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Thread: Solving an equation by iteration

  1. #1
    Apr 2011

    Solving an equation by iteration


    I need to solve an equation for a course for psychology, but since I am studying psychology maths is not really my thing...

    The equation I need to solve is:

    xt+1 = f(xt) = axt(1-2xt)

    The docent provide an example in the lecture (a logistic map equation with iteration, needed for understanding complex dynamic behaviour) but I don't get the steps in between and this is a bit too complex for me This was his example:

    Yi+1= rYi(1Yi)

    i=0 Y0->Y1= rY0(1Y0)

    i=1 Y1->Y2= rY1(1Y1)

    = r2Y0= r rY0(1Y0) (1rY0(1Y0) )
    = r3Y04+2r3Y03r2(1+r)Y02+r2Y0

    Can someone help me (also writing the steps I need to solve the equation)

    I would be far mooore than thankful!

    A desperate student from the Netherlands
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  2. #2
    MHF Contributor

    Mar 2011
    what the equation says is:

    x(current) = a*(x(previous))*(1 - 2x(previous)). apparently "a" here is some constant that represents some feature of the system.

    so we start with t = 0, to get an initial value of x0. the next value x1, is:

    x1 = ax0(1 - 2x0) = ax0 - 2a(x0)^2

    the second value, x2, is:

    x2 = ax1(1 - 2x1) = a(ax0 - 2a(x0)^2)(1 - 2(ax0 - 2a(x0)^2))

    = (a^2x0 - 2a^2(x0)^2)(1 - 2ax0 + 4a(x0)^2) = a^2x0 - (2a^2 + 2a^3)(x0)^2 + 8a^3(x0)^3 - 8a^3(x0)^4

    i am not sure what you mean by "solve" the equation. i can't think of some other formula for x_n solely in terms of x0.

    i'm pretty sure that if one does exist, it's fairly complicated, and horrendous.

    to calculate any given x_n, you just need to start with x0, and work your way to it.
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