How the Mathematical Ninja approximates ln⁡(5)

“Isn’t it somewhere around $\varphi$?” asked the student, brightly. “That number sure crops up in a lot of places!”

The Mathematical Ninja’s eyes narrowed.

“Like shells! And body proportions! And arrawk!”

Hands dusted. The Mathematical Ninja stood back. “The Vitruvian student!”

The student arrawked again as the circular machine he was now attached to started to stretch his legs and arms to the diagonal and back again.

“Not really,” said the Mathematical Ninja. “I mean, you’re right that it’s fairly close to $\varphi$. But it’s not really related to $\varphi$. And the ubiquity of the so-called golden ratio… well, it’s overstated, to say the least.”

“I… I see.”

“But, if I wanted to work out $\ln (5)$, I’d probably make use of the fact that $\ln (2)$ is about 0.69315.”

“Obviously,” said the student, who knew it was a bit short of 0.7.

“Now, there are two useful things: the Taylor series for $\ln \left(\frac{1+x}{1-x}\right)$, which is $2x+\frac{2}{3}{x}^3$, plus smaller terms; and the fact that $\frac{2^7}{5^3}$ isn’t too far from 1.”

“It’s 1.024!”

“If you must use such vulgar decimals, then yes, it’s 1.024. In any event, $\ln \left(\frac{128}{125}\right)$ is $\ln \left(\frac{1+x}{1-x}\right)$ if you take $x$ to be $\frac{3}{253}$, the difference over the sum. That makes the logarithm about $\frac{6}{253}$ plus some really tiny terms.”

“Which you just ignore?”

“Which I just ignore, because ${\left(\frac{3}{253}\right)}^3$ is on the order of 2 parts in a million. I can also write this logarithm as $7\ln (2)-3\ln (5)$.”

“Aha! And you know what $\ln (2)$ is!”

“I do. $7\times 0.69315-3\ln (5)\approx \frac{6}{253}$. Dividing everything by 3 gives $7\times 0.23105-\ln (5)\approx \frac{2}{253}$.”

“So $\ln (5)$ is about 1.61735 minus $\frac{2}{253}$?” quivered the student. “And $\frac{2}{253}=\frac{8}{1012}$, which is about 0.008. Less 1% or so, that’s 0.0079.”

“Good enough for Renaissance work,” said the Mathematical Ninja. “So that gives 1.6094 or 1.6095, depending on rounding, as a value for $\ln (5)$.”

“And $\varphi$ is about 1.618!” said the student. “Someone should make a stock-trading pattern based on $\varphi$, that would be a sure-fire winner.”

“Don’t push your luck,” said the Mathematical Ninja.