@ -179,19 +179,23 @@ by the sequence of points $(z_i)_{i\in\N}$ defined by the recurrence relation
$$ z_{i+1}=z_i+h\cdot f(z_i,t_i) ,$$
where $h$ is the step size.
In our case, we have $$f(z,t)=-\left(\frac{\partial H}{\partial z}(z,t)\right)^{-1}\frac{\partial H}{\partial t}(z,t)$$ and $t_0=1$, since we track from $1$ to $0$. For the same
reason, we set $$t_i=t_{i-1}-h.$$ We will also use a variable step size, based on the output of each iteration.
reason, we set $$t_{i+1}=t_i-h.$$
\subsubsection{Corrector: Newton's method}
Since we want to solve $$H(z,t)=0,$$ we can use Newton's method to improve the approximation $\widetilde{z_i}$ obtained by Euler's method to a solution of such equation.
This is done by moving towards the root of the tangent line of $H$ at the current approximation, or in other words through the iteration
where this time $z_0=\widetilde{z}_i$, with $\widetilde{z}_i$ and $t_{i+1}$ obtained from the $i$-th Euler step.
Usually, only a few steps of Newton's method are needed; we will use a fixed maximum of $10$ steps,
stopping the iterations when the desired accuracy is reached, for instance when the norm of $H(z_i,t_i)$ is less than $10^{-8}$.
Usually, only a few steps of Newton's method are needed; we will use a fixed number of 5 iterations.
At this point, we use the final value of the Newton iteration as the starting value for the next Euler step.
\subsubsection{Adaptive step size}
In order to improve the efficiency of the algorithm, we will use an adaptive step size, which will be based on the norm of the residual of the Newton iteration.
If the desired accuracy is not reached, for instance when the norm of $H(z_i,t_i)$ is bigger than $10^{-8}$,
then we halve the step size; if instead we have 5 "successful" iterations in a row, we double the step size.
