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February 15

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

How do you solve ${\frac {n!}{(n-r)!n^{r}}}\geq 1-\epsilon$ for $n$ , given $r$ and $\epsilon$ ? The parameter values of interest are of the order $r\sim 10^{9}$ and $\epsilon \sim 10^{-5}$ . 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)[reply]

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

$\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\cdots \left(1-{\frac {r-1}{n}}\right)\geq 1-\epsilon$

$\log \left(1-{\frac {1}{n}}\right)+\log \left(1-{\frac {2}{n}}\right)+\cdots +\log \left(1-{\frac {r-1}{n}}\right)\geq \log(1-\epsilon ).$

In approximate terms, we can write this as

$-{\frac {1}{n}}-{\frac {2}{n}}-\cdots -{\frac {r-1}{n}}\geq \log(1-\epsilon )$

$-{\frac {r(r-1)}{2n}}\geq \log(1-\epsilon )$

$n\geq {\frac {r(r-1)}{-2\log(1-\epsilon )}}\approx {\frac {r^{2}}{2\epsilon }}$

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 ${\frac {\epsilon }{r}}$ . In other words, in using the formula $n\geq {\frac {r(r-1)}{-2\log(1-\epsilon )}}$ you commit an error of at most about ${\frac {\epsilon }{r}}\approx 10^{-14}$ 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 $\log(1-\epsilon )$ with $\epsilon$ , the error rises to about $\epsilon /2=5\times 10^{-6}$ times the value of n. 64.140.122.50 (talk) 05:58, 15 February 2013 (UTC)[reply]

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

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

February 15

How do you solve n ! ( n − r ) ! n r ≥ 1 − ϵ {\displaystyle {\frac {n!}{(n-r)!n^{r}}}\geq 1-\epsilon } for n?

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