Georg,
> is there an efficient way in sage to find the smallest integer k for
> which the inequality
>
> b^(k+1) / (factorial(k) * factorial(k+1)) <= 1
>
> is true (b > 0)
Stirling's expansion gives (when b goes to infinity) k ~ sqrt(b)*exp(1).
Thus it suffices to evaluates f(k) = b^(k+1) / (factorial(k) * factorial(k+1))
at this starting point. If f(k) < 1, increasing k by 1 will multiply f(k) by a
factor about b/k^2 ~ exp(-2); if f(k) < 1, decreasing k by 1 will multiply f(k)
by a factor about exp(2). Thus log(f(k))/2 at the starting point should give
you the number of corrections steps (more probably 0 or 1).
Paul Zimmermann
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