## Proof for a Post a While Back

In a post a while back, I mentioned that I’d found that:

$1-\frac{1}{\sqrt[k]{2}}\approx\frac{\ln 2}{k}$

for sufficiently large k, which relates to finding the median of the minimum of a set of k independent, identically-distributed random variables.  I didn’t have a proof at the time, but I’ve thought of a very simple “proof”:

$(1-\frac{1}{x})^x \approx \frac{1}{e}$ for large x.

$1-\frac{1}{x} \approx \frac{1}{\sqrt[x]{e}}$

$1-\frac{1}{\sqrt[x]{e}} \approx \frac{1}{x}$

$1-\frac{1}{\sqrt[k]{e^{\ln 2}}} \approx \frac{\ln 2}{k}; x = \frac{k}{\ln 2}$

$1-\frac{1}{\sqrt[k]{2}} \approx \frac{\ln 2}{k}$

It’s by no means rigorous, as things that are “approximately equal” put through certain transformations can result in values that  are not even close, but it seems to check out in this case.

1. Actually, if you perform a series expansion of the LHS of the equation around x = Infinity you will see that the first term is ln2/k. The second one is (ln2)^2/(2k^2), which for k very large is much smaller than the first term. In principle, you are looking at $1 - 2^{-1/k}$ for k –> Infinity. You could also look at $1 - 2^{-q}$ for q –> 0 which is easier to treat. Will reply to your email later…
2. Ah yes, of course, the Taylor series for $1-2^{-q}$ about $q=0$ starts with $(\ln 2)q - 0.5 (\ln 2)^2 q^2$. Thanks! 🙂