Summing a geometric series: Secrets of the Mathematical Ninja

"I don't know if you were actually working stuff out there, or if you just muttered at random for a bit and guessed," said the Mathematical Ninja's cheeky student.

Stung, the Mathematical Ninja was forced to explain how he figured out $\left(1 - \frac{1}{8}\right)^{19}$.

It's a simple enough trick that he's explained many times before: it involves knowing that $\left(1 + \frac{1}{k}\right)^{k} \simeq e.$ In fact, $\left(1 - \frac{1}{8}\right)^{8}$ is about six or seven percent away from $1/e$. As a rule of thumb, the percentage error is about $\frac{50}{n}$, and the fraction is lower than $1/e$.

That means, $\left(1 - \frac{1}{8}\right)^{19}$ is roughly $\left(\frac{1}{e} - 7\% \right)^{19/8}$. You need to multiply the error by the power ($\frac{50\times19}{8\times 8} \simeq \frac{1000}{64}$ - so that'll make it pretty much $e^{-2.4} - 16%$).

But what's $e^{-2.4}$? Well, any fule no that since $\ln(10) \simeq 2.3$ and $\ln(12) = 2\ln(2) + \ln(3) \simeq 1.4 + 1.1 = 2.5$, you can surmise that $e^{-2.4} \simeq \frac{1}{11}$.

That's a tenth, less 10% (give or take), less another 16% - somewhere in the upper 0.07s. The correct answer? 0.079.

Muttering randomly, indeed.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

This site uses Akismet to reduce spam. Learn how your comment data is processed.