Difficulty: (simple version) *** (advanced version) ***** Impressiveness: ***** Accuracy: ***

If you’re a statistician, you quite often end up working out powers of numbers just a little less than 1. What’s the probability of rolling a pair of dice ten times and never getting a double six? It’s $\left(\frac{35}{36}\right)^{10}$. And what does that work out to be? I guessed ‘a bit more than $\frac{26}{36}$, so about 3/4’. In fact, it’s 0.7545. Not too shabby. How did I do it?

Well, that one was easy. For numbers close enough to 1, the binomial expansion is a) really simple and b) pretty accurate if you just use the first two terms. The devilishly difficult formula is $(1 + x)^n = 1 + nx + O(x^2)$ - where the ‘big O’ just means the other terms are $x^2$ and higher powers. When $x$ is small, you can safely ignore them.

So, to work out $\left(\frac{35}{36}\right)^{10}$, you just say ‘it’s $\frac{1}{36}$ less than 1, so $x = \frac{1}{36}$ and $n = 10$. I get $1 - \frac{10}{36}$, which is $\frac{26}{36}$ - the next term would be a plus, so I’ll say “a bit more than…” just to look cocky.’

This works great - as long as $nx$ isn’t too big. (Once it gets more than about $\frac{1}{3}$, I’d start to worry - in fact, the $\frac{10}{36}$ from the example was big enough for me to be chewing my lip about whether I’d be close enough. So, what do you do if your power or your $x$ is too big?

For instance, I was idly wondering, what’s $0.9^{20}$, as one does. And I was walking to the station without my phone, so I had to come up with a ballpark figure in my head. Contrived? Well, maybe. But such is the life of a maths writer.

So, here’s what I did. I made use of the fact that - for small $x$ - $\ln(1+x) \simeq x - \frac{1}{2}x^2$. In fact, I ignored the second term, but it might be useful sometimes. That means, $\ln(0.9) = \ln(1 - 0.1) \simeq - 0.1 - 0.005 = -0.105$.

Then I used my log rules: $\ln(0.9^{20}) = 20\ln(0.9) \simeq -2.1$. So all I need to do is find $e^{-2.1}$ which - because I know my logs - I can say is pretty close to $\frac{1}{8}$. The actual answer? 0.122, so I’m about 3% off.

(Since you ask: I know $\ln(2)$ is about 0.7. So $\ln(8)$ is about 2.1, and $\ln(1/8)$ is about -2.1.)

So there you have it: two quick ways to figure out powers of numbers close to 1, in your head.