Hmmm, just to get more familiar to it....
\( z=z^z \)
\( 0=z^z-z \)
\( 0=z \cdot (z^{z-1} - 1) \)
\( 0= f^{\circ 2}(1) \)
where
. \( f(x)=z^x-1 \)
To look for zeros it might be helpful to list zeros of the real- and the imaginary parts separately first.
update: the Wolfram-alpha-contourplots for z^(z-1)-1 from 1+I to 10+10*I (separate for real and for imaginary parts) give zeros on continuous lines, and an overlay seems to indicate more systematical zeros at the crossings of the lines.
Some small programming in Pari/GP gave the following additional solutions
which can easily be prolonged because the pairwise distances seem to approximate a linear scheme. Of course the conjugate numbers are also roots.
This does not exclude further roots, I would especially look for them in the angle indicated by the interpolation curve through the roots (and their conjugates) with the real axis
update 2: I've computed the regression-line for the progression of the real parts of the roots and for the imaginary parts of the roots.
This reproduces the given roots for n=1 to 14 perfectly and gives guesses for the next couple of roots. Such further guesses should be improved by an application of the Newton-rootfinder algorithm, of course.
update 3: picture based on first 41 complex roots (41=1+2*20):
update 4: contourplot of imaginary zeros (black lines) and real zeros (white lines) for z^{z-1}-1 . The smallest known 9 roots from the previous picture are dotted with red color. The image suggests there are *no more* zeros besides that on the curce determined in the previous plot, because the black and white lines have pairwise only one crossing.
\( z=z^z \)
\( 0=z^z-z \)
\( 0=z \cdot (z^{z-1} - 1) \)
\( 0= f^{\circ 2}(1) \)
where
. \( f(x)=z^x-1 \)
To look for zeros it might be helpful to list zeros of the real- and the imaginary parts separately first.
update: the Wolfram-alpha-contourplots for z^(z-1)-1 from 1+I to 10+10*I (separate for real and for imaginary parts) give zeros on continuous lines, and an overlay seems to indicate more systematical zeros at the crossings of the lines.
Some small programming in Pari/GP gave the following additional solutions
Code:
2.863+3.223*I
3.727+5.318*I
4.433+7.194*I
5.057+8.947*I
5.627+10.62*I
6.160+12.23*I
6.664+13.79*I
7.144+15.31*I
7.606+16.80*I
8.051+18.27*I
8.482+19.70*I
8.902+21.12*I
9.310+22.51*I
9.710+23.89*I
...This does not exclude further roots, I would especially look for them in the angle indicated by the interpolation curve through the roots (and their conjugates) with the real axis
update 2: I've computed the regression-line for the progression of the real parts of the roots and for the imaginary parts of the roots.
Code:
Re(n) = 1.207 E-12*n^13 - 1.275 E-10*n^12 + 0.000000006121*n^11 - 0.0000001769*n^10 + 0.000003433*n^9 - 0.00004727*n^8 + 0.0004760*n^7
- 0.003558*n^6 + 0.01987*n^5 - 0.08292*n^4 + 0.2585*n^3 - 0.6192*n^2 + 1.714*n + 1.576
Im(n) = 1.340 E-12*n^13 - 1.416 E-10*n^12 + 0.000000006805*n^11 - 0.0000001968*n^10 + 0.000003824*n^9 - 0.00005274*n^8 + 0.0005321*n^7
- 0.003990*n^6 + 0.02239*n^5 - 0.09412*n^4 + 0.2979*n^3 - 0.7428*n^2 + 3.151*n + 0.5923
est_root(n)= Re(n)+ I*Im(n)This reproduces the given roots for n=1 to 14 perfectly and gives guesses for the next couple of roots. Such further guesses should be improved by an application of the Newton-rootfinder algorithm, of course.
update 3: picture based on first 41 complex roots (41=1+2*20):
update 4: contourplot of imaginary zeros (black lines) and real zeros (white lines) for z^{z-1}-1 . The smallest known 9 roots from the previous picture are dotted with red color. The image suggests there are *no more* zeros besides that on the curce determined in the previous plot, because the black and white lines have pairwise only one crossing.
Gottfried Helms, Kassel