## Kainui one year ago Weird accident. Is this just a coincidence?

1. Kainui

So I was looking at the Stirling approximation for $$\ln(x!) \approx x \ln x -x +1$$. Then I looked at the integral representation of $$x! = \int_0^\infty t^x e^{-t}dt$$ and then here I made a mistake as well as a weird bit of playing around. =D I took the logarithm of both sides: $\ln(x!) = \ln\int_0^\infty t^x e^{-t}dt$ And made the mistake of thinking I could bring the logarithm inside the sum... $\ln(x!) = \int_0^\infty \ln( t^x e^{-t}dt)$ Then did the weird thing: $\int_0^\infty x \ln( t) + \int_0^\infty -t + \int_0^\infty \ln(dt)$ Anyways then I recognized that there is an odd similarity here between this and Stirling's approximation: $x\int_0^\infty \ln( t) - \int_0^\infty t + \int_0^\infty \ln(dt)$ $x\ln x - x + 1$ Is there a reason for this?

what type of math is this???? :O

3. Kainui

Calculus :D

ohhh my lawwwd. so glad i dont take it sorry i cant help you i cant even read it haha goodluck though

5. Kainui

The derivation of Striling's approximation is what I had in mind while I was doing this, which is why I must have made this mistake in bringing the $$\ln$$ inside the sum: $\ln (x!) = \sum_{k=1}^x \ln k \approx \int_1^x \ln t dt = x \ln x -x +1$

6. freckles

so infinity is suppose to be x ?

7. freckles

wait nevermind I'm confused

8. freckles

not sure how the improper integrals turned into function of x

9. Kainui

No, these are two different things, the infinity is from the gamma function, but it looks like you're catching yourself. There are two separate integrals going on I think haha

10. Kainui

So I'll put them up here for reference: Stirling's Approximation of $$\ln (x!)$$ $\ln (x!) \approx \int_1^x \ln x dx$ Gamma function $$\Gamma(x+1)=x!$$ $x! = \int_0^\infty t^xe^{-t}dt$ Yeah this is kind of confusing I can help explain more though since I sorta skipped around.

11. Kainui

I can derive or help you derive either of these, although I put the basic steps for deriving Stirling's approximation $$\ln(x!) \approx x \ln x - x +1$$ which has that first integral as an intermediate step to deriving it.

12. freckles

ok I'm going to need a few math moments

13. freckles

$\ln (x!) = \sum_{k=1}^x \ln k \approx \int_1^x \ln t dt = x \ln x -x +1$ I do understand this... and you are saying... that weird thing you did looks similar to this right?

14. freckles

that is kind of weird

15. Kainui

Yeah, for some reason the messed up thing ends up almost identical in a way... It's just strangely similar.

16. Kainui

If we could somehow show $\int_0^\infty \ln t = \ln x$ $\int_0^\infty t = x$ $\int_0^\infty \ln(dt) = 1$ It would work. Of course, these statements are all nonsense since the first two are infinite and the last one is... well I am not sure. Maybe there's some way around this though? I guess there's something fishy going on here.

17. Kainui

I guess this isn't entirely true, I should replace all the = signs with $$\approx$$ since the left hand side are the actual terms (although wrong... #_#) while the terms on the other side are the terms of the approximation. Confusing.

18. freckles

ok I think I'm almost there to seeing why your stuff happened weird like it looks like: $\ln(x^n e^{-x})= n \ln(n)-n+\text{ blah }$ http://mathworld.wolfram.com/StirlingsApproximation.html

19. Kainui

Ohhhh it looks like they derive it using the factorial interesting thanks, this is just what I needed it looks like, just gotta read it!

20. Kainui

This is an excellent and really close approximation on that link you shared too which looks great $x! \approx \sqrt{\left(2x+\frac{1}{3}\right)\pi }*x^xe^{-x}$ https://www.desmos.com/calculator/cmiwh9gfbh

21. freckles

I wouldn't have thought of any of that you know rewriting ln(x^n*exp(-x)) like they did in lines 9-16

22. Kainui

I don't really understand the reasoning though at line 8, how does they come to this conclusion?

23. freckles

the part where they have x=n+xi where xi<<n ?

24. freckles

I know f'=0 when x=n

25. freckles

and so we have a sharp corner there for ln(exp(x)*x^n)

26. freckles

$\ln((n+\xi)^n e^{-(n+\xi)}) \\ =\ln((n+\xi)^n)+\ln(e^{-(n+\xi)}) \\ =n \ln(n+\xi)-(n+\xi)$

27. freckles

I don't know why they found the critical point

28. Kainui

Ohhhhh I think I see. They found the critical point because that's the local max. This means this is where the most contribution comes from, for instance if you integrate $$-x^2+5$$ where it's larger than 0, the most area contribution will be the integral around x=0 since that's where the max is.

29. Kainui

I think they should have just taken the derivative of $$x^ne^{-x}$$ instead to make it clearer that that's what's happening I think

30. dan815

good mistake :)

31. Kainui

ty I'm expert mistaker