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Math Help - Numerical analysis

  1. #1
    MHF Contributor Amer's Avatar
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    Numerical analysis

    I want just how to solve it give me a hint

    The question:-

    Use fixed-point iteration method to determine a solution accurate to within 10^{-2} for x^3-x-1=0 on [1,2] use p_{0}=1

    here is my solution (correct me if I was wrong )

    first we should dtermine F(x)=x so

    x^3-x-1=0 \Rightarrow x^3=x+1 \Rightarrow x=\sqrt[3]{x+1}\Rightarrow F(x)=\sqrt[3]{x+1}

    F'(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}

    \mid F'(x) \mid <1 so F(x) is converge now

    F(p_{0})=\sqrt[3]{2}=1.2599210498949=p_1

    F(p_1)=1.3122938366833=p_2

    F(p_2)=1.3223538191388=p_3

    ..................etc

    when I should stop ??
    my solution is correct or it is wrong ??

    Thanks for the help
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  2. #2
    MHF Contributor arbolis's Avatar
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    You should stop when the first 2 digits after the coma don't change with newer iterations.
    For the rest I can't help you because I have to revise my Numerical Analysis course first.
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  3. #3
    Super Member Random Variable's Avatar
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    Why not use Newton's Iteration?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Random Variable View Post
    Why not use Newton's Iteration?
    Because the question ask to use fixed-point iteration.
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Random Variable View Post
    Why not use Newton's Iteration?
    I know it but I should understand how this work to understand fixed point iteration and newton is a special function F(x)

    I can use it to find F(x) but I do not know how to work with fixed point iteration in general
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  6. #6
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    Say you decide that the answer should be x=1.32 correct to 2~decimal places. How can you check your answer? Use the sign change method.

    If 1.32 is correct to 2 d.p. then the root must lie in the interval (1.315,1.325).

    f(x)=x^3-x-1, f(1.315)=-0.0411 and f(1.325)=0.0012 (4 d.p.).

    Since f(x) changes sign and is continuous on (1.315,1.325), the root is 1.32 correct to 2 d.p.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Amer View Post
    I want just how to solve it give me a hint

    The question:-

    Use fixed-point iteration method to determine a solution accurate to within 10^{-2} for x^3-x-1=0 on [1,2] use p_{0}=1
    The iteration we will consider is generated from the rearrangement:

    x=f(x)=(1+x)^{1/3}

    then:

    x_n=f(x_{n-1})

    is our iteration.

    Now we expand f(x) as a Taylor series about the root:

    f(x+\varepsilon)=f(x)+\varepsilon f'(x)+ \varepsilon^2 R(x)

    where \varepsilon^2 R(x) is the remainder term, and as the serier is alternating from the second term onwards we know that this is bound by the absolute value of the first neglected term: |\varepsilon^2 f''(x)/2!|< 0.2a; \varepsilon^2 (please justify this last inequality yourself, or something tighter if you like)

    Hence as on [1,2] |\varepsilon|<1 we have \varepsilon^2 R(x)<0.2\;\varepsilon

    and so:

     <br />
|f(x)-f(x+\varepsilon)|<0.5 \varepsilon.<br />

    Hence in the iteration the error more than halves at each pass (in fact it does better than this I leave it to the reader to find tighter bounds here).

    Hence the error after n passes through the iteration is less than 2^{-n} and we need to find n such that:

    2^{-n}<10^{-2}

    this will guarantee that our error is less than the required limit, which looks like 6 iterations to me.

    (in fact this is pessimistic since I have been rather generous in estimating the bounds)

    CB
    Last edited by CaptainBlack; July 15th 2009 at 12:06 PM.
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