# Newton-Raphson convergence issue

I am working with the following function where r= 2^(-d) and d =[1:100]

I am comparing 5 different algorithms to compute the roots of this function but the Newton Raphson method doesn't converge anymore above a certain value of r.

Column n°5 on this screenshot How can I solve this issue?

Thank you in advance

• Hey I found the issue. The initial value is too small to let NR converge so if I set the initial value to something like 0.5 it converges no problem – Yrden May 3 '18 at 7:01

## 1 Answer

Admitting that you look for the first positive zero of equation $$f(u)=2 \sin ^2\left(\frac{\pi}{2} \left(4 u^3-3 u^4\right)\right)-\frac 1 {2^d}$$ I am rather surprised by the value $3.1803$ given in the table for $d=2$; it effectively corresponds to the $n^{th}$ root of the equation for a very large $n$ (i gave up trying to count it !).

The key problem with Newton method is the initial guess. What we know, at least from the very first case $(d=1)$ is that the root is small. So, using Taylor expansion around $u=0$, what we have is $$2 \sin ^2\left(\frac{\pi}{2} \left(4 u^3-3 u^4\right)\right)=8 \pi ^2 u^6-12 \pi ^2 u^7+O\left(u^8\right)$$ giving as an initial underestimate $$u_0=\frac 1 {\sqrt{\pi^2\, 2^{d+3}}}$$ By Darboux-Fourier theorem, we know that we shall face one overshoot of the solution since, for any $d$, $f(u_0)\times f''(u_0) <0$.

Using the method for $d=1$, the iterates are $$\left( \begin{array}{cc} n & u_n \\ 0 & 0.430127 \\ 1 & 0.542201 \\ 2 & 0.515233 \\ 3 & 0.513710 \\ 4 & 0.513705 \end{array} \right)$$ Doing the same for $d=2$ $$\left( \begin{array}{cc} n & u_n \\ 0 & 0.383200 \\ 1 & 0.459453 \\ 2 & 0.442439 \\ 3 & 0.441355 \\ 4 & 0.441351 \end{array} \right)$$ Working with illimited precision, Newton method converges without any problem (I tried up to $d=10^6$).

Working with limited precision (all numbers being declared as double precision real numbers - the $2$ in the definition of $r$ is crucial) and using as convergence criteria $\frac{|\Delta u|} u \leq 10^{-16}$, I did not face any problem at all except above $d=1017$ $(2^{-1018}\approx 3.56 \times 10^{-307})$. For $d \geq 978$, the estimate was the solution.