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Infinite tetration of the imaginary unit
#27
Hi Sheldon,

thanks for the great discussion, btw!

Now I'm looking at the problem of the integer-period, for instance, a period of four.

First I go one more step behind the ''u'' parameter. Because we want to have it with abs(u) or length =1, its real and imaginary part are just cos and sin of the same angular parameter, call it p. To have a period of four we set p=Pi/2
Now I have u,t and b dependent on the parameter p:
Code:
.
p   = Pi/2                                   ~ 1.57079632679
u_p = e^(p*I)  = cos(p) + I*sin(p)           = 1*I
t_p = exp(u)   = e^e^(p*I)                   ~ 0.540302305868 + 0.841470984808*I
b_p = exp(u/t) = e^(e^(p*I - e^(p*I))     )  ~ 1.98933207608 + 1.19328219947*I

where the expression for b_p can also be written in cos/sin-terms .

Then the powerseries for b_p^x, dependent on x, begins
Code:
b_p^x=
                                               1
              +0.841470984808 + 0.540302305868*I x^1
              +0.208073418274 + 0.454648713413*I x^2
             -0.0235200013433 + 0.164998749433*I x^3
            -0.0272351508693 + 0.0315334373045*I x^4
          -0.00799103562219 + 0.00236385154553*I x^5
         -0.00133356984257 - 0.000388077080832*I x^6
        -0.000130354483873 - 0.000149583780624*I x^7
     -0.00000360863179089 - 0.0000245376549262*I x^8
    +0.00000113568806559 - 0.00000251083074814*I x^9
  +0.000000231225619785 - 0.000000149917634174*I x^10
  +0.0000000250518630389 - 1.10873065678 E-10*I x^11
  + ...
Now we want to see more for the four-times iteration. Series having a constant term are not nice for iteration, so we recenter by the fixpoint.
We get
Code:
g(x)=b_p^(x+t)-t=                   1.00000000000*I x^1
                -0.270151152934 + 0.420735492404*I x^2
               -0.151549571138 + 0.0693578060912*I x^3
             -0.0412496873584 - 0.00588000033583*I x^4
            -0.00630668746090 - 0.00544703017386*I x^5
           -0.000393975257588 - 0.00133183927037*I x^6
          0.0000554395829760 - 0.000190509977510*I x^7
         0.0000186979725780 - 0.0000162943104841*I x^8
      0.00000272640610291 - 0.000000400959087876*I x^9
    0.000000251083074814 + 0.000000113568806559*I x^10
  0.0000000136288758340 + 0.0000000210205108896*I x^11
  ...
Let's call this function g(x), such that g(x-t)+t = b_p^x . Now the fourth iteration means the fourth power of the associated Bell-matrix and this gives then the powerseries
Code:
g[4](x)=                                    
                            1.00000000000 x^1
                                        0 x^2
                                        0 x^3
                                        0 x^4
      0.0369804674784 + 0.0209284634084*I x^5
     -0.0133357749117 + 0.0430512149364*I x^6
    -0.0314482605408 - 0.00625863317399*I x^7
     0.00325514172007 - 0.0187262354932*I x^8
     0.0111167178521 + 0.00450965240622*I x^9
  -0.00619602644164 + 0.00863859539483*I x^10
  -0.00808355752859 - 0.00636782383183*I x^11
  + ...
which immediately shows, that not all x are mapped to themselves after four iterations. But now the beginning of that series looks *very* strange to me - the first four coefficients are exactly what one would expect for a precise four-step-period but with the fifth and the following it looks, as if there were some small, but systematic, distortion in that whole formula. Perhaps I've also an error anywhere, but because we actually shall not get a clean four-step-period I think the series is ok.

Hmmm - what does this tell to us now?

Gottfried



[update 1] Well, numerically this agrees with the direct computation.

Even more strange looks the formal powerseries for the sum of g[4](x) and its inverse:
Code:
(g[4](x)+g[-4](x))/2=        1.00000000000 x^1
                                         0 x^2
                                         0 x^3
                                         0 x^4
                                         0 x^5
                                         0 x^6
                                         0 x^7
                                         0 x^8
     0.00232388598570 + 0.00386972180224*I x^9
   -0.00766787431799 + 0.00722126227129*I x^10
    -0.0112186194499 - 0.00878239357562*I x^11
     0.00780725843775 - 0.0123162125380*I x^12
     0.0108462124501 + 0.00556062630806*I x^13
   -0.00254708386547 + 0.00766320056079*I x^14
  -0.00366236379688 + 0.000506041490944*I x^15
  -0.00286712784159 + 0.000171130301212*I x^16
   -0.00308230052060 - 0.00406930549958*I x^17
  ...

Now we have the first 8 coefficients zero???



[update 2] Now things become really strange. We can build a sum of the weighted iterates which seems to asymptotically gives a powerseries where only one coefficient <>0 remains at x^1. The beginning of the sum is, where the []-brackets denote the iterate and the (x)-part is left away:
Code:
.
s1(p)=((g[4]+g[-4]) - (g[8]+g[-8])/4) - ((g[16]+g[-16])/16 - (g[32]+g[-32])/64 )/16    ...

I tried to formulate this as recursion (hope I've debugged all errors)
Code:
.
let g4[k]=g[4*k]
then
s0(p)=(g4[2^p]+g4[-2^p]) / 4^p
s1(p)=(s0(2p)-s0(2p+1)) / 4^(p*2)
s2(p)=(s1(2p)-s1(2p+1)) / 4^(p*2^2)
s3(p)=(s2(2p)-s2(2p+1)) / 4^(p*2^3)
..
lim n->inf s_n(0) = a x + 0
where the value of a is not yet clear to me.
This is a very strange sum of the iterates, I'd say...

Gottfried Helms, Kassel
Reply


Messages In This Thread
Infinite tetration of the imaginary unit - by GFR - 02/10/2008, 12:09 AM
RE: Infinite tetration of the imaginary unit - by Gottfried - 06/21/2011, 08:00 PM

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