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I'm trying to find the order of convergence of the fixed point iteration $p_{n+1} = g(p_n)$ for $n \geq 0 $ where $g(x) = (3+x-3x^2)^{1/4}$ and the point of convergence is at $x = p$

This is how I started solving the problem. $$ p - p_{n+1} = p - (3+p_n - 3p_{n}^2)^{1/4} $$ Since $\epsilon_{n} = p - p_n$,

$$ \epsilon_{n+1} = p-(3+p_n - 3p_{n}^2)^{1/4} $$ After some algebraic manipulation, I've reached (unless my math is way off), $$ -2p\epsilon_{n+1} +\epsilon_{n+1}^2 = 3 +p -3p^2 + (4p-1)\epsilon_n-2\epsilon_n^2 $$ I'm not sure how to move on from here.

pa1
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  • Something is wrong here. $p$ must be equal to $1$. As $\epsilon_n \to 0$, the LHS tends to $0$ while the RHS tends to $3+p-3p^2 = 1$. – NN2 Apr 21 '23 at 20:19
  • You subtract $p_{n+1}=g(p_n)$ and $p=g(p)$ to get $ϵ_{n+1}=g'(c_n)ϵ_n$ with some midpoint $c_n$, and thus have to find bounds for the derivative. – Lutz Lehmann Apr 22 '23 at 04:22
  • @NN2 Yes, you're correct. I've forgotten to mention that. – pa1 Apr 22 '23 at 07:43

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