When we use Newton's-Raphson method in the following equation $f(x)=x^{50}-1 =0$ for $x>0$ with $x_0=\frac{1}{2}$, there is very slow convergence for the $x=1$ root.
[Math] Newton-Raphson method very slow convergence
newton raphsonnumerical methods
newton raphsonnumerical methods
When we use Newton's-Raphson method in the following equation $f(x)=x^{50}-1 =0$ for $x>0$ with $x_0=\frac{1}{2}$, there is very slow convergence for the $x=1$ root.
Best Answer
As André Nicolas commented, there is no problem with multiplicity.
The iterative scheme is slow because $$x_{n+1}=x_n-\Delta x_ n$$ with $$\Delta x_ n=\frac{x_n^{50}-1}{50 \,x_n^{49}}=\frac 1{50}(x_n-\frac 1{x_n^{49}})$$ and, close to $x=\frac 12$, $\frac 1{x_n^{49}}=2^{49}\approx 5.63\times 10^{14}$. So basically, you start iterating at $x_0=\frac 12$ and the first iterate is $x_1\approx 1.1259\times 10^{13}$. It will effectively be a long way to go back to $1$.
Strating from below leads to an overshoot of the solution (Darboux theorem since in this case $f(x_0)f''(x_0)<0$).
Even if we start closer to the solution, the process is quite slow. Starting with $x_0=0.9$, the successive iterates are $$x_1=4.37459$$ $$x_2=4.28709$$ $$x_3=4.20135$$ $$x_4=4.11732$$ Now compare if starting with $x_0=1.1$ $$x_1=1.07819$$ $$x_2=1.05712$$ $$x_3=1.03730$$ $$x_4=1.01988$$