^3\approx 2^{10}$ so $\ln 2\approx 0.3\ln 10.$ And $e\approx 3^3/10$ so $\ln 10\approx -1+3\ln 3.$ And $3\approx 1.1e$ so $\ln 3\approx 1.1 $ Therefore $\ln 2\approx$ $ 0.3\ln 10\approx$ $ -0.3+0.9\ln 3\approx$ $ -0.3+0.9(1.1)=0.69$ :) — DanielWainfleet, Nov 10 '17 at 03:08

Bumblebee · Answer 1 · 2017-11-09T23:59:35.823

3

Consider the function $$f(x)=\ln(x)-\dfrac{x^2 - 1}{2x}$$ on $[1,\infty).$ This function is smooth and $\dfrac{df}{dx}=-\dfrac{(x-1)^2}{2x^2}$ and $\dfrac{d^2f}{dx^2}=\dfrac{1-x}{x^3}.$ Thus this function is strictly decreasing and concave down. As $f(1)=0,$ function is negative for all $x\gt 1.$

EDIT:
Best way to come up with these type of inequalities is mean value theorems. Also $\ln x$ has a very slow growth rate (look at the graph), there for any fast growing function would exceed $\ln x$ at some point. In your case that function is $\dfrac12\left(x-\dfrac1x\right)$ and the crossing point is $x=1.$

edited Nov 09 '17 at 23:59

answered Nov 09 '17 at 22:52

Bumblebee

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I'm not interested in a proof of this fact as much as how one might come up with (x^2 - 1) / 2x. – Nathan Nov 09 '17 at 23:04
The asker already proved the inequality. – Peter Nov 09 '17 at 23:04
@NMEM: Look at my edit. That's how I think about these type of upper bounds of $\log$ function. – Bumblebee Nov 10 '17 at 00:01
See this nice graph – Bumblebee Nov 10 '17 at 00:06

score 3 · Accepted Answer · edited Jun 12 '20 at 10:38

By the Cauchy-Schwarz inequality, for any $x\geq 1$ we have

$$ \log(x)=\int_{1}^{x}\frac{du}{u}\leq\sqrt{\int_{1}^{x}du\int_{1}^{x}\frac{du}{u^2}} = \sqrt{x}-\frac{1}{\sqrt{x}} $$ which is sharper than the wanted inequality.

Newton's method then allows to state $$ \log(x)\leq \sqrt{x}-\frac{1}{\sqrt{x}}-1+\frac{x}{\exp\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)} $$ and so on.
Other accurate bounds can be derived from the (generalized) Shafer-Fink inequality, for instance.

Anyway, one of the most efficient ways for approximating $\log 2$ is to exploit Beuker-like integrals: $$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x}\,dx = -\frac{621}{56}+16\log 2$$ and since the LHS is bounded between $0$ and $\frac{1}{4^4}$, we have $\log(2)\leq\frac{39}{56}$ which is both simple and pretty accurate.

Intuition for an upper bound on $\ln(x)$ for $x \geq 1$

2 Answers2