Stirling's Formula: How does n relate to n^n.

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In summary, the conversation focuses on proving Stirling's formula and how the factor e comes into place. The conversation suggests using algebraic methods and considering the relationship between n^n and n!. Ultimately, the conversation provides a step-by-step guide on how to prove the formula and suggests using the second method for clarity in the paper.
  • #1
Ryuky
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Hello all.
I am doing a school paper on Stirling's formula and I want first to show how the factor e comes into place. So I found somewhere on the net the definition n! = (1-1/2)×(1-1/3)2... × (1-1/n)n-1] × nn.

I will then use this result and the definition of e^x since we are talking for large n to proceed.

Homework Statement


Problem is, how do I prove it? Although I haven't tried induction yet, I was told that I should find an algebraic method.

I don't want a complete solution, just a hint to guide me through the proof.

Thank you
 
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  • #2
First represent all the 1-.5 and 1-1/3 by (1-1/n+1)^n, we note this is (n/(n+1))^n.
Now note what happens as we multiply this for different n's:
(1/2)^1 * (2/3)^2 * (3/4)^3... = (1^1 * 2^2 * 3^3...)/(2^1 * 3^2 * 4^3...).
Do you notice what is happening? Try canceling things up, in problems like this, I recommend you put things in variables like I did (n/(n+1)) in our case.Thanks, Bonaparte
 
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  • #3
Bonaparte said:
First represent all the 1-.5 and 1-1/3 by (1-1/n+1)^n, we note this is (n/(n+1))^n.
Now note what happens as we multiply this for different n's:
(1/2)^1 * (2/3)^2 * (3/4)^3... = (1^1 * 2^2 * 3^3...)/(2^1 * 3^2 * 4^3...).
Do you notice what is happening? Try canceling things up, in problems like this, I recommend you put things in variables like I did (n/(n+1))^n in our case.Thanks, Bonaparte

Thank you for your response. I got it now, the fraction reduces to n!/(n+1)^n = Product from 1 to n of (n/(n+1))^n which leads to n!=(n+1)^n * the Product, but then the last term of the product cancels out with (n+1)^n and so comes the required result.

However, I am still interested on how you derived it. I mean, suppose we had absolutely no clue on how does n^n relate to n!. How would we then, think of this relationship? What steps would we have to take?
 
  • #4
The relationship between n^n and n! is ruled by their definition. Now that I think about it a clearer way to have stated this is instead of n/n+1 (where we have to change n^n to (n+1)^n) is to state it in their way (1-1/n)^(n-1).
Then we have ((n-1)/n)^(n-1). Then proceed as before, although now it will be clearer. Considering your writing a paper about it, I recommend the second way.

As for your question, I derived it by look at the numbers, seeing the pattern (which is not even necessary considering its given) and implementing it into numbers.

Thanks and good luck, Bonaparte
 

1. What is Stirling's Formula?

Stirling's Formula is an approximation formula that is used to estimate the value of factorials for large numbers. It was derived by Scottish mathematician James Stirling in the 18th century.

2. How does Stirling's Formula work?

Stirling's Formula approximates the factorial of a large number n by using the constant e (2.71828...) and the square root of 2π. The formula is n! ≈ √(2πn)(n/e)^n.

3. What is the relationship between n and n^n in Stirling's Formula?

In Stirling's Formula, n^n refers to the exponential function with base n. This means that as n gets larger, n^n also gets larger at an increasingly faster rate.

4. Can Stirling's Formula be used for all values of n?

No, Stirling's Formula is most accurate for large values of n. As n gets smaller, the approximation becomes less accurate.

5. How accurate is Stirling's Formula compared to the actual value of n factorial?

Stirling's Formula is a very good approximation for large values of n, but it is not exact. The larger the value of n, the closer the approximation will be to the actual value of n factorial. However, for smaller values of n, the approximation may be off by a larger margin.

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