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From: Gottfried Helms on 22 Jun 2010 08:46
Hi -

I am very unfimiliar with integrals, but fiddling with
some powerseries-inversion I came to a set of coefficients
which asymptotically can be described by the formula

x 1 - (1-t)^n
ln (x) = lim integral ( ------------ )dt - abs(S[n,1])/n!
n->inf, eps t
eps->0

where S[n,1] is the Stirlingnumber 1st kind of the second column of the
Stirling-matrix. (compare: for n=0,1,2,3,... = {0, 1, -3, 11, -50,...} )

For example using x=0.5 and only small n=10 this gives the value -0.693064856151
which has an absolute error of about 0.0000823 .

Consequently
x 1 - (1-t)^n
ln(x) - ln(y) = lim integral ( ------------ )dt
n->inf, y t


I didn't see that integral-formula before. Is it just a simple reformulation
of the usual integral formula (as online for instance in wikipedia or mathworld) ?
And if, how would one translate that formulae into each other?

------------

Interestingly, the first formula allows to define continuously interpolated
Stirling-numbers of the first kind.
Just reorder the first equation

x 1 - (1-t)^n
abs(S[n,1])/n! = lim integral( ------------ )dt - log(x)
eps->0 eps t

and for x=1 we get for each n=0,1,2,3,4,... the appropriate Stirlingnumber
S[n,1].

If I simply write fi(n,x) for the limit in the above formula, then
I have

abs(S[n,1])/n! = fi(n,x) - log(x)

and using x=1, eps~1e-100

abs(S[n,1]) = fi(n,1) *n!

I get the Stirling-numbers accurate already to dozens of digits.

Now I generalize n! = gamma(1+n) and use a continuous "s" instead of "n" :

abs(S[s,1]) = fi(s,1) * gamma(1+s)

For the half-step I get then S[0,1],S[0.5,1], S[1,1], ...

S[ 0 ,1] 0.,
S[0.5,1] 0.543882461478, = ( 1 - 1*log(2))*gamma(0.5)
S[ 1 ,1] 1.
S[1.5,1] 1.70205061767, = ( 4 - 3*log(2))*gamma(1.5)
S[ 2 ,1] 3.
S[2.5,1] 5.58446693235, =( 23/3 - 5*log(2))*gamma(2.5)
S[ 3 ,1] 11.
S[3.5,1] 22.8689852337, =( 176/3/5 - 7*log(2))*gamma(7/2)
S[ 4 ,1] 50.
S[4.5,1] 114.5421619481 = ( 1689/3/5/7 - 9*log(2))*gamma(9/2)
S[ 5 ,1] 274.
S[5.5,1] 682.3246684993 = (19524/3/5/7/9 - 11*log(2))*gamma(11/2)
...

I remember that there are discussions on and proposals for the
interpolation of that Stirling numbers but didn't find this
online at the moment, so I couldn't compare my values with
that of existing generalizations. Does someone around here
has a clue whether this meets something reasonable and/or
the existing interpolations ?

TIA -

Gottfried Helms
From: Gottfried Helms on 23 Jun 2010 02:54
Am 22.06.2010 14:46 schrieb Gottfried Helms:
>

In my previous post I used a formula for interpolation
of the unsigned Stirling-numbers 1st kind.
This was the given formula:
>
> x 1 - (1-t)^n
> abs(S[n,1])/n! = lim integral( ------------ )dt - log(x)
> eps->0 eps t
>
> and for x=1 we get for each n=0,1,2,3,4,... the appropriate Stirlingnumber
> S[n,1].

If I adapt the formula to give the original (signed) numbers (and also
insert 1 for x getting the log(x) away) I get the signed numbers

1 (t-1)^s - (-1)^s
S[s,1]) = lim integral( ----------------- )dt * gamma(1+s)
eps->0 eps t

I get the signed Stirlingnumbers of the first kind (second column)
at s=0 and at positive integer s.

The interpolation for the half-integer indexes shows now imaginary
value (however with the same values as given before only on the
imaginary axes now). Here I just inserted the corrections:

For the half-step I get then S[0,1],S[0.5,1], S[1,1], ...

S[ 0 ,1]= 0.,
S[0.5,1]= -I*0.543882461478, = -I*( 1 - 1*log(2))*gamma(0.5)
S[ 1 ,1]= 1.
S[1.5,1]= I*1.70205061767, = I*( 4 - 3*log(2))*gamma(1.5)
S[ 2 ,1]= -3.
S[2.5,1]= -I*5.58446693235, = -I*( 23/3 - 5*log(2))*gamma(2.5)
S[ 3 ,1]= 11.
S[3.5,1]= I*22.8689852337, = I*( 176/3/5 - 7*log(2))*gamma(7/2)
S[ 4 ,1]= -50.
S[4.5,1]= -I*114.5421619481 = -I*( 1689/3/5/7 - 9*log(2))*gamma(9/2)
S[ 5 ,1]= 274.
S[5.5,1]= I*682.3246684993 = I*(19524/3/5/7/9 - 11*log(2))*gamma(11/2)
...



Here is a list in smaller steps. Left hand the "original" values (real, imaginary),
right hand divided by gamma(1+n). The plot of the gamma-divided values shows
a nice and smooth spiral in the complex plane.

real imaginary real imaginary
S[1 ,1]= 1 0 1 0
S[1.1,1]= 1.0575 0.3436 1.0105 0.3283
S[1.2,1]= 0.9997 0.7263 0.9073 0.6592
S[1.3,1]= 0.8073 1.1112 0.6920 0.9524
S[1.4,1]= 0.4722 1.4532 0.3801 1.1699

S[1.5,1]= 0 1.7021 0 1.2804
S[1.6,1]= -0.5868 1.8060 -0.4105 1.2633
S[1.7,1]= -1.2475 1.7171 -0.8076 1.1116
S[1.8,1]= -1.9229 1.3971 -1.1470 0.8333
S[1.9,1]= -2.5369 0.8243 -1.3883 0.4511

S[2 ,1]= -3 0 -1.5 0
S[2.1,1]= -3.2161 -1.0450 -1.4634 -0.4755
S[2.2,1]= -3.0907 -2.2455 -1.2751 -0.9264
S[2.3,1]= -2.5426 -3.4996 -0.9475 -1.3042
S[2.4,1]= -1.5171 -4.6691 -0.5089 -1.5662

S[2.5,1]= 0 -5.5845 0 -1.6804
S[2.6,1]= 1.9675 -6.0552 0.5293 -1.6290
S[2.7,1]= 4.2762 -5.8857 1.0253 -1.4112
S[2.8,1]= 6.7405 -4.8973 1.4359 -1.0433
S[2.9,1]= 9.0948 -2.9551 1.7162 -0.5576

S[3 ,1]= 11 0 1.8333 0
S[3.1,1]= 12.0599 3.9185 1.7702 0.5752
S[3.2,1]= 11.8512 8.6104 1.5279 1.1101
S[3.3,1]= 9.9680 13.7198 1.1256 1.5493
S[3.4,1]= 6.0793 18.7103 0.5998 1.8459

S[3.5,1]= 0 22.8690 0 1.9661
S[3.6,1]= -8.2315 25.3338 -0.6151 1.8932
S[3.7,1]= -18.2735 25.1513 -1.1842 1.6299
S[3.8,1]= -29.4116 21.3688 -1.6488 1.1979
S[3.9,1]= -40.5098 13.1624 -1.9601 0.6369

S[4 ,1]= -50 0 -2.0833 0

-----------------------------------------------
Don't know, whether it is in some way significan, that the term in the
integral equals the s'th cyclotomic polynomial with two bases (t-1 and -1)

1 (t-1)^s - (-1)^s
S[s,1]) = lim integral( ----------------- )dt * gamma(1+s)
eps->0 eps (t-1) - (-1)




Gottfried Helms
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