<s>[INST] Can you please help me whether $N-R$ method or other methods converges? Consider the system of equations
\begin{align}
&f(x,y)=x+\frac{y^4}{2}+\frac{x^{32}}{4}+\frac{y^{128}}{8}=0 \\
&g(x,y)=y+\frac{x^8}{2}+\frac{y^{32}}{4}+\frac{x^{256}}{8}=0.
\end{align}
I want to solve it using Newton-Raphson process or any other methods
Consider the initial guess $(x_0,y_0)=((-2)^{5/31},(-2)^{9/31})$, which is either $( 1.118,-1.223 )$ or
$(x_0,y_0)=(-1.118,1.223)$.
I want to see whether the Newton-Raphson method converges with this
initial guess.
Note that the initial guess $(x_0,y_0)$ is a simultaneous zero of the
following truncated system obtained from the original system
\begin{align}
&x+\frac{y^4}{2}=0\\
&y+\frac{x^8}{2}=0.
\end{align}
By hand calculation, it is laborious. For the jacobian  is $$J
=\begin{pmatrix} 1+8x^{31} & 2y^3+16y^{127} \\ 2x^3+32x^{255} &
1+8y^{31} \end{pmatrix} \Rightarrow J((x_0,y_0)) \approx
\begin{pmatrix} 255 & -2.03 \times 10^{12} \\ 7.21 \times 10^{13} &
-4104\end{pmatrix}$$
So the 2nd iteration is given by
\begin{align} \begin{pmatrix} x_1 \\ y_1
\end{pmatrix}&=\begin{pmatrix}1.118 \\ 1.223 \end{pmatrix}
-J((x_0,y_0))^{-1} \begin{pmatrix}f((x_0,y_0)) \\ g((x_0,y_0))
\end{pmatrix}
\end{align}
which seems difficult to calculate because I can not invert the huge matrix.
The zeros of the truncated system should converge to the solutions
of the original system.
Can you please help me whether N-R method or any other numerical methods converges ?

More, specifically, how to show the simultaneous zeroes of the truncated system $
x+\frac{y^4}{2}=0=y+\frac{x^8}{2}$ converges to the simultaneous zeroes of the original system $f(x,y)=0=g(x,y)$ ?

Thanks
 [/INST] By performing the contour plot for some truncated approximations, we can infer for which the NR algorithm will work.
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9}=0 \\
}
$$
Here NR works with one solution.

$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9} + \frac{y^{2187}}{27}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9} + \frac{x^{6561}}{27}=0 \\
}
$$
Here NR succeeds with three solutions

$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9} + \frac{y^{2187}}{27}+\frac{x^{59049}}{81}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9} + \frac{x^{6561}}{27}+\frac{y^{59049}}{81}=0 \\
}
$$
Here NR only gives one solution. The two contours only cross one time.

</s>