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Here is the dot


And here is how I arrived at it »


Pretty simple IFS using a 2 x 2 grid as the base for the iterations. Is there a way to describe this point as well as its siblings, all of which lie on the dotted red line(pic 2), using an equation?

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Well, it's a limit. Exactly what equations were your IFS using? – J. M. Sep 30 '10 at 23:26
-1: too little information – Jyotirmoy Bhattacharya Oct 1 '10 at 3:07
There are no equations involved in its construction as my program merely realizes the steps in pic 2 - one subroutine constructs a 2 x 2 grid seen in step 1 of pic 2 and then builds out the IFS through recursion as shown in steps 2,3 4 and 5. Number of steps in the recursion as well as the location for the next iteration is the program input, which in this case is 2,1,3 (2nd Quad, !st Quad(of 2nd Quad) and 3rd Quad (of 1st Quad (of 2nd Quad)). – saber soap Oct 1 '10 at 7:05
While it's trivial to throw on a scale for the axes and store the point as coords, for larger iterations the coords are unwieldy. No matter what the scale of the axes, for the nth iteration the dot will move to a x,y coord each having n digits. After 5 x 10^7 iterations, that x,y gets massive. While looking to solve this, I chanced upon the Taylor series where any term in the series is simply » That gave me the idea that instead storing the dot as coords, I'm better off storing the dot as the nth term of a general equation. How do I cook up that equation,is my question. – saber soap Oct 1 '10 at 7:39
But you have an algorithm... that would give information that can be just as valuable as (if not more so) the series of pictures you have. – J. M. Oct 1 '10 at 16:23

It seems like you are applying $f_1,f_2,f_3$ where 1,2,3 maps the entire square to down right, up right, and down left respectively, and your red dot is the solution to $f_1(f_2(f_3(x))) = x.$ Now, you just have to find these functions, which is rather easy and left as an exercise. Now, the composition of these 3 functions (in the order above) is:

$$f(x,y) = ((x+5)/8,(y-3)/8)$$

Now, solve $(x+5)/8 = x$, giving $x=5/7.$ Now, solve $(y-3)/8 = x$, giving $x=-3/7.$

The red dot should be $(5/7,-3/7).$

I have assumed that the corners of the original square is $(\pm 1,\pm 1)$.

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It's not very clear what you're trying to ask, but what you're describing looks like a region quadtree.

Each step in your IFS corresponds to a choice of child node (or subquadrant), and the tree path of any given point is the same as the list of numeric input steps in your question.

It does not really matter whether you store the point as an arbitrary-precision coordinate pair or tree traversal: the binary digits required to store its coordinates correspond exactly to its quadtree traversal path.

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