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Limit of (x! e^x) / (x^x *x^1/2)

  1. Jul 28, 2005 #1
    hi ,

    1.)how do I find the limit of (x! e^x) / (x^x *x^1/2) as x tends to infinity ?

    2.)and is f(x)= x! a function ? if so, how do I find the derivative ?

    thanks for any help

    Roger
     
  2. jcsd
  3. Jul 28, 2005 #2
    Sterling's formula says limit [tex]n!=\frac{n^n}{e^n}\sqrt{2n\pi}[/tex] We need only substitute X for n, since the function is continuous for positive X, to work your problem.
     
  4. Jul 28, 2005 #3

    George Jones

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    I thought about saying this, but then I wondered whether the point of Roger's problem is to derive Stirling's formula.

    Regards,
    George
     
  5. Jul 28, 2005 #4

    George Jones

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    A standard extension of factorial to all real numbers except the negative integers is by way of the gamma function. Then [itex]\Gamma (x + 1) = x![/itex]. One way to derive Stirling's formula is by using the standard integral representation of the gamma function. A couple of the steps are, however, not completely obvious.

    Regards,
    George
     
  6. Jul 29, 2005 #5
    thanks for your help

    Can I find the limit without using sterlings formula ?
     
  7. Jul 29, 2005 #6
    When I learned about Sterling's formula it was a graduate course and the professor put the derivation on the board. It is not that simple. Note the presence of [tex]\sqrt(2n\pi)[/tex]. This frquently means the use of complex integration, but not here: http://courses.ncssm.edu/math/Stat_Inst/PDFS/appndx_1.pdf
     
    Last edited: Jul 29, 2005
  8. Jul 29, 2005 #7
    interesting formula
     
  9. Jul 30, 2005 #8
    Thanks for the information.

    Does the extension to real numbers excluding negative, mean that the factorial becomes 'continuous' so that a derivative exists ?
     
  10. Jul 30, 2005 #9
    Yes and its usually expressed in terms of digamma function.

    -- AI
     
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