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

  • Thread starter roger
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  • #1
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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
 

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  • #2
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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.
 
  • #3
George Jones
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robert Ihnot said:
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.
I thought about saying this, but then I wondered whether the point of Roger's problem is to derive Stirling's formula.

Regards,
George
 
  • #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
 
  • #5
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thanks for your help

Can I find the limit without using sterlings formula ?
 
  • #6
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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
 
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  • #7
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interesting formula
 
  • #8
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George Jones said:
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
Thanks for the information.

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

-- AI
 
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