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Newton's cubic Math Modeling

asked 2011-12-12 15:05:24 -0500

rower9911 gravatar image

updated 2011-12-13 02:51:12 -0500

I need to find the "black points" (where the function becomes undefined) of an equation x^3 -1 using newtons method. So the function turns out to be F(x) = x - (x^3-1)/(3x^2). Is there a way to find a majority of the points (where x is undefined as in the Newton cubic fractal photo aka the black points) using Sage?

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Can you be a little more specific? `x^3-1` is defined for all complex `x`, so I assume you are referring to something else.

kcrisman gravatar imagekcrisman ( 2011-12-13 02:35:58 -0500 )edit

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answered 2011-12-13 03:25:47 -0500

kcrisman gravatar image

I don't think this is a built-in behavior. I'd be interested in a reference for an algorithm, even! Basically, this would just be 0, anything whose tangent line sends it to zero, anything whose tangent line sends it to one of those, etc., right?

If you just wanted to visualize this, you could try using some of the ideas at this ticket about Newton basins, though it wouldn't give the "black points".

For a graphical way of exploring this, I would search sagenb.org for implementations of Newton's method in interactive mode.

You could use solve to try to get some of these points, though I'm not sure how useful this would be.

sage: LA(x) = x - (x^3-1)/(3*x^2)
sage: LA(0)
---------------------------------------------------------------------------
RuntimeError: power::eval(): division by zero
sage: solve(LA==0,x)
[x == 1/4*(I*(-1)^(1/3)*sqrt(3) - (-1)^(1/3))*2^(2/3), x == 1/4*(-I*(-1)^(1/3)*sqrt(3) - (-1)^(1/3))*2^(2/3), x == 1/2*(-1)^(1/3)*2^(2/3)]
sage: S = solve(LA==0,x)
sage: for s in S:
    solve(LA == s.rhs(),x)
....:     
[x == -1/576*(-18*I*sqrt(3) - 18)*(-I*sqrt(3) + 1)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) + 1/24*(3*I*sqrt(3) - 3)*(-1)^(1/3)*2^(2/3) - 1/2*(1/8*sqrt(6) - 5/16)^(1/3)*(I*sqrt(3) + 1), x == -1/576*(-18*I*sqrt(3) - 18)*(I*sqrt(3) + 1)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) + 1/24*(3*I*sqrt(3) - 3)*(-1)^(1/3)*2^(2/3) - 1/2*(1/8*sqrt(6) - 5/16)^(1/3)*(-I*sqrt(3) + 1), x == 1/24*(3*I*sqrt(3) - 3)*(-1)^(1/3)*2^(2/3) + 1/288*(-18*I*sqrt(3) - 18)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) + (1/8*sqrt(6) - 5/16)^(1/3)]
[x == -1/576*(-I*sqrt(3) + 1)*(18*I*sqrt(3) - 18)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) - 1/24*(3*I*sqrt(3) + 3)*(-1)^(1/3)*2^(2/3) - 1/2*(1/8*sqrt(6) - 5/16)^(1/3)*(I*sqrt(3) + 1), x == -1/576*(I*sqrt(3) + 1)*(18*I*sqrt(3) - 18)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) - 1/24*(3*I*sqrt(3) + 3)*(-1)^(1/3)*2^(2/3) - 1/2*(1/8*sqrt(6) - 5/16)^(1/3)*(-I*sqrt(3) + 1), x == -1/24*(3*I*sqrt(3) + 3)*(-1)^(1/3)*2^(2/3) + 1/288*(18*I*sqrt(3) - 18)*2^(1/3)/(1/8*sqrt(6) - 5/16)^(1/3) + (1/8*sqrt(6) - 5/16)^(1/3)]
[x == -1/2*(I*sqrt(3) + 1)*(1/8*sqrt(2)*sqrt(3) - 5/16)^(1/3) - 1/16*(-I*sqrt(3) + 1)*2 ...
(more)
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thanks. i appreciate your help. I'm still looking for an algorithm but cant seem to come up with anything!

rower9911 gravatar imagerower9911 ( 2011-12-13 13:23:42 -0500 )edit
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You may find `http://en.wikipedia.org/wiki/Newton_fractal` useful as well - basically, you're looking for a Julia set, and you might find algorithms for finding those out there.

kcrisman gravatar imagekcrisman ( 2011-12-13 14:34:21 -0500 )edit

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Asked: 2011-12-12 15:05:24 -0500

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Last updated: Dec 13 '11