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.

~ by Neil Dickson on December 14, 2009.

2 Responses to “Proof for a Post a While Back”

  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! 🙂

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: