Weird accident. Is this just a coincidence?

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- Kainui

Weird accident. Is this just a coincidence?

- katieb

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- 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?

- adajiamcneal

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

- Kainui

Calculus :D

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## More answers

- adajiamcneal

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

- 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:
\[\ln (x!) = \sum_{k=1}^x \ln k \approx \int_1^x \ln t dt = x \ln x -x +1\]

- freckles

so infinity is suppose to be x ?

- freckles

wait nevermind I'm confused

- freckles

not sure how the improper integrals turned into function of x

- 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

- 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.

- 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.

- freckles

ok I'm going to need a few math moments

- 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?

- freckles

that is kind of weird

- Kainui

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

- 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.

- 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.

- 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

- 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!

- 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

- 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

- Kainui

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

- freckles

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

- freckles

I know f'=0 when x=n

- freckles

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

- freckles

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

- freckles

I don't know why they found the critical point

- 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.

- 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

- dan815

good mistake :)

- Kainui

ty I'm expert mistaker

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