Existence of solution to f(x)=0

**Mollier** · July 18th 2010, 09:00 AM

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.

**roninpro** · July 18th 2010, 09:43 AM

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 $g(x_0)=x_0$ ).

**Mollier** · July 18th 2010, 08:44 PM

Originally Posted by roninpro

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.

**earboth** · July 18th 2010, 09:24 PM

Originally Posted by Mollier

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 $x-g(x)=0$

**roninpro** · July 18th 2010, 09:28 PM

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 $g(x)=\log(2x+1)$ .

**Mollier** · July 18th 2010, 10:12 PM

Thank you very much!

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