1-exp(1)+exp(exp(1))-exp(exp(exp(1))) + - ... =?
Gottfried Helms
I asked this in sci.math, but without useful success.
Perhaps the subject is more appropriate here...
I'm considering divergent summation, and wouldn't
have a clue, with which technique one would access
this problem.
Recently I found a possible solution, which however
depends on proofs for convergence or at least
summability of intermediate result.
For a start, a current approximation was
0.246960868[386...
Here the explanation which I sent to sci.math,
inserted comments marked by "**>".
Is someone out here, who could comment on the following?
TIA -
Gottfried Helms
---------------------------------------------
(posting to sci.math, answering to a respondent)
Well, to some divergent series a value can be
assigned by Cesaro-,Euler-, or Borel-summation.
To others these summation-methods do not suffice
to "limit" them in terms of their partial sums.
Now this is a very badly diverging series...
It occurs in a certain "experimental" context, where
I look at powerseries.
Assume, I have the coefficients of the powerseries
of x in a vector
V(x) = [1,x,x^2,x^3,...]
Now assume another vector A0, containing
A0 = [1/0!, 1/1!, 1/2!, ...]
then the rowvector V(x)~ by A0 is
V(x)~ * A0 = exp(x)
Now I have another vector A1, and this performs
V(x) ~ * A1 = exp(x)^2
and assume, I have a set of vectors A0,A1,...
collected in a matrix A, giving
V(x)~ * A = [exp(x),exp(x)^2,exp(x)^3,...]
or more precisely a matrix B ( like A with an additional
left column) which performs
V(x)~ * B = [1, exp(x),exp(x)^2,exp(x)^3,...]
then I can say:
V(x) ~ * B = V(exp(x))~
and B transforms a powerseries in x in one of exp(x).
------------------------------------
Now, if I consider iterated application, then I have
V(x)~ * B = V(exp(x))~
V(exp(x))~ * B = V(exp(exp(x)))~
V(exp(exp(x)))~ * B = V(exp(exp(exp(x))))~
...
and so on.
I may write this in terms of powers of B:
V(x)~ * B^0 = V(x)~
V(x)~ * B^1 = V(exp(x))~
V(x)~ * B^2 = V(exp(exp(x)))~
V(x)~ * B^3 = V(exp(exp(exp(x))))~
...
V(x)~ * B^k = V(...)~
...
B is not triangular, so its powers are not defined for
infinite dimension - but it seems, that comparing the
powers of B with increasing finite dimensions, there
is a convergence in the sum in the dominating terms.
So assume a reasonable behaviour for the beginning.
---------------------------
The alternating sum of theses values could then be
written in terms of a geometric series of B:
V(x)~ * (B^0 -B^1 + B^2 - B^3 + B^4 ...)
= V(x)~ - V(exp(x))~ + V(exp(exp(x))) - ...
For a geometric series the shorthand formula is also
valid for matrices,
(I - B + B^2 - B^3 + B^4 -+...) * ( I + B) = I
(I - B + B^2 - B^3 + B^4 -+...) = I * (I + B)^-1
writing M for the reciprocal of the term
M = (I + B)^-1
*if* the powers can be used *and* the parenthese is invertible.
Now for finite dimensions d, the matrices M_d seems to stabilize
with higher d, and a convergent for M_d with d->oo
seems reasonable, so practically I assume the finite
version of M_d with d=64 as sufficient precise, where
I calculate with float-precision of about 200 digits (Pari/Gp),
and final approximations to ~ 12 significant digits for display.
--------------------------------------
The top-left-submatrix of M is then about:
1/2 -0.24696 -0.036280 0.063037 ....
0 0.60330 -0.38514 -0.12045
0 -0.11631 0.61686 -0.42321
0 -0.00013547 -0.16041 0.59960
0 0.0081171 -0.013595 -0.18060
0 0.000074344 0.014385 -0.021246
0 -0.0012697 0.0031833 0.017801
0 -0.00010936 -0.0023863 0.0059199
....
where the sequences of the absolute values in the columns
seem to converge to zero.
-----------------------------------------
For the sum of the first column of M I get now
lim V(1)~ * M[,0] = 1/2
which is obvious since
V(1)~ * column([1/2,0,0,0,...]) = 1/2
and this is implying, or a shorthand for, the first
column of the explicite and divergent rowvectorsum
of all V(x)~ * B^k : 1 - 1 + 1 - 1 + ...
[1, x, x^2, x^3, x^4,...]
- [1, exp(x), exp(x)^2, ...]
+ [1, exp(exp(x)), exp(exp(x))^2 , ...]
- ....
+ ....
with x=1, in the sense of Cesaro or Euler-summation.
giving also 1/2 this explicte way.
--------------------------------------------
Analoguously I apply this to the second column, M[,1] ,
which -different from the column-sum of B_k's- still
seems to be convergent with x=1.
lim V(1)~ * M[,1] = 0.246960868[386...
(Again implicitely) this expresses the replacement
for the alternating sum
x - exp(x) + exp(exp(x)) - ...
of the first columns of all V(x)~ * B^k by its construction.
--------------------------------------------
The intention of my question was, -besides of the
amazing formulation of the problem - to get counter-
checks with other methods, and still would like to
see, whether this result is conform with other
methods.
**>
I *assumed* from heuristics, that M_d stabilizes with
higher d and thus in principle we would expect some
summability for the alternating sums of V(x)~ * B^k,
but I didn't work this out so far.
Note, that an equivalent, but somehow inverse idea,
can be checked easily by low order summation-techniques:
x - log(1+x) + log(1+log(1+x)) - log(1+log(1+log(1+x))) +... - ...
which has oscillating divergence, which one however can
handle by Eulersummation, for instance.
-----
Gottfried Helms
Oscar Lanzi III
What does your method give when the base of your iterated exponentiation
is changed from e to sqrt(2)? When the latter base is used the terms
are bounded and you will get an Euler-summable expression, whose Euler
sum can be compared with what your method gives.
--OL
Gottfried Helms
Am 24.06.2007 00:03 schrieb Gottfried Helms:
>
> writing M for the reciprocal of the term
>
> M = (I + B)^-1
>
> *if* the powers can be used *and* the parenthese is invertible.
>
> Now for finite dimensions d, the matrices M_d seems to stabilize
> with higher d, and a convergent for M_d with d->oo
> seems reasonable, so practically I assume the finite
> version of M_d with d=64 as sufficient precise, where
> I calculate with float-precision of about 200 digits (Pari/Gp),
> and final approximations to ~ 12 significant digits for display.
>
B1 = (I + B)
and
M = B1^-1
which support the assumption of stabilizing of M and convergence/
summability of the column-sums of M according to
V(1)~ * M
---------------------------------------------------------
To invert B1 I decompose it into LDU - components, where
L is a lower left triangular matrix, D is diagonal, U is an
upper right triangular matrix.
With increasing dimension d the coefficents in L,D,U are
constant, only new rows/columns are added.
Here I give the images for d = 12, where for display I reduce
the rows to 8 and columns to 6
The left component L:
VE(tL,8,6):
1 . . . . .
0 1 . . . .
0 1/4 1 . . .
0 1/12 7/15 1 . .
0 1/48 1/4 19/28 1 .
0 1/240 31/300 13/28 103/115 1
0 1/1440 7/200 211/840 5774/7935 51775/46483
0 1/10080 127/12600 19/168 7499/15870 2028343/1952286
-----------------------------------------------
in float display:
1.0*VE(tL,8,6):
1.00000 . . . . .
0 1.00000 . . . .
0 0.250000 1.00000 . . .
0 0.0833333 0.466667 1.00000 . .
0 0.0208333 0.250000 0.678571 1.00000 .
0 0.00416667 0.103333 0.464286 0.895652 1.00000
0 0.000694444 0.0350000 0.251190 0.727662 1.11385
0 0.0000992063 0.0100794 0.113095 0.472527 1.03896
=======================================================================
The right component R is:
VE(tR,8,6):
1 1/2 1/2 1/2 1/2 1/2
. 1 1 3/2 2 5/2
. . 1 3/2 14/5 9/2
. . . 1 212/105 13/3
. . . . 1 2695/1058
. . . . . 1
. . . . . 0
. . . . . 0
-----------------------------------------------
in float display:
1.0*VE(tR,8,6):
1.00000 0.500000 0.500000 0.500000 0.500000 0.500000
. 1.00000 1.00000 1.50000 2.00000 2.50000
. . 1.00000 1.50000 2.80000 4.50000
. . . 1.00000 2.01905 4.33333
. . . . 1.00000 2.54726
. . . . . 1.00000
. . . . . 0
. . . . . 0
=======================================================================
The diagonal is
VE(tD,8):
2 2 5/2 7/2 529/105 2021/276 497335/46483 77937481/4973350
1.0*VE(tD,8):
2.00000 2.00000 2.50000 3.50000 5.03810 7.32246 10.6993 15.6710
Note, that the entries of this diagonal do not change regardless
of the dimension; increasing the dimension simply adds new values.
=======================================================================
Now, to invert B1 means to use the inverse of the three
components, where because of their triangularity plus
the consistency of coefficients regarding the size of
dimension the coefficients are again consistent:
The inverse of the diagonal D in rational and float display
1.0*tDinv,tDinv:
1/2 0.500000
1/2 0.500000
2/5 0.400000
2/7 0.285714
105/529 0.198488
276/2021 0.136566
46483/497335 0.0934642
4973350/77937481 0.0638120
245503065150/5644742532671 0.0434923
18966334909774560/640455736601965177 0.0296138
322789691247390449208/16020371227049741794583 0.0201487
212540844137878775546487/15511826733078660216267418 0.0137019
-----------------------------------------------
The inverse of the right component R
VE(tRInv,8,8):
1 -1/2 0 1/4 -1/210 -2039/6348 2831/92966 2597513/4774416
. 1 -1 0 4/5 -20/529 -55230/46483 70259/397868
. . 1 -3/2 8/35 750/529 -20610/46483 -899619/397868
. . . 1 -212/105 1285/1587 90492/46483 -1496523/994670
. . . . 1 -2695/1058 78864/46483 4426723/1989340
. . . . . 1 -6210/2021 5714667/1989340
. . . . . . 1 -42932869/11936040
. . . . . . . 1
1.0*VE(tRInv,8,8):
1.00000 -0.500000 0 0.250000 -0.00476190 -0.321204 0.0304520 0.544048
. 1.00000 -1.00000 0 0.800000 -0.0378072 -1.18818 0.176589
. . 1.00000 -1.50000 0.228571 1.41777 -0.443388 -2.26110
. . . 1.00000 -2.01905 0.809704 1.94678 -1.50454
. . . . 1.00000 -2.54726 1.69662 2.22522
. . . . . 1.00000 -3.07274 2.87264
. . . . . . 1.00000 -3.59691
. . . . . . . 1.00000
-----------------------------------------------
Inverse of the left-component
VE(tLInv,8,8):
1 . . . . . . .
0 1 . . . . . .
0 -1/4 1 . . . . .
0 1/30 -7/15 1 . . . .
0 2/105 1/15 -19/28 1 . . .
0 -1/92 37/690 33/230 -103/115 1 . .
0 -385/185932 -3628/139449 153891/1859320 188227/697245 -51775/46483 1 .
0 38677/10444035 -513748/52220175 -4511833/87033625 25638092/261100875 1861897/4177614 -3314476/2486675 1
1.0*VE(tLInv,8,8):
1.00000 . . . . . . .
0 1.00000 . . . . . .
0 -0.250000 1.00000 . . . . .
0 0.0333333 -0.466667 1.00000 . . . .
0 0.0190476 0.0666667 -0.678571 1.00000 . . .
0 -0.0108696 0.0536232 0.143478 -0.895652 1.00000 . .
0 -0.00207065 -0.0260167 0.0827674 0.269958 -1.11385 1.00000 .
0 0.00370326 -0.00983811 -0.0518401 0.0981923 0.445684 -1.33289 1.00000
=======================================================================
Now M_d for a certain dimension d is
M_d = B1_d^-1 = R_d^-1 * D_d^-1 * L_d^-1
and one had to show, that the occuring partial sums for the
entries of M_d, for d = 1 to inf are either convergent or
can be limited by a summation technique.
For this analytical expressions for the entries in R^-1, D^-1,L^-1
are required which I don't see at the moment. But since the entries
have alternating signs, and the values in the inverted diagonal seem
to decrease, I think, asymptotics should be sufficient for a first
estimate of the validity of the general idea and the given provisorial
result.
Gottfried Helms
Gottfried Helms
What I get is, even with (my) Eulersum of order 1 (direct summing)
dim = 64
%pri B = 1.0*dFac(-1)*PInv*St2F * P~ \\ the initial version of B, performing summing with parameter exp(1)
%pri B_test = dV(log(2)/2) * B \\ test-variant which performs with parameter sqrt(2)
1) Check appropriateness of B_test
----------------------------------------------------
%pri ESum(1.0)*dV(1)*B_test \\ example-summation , Eulersum of low order 1 is possible
1.00000 1.41421 2.00000 2.82843 4.00000 5.65685 8.00000 11.3137 16.0000 22.6274 32.0000 45.2548 64.0000 90.5097 128.000 181.019 256.000 362.039 512.000 724.077 1024.00 1448.15 2048.00 2896.31 4096.00 5792.62 8192.00 11585.2 16384.0 23170.5 32768.0 46341.0 65536.0 92681.9 131072. 185364. 262144. 370728. 524288. 741455. 1.04858E6 1.48291E6 2.09715E6 2.96582E6 4.19430E6 5.93164E6 8.38861E6 1.18633E7 1.67772E7 2.37266E7 3.35544E7 4.74531E7 6.71089E7 9.49063E7 1..34218E8 1.89813E8 2.68435E8 3.79625E8 5.36871E8 7.59250E8 1.07374E9 1.51850E9 2.14748E9 3.03700E9
This means:
[1,1,1,1...] ---> [1, sqrt(2), sqrt(2)^2 , sqrt(2)^3, ...]
----------------------------------------------------
%pri ESum(1.0)*dV(1) * B_test^2 \\ example-summation , Eulersum of low order 1 is possible
1.00000 1.63253 2.66514 4.35092 7.10299 11.5958 18.9305 30.9046 50.4525 82.3651 134.463 219.515 358.364 585.039 955.091 1559.21 2545.46 4155.53 6784.01 11075.1 18080.4 29516.7 48186.8 78666.2 128425. 209657. 342270. 558765. 912200. 1.48919E6 2.43114E6 3.96891E6 6.47935E6 1.05777E7 1.72684E7 2.81911E7 4.60228E7 7.51334E7 1.22657E8 2.00241E8 3.26899E8 5.33672E8 8.71234E8 1.42231E9 2.32196E9 3.79067E9 6.18837E9 1.01027E10 1.64929E10 2.69251E10 4.39559E10 7.17593E10 1.17149E11 1.91249E11 3.12219E11 5.09705E11 8.32108E11 1.35844E12 2.21769E12 3.62043E12 5.91046E12 9.64898E12 1.57522E13 2.57159E13
This means:
[1,1,1,1...] ---> [1, sqrt(2)^sqrt(2), (sqrt(2)^sqrt(2))^2 , (sqrt(2)^sqrt(2))^3, ...]
----------------------------------------------------
%pri ESum(1.0)*dV(1) * B_test^3
1.00000 1.76084 3.10056 5.45958 9.61345 16.9277 29.8070 52.4854 92.4184 162.734 286.548 504.566 888.459 1564.43 2754.72 4850.61 8541.15 15039.6 26482.3 46631.1 82109.9 144582. 254586. 448286. 789360. 1.38994E6 2.44745E6 4.30957E6 7.58847E6 1.33621E7 2.35285E7 4.14298E7 7.29513E7 1.28456E8 2.26190E8 3.98284E8 7.01314E8 1.23490E9 2.17446E9 3.82888E9 6.74204E9 1.18717E10 2.09041E10 3.68087E10 6.48143E10 1.14128E11 2.00960E11 3.53859E11 6.23088E11 1.09716E12 1.93192E12 3.40180E12 5.99003E12 1.05475E13 1.85724E13 3.27030E13 5.75848E13 1.01398E14 1.78545E14 3.14389E14 5.53587E14 9.74776E14 1.71642E15 3.02233E15
This means:
[1,1,1,1...] ---> [1, sqrt(2)^sqrt(2)^sqrt(2), (sqrt(2)^sqrt(2)^sqrt(2))^2 , (sqrt(2)^sqrt(2)^sqrt(2))^3, ...]
----------------------------------------------------
2) Now check results of infinitely iterated use of B_test by
use of M_test
----------------------------------------------------
B1_test = matid(n) + B_test ;
tmp = CV_LR(B1_test); \\ compute L-D-U - components of tst
\\ result in matrices CV_L CV_D CV_R
dim = 64 \\ compute the Inverse of I + B_test using different dimensions
M_test = EMMul(VE(CV_RInv,dim), VE(CV_DInv*CV_LInv,dim), 1.0)
\\ compute the inverse matrix M_test of B1_test by L-D-U-components
\\ with dim as selectable dimension,
\\ implicte euler-summation (here order 1.0)
\\ in the matrix-multiplications
\\ results
%pri ESum(1.0)*dV(1)*M_test \\ Eulersum(1.0)
0.500000 0.362302 0.201330 0.0223437 -0.161780 -0.326291 -0.429142 ...
where in the second column is the interesting result.
Note, that for summation of the more right columns higher orders
of the Euler-summation may be needed, so these numbers need be
crosschecked.
---------------------------------------------------------
The result of the above operations is now:
x= lim 1 - sqrt(2)^1 + sqrt(2)^sqrt(2)^1 - sqrt(2)^sqrt(2)^sqrt(2)^1 + ... - ...
x ~ 0.3623015472668659
Curious, whether this result makes sense...
Note, that using V(2) as initial powerseries we get
%pri ESum(1.0)*dV(2)*M_test \\ Eulersum(1.0)
0.500000 1.00000 2.00000 4.00000 8.00000 16.0000 32.0000 64.0000 128.000 256.000 512.000 1024.00 2048.00 4096.00 8192.00 16384.0 32768.0 65536.0 131072. 262144. 524288. 1.04858E6 2.09715E6 4.19430E6 8.38861E6 1.67772E7 3.35544E7 6.71089E7 1.34218E8 2.68435E8 5.36871E8 1.07374E9 2.14748E9 4.29497E9 8.58993E9 1.71799E10 3.43597E10 6.87195E10 1.37439E11 2.74878E11 5.49756E11 1.09951E12 2.19902E12 4.39805E12 8.79609E12 1.75922E13 3.51844E13 7.03688E13 1.40738E14 2.81476E14 5.62953E14 1.12591E15 2.25184E15 4.50372E15 9.00763E15 1.80158E16 3.60334E16 7.20723E16 1.44161E17 2.88372E17 5.76888E17 1.15419E18 2.30953E18 4.62225E18
and in the second column we have the implicte computation of:
x= lim 2 - sqrt(2)^2 + sqrt(2)^sqrt(2)^2 - sqrt(2)^sqrt(2)^sqrt(2)^2 + ... - ...
= 1.000
Gottfried Helms
Gottfried Helms
Am 24.06.2007 14:48 schrieb Oscar Lanzi III:
y = lim sqrt(2) - sqrt(2)^sqrt(2) + ... \\ Eulersummation
then
y ~ 0.6376984527331341...
Let x be my previous result
x ~ 0.3623015472668659...
then we have
y = 1.0000000000... - x
I think I should look for a small correction...
Thanks for the hint concerning sqrt(2).
Gottfried Helms
Gottfried Helms
Am 24.06.2007 14:48 schrieb Oscar Lanzi III:
I had a different series, so both results are
perfectly compatible:
I had
> The result of the above operations is now:
>
> x= lim 1 - sqrt(2)^1 + sqrt(2)^sqrt(2)^1 - sqrt(2)^sqrt(2)^sqrt(2)^1 + ... - ...
> x ~ 0.3623015472668659
>
and by direct summation I was missing the first term 1:
> y = lim sqrt(2) - sqrt(2)^sqrt(2) + ... \\ Eulersummation
>
> then
>
> y ~ 0.6376984527331341...
and indeed
x = 1 - y
So this seems to be a one-example confirmation of my method.
Gottfried
-------
Gottfried Helms
Oscar Lanzi III
Good.
Now test this thing by approaching and then breaking through the
singularity (with respect to the boundedness of the terms and thus Euler
summation) at b(ase) = exp(1/e) = 1.444+. I would recommend proceeding
in two stages: first examine the behavior as you come up to the
Euler-summation singularity; then, if that is satisfactory, step beyond
it with your method.
Part 1: The approach
We have seen what the sum (both Euler's and yours) is at b = sqrt(2) =
1.414+. Examine the Euler sum for b = 1.43, 1.44, 1.444. You can also
try your method if you wish for these bases, although the result for b =
sqrt(2) seems to indicate that you'll just get back the Euler sum.
What you need to see is evidence that the Euler sum does _not_ duplicate
the square root singularity with b that we get in the individual terms.
To wit, as b approaches e^(1/e) from below the n-th term apporaches
a*((-1)^(n-1)) with a defined by b = a^(1/a) and a <= e. This
asymptotic behavior contributes a/2 to the Euler sum, where a and
therefore a/2 will contain a square-root singularity. The deviations of
the terms from this behavior for finite n must be such as to cancel out
this square root singularity. _If they don't, then an analytic
continuation across b = e^(1/e) will surely go into complex numbers,
turning your real sums into sauce_. To check this, plot your sums
against a (or against sqrt[(exp(1/e)-b]) and against b, and see which
relation appears (essentially) linear. What you want, of course, is
linear behavior versus b. This method is not foolproof, but it will at
least screen out the most obvious possible pitfall in your scheme.
Part 2: Over the edge
If Part 1 appears satisfactory, examine the sum by your method (Euler's
will no longer work) for b = 1.45, 1.46, 1.47. Compare these results
with the earlier ones for b = 1.43-1.444, to see if there is reasonably
smooth behavior. If yes, you have an apparently suitable analytic
continuation that can be applied for "large" values of b such as e.
Have fun! :-)
--OL
Gottfried Helms
Am 25.06.2007 13:27 schrieb Oscar Lanzi III:
> Good.
>
> Now test this thing by approaching and then breaking through the
> singularity (with respect to the boundedness of the terms and thus Euler
> summation) at b(ase) = exp(1/e) = 1.444+.
:-) Just came across it. How did you get the idea, that this point
would produce a singularity, btw?
Also I used different notation for indication of the base. The indicated
bases would be 1/e = 0.3678794411714423... in my notation below.
(For exp(1/e) in my notation I get ~ 0.3562483198426627...)
> I would recommend proceeding
> in two stages: first examine the behavior as you come up to the
> Euler-summation singularity; then, if that is satisfactory, step beyond
> it with your method.
>
> Part 1: The approach
>
> We have seen what the sum (both Euler's and yours) is at b = sqrt(2) =
> 1.414+. Examine the Euler sum for b = 1.43, 1.44, 1.444. You can also
> try your method if you wish for these bases, although the result for b =
> sqrt(2) seems to indicate that you'll just get back the Euler sum.
>
> What you need to see is evidence that the Euler sum does _not_ duplicate
> the square root singularity with b that we get in the individual terms.
> To wit, as b approaches e^(1/e) from below the n-th term apporaches
> a*((-1)^(n-1)) with a defined by b = a^(1/a) and a <= e. This
> asymptotic behavior contributes a/2 to the Euler sum, where a and
> therefore a/2 will contain a square-root singularity. The deviations of
> the terms from this behavior for finite n must be such as to cancel out
> this square root singularity. _If they don't, then an analytic
> continuation across b = e^(1/e) will surely go into complex numbers,
> turning your real sums into sauce_. To check this, plot your sums
> against a (or against sqrt[(exp(1/e)-b]) and against b, and see which
> relation appears (essentially) linear. What you want, of course, is
> linear behavior versus b. This method is not foolproof, but it will at
> least screen out the most obvious possible pitfall in your scheme.
Well, I'll try to configure my testenvironment according to this...
>
> Part 2: Over the edge
>
> If Part 1 appears satisfactory, examine the sum by your method (Euler's
> will no longer work) for b = 1.45, 1.46, 1.47. Compare these results
> with the earlier ones for b = 1.43-1.444, to see if there is reasonably
> smooth behavior. If yes, you have an apparently suitable analytic
> continuation that can be applied for "large" values of b such as e.
>
> Have fun! :-)
>
> --OL
Here I have some results; I'll see, whether I can go more into detail
tomorrow to examine it in the near of s = 1/e +- eps and to do some plots.
(I'm using "s" in the following to prevent confusion with your notation of
"b")
I checked the intermediate summation-terms of the matrix-
multiplication for apparent divergence. The following
entries seem to be good approximations without need of
high orders of Eulersummation:
1 - 5 + 5^5 - 5^5^5 + ... %445 = 0.1976203324895644
1 - pi + pi^pi - pi^pi^pi + ... %432 = 0.2324670739643584
1 - e + e^e - e^e^e + ... %457 = 0.2469608978527557
1 - 2 + 2^2 - 2^2^2 + ... %469 = 0.2874086990716591
s = phi+1 (= 1.6180339887498948..)
1 - s + s^s - s^s^s + ... %507 = 0.3278048495149163
s=sqrt(2)
1 - s + s^s - s^s^s + ... %481 = 0.3623015472668659
s = phi (= 0.6180339887498948..)
1 - s + s^s - s^s^s + ... %494 = 0.7861534997409138
s = 1/2
1 - s + s^s - s^s^s + ... %519 = 0.9382530028218765
\\ the following are questionable --------------
\\ 1/e ~ 0.3678794411714423
s = 1/e + 0.01
1 - s + s^s - s^s^s + ... %531 = 1.172056144413609
s = 1/e
1 - s + s^s - s^s^s + ... --- division by zero ---
s = 1/e-0.01
1 - s + s^s - s^s^s + .. %551 = 1.223179007248831
s = 0.01
1 - s + s^s - s^s^s + .. --- seems to diverge ---
\\ the following is specifically questionable, since my Euler-sum
\\ procedure is not configured well for complex summation,
\\ although the summation-terms diverge "not too strong"
\\
\\ the implicite value log(-1/2) is used by the Pari-convention as
\\ ~ -0.6931471805599453 + 3.141592653589793*I
s = -1/2
1 - s + s^s - s^s^s + .. %592 = 0.3755075014152616 - 1.118700997443251*I
Next I'll try your suggestion to get more results near the singularity
and speed up the computation to get sufficient coordinates for a plot.
Thanks for the comment!
Gottfried
---
Gottfried Helms, Kassel
Gottfried Helms
I just added a documentation about the terms for
the Euler-summation on the real axis near the singularity.
It looks very good and is at
http://go.helms-net.de/math/binomial_new/PowertowerproblemDocSummation.htm
Gottfried
----
Gottfried Helms, Kassel
Oscar Lanzi III
The "singularity" I was referring to has to do with the terms of the
series that suddenly go from bounded to increasing extremely fast.
Therefore there is a catastrophic breakdown of the Euler summation.
But it turns out that the iterated Shanks extrapolation continues to
converge, and that its results indicate continuous behavior across the
Euler-summation singularity. Following are sums I obtained using
Microsoft Excel:
Base Sum
1.43 0.359117
1.44 0.357152
-------Euler-summation Singularity--------
1.45 0.355226
1.46 0.353346*
1.47 0.351477*
*Slight uncertainty since I could get only a few partial sums before the
terms overflowed.
The implication is that sums are definable across the Euler-summation
sigularity; that is, the singularity is removable. But for bases
substantially larger than e^(1/e) in my notation, the Shanks mthod
becomes computationally unfeasible and we would have to rely on your
method.
--OL
Oscar Lanzi III
I just saw your result for low bases. There you run into a singularity
not only with respect to Euler summation, but with respect to the actual
sum obtained by any method.
The singularity for s < 1 occurs at s = exp(-e), not at s = 1/e. The
quantity that's 1/e at this point is the limit of the absolute values of
the terms; to wit, given s = exp(-e), the sequence 1, s, s^s, etc.
converges to 1/e. Above this, up to s = exp(1/e), the Euler sums
converge; and as we have seen, for still larger s we can achieve
convergence with nonlinear methods (both the iterated Shanks
trnasformation and yours). But at s = exp(-e) the Euler sum becomes
divergent. The divergent behavior is understood by decomposing
rendering the summation terms as
a*(-1)^(n-1) + d_n
where a is the limiting absolute values of the terms as described
preciously. The deviations d_n asymptotically satisfy d_(n+1)/d_n = -ln
a, so the boundaries of convergence are given by a = e and a = 1/e. For
the former case (s = exp(1/e)) the signs of the d_n terms altternate,
which enables convergence simply because the absolute values of the d_n
slowly die out. But for s < 1 the signs of the d_n terms are all
positive, and at s = exp(-e) where the limiting term ratio hits 1, they
go to zero too slowly for convergence.
Below s = exp(-e) the sequence 1, s, s^s, etc. does not blow up. It
bifurcates; as s --> 0+ the terms alternately approach 1 and 0. In your
alternating series this produces a positive bias; the series diverges
linearly. Such a linear diovergence cannot be removed with standard
summation methods, so I am not surprised to see that your method
likewise breaks down at s = 0.01 (which is less than 1/e).
What is the nature of the singularity at s = exp(-e)? Given the
behavior of the d_n above we would expect a first-order pole. But a
function should be continuable across (or around) a pole. There must be
some hidden structure that prevents such contiunuation. Curiouser and
curiouser ... .
Thanks for getting back to me.
--OL
Gottfried Helms
I compiled an overview about my summing-method in
http://go.helms-net.de/math/binomial_new/10_4_Powertower.pdf
where the method is explained in more detail, and is reflected
regarding the discussion I had with Oscar Lanzi III .
Also I rewrote the study of the singularity point in
http://go.helms-net.de/math/binomial_new/PowertowerproblemDocSummation.htm
------
@ Oscar :
Oscar -
thanks again for your comments, they are helpful to make
me review open questions in my approach. Meanwhile I searched
for material about tetration and get additional caveats concerning
the range of validity of the method.
I'll answer to your recent letters in more detail tomorrow,
I'm having doubts about the validity of Euler-summation
at and beyond the critical point exp(-1), which you mentioned.
Gottfried
-------
Gottfried Helms, Kassel
Gottfried Helms
Hmm - I'm still not prepared to comment on this precisely.
However, considering an email which I got today urged me, to
try to differentiate two things, whose mixture may be a
source of misconception and may confuse (at least my) part of
discussion.
It appears now to me, that we possibly discuss two essential
different things here, which surely cover overlapping aspects.
I'm discussing an infinite sum of finite entities -
each term in the sum is finite. The other side is
the infinite powertower itself. This construct does not occur
in the given sum - however attracted and partly dominated
unconsciously my attention and focus over the last days.
Surely the bounds, given for the infinite powertower
are relevant for the mentioned sum and may or may not
exactly establish bounds for the base-parameter of the
given infinite sum.
It reminds me to the difference between formulating
a lemma for the geometric series: an infinite sum of
finite entities(powers of a certain parameter) and
for the infinite power of a parameter itself.
Well - for the infinite power there is no expression,
so let's assume an infinite one-parameter continued
fraction, for which a limiting value can be found
but which would be essentially different from an
infinite sum of rational continued fractions.
Another analogy may be given by the zeta-series. For
each real parameter >1 it is conventionally convergent,
and for each real parameter <=1 it is divergent.
But also here we deal with an infinite sum of
finite entities, in contrast to a power of an
infinite entity itself.
So I feel, this aspect should be considered more
explicitely...
With the "alternating sum of increasing powertowers"
of a certain base I think now, the best formulation
is, that my process gives coefficents a_k for a
polynomial, which then replaces the term-by-term
summation of the reflected powertower-series
and also expands the domain for the base-parameter,
which are defined by the bounds for the base-parameter
for an individual infinite powertower.
Also the coefficients a_k are parameterspecific, where
the parameter s is the base.
The variable x in the polynomial with coefficients a_k,
(which are dependend on s), allows the computation of
S(s,x) = x - s^x + s^s^x - s^s^s^x + ... -
in terms of a polynomial in x instead. The x at the top of
the powertowers are an aspect of parametrizing such a sum
different from what we discussed before (and this may
also be of less interest - we had simply set it to 1 in our
discussion).
It is comparable -for another instance- with the Stieltjes-
constants as coefficients a_k of a polynomial in x, which
allow then to compute the zeta-series zeta(x) in terms of
a polynomial.
I'm not sure, how far I got this right and formulated well,
I appreciate comments on this, too.
Gottfried Helms
Oscar -
putting some aspects of your previous posts together.
The best presentation of my method is to say that it provides for
S(s,x) = x - s^x + s^s^x - s^s^s^x + ... - (*1a)
here for x=1
S(s):= S(s,1) = 1 - s + s^s - s^s^s + ... - (*1b)
a polynomial representation in terms of coefficients as_k depending on s
S(s,x) = as_0 + as_1 x + as_2 x^2 + as_3 x^3 + ... (*2a)
S(s,1) = as_0 + as_1 + as_2 + as_3 + ... (*2b)
S(s,0) = as_0 (*2c)
For the documentation of the behaviour of S(s) in my method, I make use of
the following identity:
It is according to (*1a) and (*1b)
S(s,1) = 1 - s^1 + s^s^1 - s^s^s^1 + ... -
S(s,0) = 0 - 1 + s^1 - s^s^1 + s^s^s^1 + ... -
from where in the limit it should be
S(s,1) = - S(s,0) (*3)
By (*2b) and (*2c) it is
S(s,1) = as_0 + as_1 + as_2 + as_3 + ...
S(s,0) = as_0
and since by (*3) S(s,1) = - S(s,0) we have, that S(s,1) is completely
explained by the first of the coefficients as_0 and thus
S(s) = - as_0
Now as_0 is itself the result of a vector-product, and to check
its behaviour I document its individual additive components
as_0 = bs_0 + bs_1 + bs_2 + ...
so that the best simplificated result of my method is then
(assuming x=1)
S(s) = - ( bs_0 + bs_1 + bs_2 + bs_3 + ... ) (*4)
---------------------------------------------------------------------
> Part 1: The approach
>
> We have seen what the sum (both Euler's and yours) is at b = sqrt(2) =
> 1.414+. Examine the Euler sum for b = 1.43, 1.44, 1.444. You can also
> try your method if you wish for these bases, although the result for b =
> sqrt(2) seems to indicate that you'll just get back the Euler sum.
Let T(s) denote the infinite powertower s^s^s^s^s... .
For s providing the convergent cases T(s) I crosschecked my
method (*4) for computing S(s) by usual Eulersummation of the terms (*1b),
where I got identity in terms of float-arithmetic approximation.
That means all s in e^-e +eps < s < e^(1/e), where eps ~ 2^-12;
this includes also s=sqrt(2).
>
> Part 2: Over the edge
>
> If Part 1 appears satisfactory, examine the sum by your method (Euler's
> will no longer work) for b = 1.45, 1.46, 1.47. Compare these results
> with the earlier ones for b = 1.43-1.444, to see if there is reasonably
> smooth behavior. If yes, you have an apparently suitable analytic
> continuation that can be applied for "large" values of b such as e.
For the coefficients bs_k in (*4) I get nicely decreasing absolute values
of alternating signs, where however the alternating behaviour is not strict.
a small table for various s
param ! coefficients sum(- bs_k)
s ! bs_0 bs_1 bs_2 bs_3 bs_4 ... S(s)
-------+---------------------------------------------------------------------
1 ! 0 -0.5 0 0 0 ... 1/2
1.0625 ! 0 -0.47 0.00076 1e-6 2e-8 ... 0.4706
1.4375 ! 0 -0.367 0.009 -6e-6 1e-5 .(decreasing) 0.3576
1.5 ! 0 -0.356 0.010 -1e-4 5e-5 .(decreasing) 0.3461
3 ! 0 -0.238 -0.002 -2e-4 6e-4 .(decreasing) 0.2368
7 ! 0 -0.170 -0.017 4e-3 2e-3 .(decreasing) 0.1795
e^2 ! 0 -0.167 -0.018 4e-3 2e-3 0.1769
e^3 ! 0 -0.125 -0.023 2e-4 3e-3 0.1421
where for higher s the dominant term is further decreasing in absolute value
and with a vague floating the subsequent terms are still decreasing, while
the rate of decrease seems to diminuish. But this may be due to the finite
dimension of my matrices and I'll check this later.
For s = exp(k) I seem to get nice approximations to rationals for bs_1, saying
bs_1 = -1/2 * 1/(k+1)
which explains then the method-internal singularity at s = exp(-1).
With decreasing s I get less bs_k which are negative, so the alternating
character disppears and at s = e^(-e) I get a series with only positive
terms, which also increase, and thus cannot be summed to a finite
value.
param ! coefficients sum(- bs_k)
s ! bs_0 bs_1 bs_2 bs_3 bs_4 ... S(s)
-------+---------------------------------------------------------------------
e^(-e) ! 0 0.291 1.332 2.369 3.210 -inf-
I couldn't reproduce the behave of the Euler-summation.
The terms to be summed are for the method (*1b), s= sqrt(2)
1 -1.4142136 2.0000000 -2.8284271 4.0000000 -5.6568542 8.0000000 -11.313708
If I apply my implementation of Eulersummation I get the following
smooth sequence of partial sums:
0.50000000 0.39644661 0.41789322 0.41345148 0.41437139 0.41418087
which stabilizes at about the 10'th term in the last displayed digit.
Here I don't know, what's going on compared to your report?
Regards -
Gottfried Helms
Am 29.06.2007 15:46 schrieb Gottfried Helms:
This was crap. Wrong parameter setting:
>
> I couldn't reproduce the behave of the Euler-summation.
> The terms to be summed are for the method (*1b), s= sqrt(2)
>
> 1 -1.4142136 2.0000000 -2.8284271 4.0000000 -5.6568542 8.0000000 -11.313708
>
> If I apply my implementation of Eulersummation I get the following
> smooth sequence of partial sums:
>
> 0.50000000 0.39644661 0.41789322 0.41345148 0.41437139 0.41418087
>
> which stabilizes at about the 10'th term in the last displayed digit.
>
It must be:
---------------------------------------------------
The terms to be summed are for the method (*1b), s= sqrt(2)
1, -1.4142136, 1.6325269, -1.7608396, 1.8409109, -1.8927127, 1.9269997, ...
If I apply my implementation of Eulersummation I get the following
smooth sequence of partial sums:
0.50000000 0.39644661 0.37195908 0.36534037 0.36333599 0.36267423
which stabilizes at about the 20'th term in the last displayed digit
to 0.36230155
-----------------------------------------------------
Oscar Lanzi III
Based on the numerical values from your table it ap[pears that the
Shanks and Helms methods agree across s = exp(1/e) as was originally
demanded. Thus the Helms method appears to be valid for the original
problem with s = e.
We seem to agree that s = exp(-e) represents a "true" singularity where
convergence breaks down. Numerical investigation of the sums as s -->
exp(-e)+ indicates not a pole as I had first conjectured, but a branch
point; if s = exp(-e)+t then the sum grows in proportion to t^(-1/2)
(try it!). This accounts for the lack of analytic continuability. In
the other direction your bs_0 values point to an asymptotic approach to
1/(2*ln s+2) for large s. Already for s = e we are rather close to this
limit.
--OL
Gottfried Helms
Very good! I'll study these informations to add them
to the article.
So it seems, that this method has a resonable degree
of verification, to consider its emission to the
"scientific knowledgebase".
How to proceed?
a) I feel not competent to prove the appropriateness
of equivalence of the sum of powers of B and the
inversion of (I+B). If this is indeed needed (what
I assume) then proving the general summability of
B^0 - B^1 + B^2... must be done in terms of growthrate-
analyses for the entries of such a sum of matrices,
and already for the individual powers of B and its
parametrized versions(*1).
Although I've read the books of G.H.Hardy and K.Knopp
about infinite, divergent series, I feel not to be
experienced enough to apply the required limiting
criteria here to a sufficient extent.
Here I needed help, or I have to leave that whole
part open for further survey.
b) I don't have a clue about the requirements for
submitting articles, and even about journals, which
may be interested in that subject.
Also here I needed hints and advice.
c) ???
Regards -
Gottfried
*) For the interesting second column of B^2 I found
it consisting of Bell-numbers with inverse factorial
scaling, which is meaningful, since I know that
the Bell-numbers define the powerseries in x for
the computation of exp(exp(x)).
Since these Bell-numbers can also be computed by
exponentiaton of the pascal matrix P, this indicates,
that iterated exponentiation of P (which is itself
the first iterate of such an exponentiation then)
may describe the behave of this critical column in
the powers of B as well.
----
Gottfried Helms, Kassel
Gottfried Helms
>
> a) I feel not competent to prove the appropriateness
> of equivalence of the sum of powers of B and the
> inversion of (I+B). If this is indeed needed (what
> I assume) then proving the general summability of
> B^0 - B^1 + B^2... must be done in terms of growthrate-
> analyses for the entries of such a sum of matrices,
> and already for the individual powers of B and its
> parametrized versions(*1).
I seem to have a solution for this problem. The
entries of powers of B (as well as of the parametrized
version) can, though B itself is a non-triangular-matrix
of infinite size, apparently be described by exponentials
of nilpotent triangular pascal-matrices (P - I), so
they seem to be provable of finite size. If that holds,
then one can say that they all are expressible as
finite combinations of logarithms and exponentials,
and the equivalence of M and the alternating sum of
powers of B and Bs can (principally) be settled.