Given is the iteration $x_{k+1}=\frac{1}{11}(1-\cos(x_{k}))$ with $x_{0}\in (-\frac{\pi }{2},\frac{\pi }{2})$ without $0$. Check if the sequence converges to $x^{*}=0$ and find its convergence rate.
Well, i am stuck in this obviously very easy first part of the problem. Wolfram Alpha shows that $\lim_{x\rightarrow \infty }\frac{1}{11}(1-\cos(x))=0$, so the sequence converges to the given $x^{*}=0$. Can anyone give me a hint how to solve this limit, please?
Second, the convergence rate. I know the definitions of sublinear convergence, superlinear convergence etc. I need to calculate the $\lim_{x\rightarrow \infty }\frac{\left \| x_{k+1} -x^{*}\right \|}{\left \| x_{k}-x^{*} \right \|}$, so $lim _{x\rightarrow \infty } \frac{\left | \frac{1}{11}(1-\cos(x)) \right |}{\left | x \right |}=lim_{x\rightarrow \infty }\frac{1-\cos(x)}{11x}= 0$. So, the sequnce converges superlinearly. Is the idea correct at all? Can anybody help me with this problem, please? Thank you in advance!
I'm not so sure how you managed to obtain that $\lim_{x→∞} (1-\cos(x))/11 = 0$ because this limit doesn't exist. The graph is forever waving about between $1/11$ and $0$:
But more importantly, the behaviour of this function at infinity is not really relevant when we are talking about starting with $|x_0| < \pi/2$ and trying to get a null sequence.
Hint: you can try showing that $1-\frac{x^2}{2} < \cos x $ and use this to get a bound on $|x_k|$ which should tend to $0$ irrespective of the starting point $x_0 ∈ (-π/2,π/2)\setminus \{0\}$.
This will also give you a bound up to a constant on the convergence rate, so if a reasonable Big-O is all you want, you're done at this point. If you want to show that there is a quadratic lower bound as well, try scaling $p(x):=-(x-π/2)(x+π/2)$ to have the same maximum height in $(-\pi/2 , \pi/2)$ as $\cos(x)$ and use this to bound $cos(x)$ from above now.