Hi -

I plotted a graph, which shows y = x^(1/x) , where y has arbitrary, but purely real value, irrespectively of the bounds e^(-e)..e^(1/e).

Surely plots like this exist elsewhere but I just tried...

Each line has a fixed real value in x, and increasing values in the imaginary part. The pure real values y of the function were found by binary search.

The y-values are strongly scaled by the asinh-function, so the real values would be sinh( <plotted-y> ) .

The lines may be a bit misleading: along the lines there is *no* connection of real values (the function value is complex); only the dots are purely real. Looking at neighboured dots over the different lines, however, this seem to indicate a continuous trace this way.

The plot seems to indicate, that all real values occur infinitely often, always at a continuous line of the contours (but possibly each contour is bounded, don't know yet)

I would like to implement a tracer, which finds the real-valued contours, but I don't see at the moment, how to do that.

What do you think?

Gottfried

I plotted a graph, which shows y = x^(1/x) , where y has arbitrary, but purely real value, irrespectively of the bounds e^(-e)..e^(1/e).

Surely plots like this exist elsewhere but I just tried...

Each line has a fixed real value in x, and increasing values in the imaginary part. The pure real values y of the function were found by binary search.

The y-values are strongly scaled by the asinh-function, so the real values would be sinh( <plotted-y> ) .

The lines may be a bit misleading: along the lines there is *no* connection of real values (the function value is complex); only the dots are purely real. Looking at neighboured dots over the different lines, however, this seem to indicate a continuous trace this way.

The plot seems to indicate, that all real values occur infinitely often, always at a continuous line of the contours (but possibly each contour is bounded, don't know yet)

I would like to implement a tracer, which finds the real-valued contours, but I don't see at the moment, how to do that.

What do you think?

Gottfried

Gottfried Helms, Kassel