A friend told me his colleague estimated 0.95^32 using nothing, just approximating it in her head to be about 0.2. My calculator gives the answer 0.1937114844585.
How would one go about doing something like that? I've been burried in contemplation all morning.
Thanks in advance for any insight!
On
One possibility would be to take the logarithm: $$ \ln(0.95^{32}) = 32 \cdot \ln(1-0.05) \approx -1.6 $$ because $\ln(1+x)\approx x$ when $x$ is small.
Perhaps the colleague has memorized the natural logarithms of a handful of small integers. In particular, $\ln 5=1.609...$ (you can remember that because an English/American mile is very nearly $\ln 5$ kilometers; or perhaps because $\ln 5\approx \phi$, the golden ratio which involves $\sqrt 5$ and appears in pentagrams), so $e^{-1.6}$ is about $1/5$.
Alternatively, with common (base-10) logarithms:
$\log(1+x) \approx \frac{10}{23} x $ (one needs to memorize this conversion factor), so $$\log(0.95^{32})\approx -\frac{10}{23} \cdot\frac{1}{20} \cdot 32 = -\frac{16}{23} = -1+\frac{7}{23}$$ But $3\cdot 23=69$ so $\frac7{23}\approx 0.3 \approx \log 2$ (another remembered value). Again, the result must be close to $0.2$.
On
$$\left(0.95^2\right)^{16}\approx\left(0.9^2\right)^{8}\approx\left(0.8^2\right)^{4}\approx\left(0.65^2\right)^{2}\approx0.4^2\approx0.16\approx0.2$$
Note that in most of the steps, the number was rounded down, so an estimate of $0.2$ is reasonable here. I'd argue this is still a 'reasonable' approximation, since the rounding is relatively small and in the end, we round up to $1$dp to compensate which feels 'about right' (note that the error gets bigger the larger the power you take).
I'm no genius but $0.95^{32}\tilde{} 0.9^{16}\tilde{}0.8^8\tilde{}0.65^4\tilde{}0.42^2\tilde{}0.176$, purely mental.