--- a hazardous summation-method for 0!-1!+2!-3!+...-...
Gottfried Helms
the series in the header
su = 0! - 1! + 2! - 3! + 4! - ... + ...
L Euler found a meaningful interpretation using integrals assigning
it a value of about 0.596347�
Also the Borel-summation assigns the same value to this.
Studying the triangle of Eulerian numbers I came across the
idea to use this matrix for a summation, decomposing the entries
of the matrix into geometric series and derivatives.
Not much sophisticated reasoning about range of convergence
included, but it finds the correct value.
See:
http://go.helms-net.de/math/binomial_new/01_12_Eulermatrix.pdf
Chap. 2.2 and 2.3
How could this made waterproof?
TIA -
Gottfried Helms
Gottfried Helms
> What do you think about the approach to the divergent summation of
> the series in the header
>
> su = 0! - 1! + 2! - 3! + 4! - ... + ...
>
> L Euler found a meaningful interpretation using integrals assigning
> it a value of about 0.596347�
> Also the Borel-summation assigns the same value to this.
>
fsum(x) = 0! + 1! x + 2! x^2 + 3! x^3 +�
fsum(-1) = 0.596347362323
fsum(-2) = 0.461455316242
fsum(-3) = 0.385602012137
fsum(-4) = 0.335221361210
fsum(-5) = 0.298669749329
fsum(-6) = 0.270633013639
fsum(-7) = 0.248281352547
fsum(-8) = 0.229947781627
fsum(-9) = 0.214577094581
fsum(-10) = 0.201464233646
fsum(-11) = 0.190117766778
fsum(-12) = 0.180183310425
� �
Could someone check this using the integral-formula?
Gottfried Helms
Gottfried Helms
> � �
>
> Could someone check this using the integral-formula?
>
of summation here? Because in each column are only
finitely many geometric series involved?
Hmm...
Gottfried Helms
G. A. Edgar
<he...@uni-kassel.de> wrote:
> Am 06.11.2009 23:43 schrieb Gottfried Helms:
> > What do you think about the approach to the divergent summation of
> > the series in the header
> >
> > su = 0! - 1! + 2! - 3! + 4! - ... + ...
> >
> > L Euler found a meaningful interpretation using integrals assigning
> > it a value of about 0.596347�
> > Also the Borel-summation assigns the same value to this.
> >
> I propose the following further values for variations of that sum:
> fsum(x) = 0! + 1! x + 2! x^2 + 3! x^3 +�
>
> fsum(-1) = 0.596347362323
> fsum(-2) = 0.461455316242
> fsum(-3) = 0.385602012137
> fsum(-4) = 0.335221361210
> fsum(-5) = 0.298669749329
> fsum(-6) = 0.270633013639
> fsum(-7) = 0.248281352547
> fsum(-8) = 0.229947781627
> fsum(-9) = 0.214577094581
> fsum(-10) = 0.201464233646
> fsum(-11) = 0.190117766778
> fsum(-12) = 0.180183310425
> � �
>
> Could someone check this using the integral-formula?
>
> Gottfried Helms
According to Borel summation, we should have
-exp(1)*Ei(-1) = .5963473622
-(1/2)*exp(1/2)*Ei(-1/2) = .4614553164
-(1/3)*exp(1/3)*Ei(-1/3) = .3856020120
-(1/4)*exp(1/4)*Ei(-1/4) = .3352213612
-(1/5)*exp(1/5)*Ei(-1/5) = .2986697494
-(1/6)*exp(1/6)*Ei(-1/6) = .2706330136
-(1/7)*exp(1/7)*Ei(-1/7) = .2482813514
-(1/8)*exp(1/8)*Ei(-1/8) = .2299477818
-(1/9)*exp(1/9)*Ei(-1/9) = .2145771028
-(1/10)*exp(1/10)*Ei(-1/10) = .2014642544
-(1/11)*exp(1/11)*Ei(-1/11) = .1901177930
-(1/12)*exp(1/12)*Ei(-1/12) = .1801833179
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Gottfried Helms
>
> -exp(1)*Ei(-1) = .5963473622
> -(1/2)*exp(1/2)*Ei(-1/2) = .4614553164
> -(1/3)*exp(1/3)*Ei(-1/3) = .3856020120
> -(1/4)*exp(1/4)*Ei(-1/4) = .3352213612
> -(1/5)*exp(1/5)*Ei(-1/5) = .2986697494
> -(1/6)*exp(1/6)*Ei(-1/6) = .2706330136
> -(1/7)*exp(1/7)*Ei(-1/7) = .2482813514
> -(1/8)*exp(1/8)*Ei(-1/8) = .2299477818
> -(1/9)*exp(1/9)*Ei(-1/9) = .2145771028
> -(1/10)*exp(1/10)*Ei(-1/10) = .2014642544
> -(1/11)*exp(1/11)*Ei(-1/11) = .1901177930
> -(1/12)*exp(1/12)*Ei(-1/12) = .1801833179
>
With the Pari/GP-intnum-formula applied to the integralformula
I got now nearly the same numbers (last digit differs sometimes
due to rounding(?))
Also I see that the coefficients in my example (the colsums of
the Euler-triangle) are not well suited for Cesaro/Euler-summation,
and I assume that the differences from the 8'th digit on are due
to weak performance of the "sumalt"-procedure in Pari/GP for this
problem.
The results with my implementation of Eulersummation agree with the
sumalt-values, and because I have control over the partial sums
I can also explicitely see the difficulties with the convergence
of that partial sums using that summation: convergence is simply
poorly accelerated. So I conclude this is the source of the problem
in Pari/GP' "sumalt".
Well, meanwhile I have some examples, where such a decomposition
into divergent geometric series and reordering summation works
for the divergent case, and also another example, where it does
not. What's the critical point? I've read G.H.Hardy and K.Knopp
few years ago and may not have realized the relevance of some
related chapters there.
Could someone give some more hint?
Gottfried Helms
----------------
where it worked: (for instance) summing of Bell-numbers/
alternating sum of columns of Stirling-matrices,
see http://go.helms-net.de/math/binomial/04_5_SummingBellStirling.pdf
where it didn't work: q-binomial-matrices (occuring in my
discussion of "exponential polynomial interpolation" for
tetration, not yet rewritten with focus on the q-binomial-
matrices)
Gottfried Helms
> the series in the header
>
> su = 0! - 1! + 2! - 3! + 4! - ... + ...
>
> L Euler found a meaningful interpretation using integrals assigning
> it a value of about 0.596347�
> Also the Borel-summation assigns the same value to this.
>
> Studying the triangle of Eulerian numbers I came across the
> idea to use this matrix for a summation, decomposing the entries
> of the matrix into geometric series and derivatives.
> Not much sophisticated reasoning about range of convergence
> included, but it finds the correct value.
>
of transformations to allow divergent summation.
I'll write it in matrix-notation
Let in an algebraical matrix-formula
V(x) represent a columnvector of consecutive powers of x
(a "Vandermondevector")
F the vector of factorials [0!,1!,2!,...], dF when used
as diagonalmatrix,
E the Eulermatrix in lower triangular form,
~ the symbol for transposition
then first, we have according to the introductional example
E*V(1) = F
the factorials as results of rowsums.
If we premultiply that with the inverse factorial, then this gives
the unit-vector:
dF^-1 * F = V(1)
and the unit-vector premultiplied by a vandermonde-row-vector
with the quotient q of a geometric series evaluates to just that
geometric series in closed form; let's use q=1/2 first:
V(1/2)~ * V(1) = 2
If we use q=-1 then this is a divergent expression
V(-1)~ * V(1) = 1/2 // Cesaro/Euler-summation
But if we dissolve the V(1)-vector we get - formally:
V(-1)~ * ( dF^-1 * F ) = ???
V(-1)~ * ( dF^-1 * (E * V(1)) ) = ???
and change order of summation
( V(-1)~ * dF^-1 * E ) * V(1) = ???
( AS(-1) ~ ) * V(1) = ???
Now let's look at the lhs; the inverse factorials
premultiplied to the Eulerian triangle gives strongly
decreasing values in the intermediate result-triangle and
the premultiplication by the V(-1)-vector has nearly the
same rate of convergence as the exponential series - at
least in the first few columns, obviously.
The first few coefficients of AS(-1) are then
[0.36787944, 0.13533528, 0.0011826310, -0.0047367048,
0.00015701391, 0.00020692553, -0.000017334505, -0.0000087610541,
0.0000012416906, 0.00000034713099, -0.000000075195503,
-0.000000012470560,...]
and postmultiplied with the V(1)-vector we get the partial sums for
up to 13 terms:
0.36787944
0.50321472
0.50439736
0.49966065
0.49981766
0.50002459
0.50000726
0.49999849
0.49999974
0.50000008
0.50000001
0.50000000
0.50000000
...
For positive q we get ahead with q=0.75 and arrive at 4.000000 with 8 decimals
in the 30'th partial sum;
(V(0,75) * dF^-1 * E ) * V(1) -> 4.000
and surely for q=1 we get unresolvable divergence. The first
few terms of AS(1) are (using "sumalt" in Pari/GP)
[2.7182818, 1.9524924, 1.9957914, 2.0000389, 2.0000576,
2.0000051, 1.9999996, 1.9999999, 2.0000000, 2.0000000,
2.0000000, 2.0000000]
which very likely continues for the following terms and the
sum of all terms in AS(1) diverges then to infinity.
But for negative q we can do well: for q=-2
(V(-2) * dF^-1 * E ) * V(1)
we get the first few terms in AS(-2)
[0.13533528, 0.15365092, 0.057425669, -0.0042317431, -0.0092431540,
-0.0010507275, 0.0012603255, 0.00038236770, -0.00013302980, -0.000082748221,
0.0000069120015, 0.000014186455]
and the first few partial sums are
0.13533528
0.28898621
0.34641187
0.34218013
0.33293698
0.33188625
0.33314658
0.33352894
0.33339591
0.33331316
0.33332008
0.33333426
0.33333558
0.33333357
0.33333302
0.33333324
0.33333337
0.33333335
0.33333333
-----------------------------------------------------------
Now it would be good to have the exact range of convergence
for the column-sums. The first two columns are easy: for a
left-multiplication with a vandermonde-vector they provide
infinite range of convergence because of the reciprocal factorials.
But even if the range of convergence for a single column
would be infinite, then the (row-) sum of the column-sums need not
be convergent. This seem to happen at least for q>=1
So I guess with that rough sketch, that we have convergence/summability
for the whole range -inf< q <1 and this agrees also with the ability
to sum the alternating factorial series
Nice exercise/example for the divergent summation stuff - isn't it?
Could this be put to more precision?
Gottfried Helms
Gottfried Helms
>
> and surely for q=1 we get unresolvable divergence. The first
> few terms of AS(1) are (using "sumalt" in Pari/GP)
>
> [2.7182818, 1.9524924, 1.9957914, 2.0000389, 2.0000576,
> 2.0000051, 1.9999996, 1.9999999, 2.0000000, 2.0000000,
> 2.0000000, 2.0000000]
>
> which very likely continues for the following terms and the
> sum of all terms in AS(1) diverges then to infinity.
>
fractional parts I get
sum k=0..inf ( AS(1)[k] - 2 ) -> 2/3
If I do alternating summation (with Euler-sum) I get
sum k=0..inf ( (-1)^k*AS(1)[k] ) -> 1.761594156 // Euler-summation
which looks like
0.7615941559557648 = (h042) tanh(1)
(by Plouffe's inverter)
AS(1)*V(-1) = 1 + tanh(1) // Euler-summation
Strange...
Gottfried Helms