1

When doing financial, actuarial or probability calculations, often a person is faced with computing exponentiations.

For example, let's say the chance of a particular ship sinking is 1% per year. Then to compute the chance of the ship sinking over the course of a 30 year lifetime is 1 - 99% × 99% ... × 99% (30 times) or $1-0.99^{30}$. So, we have the problem of computing $\frac1{99}$ raised to the 30th power. Is there a fast way to do this kind of calculation manually in the head or using a pencil and paper?

MJD
  • 65,394
  • 39
  • 298
  • 580
Tyler Durden
  • 954

1 Answers1

2

I am not sure about anything precise, but I can give the following approximate rule which turns out to be rather useful: if $n \geq 10$, $(1-\frac{1}{n})^n \approx \frac{1}{e} \approx 0.37$.

So in your example, you want to approximate now $0.37^{3/10} \approx 0.37^{1/3}$. Now, $0.7^3 =0.343$, $0.8^3=0.512$ so, since the cube function grows faster as the argument increases (eg: a linear function has constant growth, so if the cube was linear, you would approximate $0.37 \approx 0.715^3$ but it is convex, albeit not too strongly), you are likely to have some sort of $0.73^3 \approx 0.37$.

So your quantity can be guessed at $0.73$. The actual value is $\sim 0.739$...

Of course, the hundredth-precision is a stroke of luck, but the tenth-precision is not.

Aphelli
  • 34,439
  • You've left a couple of dollar signs out near the beginning. – timtfj Dec 15 '18 at 16:11
  • Thanks, corrected! – Aphelli Dec 15 '18 at 16:12
  • Maybe it's also worth mentioning that the $0.37$ is an approximation to $\frac1e$. – timtfj Dec 15 '18 at 17:09
  • You are right, I added it. – Aphelli Dec 15 '18 at 17:11
  • I tried something a little different. Arriving at $\log x\approx -\frac3{10}$, which is almost the same as where you got, we now need $e^{-3/10}$. This is approximately $$1-\frac3{10}+\frac9{200}-\frac{27}{6000}.$$ Doing just the first three terms is not too hard and gets us $x\approx \frac{149}{200}=7.45$, which is not too far off. We are lucky with the next term, which is equal to $\frac9{2000}$, and has an easy denominator. So it is not too hard to include that and get $7.415$, which is quite close. – MJD Dec 16 '18 at 15:28
  • I'm not sure how helpful this is to OP, so I didn't write up the details. But it's how I would have done it. I will write it up in detail if so requested. – MJD Dec 16 '18 at 15:31