Testwiki:Reference desk/Archives/Mathematics/2013 February 15

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February 15

How do you solve $\frac{n!}{(n-r)!n^r}\ge 1-ϵ$ for n?

How do you solve $\frac{n!}{(n-r)!n^r}\ge 1-ϵ$ for $n$, given $r$ and $ϵ$? The parameter values of interest are of the order $r\sim 10^9$ and $ϵ\sim 10^{-5}$. Can someone give a good approximation formula that can be evaluated without overflows or gross loss of precision during computation? Thanks. --173.49.13.216 (talk) 03:56, 15 February 2013 (UTC)

I'm not a specialist of this kind of thing, but this can be rewritten (exactly) as

$(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{r-1}{n})\ge 1-ϵ$

$\log (1-\frac{1}{n})+\log (1-\frac{2}{n})+\cdots +\log (1-\frac{r-1}{n})\ge \log (1-ϵ).$

In approximate terms, we can write this as

$-\frac{1}{n}-\frac{2}{n}-\cdots -\frac{r-1}{n}\ge \log (1-ϵ)$

$-\frac{r(r-1)}{2n}\ge \log (1-ϵ)$

$n\ge \frac{r(r-1)}{-2\log (1-ϵ)}\approx \frac{r^2}{2ϵ}$

In linearizing the logarithms, the error in the terms on the left is at most about $\frac{r}{2n}$ times their value. Given the approximate value of n, the error on the left is, in proportion, about $\frac{ϵ}{r}$. In other words, in using the formula $n\ge \frac{r(r-1)}{-2\log (1-ϵ)}$ you commit an error of at most about $\frac{ϵ}{r}\approx 10^{-14}$ times the vaue of n. (In fact, a more careful calculation shows that the error is about 2/3 of this, and that it results in an underestimation of the minimum possible n.) If you replace r - 1 with r, this rises to about 1/r, or 10^-9. If you replace $\log (1-ϵ)$ with $ϵ$, the error rises to about $ϵ/2=5\times 10^{-6}$ times the value of n. 64.140.122.50 (talk) 05:58, 15 February 2013 (UTC)

I forgot to mention this - I'm using natural logarithms here. 64.140.122.50 (talk) 06:08, 15 February 2013 (UTC)

Thanks. Your formula is more than good enough. --173.49.13.216 (talk) 12:15, 15 February 2013 (UTC)