### (Question) Help with .ucl files

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#### JimHedron

• Fractal Freshman
• Posts: 3

#### Help with .ucl files

« on: February 05, 2019, 05:44:21 AM »
Hi.  I'm trying to create a ucl file to direct colour 'colour-by-root' on a Newton Raphson iteration.    Basically I do a normal iteration until z stabilises then in the Final section of a ucl file I examine z against each root and set a #color accordingly.   Whatever I do though the resulting fractal is totally transparent (in both UF5 and UF6).  I'm missing something.  Any help greatly appreciated.

Jim.

• 3f
• Posts: 1915

#### Re: Help with .ucl files

« Reply #1 on: February 05, 2019, 02:29:47 PM »
a) Make sure the alpha in your #color isn't set to transparent.

b) Rather than check for exactly equaling a root check for hitting a small radius around it with |#z - root| < 0.001 or similar.

#### JimHedron

• Fractal Freshman
• Posts: 3

#### Re: Help with .ucl files

« Reply #2 on: February 06, 2019, 06:25:24 AM »
Thank-you pauldebrot.  I'm still not sure what went wrong, but I went back to basics and wrote a .ucl file to colour the entire screen red and then worked up from there.   Image of the finished product attached.

• 3f
• Posts: 1915

#### Re: Help with .ucl files

« Reply #3 on: February 06, 2019, 03:47:42 PM »
Interesting. Looks like your iteration isn't starting on a critical value, though.

#### JimHedron

• Fractal Freshman
• Posts: 3

#### Re: Help with .ucl files

« Reply #4 on: February 07, 2019, 05:38:51 AM »
The image is a heavily modified Newton/Halley iteration.   It started with an experiment.  As the Halley method is faster than the Newton, I wondered if the second order term in the Halley iteration was capable of finding the roots by itself.  Basic formula  z(new) = z(old) + Halley - Newton.    I term this an 'incremental Halley'.   I then explored the effect of adding a relaxation factor R,  z(new) = z(old) + R(Halley - Newton).   The iteration proved robust over quite a wide range of R.

I then tried varying R with pixel.  This resulted in the right hand side of the above image.  It is dominated by the root at (1,0) so to resolve this I also tried R = -pixel.  This is then the left side.  There is some overlap.  The three roots can be seen in both images.

I've also taken things further by looking at R = fn(pixel) which gives some nice results.

Jim

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