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I've recently discovered that modifying the standard Newton-Raphson iteration by "squashing" $\frac{f (t)}{\dot{f} (t)}$ with the hyperbolic tangent function so that the iteration function is

$$N_f (t) = t - \tanh \left( \frac{f (t)}{\dot{f} (t)} \right)$$

results in much larger (Fatou) convergence domains, also known as (immediate) basins of attraction.

The convergence theorems of the Newton-Kantorivich type , such as those in On an improved convergence analysis of Newton’s method can be modified in a relatively straightforward way so that they apply to this modified iteration. For instance, the parameter $\eta$ defining the inequality in the referenced paper,

$$\left\| \frac{F (x_0)}{F' (x_0)} \right\| \leqslant \eta$$

can be replaced with

$$\left\| \tanh \left( \frac{F (x_0)}{F' (x_0)} \right) \right\| \leqslant 1$$

so that $\eta=1$ at most, no matter the (differentiable) function $F(x)$ , since the range of $\tanh(x)$ is $(- 1, 1)$

For the function I am applying this to, the Hardy Z function, $$\begin{array}{ll} Z (t) & = e^{i \vartheta (t)} \zeta \left( \frac{1}{2} + i t \right) \end{array}$$ where $$\vartheta (t) = - \frac{i}{2} \left( \ln \Gamma \left( \frac{1}{4} + \frac{i t}{2} \right) - \ln \Gamma \left( \frac{1}{4} - \frac{i t}{2} \right) \right) - \frac{\ln (\pi) t}{2}$$ this tanh modification has much better convergence properties than the regular Newton iteration when using the starting points, indexed by n, $$x_0 (n) = \frac{2 \pi \left( n - \frac{11}{8} \right)}{W \left( \frac{n - \frac{11}{8}}{e} \right)}$$ where $W(x)$ is the Lambert W function, which is the exact solution to the approximation equation for the Riemann zeros obtained by replacing $\vartheta$ with its Stirling approximation $$\tilde{\vartheta} (t) = \frac{t}{2} \ln \left( \frac{t}{2 \pi e} \right) - \frac{\pi}{8}$$ and solving $$\tilde{\vartheta} (x_0 (n)) = \left( n - \frac{3}{2} \right) \pi$$

However, even with this tanh modification, only about 98.02% of the first 100,000 zeta zeros converged to the n-th zero when started from the n-th approximation zero. The problem for many of these points, is that derivative of the map $$M_f (t_{}) = \frac {d}{d t} N_f (t) = {\frac {1}{ \left( \mbox {D} \left( f \right) \left( t \right) \right) ^{2}} \left( \left( \cosh \left( {\frac {f \left( t \right) }{\mbox {D} \left( f \right) \left( t \right) }} \right) \right) ^{2 } \left( \mbox {D} \left( f \right) \left( t \right) \right) ^{2}+f \left( t \right) \left( D^{ \left( 2 \right) } \right) \left( f \right) \left( t \right) - \left( \mbox {D} \left( f \right) \left( t \right) \right) ^{2} \right) \left( \cosh \left( {\frac {f \left( t \right) }{\mbox {D} \left( f \right) \left( t \right) }} \right) \right) ^{-2}} $$, which is the multiplier of the fixed point when evaluated at a zero of $f(t)$, is quite large when it happens to be near the edge of the basin of attraction, causing the trajectory to jump out of its immediate basin and land in a neighboring one.

I postulated that using the relaxed/damped Newton's iteration with a damping/relaxation factor $h_n$ which was inversely proportional to the multiplier $M_f(x_n)$ would result in much better convergence properties, so I defined the relaxation/damping factor at the m-th step to be $$h_m = \tanh (| M_f (x_m)^{- 1} |)$$ so that the iteration becomes $$N_f (t) = t - \tanh (| M_f (x_m)^{- 1} |)\tanh \left( \frac{f (t)}{\dot{f} (t)} \right)$$

and came to discover that this method of choosing the relaxation parameter results in quite excellent convergence properties.. convergence for about 99.976% of the zeros up to $n=200000$ . I should mention here that I actually use 3 starting points, if an iteration does not convergence from $x_0(n)$ then I try starting from $$\frac{(x_0 (n - 1) + x_0 (n))}{2}$$ and if that doesn't converge, starting from $$\frac{(x_0 (n) + x_0 (n + 1))}{2}$$, and finally if that doesn't converge, declaring the index n to be "not yet convergent" .

My question is, does this unrelaxed tanh iteration have a name or has it been studied? If so, does the method of choosing the relaxation parameter to be inversely proportional to the multipler have a name or has been studied before?

I have no doubt that good starting points exist for every root of $Z$ , and that extending this starting point selection method to do some sort of halving method similar to the bisection method of finding roots would do able to find a good starting point without having to do too much recursive subdividing of the interval $(\frac{(x_0 (n - 1) + x_0 (n))}{2},\frac{(x_0 (n) + x_0 (n + 1))}{2})$

p.s. I'm using the Hardy Z function as an example of a non-polynomial transcendental function with multiple zeros but it should work with any other function that Newton's method works on. If anyone is interested in the code I wrote to do this convergence, then send me a message. It is written in Java using JNA to wrap arblib which is way faster than Maple and does parallel processing to speed up the search. There are 47 indices <200,000 for which the method above converges to the wrong zero, i.e. starting from x0(n) it doesnt converge to x(n) but to some x(m) where m!=n, they are $$[21250, 22700, 28569, 35127, 35388, 48745, 54528, 55646, 62572, 67652, 73777, 80093, 86530, 86914, 88082, 89641, 92548, 96858, 98150, 100494, 103940, 104782, 115986, 120067, 128690, 130098, 132607, 133231, 141441, 141695, 147020, 149253, 152984, 155057, 156849, 158729, 163891, 164016, 165178, 171748, 175284, 180607, 182832, 187463, 190452, 192768, 196874]$$

For example, here is a graph of the iteration near the 1st non-convergent point n=21250

x0 := 19029.571920834291724

h1 := .15057973742585188810

x1 := 19029.557342014086776

h2 := 0.63772212783089483930

x2 := 19029.533810633781735

h3 := 1.

x3 := 19029.532171164322700

h4 := 1.

x4 := 19029.532198342701094

converges to wrong zero around 21250

p.s. this type of convergence analysis applied to the Grampoints was more straightforward . See Convergence of the Newton-Kantorovich Method for the Gram Points

crow
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  • I had never heard of this method before. – Jean Marie Feb 13 '17 at 19:43
  • Have you tried Newton's method for comparison? Does Newton's method converge for the 47 indices you listed? – user25406 Feb 13 '17 at 20:38
  • Yes, apparently the method I described has a convergence domain which is a superset of the regular Newton methods convergence domain. I'll post an image of the function around the first divergent point at n=21250 . The reason it diverges at this point is because the starting point is just to the left of a zero of Z' ,and doing the alternative left/right search doesn't work starting there either – crow Feb 13 '17 at 20:54
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    It's an interesting modification. It is easy to see why it works as $\tanh(x) \sim x$ when $x$ is small. When we are far away then it only allows for small $\leq 1$ changes in $x_n$ per step avoiding large overshooting (which can be further modified to $t-a\tanh(f/af')$ for some number $a$ if needed). This is probably very useful for functions with a large number of roots. – Winther Feb 13 '17 at 20:57
  • I tried this and all methods usually finds a root, however Newton's method seem to sometimes find a different root than the one you are interested in (due to overshooting). However I find convergence for all methods for the points you specify here. Is this convergence to the wrong point? E.g. for $n=22700$ then Newton's, tanh-Newton and your method all converges to $\sim 20161.9$ which is a root. – Winther Feb 13 '17 at 20:58
  • yes thats correct. they all converge but not necessarily to the point to which the starting point corresponds. i also applied some formulas of Argyros to determine the radii of the convergence balls and they agree.. the thing thats needed. i suppose one could invert these relations to show that the uniqueness ball is large enough to contain its previously found neighbor and its convergence ball small enough so that it doesnt contain any neighboring roots – crow Feb 13 '17 at 21:05
  • Just to clarify: $t = 19029.53$ is a root of $Z(t)$. What root is it supposed to converge to in this case? – Winther Feb 13 '17 at 21:19
  • 19029.53... is the 21249-th root . If the iteration doesn't converge to 19029.53 starting from x0(21249) then try starting from (x0(21248)+x0(21249))/2 or (x0(21249)+x0(21250))/2 . I'm checking convergence by using the tables found at http://www.dtc.umn.edu/~odlyZko/zeta_tables/index.html if you start at a root, you will stay there, since its a fixed-point.. I believe the modification im using somehow guarantees that it will converge to the root inside the immediate basin of attraction in which the starting point is found.. although I have not yet derived a proof of that. – crow Feb 13 '17 at 21:26
  • Winther, that makes sense .. I applied this method to Wikipedias prototypical example of a function that Newtons method suffers overshooting with, |x|^a with 0<a<0.5... it converges quite rapidly whereas neither Newtons iteration or my "unrelaxed tanh Newton method" does not but the "auto-relaxed tanh Newton iteration" does. https://en.wikipedia.org/wiki/Newton%27s_method#Overshoot – crow Feb 14 '17 at 16:30

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