What is the area of the m-set.

[TheRedshiftRider]

Let's assume we have a bailout radius of 4 (-2 to 2 and -2i to 2i). And that it has infinite iterations.

I suppose we would first have to find the size of the cardoid, then the circles around it, and then the circles around the circles (and repeat it infinitely many times). And finally we need to find out sizes of minibrots and add those to the total.

Is there a way to find this out?

[Sockratease]

What is the actual size of the m-set?

Forty Two!   

I'm not sure this question can be answered.  Size in what sense?

...we need to find out sizes of minibrots and add those to the total.

My first instinct is to say that each minibrot is the same size as the whole set.  Aren't there infinite many minibrots?  They are impossible to count, let alone measure their size.

That's why I ask what you mean by size?  As a Mathematical Construct it is Infinite by definition.  If you mean it's boundaries, then it depends on your screen resolution, I think.

[PieMan597]

Info about guesses of area here: http://mrob.com/pub/muency/pixelcounting.html

[TheRedshiftRider]

Forty Two!  

I'm not sure this question can be answered.  Size in what sense?

My first instinct is to say that each minibrot is the same size as the whole set.  Aren't there infinite many minibrots?  They are impossible to count, let alone measure their size.

That's why I ask what you mean by size?  As a Mathematical Construct it is Infinite by definition.  If you mean it's boundaries, then it depends on your screen resolution, I think.

I would say the surface area in units of 1*1i?

In the case of the minibrots I would say relative to the main set so in that case they would be smaller. And yes, we would also have to figure out the number of minibrots for every magnification level.

Info about guesses of area here: http://mrob.com/pub/muency/pixelcounting.html

Useful link, thank you.

[valera_rozuvan]

If we are talking about the area of the Mandelbrot set [1] ... should we first try to prove that the set is locally connected [2][3]?

[1] http://mrob.com/pub/muency/areaofthemandelbrotset.html
[2] https://en.wikipedia.org/wiki/Mandelbrot_set#Local_connectivity
[3] https://en.wikipedia.org/wiki/Locally_connected_space

[lycium]

I think a better approach would be to use interval arithmetic to get exact upper and lower bounds.

Some refs:

https://webserver2.tecgraf.puc-rio.br/~lhf/ftp/doc/oral/oktobermat.pdf
http://intervalmandelbrot.codeplex.com/

[TheRedshiftRider]

If we are talking about the area of the Mandelbrot set [1] ... should we first try to prove that the set is locally connected [2][3]?

[1] http://mrob.com/pub/muency/areaofthemandelbrotset.html
[2] https://en.wikipedia.org/wiki/Mandelbrot_set#Local_connectivity
[3] https://en.wikipedia.org/wiki/Locally_connected_space

Yes, area is the correct term, I forgot that. I have not thought of local connectivity. I think that wpuld be required.

I think a better approach would be to use interval arithmetic to get exact upper and lower bounds.

Some refs:

https://webserver2.tecgraf.puc-rio.br/~lhf/ftp/doc/oral/oktobermat.pdf
http://intervalmandelbrot.codeplex.com/

That would also be possible. But wouldn't it be easier to use a formula to add up all individual details of the mandelbrot set?

[lycium]

But wouldn't it be easier to use a formula to add up all individual details of the mandelbrot set?

1. That's exactly what this is (you add up rectangular areas where it's definitely contained).
2. "Use a formula" - sure, which formula?

[Chillheimer]

I'm no math-expert, but isn't the result an irrational number, just like pi, with never ending digits? Just like in PieMans link(thx). While calculation time for the next digit rises exponentially(?)..
There is no "last digit" so you never will find the real size.

What exactly do you mean by "actual size"?

[Sockratease]

I'm no math-expert, but isn't the result an irrational number, just like pi, with never ending digits? Just like in PieMans link(thx). While calculation time for the next digit rises exponentially(?)..
There is no "last digit" so you never will find the real size.

What exactly do you mean by "actual size"?

Maybe a better question is to ask how many "Significant Figures" you want in your answer.

Measurements below "The Planck Length" could be considered overkill if you think it represents a true physical limitation to real world measurements.

[TheRedshiftRider]

1. That's exactly what this is (you add up rectangular areas where it's definitely contained).
2. "Use a formula" - sure, which formula?

Oh.. okay. Well, I do not know.

I'm no math-expert, but isn't the result an irrational number, just like pi, with never ending digits? Just like in PieMans link(thx). While calculation time for the next digit rises exponentially(?)..
There is no "last digit" so you never will find the real size.

What exactly do you mean by "actual size"?

If that is true we will have to do an approximation.

Ill change the title to the correct term: the area of the mandelbrot set.

Maybe a better question is to ask how many "Significant Figures" you want in your answer.

Measurements below "The Planck Length" could be considered overkill if you think it represents a true physical limitation to real world measurements.

It could simply be a trancendental number like chillheimer. I don't know. It depends on how you want to use it.

How are we going to represent the unit 1*1i? We can not say how large 1i is.

[lycium]

Have you looked at that PDF I linked? They have some (analytic) bounds on it already.

Regarding not knowing "how large 1i is", um, it's just a unit. If you ask someone, what's the area of a circle of radius 1, it's pi, in units of 1*1. In the complex plane, it's still pi, but in units of 1*i; if this bothers you, simply disregard it - the area is just a number and still in unit lengths, not Furlong-Miles or something

Anyway, returning to the question (area of the M-set), it seems pretty trivial to make just about *any* kind of approximation (doesn't need to be fancy), even just regularly spaced / grid samples, and any modern computer can give you an answer accurate to as many decimals as you'd like...