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**Harold****Guest**

What is a mandelbrot set?I tried reading in wikipedia,but it wasn't pretty clear to me ,it'd be great if you explain the whole concept.

**zetafunc.****Guest**

Which part did you have trouble with? Here is the part of that article which might be most relevant to you:

More precisely, the Mandelbrot set is the set of values of c in the complex plane for which the orbit of 0 under iteration of the complex quadratic polynomial z[sub]n+1[/sub] = z[sub]n[/sub][sup]2[/sup] + c remains bounded. That is, a complex number c is part of the Mandelbrot set if, when starting with z[sub]0[/sub] = 0 and applying the iteration repeatedly, the absolute value of z[sub]n[/sub] remains bounded however large n gets.

For example, letting c = 1 gives the sequence 0, 1, 2, 5, 26, , which tends to infinity. As this sequence is unbounded, 1 is not an element of the Mandelbrot set. On the other hand, c = i (where i is defined as i[sup]2[/sup] = −1) gives the sequence 0, i, (−1 + i), −i, (−1 + i), −i, ..., which is bounded, and so i belongs to the Mandelbrot set.

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

hi Harold,

So if you take points in the complex plane, carry out the iteration using a computer and then colour the point according to whether it is in the set or not, you get these fantastic pictures.

What is really amazing is, if you explore a region of the plane in close up, you find the same shape appearing at any magnification (that's what is meant by fractal shape) and points very close to each other in the plane may be 'one in' and one out' of the set.

This is nicely demonstrated at

http://www.youtube.com/watch?v=gEw8xpb1aRA

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Harold****Guest**

So,I colour the point c that will give me bounded sequence in the sequence 0,qc(0),qc(qc(0)),qc(qc(qc))),....

[qc(z)=z^2+c]

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,508

I think that should do it.

If you just divided the set of complex numbers in two {c : sequnce is bounded} and {the rest} then you'd just have black and white (or whatever two colours you chose)

And how many iterations will you use to determine whether the sequence is bounded?

So I think (but this may be wrong) that a range of colours are used, chosen according to how long the sequence appears to be bounded.

If you are going to try this and need more help, I'll ask my son who did this some years ago.

If it works please post your pictures.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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