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Thread: Fixed point problem

  1. #1
    MHF Contributor arbolis's Avatar
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    Fixed point problem

    The problem states : Let $\displaystyle f(x)=x^3+4x^2-10$. The equation $\displaystyle f(x)=0$ has a unique root $\displaystyle r$ approximately equals to $\displaystyle 1.365230013$ in $\displaystyle [1,2]$.
    Show that the following function has a fixed point in $\displaystyle r$ : $\displaystyle g(x)=\frac{\sqrt{10-x^3}}{2}$.
    I'm trying to understand the fixed point signification since about 3 days but I really don't get the concept, and much less how to do exercises related to it. So if you could do this problem explaining, I will be very very grateful.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by arbolis View Post
    The problem states : Let $\displaystyle f(x)=x^3+4x^2-10$. The equation $\displaystyle f(x)=0$ has a unique root $\displaystyle r$ approximately equals to $\displaystyle 1.365230013$ in $\displaystyle [1,2]$.
    Show that the following function has a fixed point in $\displaystyle r$ : $\displaystyle g(x)=\frac{\sqrt{10-x^3}}{2}$.
    I'm trying to understand the fixed point signification since about 3 days but I really don't get the concept, and much less how to do exercises related to it. So if you could do this problem explaining, I will be very very grateful.
    This needs only the definition of fixed point... why are you getting stuck?

    Essentially you are asking us to show there exists an x such that $\displaystyle g(x) = x$
    $\displaystyle \frac{\sqrt{10-x^3}}{2} = x \Rightarrow 10 - x^3 = 4x^2 \Rightarrow x^3+4x^2-10 = 0$

    But you have already found out a value for this! Yes, x=1.365230013 is the fixed point of g(x).
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  3. #3
    MHF Contributor arbolis's Avatar
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    Essentially you are asking us to show there exists an x such that g(x)=x
    I didn't know that. So all what I had to do is to replace $\displaystyle x$ by $\displaystyle r$ in $\displaystyle g(x)-x$?

    This needs only the definition of fixed point... why are you getting stuck?
    Because I don't understand it well.
    But you have already found out a value for this! Yes, x=1.365230013 is the fixed point of g(x).
    I didn't! It was a given information.
    Thanks a lot, if I could I'd push twice on the thanks button!
    Correct me if I'm wrong : a fixed point of a function is the point where the function cross the y=x line?
    The b part of the exercise ask me to do 4 iterations using the fixed point method starting with $\displaystyle x_0=1.5$. I think I know how to do so, but I don't understand why it will converges to the root. Is this method fast, compared to the Newton's one?
    And a big question : if $\displaystyle g(x)$ wasn't given, how could I have guessed a function that would work for the fixed point method? (I guess the method to find such a function is more or less as complicated as to approximate the root of g!)
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by arbolis View Post
    I didn't know that. So all what I had to do is to replace $\displaystyle x$ by $\displaystyle r$ in $\displaystyle g(x)-x$?
    What is r?

    Correct me if I'm wrong : a fixed point of a function is the point where the function cross the y=x line?
    Yes

    The b part of the exercise ask me to do 4 iterations using the fixed point method starting with $\displaystyle x_0=1.5$. I think I know how to do so, but I don't understand why it will converges to the root. Is this method fast, compared to the Newton's one?
    I dont know actually, since I have not studied these numerical methods. But Wiki says that Newtons method is equivalent to fixed point. I am not sure though but I am sure CaptainBlack can clarify. He is very good at such Numerical Algorithms.

    And a big question : if $\displaystyle g(x)$ wasn't given, how could I have guessed a function that would work for the fixed point method?
    Just a thought: Cant you just take g(x) = f(x) - x ?
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  5. #5
    MHF Contributor arbolis's Avatar
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    To
    What is r?
    , I answer
    The equation f(x)=0 has a unique root r
    from my first post.
    Just a thought: Cant you just take g(x) = f(x) - x ?
    Hmm, I take this as a yes, without understanding well! But I know that many functions can do the job, it's the case of g in the exercise, which is not $\displaystyle f(x)-x$.
    Thanks very much, I feel I've understood at least a little bit about this method that was totally unknown some hours ago.
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