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Math Help - Fixed point iteration doubt

  1. #1
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    Fixed point iteration doubt

    Hi! I was asked to find an approximation for a root of f(x)=2sin(x\pi)+x=0 on the interval [1, 2] using the fixed point iteration method, with p0=1.

    So, I got g(x)=x=(\arcsin(-x/2))/\pi

    However, that way I just get numbers very close to zero (which is a solution for the equation, but it's not on the interval I want).

    Anyway, I checked the answer on the back of the book and they used g(x)=x=(\arcsin(-x/2))/\pi+2 and got x \simeq 1.68

    I just don't know from where does that +2 come? Could anyone explain me, please?
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  2. #2
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    Re: Fixed point iteration doubt

    It's because the inverse sine is multivalued. When you attempt to evaluate \arcsin (-1/2), your calculator will return a negative value whereas what you want is the value that is 2\pi further on (if that makes sense).

    The iteration that you need is

    \pi x = \arcsin(-x/2) + 2\pi,

    from which

    x = \frac{1}{\pi} \arcsin(-x/2) + 2.
    Thanks from Cesc1
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  3. #3
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    Re: Fixed point iteration doubt

    I get it now. Thanks!
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