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ContinuousNewtonMethod

Stephen Crowley edited this page Aug 29, 2023 · 18 revisions

The continuous Newton method involves solving a differential equation of the form:

$$ y'(t) = -\frac{f(y(t))}{f'(y(t))} $$

where $f$ is a differentiable function and whose solution $y(t)$ is a continuously varying function of a nonnegative real-valued parameter $t$ called the NewtonFlow of f which converges to a root of f as t goes to $\infty$ starting from any given point $y_0$ as the initial condition $y(0)=y_0$ at time t=0, that is,

$$f (\lim_{t \rightarrow \infty} y (t)) = 0$$

. While closed-form solutions for the continuous Newton method are known for some special cases, such as when $f$ is a quadratic polynomial or the exponential function, for most functions $f$ no closed-form solutions are available. Therefore, other methods such as numerical methods must be used to calculate or approximate the solution.

An exception to this is the SFunction ...

The Newton flow of $S(t)=\tanh(\ln(1+t^2))$

The solutions corresponding to the four roots of $S(t)=\tanh(\ln(1+t^2))$ are the limits of :

$$\pm\sqrt{-\frac{1}{-1+\frac{2}{2+a^2+a^4}+e^t+\sqrt{-\frac{a^4 (2+a^2)^2}{(2+a^2+a^4)^2}+e^{2t}}}}$$

and

$$\pm\sqrt{\frac{1-\frac{2}{2+a^2+a^4}-e^t+\sqrt{-\frac{a^4 (2+a^2)^2}{(2+a^2+a^4)^2}+e^{2t}}}{-1+\frac{2}{2+a^2+a^4}+e^t}}$$

as $t$ goes to $\infty$.

TODO: factor out the common expressions in the solutions

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