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Kainui

  • one year ago

Weird accident. Is this just a coincidence?

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  1. Kainui
    • one year ago
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    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?

  2. adajiamcneal
    • one year ago
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    what type of math is this???? :O

  3. Kainui
    • one year ago
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    Calculus :D

  4. adajiamcneal
    • one year ago
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    ohhh my lawwwd. so glad i dont take it sorry i cant help you i cant even read it haha goodluck though

  5. Kainui
    • one year ago
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    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
    • one year ago
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    so infinity is suppose to be x ?

  7. freckles
    • one year ago
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    wait nevermind I'm confused

  8. freckles
    • one year ago
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    not sure how the improper integrals turned into function of x

  9. Kainui
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ok I'm going to need a few math moments

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

  14. freckles
    • one year ago
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    that is kind of weird

  15. Kainui
    • one year ago
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    Yeah, for some reason the messed up thing ends up almost identical in a way... It's just strangely similar.

  16. Kainui
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    I don't really understand the reasoning though at line 8, how does they come to this conclusion?

  23. freckles
    • one year ago
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    the part where they have x=n+xi where xi<<n ?

  24. freckles
    • one year ago
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    I know f'=0 when x=n

  25. freckles
    • one year ago
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    and so we have a sharp corner there for ln(exp(x)*x^n)

  26. freckles
    • one year ago
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    \[\ln((n+\xi)^n e^{-(n+\xi)}) \\ =\ln((n+\xi)^n)+\ln(e^{-(n+\xi)}) \\ =n \ln(n+\xi)-(n+\xi)\]

  27. freckles
    • one year ago
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    I don't know why they found the critical point

  28. Kainui
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    good mistake :)

  31. Kainui
    • one year ago
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    ty I'm expert mistaker

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