Title: Julia Set of Riemann Zeta Function
Post by: Teenie on April 10, 2015, 03:23:14 AM
Does anyone know how to plot the Julia Set of the Riemann Zeta Function in Mandelbulber?
I figured out how to plot a still picture in Matlab but I need something a little more detailed and interactive for my project. I found several fractal programs but they either don't recognize the zeta function or they are not compatible with Windows 7. Any suggestions would be great.
I figured out how to plot a still picture in Matlab but I need something a little more detailed and interactive for my project. I found several fractal programs but they either don't recognize the zeta function or they are not compatible with Windows 7. Any suggestions would be great.
Title: Re: Julia Set of Riemann Zeta Function
Post by: DarkBeam on April 10, 2015, 09:33:40 AM
Dear Teenie,
Riemann Zeta is not a "Mandelbrot Fractal", unless you are able to completely rewrite it in a proper iterative manner, which it's not an "easy" task, still supposing it really exists.
It's just a function over complex numbers, like atan(a+i*b), with peculiar features.
As far as I know of course
:)
Luca
Riemann Zeta is not a "Mandelbrot Fractal", unless you are able to completely rewrite it in a proper iterative manner, which it's not an "easy" task, still supposing it really exists.
It's just a function over complex numbers, like atan(a+i*b), with peculiar features.
As far as I know of course
:)
Luca
Title: Re: Julia Set of Riemann Zeta Function
Post by: kram1032 on April 10, 2015, 10:57:30 AM
The Riemann Zeta function simply is the (analytic continuation of the) series
.
If you want to only look at values for
, then simply sum this up to an arbitrary point. The value will be very close very soon.
If you want to do the complex continuation, there are several formulae you can use for that purpose here http://mathworld.wolfram.com/RiemannZetaFunction.html
None of this is a Mandelbrot set though.
I tried in Mathematica once what would happen if you were to do the well known iteration for the M-Set with the Zeta function and if I recall correctly the results were not particularly interesting (and also rather slow).
If you want to only look at values for
If you want to do the complex continuation, there are several formulae you can use for that purpose here http://mathworld.wolfram.com/RiemannZetaFunction.html
None of this is a Mandelbrot set though.
I tried in Mathematica once what would happen if you were to do the well known iteration for the M-Set with the Zeta function and if I recall correctly the results were not particularly interesting (and also rather slow).
Title: Re: Julia Set of Riemann Zeta Function
Post by: DarkBeam on April 10, 2015, 11:09:57 PM
Another approach can be to use an approximation of Zeta using a suitable polynome, but you get a fractal that tends to fade outside the convergent area... Just a speedup tip by the way - and you need to find it first :)
Btw you find more details here...
http://math.stackexchange.com/questions/3271/how-to-evaluate-riemann-zeta-function
Btw you find more details here...
http://math.stackexchange.com/questions/3271/how-to-evaluate-riemann-zeta-function
Title: Re: Julia Set of Riemann Zeta Function
Post by: laser blaster on April 10, 2015, 11:35:46 PM
Th Riemann Zeta function does indeed have Mandelbrot and Julias sets with some pretty interesting patterns, see here:
http://www.dhushara.com/DarkHeart/geozeta/zetageo.htm (http://www.dhushara.com/DarkHeart/geozeta/zetageo.htm)
But I don't know if rendering these sets is a simple as the quadratic Mandelbrot. It could have multiple critical points, or even infinite critical points, and the "escape" criteria can become much more complicated than just checking if a point escapes beyond a certain radius, for instance, you might have to check for convergence to a cycle of points. I haven't studied it much, so I don't know the details.
http://www.dhushara.com/DarkHeart/geozeta/zetageo.htm (http://www.dhushara.com/DarkHeart/geozeta/zetageo.htm)
But I don't know if rendering these sets is a simple as the quadratic Mandelbrot. It could have multiple critical points, or even infinite critical points, and the "escape" criteria can become much more complicated than just checking if a point escapes beyond a certain radius, for instance, you might have to check for convergence to a cycle of points. I haven't studied it much, so I don't know the details.
Title: Re: Julia Set of Riemann Zeta Function
Post by: kram1032 on April 10, 2015, 11:39:48 PM
DarkBeam
Where do you see a polynomial?
The Riemann Zeta function is not defined via a Taylor Series. None of those serials in that math.stackexchange answer are Taylor. They are all exponential serials.
This is likely also the reason that the MSet of the Zeta function doesn't look particularly interesting: It features a very different kind of convergence to a typical Taylor series.
A Taylor series would be convergent around a point up to a certain radius (the radius of convergence)
An exponential series instead converges in a half plane. - For the Zeta function, that's the entire half plane past}\gt 1)
This convergence property is what makes the Zeta function not work with the normal algorithm.
Presumably, you'd have to use a bailout method that takes into account this changed convergence geometry, i.e. one that decides that points which are past a certain distance from}= 1)
will definitely diverge.
I assume that's what was done to get the pictures laser blaster showed.
Where do you see a polynomial?
The Riemann Zeta function is not defined via a Taylor Series. None of those serials in that math.stackexchange answer are Taylor. They are all exponential serials.
This is likely also the reason that the MSet of the Zeta function doesn't look particularly interesting: It features a very different kind of convergence to a typical Taylor series.
A Taylor series would be convergent around a point up to a certain radius (the radius of convergence)
An exponential series instead converges in a half plane. - For the Zeta function, that's the entire half plane past
This convergence property is what makes the Zeta function not work with the normal algorithm.
Presumably, you'd have to use a bailout method that takes into account this changed convergence geometry, i.e. one that decides that points which are past a certain distance from
I assume that's what was done to get the pictures laser blaster showed.