What is the surface area of the Mandelbrot set?

[matsoljare]

Or any given M or J set, for that matter... perhaps what i should ask is, how can it be computed?

[Krumel]

As far as I know, the only way is pixel-counting.
(except for the julia set for 0, which has obviously an surface of 2pi)

[stigomaster]

Because they are fractals, I believe the boundary is not measurable unless you measure in fractional dimensions. The boundary of the Mandebrot set has fractal dimension 2 and so you can not measure the length but rather the area

[Tglad]

1.50659177 roughly, its in wikipedia.
As for calculating them, you could box count, or calculate the area of the level sets at ever larger iterations.

[mrob]

I am Robert Munafo, and I did the work back in the 1990's and in 2003 to arrive at the estimate 1.50659177 +- 0.00000008. The error term is the standard error of the mean based on 20 independent surveys on grids of approximately 262144x524288 pixels, with max iterations 67108864.

To get useful statistical data, the grids are all staggered slightly and use slightly different row spacings. More details on my pixel counting page: http://mrob.com/pub/muency/pixelcounting.html A brief summary of early work on the area estimate is here: http://mrob.com/pub/muency/areahistory.html

I am currently running the program again on modern equipment after updating it for the 64-bit runtime model and adding multithreading capability. The program saves its work to protect against power failures. It is currently working on the 524288x1048576 grid size, using max iterations of 268435456, and I expect those results by about 10 days from now (Halloween!)

[lkmitch]

Back in 2001, I got 1.506484:

http://www.kerrymitchellart.com/articles/area/mandelbrot-area.html

[mrob]

lkmitch --

It's great to see you here on the forum!

I realize now that I left you out of my area history for years (because it's out of date  ) so I will rectify that situation now ...