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I'm trying to see if the iterative method $x_n=g(x_{n-1})$ where $g(x)=2\sqrt{x-1}$ will converge to $2$, if I take $x_0$ that is sufficiently close to $2$.

Indeed notice that $g(2)=2$. and we have a theorem that states that if $|g'(2)|<1$ then there is a neighborhood of $2$ such that if we take an $x_0$ from that neighborhood, the method will converge to $2$.

We also know that if $|g'(2)|>1$ then there is no such neighborhood.

But notice now that $g'(x)=\frac{1}{\sqrt{x-1}}$, and so $g'(2)=1$.

What can we say about this situation? will it converge?

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  • $\begingroup$ That should diverge, the absolute value needs to be less than 1. $\endgroup$ – bobbym Jun 26 '14 at 13:16
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    $\begingroup$ @bobbym, that's only a sufficient condition for convergence. Oria, try looking at a plot of the function. It converges if $2 \leq x_0 < 2+\epsilon$ but diverges if $2-\epsilon < x_0 < 2$. $\endgroup$ – Antonio Vargas Jun 26 '14 at 13:19
  • $\begingroup$ I think you are correct. if we take a number that is very very close to $2$, but smaller than $2$, than the derivative will be larger than one. There can be no neighborhood of $2$, since any value smaller than $2$ will yield derivative larger than 1 $\endgroup$ – Oria Gruber Jun 26 '14 at 13:19
  • $\begingroup$ Isn't it mandatory for convergence Antonio? $\endgroup$ – Oria Gruber Jun 26 '14 at 13:20
  • $\begingroup$ No, take the system given by $g(x) = 2 + (x-2) - (x-2)^3$. Here $g'(2) = 1$ but the dynamical system converges in a neighborhood of $x=2$. If $|g'(x_\infty)| = 1$ it usually only means that the first derivative doesn't give you enough information to determine whether the system converges or diverges. $\endgroup$ – Antonio Vargas Jun 26 '14 at 13:22

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