[math-fun] convergent Stirling's formula
On 2016-06-18 23:26, rcs@xmission.com wrote:
I'm hoping for something like a recurrence with a bounded number of terms. The goal is a tolerable formula for arbitrary precision gamma. The wikipedia page gives/linksto several convergent formulas, but they all have disabling features that spoil their use in long summations.
Rich
For stuff like sum 1/(n+a)! you can just use a matrix product. So you're thinking more arcane. Something like sum 1/f(n)! ? --rwg
----------- Quoting rwg <rwg@sdf.org>:
On 2016-06-18 17:50, rcs@xmission.com wrote:
Do you have a formula, or a recurrence, for the Poch form?
Ouch! I had this in the clipboard but forgot to paste it:
https://en.wikipedia.org/wiki/Stirling%27s_approximation#A_convergent_versio...
--rwg
Converting the denominators to factorials/2, starting at 12, I find the numerators are 1, 5, 59, 1218, 38376, ... When these are divided by 12, 60, 360, 2520, 20160, we get the fractions below.
I tried OEIS and Superseeker, no joy.
It would be nice to have a better formula for X!
Rich
--------- Quoting Bill Gosper <billgosper@gmail.com>:
... But with "rising factorials" (Pochhammers) instead of x^n, and weighted row sums of Stirling's Triangle instead of Bernoullis, Out[839]= 1 1 ---------- + ------------------ + 12 (1 + x) 12 (1 + x) (2 + x)
59 --------------------------- + 360 (1 + x) (2 + x) (3 + x)
29 ---------------------------------- + 60 (1 + x) (2 + x) (3 + x) (4 + x)
533 ------------------------------------------- + 280 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x)
1577 --------------------------------------------------- 168 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x)
+ 280361 /
(5040 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) (7 + x))
Both of these series get added to the usual Stirling approximation ln(x^x/e^x ?(2?x)). The coefficients of this convergent series actually outgrow the Bernoullis, but the Pochhammers grow faster still. ("Pochhammers" is so appealing compared to "rising factorial powers". Too bad that, as usual, Pochhammer didn't invent them.)
I also need the coefficients, the 533/280 and 1577/268 etc. I tried multiplying these by n!, getting to my 1, 5, 59, 1218, 38376, ... but nothing jumps out. If these can reasonably be derived from Bernoulli #s, it might be possible to use an approximate recurrence for zeta(n) values, and polish to an exact B(n) with Von Staudts formula for the fractional part. Maybe. Rich ---------- Quoting Bill Gosper <billgosper@gmail.com>:
On 2016-06-18 23:26, rcs@xmission.com wrote:
I'm hoping for something like a recurrence with a bounded number of terms. The goal is a tolerable formula for arbitrary precision gamma. The wikipedia page gives/linksto several convergent formulas, but they all have disabling features that spoil their use in long summations.
Rich
For stuff like sum 1/(n+a)! you can just use a matrix product. So you're thinking more arcane. Something like sum 1/f(n)! ? --rwg
----------- Quoting rwg <rwg@sdf.org>:
On 2016-06-18 17:50, rcs@xmission.com wrote:
Do you have a formula, or a recurrence, for the Poch form?
Ouch! I had this in the clipboard but forgot to paste it:
https://en.wikipedia.org/wiki/Stirling%27s_approximation#A_convergent_versio...
--rwg
Converting the denominators to factorials/2, starting at 12, I find the numerators are 1, 5, 59, 1218, 38376, ... When these are divided by 12, 60, 360, 2520, 20160, we get the fractions below.
I tried OEIS and Superseeker, no joy.
It would be nice to have a better formula for X!
Rich
--------- Quoting Bill Gosper <billgosper@gmail.com>:
... But with "rising factorials" (Pochhammers) instead of x^n, and weighted row sums of Stirling's Triangle instead of Bernoullis, Out[839]= 1 1 ---------- + ------------------ + 12 (1 + x) 12 (1 + x) (2 + x)
59 --------------------------- + 360 (1 + x) (2 + x) (3 + x)
29 ---------------------------------- + 60 (1 + x) (2 + x) (3 + x) (4 + x)
533 ------------------------------------------- + 280 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x)
1577 --------------------------------------------------- 168 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x)
+ 280361 /
(5040 (1 + x) (2 + x) (3 + x) (4 + x) (5 + x) (6 + x) (7 + x))
Both of these series get added to the usual Stirling approximation ln(x^x/e^x ?(2?x)). The coefficients of this convergent series actually outgrow the Bernoullis, but the Pochhammers grow faster still. ("Pochhammers" is so appealing compared to "rising factorial powers". Too bad that, as usual, Pochhammer didn't invent them.)
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