I just learned about Fixed points from our numerical methods class, and from my understanding a fixed point of a function is all the $x$'s of a given $f(x)$ that intersect with $y=x$, which means its image is equal to itself. my question is, what is so special about these points and how they are used to find roots of a function?

  • $\begingroup$ see e.g. en.wikipedia.org/wiki/Fixed_point_(mathematics)#Applications to get an idea $\endgroup$ Jul 31, 2016 at 18:08
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    $\begingroup$ If you want x such that f(x)=0, define the function g(x)=f(x)+x and search a fixed point of g. $\endgroup$ Jul 31, 2016 at 18:10
  • $\begingroup$ @vvnitram is finding fixed points easier that finding roots directly? is that why I use them? $\endgroup$ Jul 31, 2016 at 18:22
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    $\begingroup$ No, but the iteration is easy using fix-points. For example, to find the solution of $x=cos(x)$ with a table calculator, you just enter some number and press "cos" until the result does not change. If the conditions of the banach-fixpoint-theorem are satisfied, the iteration $x_{n+1}=f(x_n)$ converges and gives a root. If you try this, be sure that the calculator is set on "rad". $\endgroup$
    – Peter
    Jul 31, 2016 at 18:24
  • $\begingroup$ The point is, you have an equation $f(x)=0$, you rewrite this equation (by doing some algebra that depends on what $f$ is) as $g(x)=x$. This algebra converts the problem of finding roots of $f$ into finding fixed points of $g$. Under an assumption about $g$, fixed point iteration will converge if your initial guess is good enough. This assumption is not trivial. For example, suppose you are trying to find $\sqrt{2}$. You can think of this as the positive root of $f(x)=x^2-2$. For this problem, fixed point iteration with $g(x)=x^2+x-2$ will not converge, whereas it will with $g(x)=x/2+1/x$. $\endgroup$
    – Ian
    Jul 31, 2016 at 19:07


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