Maths Dopers - Question for you!

If you consider the formula:

(x+1/2)^(x+1/2) = k. x!.e^x

As x gets large k quickly moves towards a value of around 0.658 something. Is it possible to say what value k takes as x tends to infinity? Is it a special number, like pi/4 or what have you?

Are the dots supposed to indicate multiplication?

yes, the dots are for multiplication.

I don’t know the answer, but this should make it easier to read:

(x + ½)[sup](x + ½)[/sup] = (k)(x!)(e[sup]x[/sup])
k = (x + ½)[sup](x + ½)[/sup]/[(x!)(e[sup]x[/sup])]

Stirling’s formula has the usual form:

x! ~ x^(x+1/2) e^(-x) (2 pi)^(1/2)

where the tilde (~) indicates an asymptotic relationship. So if you plot the function

y(x) = x! e^(x) / x^(x + 1/2)

it will approach sqrt(2 pi) as x increases without bound.

biqu has the right approach using Stirling’s formula, but I believe s/he has a mistake in the limit calculation. I’m getting sqrt[e/(2pi)].

The local version of Maple agrees with biqu: it gives me sqrt(2*pi) as well.

Yes, that’s precisely the limit that answers the OP. The limit I gave was for the ratio x! e^(x) / x^(x + 1/2), which follows directly from Stirling’s formula without additional manipulation of limits with logarithms and L’Hospital’s rule. When I went back and applied those techniques to the ratio in the OP, the limit quoted by Cabbage fell out immediately.

I just plain goofed up. Ignore me and listen to biqu and Cabbage. I’ll just slink away now.

:slight_smile: Thanks for the very good answers.