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#1 2007-12-11 17:03:20
- ganesh
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James Stirling Formula
This is a brilliant approximation for factorials, particularly, factorials of higher order numbers.
For example, 1000! as per this formula is 4.023537292 x 10^2567. The actual value as per the calculator in the scientific mode is 4.0238726 x 10^2567. 
Character is who you are when no one is looking.
#2 2007-12-12 03:44:40
- mikau
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Re: James Stirling Formula
sweetness!
A logarithm is just a misspelled algorithm.
#3 2007-12-12 03:56:04
- Identity
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Re: James Stirling Formula
mikau wrote:
sweetness!
Very much so, if you think an error of
is ok.
#4 2007-12-12 04:18:01
- mikau
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Re: James Stirling Formula
I just like how it contains both pi and e.
A logarithm is just a misspelled algorithm.
#5 2007-12-12 04:31:17
- TheDude
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Re: James Stirling Formula
Identity wrote:
mikau wrote:
sweetness!
Very much so, if you think an error of
is ok.
Which comes out to a relative error of 0.008%. I'll take that.
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#6 2007-12-12 04:36:09
- Daniel123
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Re: James Stirling Formula
.. not exactly big!Identity wrote:
mikau wrote:
sweetness!
Very much so, if you think an error of
is ok.
EDIT: Aah post collison
Last edited by Daniel123 (2007-12-12 04:37:01)
#7 2007-12-12 05:17:37
- mikau
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Re: James Stirling Formula
depends on what your priorities are i guess.
A logarithm is just a misspelled algorithm.
#8 2007-12-12 05:19:17
- Identity
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Re: James Stirling Formula
ganesh wrote:
This is a brilliant approximation for factorials, particularly, factorials of higher order numbers.
For example, 1000! as per this formula is 4.023537292 x 10^2567. The actual value as per the calculator in the scientific mode is 4.0238726 x 10^2567.
So does this actually converge on the factorial value as n goes to infinity?
#9 2007-12-12 05:22:16
- luca-deltodesco
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Re: James Stirling Formula
well both n! and the approximation both diverge to infinity as n goes to infinity 
The Beginning Of All Things To End.
The End Of All Things To Come.
#10 2007-12-12 05:35:06
- TheDude
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Re: James Stirling Formula
Yes.
http://en.wikipedia.org/wiki/Stirling%27s_approximation
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#11 2007-12-12 05:35:32
- mikau
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Re: James Stirling Formula
i think he meant, does
Last edited by mikau (2007-12-12 05:36:33)
A logarithm is just a misspelled algorithm.
#12 2007-12-12 06:39:54
- mathsyperson
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Re: James Stirling Formula
I agree that percentage difference is more important than absolute difference.
If you consider the absolute argument the other way, you could say that 10mg of poison on your food is only 10mg more than the recommended amount and so not worth worrying about.
On the same theme, I would guess that this is false:
, but this is true:
Why did the vector cross the road?
It wanted to be normal.
#13 2007-12-12 09:01:51
- luca-deltodesco
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Re: James Stirling Formula
isnt that identical? the only time that that would converge to 1, is if the first converged to 0?
The Beginning Of All Things To End.
The End Of All Things To Come.
#14 2007-12-12 10:18:16
- MathsIsFun
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Re: James Stirling Formula
Because the error may grow, but not as fast as n! grows.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#15 2007-12-12 12:19:35
- mikau
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Re: James Stirling Formula
yeah. Note
but
Last edited by mikau (2007-12-12 12:20:49)
A logarithm is just a misspelled algorithm.
#16 2007-12-12 13:35:00
- Ricky
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Re: James Stirling Formula
What additional requirement can we impose so that Luca's statement holds?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#17 2007-12-13 01:05:42
- mathsyperson
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Re: James Stirling Formula
Equality? That is, instead of
just getting arbitrarily close to 1, it actually has to get there.Why did the vector cross the road?
It wanted to be normal.
#18 2007-12-13 02:53:20
- Ricky
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Re: James Stirling Formula
Certainly you can come up with a restriction far less restricting than that. Remember, this restriction can't apply to Mikau's example.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#19 2007-12-13 04:07:28
- mikau
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Re: James Stirling Formula
if the two limits each converge to the same finite number?
Last edited by mikau (2007-12-13 04:08:17)
A logarithm is just a misspelled algorithm.
#20 2007-12-13 04:21:43
- Ricky
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Re: James Stirling Formula
Bingo.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#21 2007-12-13 04:32:25
- TheDude
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Re: James Stirling Formula
It is also possible for the absolute error to approach 0 while the limits themselves diverge to infinity. As a trivial example, let f(x) = x^2 and g(x) = x^2 + 1/x. Then
and
but
Last edited by TheDude (2007-12-13 07:09:28)
Wrap it in bacon
#22 2007-12-13 06:55:02
- mikau
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Re: James Stirling Formula
Ricky wrote:
Bingo.
awesome! But are there any other restrictions that would do it?
A logarithm is just a misspelled algorithm.