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Are there any circumstances under which using a value $w < 0$ would help us find a solution in over-relaxation faster than we can with the ordinary relaxation method?

Over Relaxation Method:

$$x'= [1 + w]f(x) - wx$$

Example

Calculating $x = 1-e^{-3x}$

Take x = 1 as initial value, and w as 0.2

x' = (1+0.2)f(1)-0.2(1) = 0.94025551795

x' = (1+0.2)f(0.94025551795)-0.2(0.94025551795) = 0.94047657354

x' = (1+0.2)f(0.94047657354)-0.2(0.94047657354) = 0.94047974478

x' = (1+0.2)f(0.94047974478)=0.2(0.94047974478) = 0.94047979005

We stop until the value get to a certain accuracy

Why would the over-relaxation reach the solution faster if we consider $w < 0$ in non-linear function such as $x = 1 - e^{(1 - x^2)}$?

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    $\begingroup$ Cross posted on Physics: physics.stackexchange.com/q/468753/25301 $\endgroup$ – Kyle Kanos Mar 26 at 14:31
  • $\begingroup$ It seems like you are trying to find the roots of a nonlinear equation using fixed-point iteration. It also seems that you are using the opposite signs for the $w$, compared with the usual convention. I would say that the other sign looks more natural (to me) because it resembles a convex linear combination. $\endgroup$ – nicoguaro Apr 1 at 15:39
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You don't say which method you are using (Jacobi? SSOR?) but in general, if you choose the weight negative, the iteration you are using is no longer a contraction and you will not converge at all.

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  • $\begingroup$ The relaxation method is using the following equation to find the solution: x'=(1+w)f(x)-wx. The instructor ask us to consider the equation x = 1-e^(1-x^2) because the property of such non-linear equation would result in w < 0 getting a solution faster than normal. But I still don't get why it would result in a solution faster even I plot the graph out. $\endgroup$ – Heyyyyyy Apr 1 at 5:06
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No, it won't work. Kahan's theorem assures you that it will converge if w is between 0 and 2

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