Proving Stirling's formula help

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In summary: Hors?In summary, the conversation is about proving Stirling's formula and using a substitution method with the hint of t=ny and ln(1+y)=y-0.5y^2. The conversation also mentions using the method of steepest descent and suggests resources for further help.
  • #1
Physicist
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proving Stirling's formula.. help please

How can I prove Stirling's formula?

n!= integral from 0 to inf. exp(-t) t^n dt= n^n exp(-n) (2 pi n)^0.5

there's a hint to use the substitution t=ny & ln(1+y) = y- 0.5 y^2

I tried to use it but I couldn't intgrate.. I tried integrating by parts but it became more complicated.. :frown:

Can anyone help?

(How can I write the mathematical symbols here?)

Thanks
 
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  • #2
Physicist said:
How can I prove Stirling's formula?

n!= integral from 0 to inf. exp(-t) t^n dt= n^n exp(-n) (2 pi n)^0.5

there's a hint to use the substitution t=ny & ln(1+y) = y- 0.5 y^2

I tried to use it but I couldn't intgrate.. I tried integrating by parts but it became more complicated.. :frown:

Can anyone help?

(How can I write the mathematical symbols here?)

Thanks
Try:
http://www.sosmath.com/calculus/sequence/stirling/stirling.html

AM
 
  • #3
Another approach would be to use the method of steepest descent. Basically, you can find where [itex]t^n e^{-t}[/itex] is a maximum and observe that the most significant contribution to the integral comes from near that maximum.
 
  • #4
Thanks for helping.. but I should uuse the substitution t=ny..

HELP PLZ
 
  • #6
Thanks alot..
 
  • #7


How can I find equivalent Frenkel defects in the crystal through the equivalent Stralink
 

1. What is Stirling's formula?

Stirling's formula is an asymptotic approximation for the factorial function, which is often used in mathematics and statistics to estimate large factorials without having to compute them directly.

2. How does Stirling's formula help in proving other mathematical theorems?

Stirling's formula can be used in many mathematical proofs and calculations, especially those involving factorials, binomial coefficients, and logarithms. It provides a useful approximation and simplifies complex expressions, making it easier to prove certain theorems.

3. Can Stirling's formula be used for all factorials?

No, Stirling's formula is an approximation and is most accurate for large values of n. It becomes less accurate as n decreases, and at n=1 it is not applicable.

4. Why is Stirling's formula important in statistics?

In statistics, Stirling's formula is used to approximate the probability of rare events and to simplify complex calculations involving large sample sizes. It is also used in probability distributions such as the normal distribution and the Poisson distribution.

5. Are there any limitations to using Stirling's formula?

Yes, Stirling's formula is an approximation and may introduce errors in certain calculations. It is also not suitable for all types of factorials and may not give accurate results for small values of n. Additionally, it does not take into account any other variables or factors that may affect the final result.

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