 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Root factorial (Posted on 2015-11-25) A graph of y=x√(x!) shows a figure that appears to be approaching a linear asymptote.

Find an approximation for the equation of this line.

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) thoughts Comment 1 of 1
Stirling's approximation for ln(n!) is n * ln(n) - n.

Taking the antilog: n! ~= n^n / e^n = (n/e)^n

Substituting x for n, y would seem to approach x/e.

However, the difference (n!)^(1/n) - x/e continues to get larger and larger with larger x.

Going to Wolfram MathWorld, a better version of a Stirling approximation is n * ln(n) - n + 1

This would not seem to change things as the nth (or xth) root of e would get closer and closer to 1.  But this would only make the ratio between the approximation and the actual value approach 1, and not necessarily the difference to approach zero.

Complicating this is that on the MathWorld page, we see that the integral which this evaluates is only an approximation to the summation of the natural logs of the numbers being multiplied.

Perhaps (likely?) the difference approaches some value, so we could say something like y = x/e - k, where k is some number below, say, 10. The formula n * ln(n) - n + 1 is shown by MathWorld to be an exact value for the integral.

However, the Wikipedia article on Stirling's approximation lists

n! ~ sqrt(2*pi*n) * (n/e)^n

which makes the asymptote of nth-root(n!) to be (n/e) * (2*n)th-root(2*pi*n), but this is not linear.

 Posted by Charlie on 2015-11-25 15:01:26 Please log in:

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