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

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

    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 $\displaystyle g$ is a certain real-valued function". If you define $\displaystyle g(x)=x-f(x)$, 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. $\displaystyle f(x_0)=0$ if and only if $\displaystyle g(x_0)=x_0$).
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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 $\displaystyle g$ is a certain real-valued function". If you define $\displaystyle g(x)=x-f(x)$, 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. $\displaystyle f(x_0)=0$ if and only if $\displaystyle g(x_0)=x_0$).
    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 $\displaystyle f(x)=e^x - 2x - 1$ for $\displaystyle x\in [1,2]$, how do I then find $\displaystyle g(x)$?
    If I let $\displaystyle g(x)=x-f(x)$ I get $\displaystyle g(x)=-e^x+3x+1$. In my book they get $\displaystyle g(x)=ln(2x+1)$ and I don't see how they get that.
    Last edited by Mollier; Jul 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 $\displaystyle f(x)=e^x - 2x - 1$ for $\displaystyle x\in [1,2]$, how do I then find $\displaystyle g(x)$?
    If I let $\displaystyle g(x)=x-f(x)$ I get $\displaystyle g(x)=-e^x+3x+1$. In my book they get $\displaystyle g(x)=ln(2x+1)$ and I don't see how they get that.
    I'll give you only the steps leading to the given result:

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

    So

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

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