A hazardous summation-method for 0!-1!+2!-3!+...-...
Newsgroups: sci.math

Editorial note: I retrieved the following discussion using the goggle-archive. I changed the formatting for better readability, removed some citations of the foregoing msgs where this seemed to be meaningful and made a few small corrections. The original exchange can be seen at the online-link above. Gottfried Helms, 18.7.2010

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Fri, 06 Nov 2009 23:43:08 +0100

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 su ~ 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.

Chap. 2.2 and 2.3

How could this be made waterproof?

TIA -

Gottfried Helms

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Sat, 07 Nov 2009 08:09:20 +0100

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

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Sat, 07 Nov 2009 11:28:49 +0100

Am 07.11.2009 08:09 schrieb Gottfried Helms:
>    …     …
> Could someone check this using the integral-formula?

Ok, that was an easy one. But why could I change order of summation here? Because in each column are only finitely many geometric series involved?

Hmm...

Gottfried Helms

G. A. Edgar" <ed...@math.ohio-state.edu.invalid>
Sat, 07 Nov 2009 05:52:41 –0500

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/

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Sat, 07 Nov 2009 12:43:34 +0100

Am 07.11.2009 11:52 schrieb G. A. Edgar:

> -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

Yes, thanks! 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's "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

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)

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Mon, 09 Nov 2009 21:50:03 +0100

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.

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.

Interestingly, that triangle can be used for a wider class 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 with q=0.75 we arrive at the expected value of 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

Von: Gottfried Helms <he...@uni-kassel.de>
Datum: Mon, 09 Nov 2009 22:13:38

Am 09.11.2009 21:50 schrieb 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.

Just being curious: if I remove the constant 2 and sum the fractional parts I get If I do alternating summation (with Euler-sum) I get // Euler-summation

which looks like

0.7615941559557648 =  tanh(1)                         // (by Plouffe's inverter)

so

AS(1)*V(-1) = 1 + tanh(1)   // Euler-summation

Strange...

Gottfried Helms