Question

Weird accident. Is this just a coincidence?

Answer 1

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?

Answer 2

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

Answer 3

Calculus :D

Answer 4

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

Answer 5

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:

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

Answer 6

so infinity is suppose to be x ?

Answer 7

wait nevermind I'm confused

Answer 8

not sure how the improper integrals turned into function of x

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

ok I'm going to need a few math moments

Answer 13

\[\ln (x!) = \sum_{k=1}^x \ln k \approx \int_1^x \ln t dt = x \ln x -x +1\]
I do understand this...

and you are saying... that weird thing you did looks similar to this
right?

Answer 14

that is kind of weird

Answer 15

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

Answer 16

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.

Answer 17

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.

Answer 18

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

Answer 19

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!

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

I know f'=0 when x=n

Answer 25

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

Answer 26

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

Answer 27

I don't know why they found the critical point

Answer 28

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.

Answer 29

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

Answer 30

good mistake :)

Answer 31

ty I'm expert mistaker