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Here is the dot

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And here is how I arrived at it »

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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?2010-09-30
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    -1: too little information2010-10-01
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    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)).2010-10-01
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    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 » http://imgur.com/ybs7U. 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.2010-10-01
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    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.2010-10-01
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    The algorithm is pic 2. Here's the code » http://www.mediafire.com/?lazew4xvyc08wpp It uses 2, 1, 3 as the default input; enter your own sequence and see the results. More info in the file. There's no math involved as you can see from the file's html code. Tested in Chrome but should work in other browsers. The center of the grids that are affected by the sequence is the protagonist. The equation I'm looking for is its path as it moves with each iteration so I can retrieve, at a later date, it as well as its predecessors who were on the path, simply by plugging in simple integers into the eqn.2010-10-01

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