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

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Kainui
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.0what type of math is this???? :O

adajiamcneal
 one year ago
Best ResponseYou've already chosen the best response.0ohhh my lawwwd. so glad i dont take it sorry i cant help you i cant even read it haha goodluck though

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.2so infinity is suppose to be x ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2wait nevermind I'm confused

freckles
 one year ago
Best ResponseYou've already chosen the best response.2not sure how the improper integrals turned into function of x

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1No, 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2ok I'm going to need a few math moments

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\ln (x!) = \sum_{k=1}^x \ln k \approx \int_1^x \ln t dt = x \ln x x +1\] I do understand this... and you are saying... that weird thing you did looks similar to this right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is kind of weird

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, for some reason the messed up thing ends up almost identical in a way... It's just strangely similar.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2ok 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
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhh 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
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year ago
Best ResponseYou've already chosen the best response.2I wouldn't have thought of any of that you know rewriting ln(x^n*exp(x)) like they did in lines 916

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1I don't really understand the reasoning though at line 8, how does they come to this conclusion?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the part where they have x=n+xi where xi<<n ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and so we have a sharp corner there for ln(exp(x)*x^n)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\ln((n+\xi)^n e^{(n+\xi)}) \\ =\ln((n+\xi)^n)+\ln(e^{(n+\xi)}) \\ =n \ln(n+\xi)(n+\xi)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I don't know why they found the critical point

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Ohhhhh 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
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