Hi -

although I wanted to be a bit detached for some weeks and look into the basics of the divergent summation thing, I've just made a big step to backup my eigensystem-decomposition by a compeletely elementary iterative method, and I want to share this.

The symbolic eigensystem-decomposition is enormous resources-consuming and also not yet well documented and verified in a general manner. Here I propose a completely elementary (while still matrix-based) way to arrive at the same result as by eigensystem-analysis.

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

As you might recall from my "continuous iteration" article I've found a matrix representation for the coefficients of the powerseries for the U-tetration

, where

.

Here I'll extend the notation for the powerseries a bit, compared to the text in the article:

Let me, as earlier, use the notation u=log(t) for this text.

Then

* the b_k are powers of u (but of no special interest here),

* d_k are products of (u^j-1), like

and

* a_k(t,h) are bivariate polynomials, whose numerical coefficients are the sole mysteries in this powerseries.

Since we have two formal variables in each a_k, the coefficients can be arranged in matrices, whose row- and column-multiplieres are the consecutive powers of the parameters.

For instance a3(t,h) looks like the following:

The 3! removed from denominator makes this

d3 is the common denominator, it is

This removed it is

Also b3 = u^2 removed, this is

Now the numerical coefficients of this polynomial can be arranged in a matrix

Code:

```

A3 = 2 -3 1

1 -2 2

where we understand, that the columns are to be multiplied by consecutive powers of v=u^h, beginning at 1 and the rows by consecutive powers of u, beginning at 0:

Code:

```

A3 = [ 2 -3 1 ]*u^0

[ 1 -3 2 ]*u^1

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

*v *v^2 *v^3

The value of this term, given u and h, is then simply the appropriate built rowsums added.

Now v is u^h and we have the interesting relation, depending on h that any integer height h provides a columnshift by h rows. So if h=0 we have no shift

Code:

```

A3 = [ 2 -3 1 ]*u^0

[ 1 -3 2 ]*u^1

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

*1 *1 *1

and we have simple rowsums of zero (and also all A matrices provide coefficients of zero to the original powerseries the same way).

If h=1 we have

Code:

```

A3 = [ 2 -3 1 ]*u^0

[ 1 -3 2 ]*u^1

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

*u *u^2 *u^3

and in effect, this provides a column-shift

Code:

```

A3 = [ 2 ]*u^1

[ 1 -3 ]*u^2

[ -3 1 ]*u^3

[ 2 ]*u^4

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

= u(2 - 2u - 2u^2 + 2u^3 )

= 2u(u - 1) *(u^2 - 1 ))

= 2u*d3

and we have another sum, where also d3 cancels the denominator.

With h=2 we have the most interesting case (concerning a_3)

Code:

```

A3 = [ 2 ]*u^2

[ 1 ]*u^3

[ -3 ]*u^4

[ -3 ]*u^5

[ 1 ]*u^6

[ 2 ]*u^7

= u^2(2 + u - 3u^2 - 3u^3 + u^4 + 2u^5)

= u^2*(u - 1)*(u^2-1)* (2*u^2 + 3*u + 2)

= u^2 * d3 * (2*u^2 + 3*u + 2)

where again d3 cancels the denominator.

From inspection of the first few A-matrices I concluded, that for each

integer h the numerical coefficients of all A-matrices cancel the denominator. This even allows u=1, t=exp(1) and via fixpointshift the tetration-bases b=e^(1/e)) for integer heights, since the zero-denominators cancel.

It is interesting concerning fractional heights, for instance if h is half-integer, all these column-shifting has a special consequence. For h=1/2 we have

Code:

```

A3 = [ 2 ]*u^0.5

[ -3 ]*u^1.0

[ 1 1 ]*u^1.5

[ -3 ]*u^2.0

[ 2 ]*u^2.5

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

where the cancelling of the denominators cannot occur. This may give a feeling for the mysteries of fractional iteration.

========================================================================

But this is only for recalling some known things. This msg is intended to cover another aspect.

The symbolic eigen-decomponsition, which gives us the A-matrices, is extremely costly in time and memory, so I was searching for another method to determine the A-matrices independently.

My idea was the striking property, that the coefficients seem to guarantee, that for integer heights the denominator is always a factor of the numerator.

So the A-matrices must be expressible in some multiples of the denominators d, let's express their polynomial-coefficients as vector D. If these multiples can be determined apriori then the costly eigenvalue-decomposition is not needed.

Yesterday I found the solution (with the help of members of the seqfan-mailing list), and this is an extremely simple routine.

We use the same example A_3 here, but since we don't want to use the eigendecomposition, all numerical coefficients in A are unknown. So we have

Code:

```

RS= // row-sums

A3 = [ a0 -a1 a2]*u^0 [ ?]*u^0

[ b0 -b1 b2]*u^1 [ ?]*u^1

[ 0 0 0]*u^2 [ 0]*u^2

[ 0 0 0]*u^3 [ 0]*u^3

//I've extended the rows to get

// compatibility with the following

// matrix-operation

and the row-sums must equal an integer composition of D3 = column([1,-1,-1,1). We know by 3'rd and 4'th row, that the row-sums RS are zero, so in

the unknown k0 must be zero.

The next condition exploits the properties at h=1 since we have

Code:

```

RS =

A3 = [a0 ]*u^1 [ ?]*u^1

[b0 -a1 ]*u^2 [ ?]*u^2

[ -b1 a2]*u^3 [ ?]*u^3

[ b2]*u^4 [ ?]*u^4

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

and we have that

but have no information about k0 here. From Eigenanalysis we know, that k0=1

The next condition, using h=2 is

Code:

```

RS=

A3 = [ a0 ]*u^2 [ a0]*u^2

[ b0 ]*u^3 [ b0]*u^3

[ -a1 ]*u^4 [-a1]*u^4

[ -b1 ]*u^5 [-b1]*u^5

[ a2]*u^6 [ a2]*u^6

[ b2]*u^7 [ b2]*u^7

and since this must again be an integer composition of the polynomial d3, we may express this by the matrix-multiplication

where MD is the matrix built from D3 with column-shift

Code:

```

MD3 = [ 1 . . ] K3= [k0]

[-1 1 . ] [k1]

[-1 -1 1 ] [k2]

[ 1 -1 -1 ]

[ . 1 -1 ]

[ . . 1 ]

and K3 must be a vector now with integer weigths for each column in MD3

If K3 would be known, the result MD3 * K3 must equal RS, and from RS we can uniquely determine the unknown coefficients a0 to b2.

So, if the analoguous K-vectors for all A were known, also all numerical coefficients in the A-matrices were immediately known.

This is the concept of my new solution.

I collected some K-vectors into a matrix (from the known eigensystem-based solutions) to find a recursive generation rule for this matrix and posted the first nontrivial K-matrix into the mailing list, and a member (Richard Mathar) indeed found a computation-rule for this first K-matrix (there called "H"-matrices) , which I could systematize and generalize to a very simple iteration-process for all K- and then subsequently A-matrices analoguously.

Here an excerpt of my concluding mail to seqfan-list (I'm a bit tired of typing, but edited it a bit for this msg...)

Code:

```

==========================================================

a) Assume a matrix-function "rowshift(M)" which

computes "M1 = rowshift(M)" in the following way:

M = [a,b,c,...]

[k,l,m,...]

[r,s,t,...]

[... ]

M1 = [a,b,c, ... ]

[0,k,l,m, ... ]

[0,0,r,s,t,...]

[ ... ]

b) Assume the lower-trianguler matrix of Stirling-numbers 2'nd kind

S = [1 0 0 0 ...]

[1 1 0 0 ...]

[1 3 1 0 ...]

[1 7 6 1 ...]

[ ... ]

c)then with

H0 = [1]

[1]

[1]

[1]

...

we have the iterative Mathar-products (*1)

d)

H1 = S * rowshift(H0) \\ which is then = S * I = S

H2 = S * rowshift(H1)

H3 = S * rowshift(H2)

...

where

H1 =

1 . . . .

1 1 . . .

1 3 1 . .

1 7 6 1 .

1 15 25 10 1

H2=

1 . . . . . . . .

1 1 1 . . . . . .

1 3 4 3 1 . . . .

1 7 13 19 13 6 1 . .

1 15 40 85 96 75 35 10 1

H3=

1 . . . . . . . . . . . .

1 1 1 1 . . . . . . . . .

1 3 4 6 4 3 1 . . . . . .

1 7 13 26 31 31 25 13 6 1 . . .

1 15 40 100 171 220 255 215 156 85 35 10 1

...

and so on

(based on the Maple-implementation of R.Mathar)

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

In my basic problem-description I also had the vector D, which I also

should index now. It contains the coefficients of the polynomials in u:

e) Dk = polcoeffs(prod(j=1,k -1, u^j - 1))

Say

D3 = columnvector([1 -1 -1 1])

f) Then define the matrix MD3 as the concatenation of shifted D3

MD3 =

1

-1 1

-1 -1 1

1 -1 -1

1 -1 ...

1

...

up to the required dimension for the following matrix-multiplication

to obtain the k'th coefficient in the original powerseries

Ut(x,h) = a1(t,h) b1/d1/1!*x + a2(t,h)*b2/d2/2!*x^2 + ...

Then the coefficients for the bivariate polynomial in u=log(t) and

v=u^h of the k'th coefficient are in the vector

g) Vk = MDk * transpose(Hj[k,])

where [k,] denotes the k'th row and the index j at Hj indicates the

j=binomial(k-1,2) 'th Mathar-iterate.

So we have three essential steps to compute one A-matrix:

1.a) compute the D_k-vector as "polcoeffs(prod(j=1..k-1,u^j - 1))"

1.b) compute the MD_k-matrix by concatenation/shifting of D_k

2) compute the m'th iterate H_m(), where m=binomial(k-1,2), use the

k'th row of the result as K-vector K_k = H_m[k,]

3.a) compute RS_k = MD_k * transpose(K_k)

3.b) reformat the vector RS_k into a matrix A_k of k columns

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

This is an eminent simple recursive scheme to obtain the coefficients

for the continuous iteration of x -> t^x-1.

However it is again enormous consumptive: the required iterations of

the Mathar-products is quadratic (precisely:binomial(index-1,2)) with

the index. So for index k=20 I need already 171 iterations with

always growing matrices... surely some shortcuts may be implemented.

The working load is in the number of columns of the H-matrices: the

final number of columns for k=20 is 1/2*k*(k^2-4k+5)=3250, and grows

cubically with the index.

However - the fact, that this simple scheme is able to mimic the

symbolic eigendecomposition of a matrix-operator is a very astonishing

aspect.

========================================================================

(end of mail to seqfan)

So, for our members, who did not understand my matrx-method with eigendecomposition (or did not trust it ;-) ) here is a completely elementary approach, simple to implement.

However - now two hypotheses are involved, which need proof:

a) that indeed the property holds, that A-matrices using integer values of height h contain the denominators d_k as factor

b) that the Mathar-process indeed provides the correct H-matrices.

I checked that up to h=20 and did not find an error.

I append also some A-matrices, computed by the above process. Note, they are transposes compared to my mentioned text about "continuous iteration"

Gottfried

========================================================================

Code:

```

D_2= (vector)

-1 1

K_2= (vector)

1

A_2= (matrix)

-1 1

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

D_3= (vector)

1 -1 -1 1

K_3= (vector)

1 3 1

A_3= (matrix)

1 -3 2

2 -3 1

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

D_4=

-1 1 1 0 -1 -1 1

K_4=

1 7 13 26 31 31 25 13 6 1

A_4=

-1 7 -12 6

-6 18 -18 6

-5 18 -18 5

-6 11 -6 1

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

D_5=

1 -1 -1 0 0 2 0 0 -1 -1 1

K_5=

1 15 40 100 186 310 490 705 921 1140 1315 1435 1481 1420

1285 1105 886 660 455 285 166 85 35 10 1

A_5=

1 -15 50 -60 24

14 -75 145 -120 36

24 -130 230 -170 46

45 -180 275 -180 40

46 -165 215 -120 24

26 -105 130 -60 9

24 -50 35 -10 1

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

D_6=

-1 1 1 0 0 -1 -1 -1 1 1 1 0 0 -1 -1 1

K_6=

1 31 121 366 861 1642 2982 4932 7727 11497 16628 23127

31277 40937 52147 64612 78297 92497 107162 121451 135002 146787

156632 163631 167871 168862 166802 161541 153616 142981 130527

116621 102186 87531 73486 60166 48101 37376 28236 20635 14656

10026 6611 4130 2440 1306 636 255 80 15 1

A_6=

-1 31 -180 390 -360 120

-30 270 -870 1290 -900 240

-89 694 -1920 2515 -1590 390

-214 1364 -3345 3905 -2190 480

-374 2025 -4440 4825 -2550 514

-416 2395 -4995 4925 -2325 416

-511 2430 -4530 4110 -1800 301

-461 2006 -3480 2885 -1110 160

-330 1336 -2055 1495 -510 64

-154 675 -960 575 -150 14

-120 274 -225 85 -15 1

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

D_7=

1 -1 -1 0 0 1 0 2 0 -1 -1 -1 -1 0 2 0 1 0 0 -1 -1 1

K_7=

1 63 364 1316 3857 8540 17522 32676 56763 92722 145565

219590 321413 457324 635782 865844 1158395 1523599 1973805

2519790 3172421 3942099 4837273 5864971 7030269 8336258 9781520

11364262 13075818 14906199 16838921 18855277 20927928 23031974

25132905 27201587 29200578 31099439 32859715 34454231 35847014

37015524 37930670 38578071 38937003 39005785 38775716 38259081

37461558 36406349 35109662 33604207 31912699 30072889 28112666

26071493 23978150 21871710 19778472 17733072 15757701 13878746

12110168 10468493 8959111 7590492 6361384 5272659 4318307 3494092

2790081 2198385 1707203 1306193 983171 727447 527625 374374

258812 173747 112567 69951 41279 22883 11698 5369 2142 686 161

21 1

A_7=

1 -63 602 -2100 3360 -2520 720

62 -903 4501 -10500 12600 -7560 1800

300 -3290 13300 -26740 28700 -15750 3480

889 -8337 29778 -53340 51590 -25830 5250

2177 -16485 52269 -86415 78050 -36624 7028

3368 -25977 78218 -121345 103040 -45360 8056

5188 -36421 102242 -148715 118615 -49161 8252

6980 -44604 117719 -161840 121800 -47481 7426

8007 -48538 120967 -156765 110985 -40635 5979

7867 -46599 110173 -134960 90160 -30849 4208

7188 -40215 89656 -102620 63525 -20076 2542

6111 -30723 63357 -67585 38885 -11340 1295

4270 -20216 39081 -38220 19600 -5019 504

2528 -11193 19355 -16695 7525 -1659 139

1044 -4872 7658 -5425 1890 -315 20

720 -1764 1624 -735 175 -21 1

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

=======================================================================