How to check if an algorithm converges in an interval, for example $x_{k+1}:=\frac{1}{11}(1-\cos(x_k))$ does it converge for any startpoint $x_0\in (-\pi/2,\pi/2)\setminus\{0\}$ ? (as hint: consider the Taylor approximation)
I know only the fixed point theorem of Banach, here if i choose $g(x)=\frac{1}{11}(1-\cos(x))$, differentiable with $g'(x)<1$ but the domain is not a closed set, is this a big problem ?
What has Taylor series to do with this ?
is our solution is $x^*$ then we are hoping our iteration is a fixed point problem so $g(x^*)=x^*$, now if $g'(x)\lt 1$, then it has lipschitz constant $L\lt1$, so:
$\|x^*-x^{k+1}\|=\|g(x^*)-g(x^k)\|\leq L\|x^*-x^k\|\lt\|x^*-x^k\|$
So we see as $k\to\infty$ $x^k\to x^*$, so the iteration converges.
Also your split and open domain does not make a difference, we have a contraction, so if $x^0\in(-\pi/2,0)$ then we will stay in that interval, likewise if $x^0\in(0,\pi/2)$.