complex iteration (complex "height")

[Gottfried]

with higher precision I recomputed the value and got in the 92..96 partial-sums the following approximations:

Code:

´
  ...
  1.210309025559961+0.5058275713618201*I
---------------------------------------------
  1.210309011      + .5058275611          *I  <--- yours

So this differs from the 7'th digit;

Haha but I just used low precision so here is my 20 digits result:

Code:

1.2103090255599614766+.50582757136182013605*I

update: once for later time, 50 Digits! You see the last some digits are always unreliable.

Code:

1.2103090255599614779588104735397176784037341102467+.50582757136182013700589565226517951794360523253345*I

The last few partial sums computed with internal precision of 400, displayed precision 50, Euler-sum of order 2.538, where order=1 means direct summation without transformation (the last few of partial sums of 128 terms now)

Code:

´
   1.21030902555996147 05286038734660188979105600443644+0.50582757136182012 502296304591137652670054111308587*I
   1.21030902555996147 44685314155510971433670516046106+0.50582757136182012 835671840643903386779427797897960*I
   1.21030902555996147 66583812490070529178600987551324+0.50582757136182013 098572869953873370073216724647955*I
   1.21030902555996147 77865734208122443540107642898914+0.50582757136182013 294937569862301423395913769512951*I
   1.21030902555996147 82987507553564562088347731401270+0.50582757136182013 435542605993548055302116773994320*I
   1.21030902555996147 84733868997136185883982264653880+0.50582757136182013 532695350500506166087181069651300*I
   1.21030902555996147 84778386937626628200500840930051+0.50582757136182013 597695153820676613393345071380896*I
   ------------------------------------------------------------------------------------------------------------------------
   1.21030902555996147 79588104735397176784037341102467+ .50582757136182013 700589565226517951794360523253345*I <--- your value

The non-vanishing increase means, I'd use more precise Euler-order of a bit higher order. I'll see, whether I can find a better transformation.

[update] A better transformation uses a Stirling kind 2 transformation first, which is also regular. Then I don't need high Euler-orders, and get for example the last few partial-sums (128 terms):

Code:

´
  1.210309025559961477958810473539717649 4269005980356+0.5058275713618201370058956522651794 9881594263547436*I
  1.210309025559961477958810473539717660 1745711120037+0.5058275713618201370058956522651795 1511709385013158*I
  1.210309025559961477958810473539717668 9726354895139+0.5058275713618201370058956522651795 2061256599110112*I
  1.210309025559961477958810473539717674 4384709020618+0.5058275713618201370058956522651795 2132025558347280*I
  1.210309025559961477958810473539717677 2354562276144+0.5058275713618201370058956522651795 2045256215306036*I
  1.210309025559961477958810473539717678 3946530537033+0.5058275713618201370058956522651795 1941438348397801*I
  1.210309025559961477958810473539717678 7249247611291+0.5058275713618201370058956522651795 1865625158284700*I
-------------------------------------------------------------
  1.210309025559961477958810473539717678 4037341102467+ .5058275713618201370058956522651795 1794360523253345*I <--- your value

Again, the change of sign of change in differences in partial sums shows, that the summation is still not optimal.