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Math Help - Existence of solution to f(x)=0

  1. #1
    Member Mollier's Avatar
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    Existence of solution to f(x)=0

    Hi,

    There's a theorem in my book that states the following;

    Let f be a real-valued function, defined and continuous on a bounded closed interval [a,b] of the real line. Assume, further, that f(a)f(b)\leq 0; then, there exists \xi in [a,b] such that f(\xi)=0.

    I feel quite comfortable with that theorem. Later on in the book the authors write:

    "An alternative sufficient condition for the existence of a solution to the equation f(x)=0 is arrived at by rewriting it in the equivalent form x-g(x)=0 where g is a certain real-valued function, defined and continuous on [a,b]. The problem of solving the equation f(x)=0 is converted into one of finding \xi such that \xi-g(\xi)=0."

    Now that I do not understand. I don't really see how x-g(x)=0 is equivalent to f(x)=0. Could someone please explain this to me, or point me to some resources?

    Thanks.
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  2. #2
    Senior Member roninpro's Avatar
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    The author does say "where g is a certain real-valued function". If you define g(x)=x-f(x), then you have an equivalent form.

    The thing that strikes me about this form is that your existence question is transformed into a fixed point question (i.e. f(x_0)=0 if and only if g(x_0)=x_0).
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by roninpro View Post
    The author does say "where g is a certain real-valued function". If you define g(x)=x-f(x), then you have an equivalent form.

    The thing that strikes me about this form is that your existence question is transformed into a fixed point question (i.e. f(x_0)=0 if and only if g(x_0)=x_0).
    Thank you. The chapter I am reading is about simple iteration. The authors discuss all the theorems involved, for example Brouwer's fixed point theorem, contraction mapping theorem etc.

    Let me be a bit more specific. If f(x)=e^x - 2x - 1 for x\in [1,2], how do I then find g(x)?
    If I let g(x)=x-f(x) I get g(x)=-e^x+3x+1. In my book they get g(x)=ln(2x+1) and I don't see how they get that.
    Last edited by Mollier; July 18th 2010 at 08:59 PM.
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  4. #4
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    Quote Originally Posted by Mollier View Post
    Thank you. The chapter I am reading is about simple iteration. The authors discuss all the theorems involved, for example Brouwer's fixed point theorem, contraction mapping theorem etc.

    Let me be a bit more specific. If f(x)=e^x - 2x - 1 for x\in [1,2], how do I then find g(x)?
    If I let g(x)=x-f(x) I get g(x)=-e^x+3x+1. In my book they get g(x)=ln(2x+1) and I don't see how they get that.
    I'll give you only the steps leading to the given result:

    e^x-2x-1=0~\implies~e^x=2x+1~\implies~x=\ln(2x+1)

    So

    x-\ln(2x+1)=0 correspond to x-g(x)=0
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  5. #5
    Senior Member roninpro's Avatar
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    Defining g(x)=x-f(x) is only one way to get an equivalent form. In this case, we can get one via algebraic manipulation.

    First, we are solving f(x)=e^x-2x-1=0. Add 2x+1 to both sides to receive e^x=2x+1. Then take the log for x=\log (2x+1). We now see that a solution to f(x)=0 is the same as a fixed point for g(x)=\log(2x+1).
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  6. #6
    Member Mollier's Avatar
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    Thank you very much!
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