Thread Rating:
  • 1 Vote(s) - 3 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#47
bo198214 Wrote:I am not sure what you mean. The method is about repeated applying of and not of . You can not pull out the .


May be I am trying to say something useless, but I am sure, that in the region where h(z) does not converge, and may be even in the region where it does converge( e.g h(sqrt(2))=2 or 4 ) e.g h(e^pi/2) You can not limit yourself to only one root of -1, but You have to use both separately, or information is lost. So the real way to get accurate results is to study both rotations anticlokwise (i) and clockwise (-i) independently. I am not sure computer programs handle it that way. I will try once more:

Knoebel (1981) gave the following series for

h(z) = 1+ lnz + 3^2/3!*(lnz)^2 .........

Now, if we take h(e^(pi/2) we know that answers will be +i, -i.

ln(e^pi/2) = pi/2+- 2*pi*i*k -> I stress +-, k >0.

If You put this in series above( I know they do not coverge, but they might give h(z) analytic continuation), you will get :

h(z,k) = SUM (( (n^n-1)/n!) * ( pi/2+-2pi*I*k)^n-1 or
h(z,k) = SUM (((n^n-1)/n!)* pi/2^(n-1)* (1+-4*I*k)^n-1

Therefore, if we get h(z) = i when k = 1, than we must get h(z) =-i when with k =-1. If k=0, we have a special case, as we can not get i in the series if k=0.

For k=2 and -2 the values of h(z) again may be complex conjugates , but may be not.

For k=3 and -3 again the imaginary values will definitely differ by sign because values for 1 and - 1 were complex conjugates.

So basically for each pair of k +- You will get 2 values of h(z) which will be complex conjugates, but sometimes without imaginary part when they converge. Then there will seem to be only 1 value, a real number.But not all cases.

As a conjecture, perhaps h (sqrt2) will have one real value at from ln(sqrt2) and another from ln(-sqrt2).
h(cube root 2) may have 3 values, h(4th root of 2) may have 4 converging values etc.
Reply


Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/13/2007, 11:01 PM
RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM
RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM
RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM

Possibly Related Threads...
Thread Author Replies Views Last Post
  b^b^x with base 0<b<e^-e have three real fixpoints Gottfried 1 2,122 11/07/2017, 11:06 AM
Last Post: sheldonison
  2 real fixpoints again ....... tommy1729 10 8,446 02/23/2016, 10:17 PM
Last Post: tommy1729
  A new set of numbers is necessary to extend tetration to real exponents. marraco 7 8,660 03/19/2015, 10:45 PM
Last Post: marraco
  Real-analytic tetration uniqueness criterion? mike3 25 19,867 06/15/2014, 10:17 PM
Last Post: tommy1729
  About real limits tommy1729 1 2,278 09/23/2013, 09:24 PM
Last Post: tommy1729
  Solutions to f ' (x) = f(f(x)) ? tommy1729 1 2,183 08/12/2013, 12:10 AM
Last Post: tommy1729
  Real and complex behaviour of the base change function (was: The "cheta" function) bo198214 39 48,976 08/13/2011, 06:33 PM
Last Post: bo198214
  The imaginary tetration unit? ssroot of -1 JmsNxn 2 5,187 07/15/2011, 05:12 PM
Last Post: JmsNxn
  Tetration and imaginary numbers. robo37 2 5,166 07/13/2011, 03:25 PM
Last Post: robo37
  Infinite tetration of the imaginary unit GFR 40 51,982 06/26/2011, 08:06 AM
Last Post: bo198214



Users browsing this thread: 1 Guest(s)