Mark McClure
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 Feb 1 awarded Nice Answer Jan 18 comment Using Solve to solve the equation $x^{1/3}=-1$ I don't think there is a solution. I believe that the image of the complex plane under $z\to z^{1/3}$ is $\{r e^{i\theta}: r\geq 0, -\pi/3 < \theta \leq \pi/3\}$, given the location of the branch cut. Jan 17 awarded Yearling Jan 11 awarded Enlightened Jan 11 awarded Nice Answer Jan 6 awarded Enlightened Jan 6 awarded Nice Answer Jan 2 awarded Nice Answer Dec 24 comment Revolution of Koch Snowflake @ubpdqn Thanks - and same to you! Dec 24 revised Revolution of Koch Snowflake deleted 4 characters in body Dec 24 answered Revolution of Koch Snowflake Dec 11 revised Numerically evaluating an integral related to Cantor's staircase added 298 characters in body Dec 11 revised Numerically evaluating an integral related to Cantor's staircase added 690 characters in body Dec 11 comment Numerically evaluating an integral related to Cantor's staircase @J.M. Glad you like it! Dec 10 answered Numerically evaluating an integral related to Cantor's staircase Dec 7 comment Basins of attraction of equilibrium points For the system $x'=V_x,y'=V_y$, an equilibrium will be attractive iff it's a local minimum of the potential function. Before proceeding further, though, you should make sure that's the system you're interested in. Perhaps, you're really interested in a second order system, which is likely in celestial mechanics. That's a more complicated question, as you've got to deal with initial conditions on $x'$ and $y'$ as well. Dec 7 comment Basins of attraction of equilibrium points As you have shown, the algebraic system $V_x=0,V_y=0$ has four numeric solutions and, if we apply Newton's method to that system, each of those solutions has an interesting basin of attraction. But that is a different question from whether those equilibrium points are attractive under a system of differential equations like $x'=V_x,y'=V_y$. Dec 7 comment Basins of attraction using Newton's method Part II I agree with Quantum_Oli - it would be nice to know where this came from. It appears there might be an interesting question here but the image looks like the Julia set of a polynomial - I can't imagine how your non-differentiable function generated it. Dec 7 comment Basins of attraction of equilibrium points I'm quite sure that the basins of attraction of Newton's method in two real variables have been well studied; there's an elementary discussion on pages 486-487 of Gil Strang's calculus text. If you're interested in studying the basins of attraction of the smooth dynamical system with a given potential, however, that's quite a different thing. Only one of your equilibrium points is attractive for that system. Dec 5 comment Basins of attraction using Newton's method The first thing my answer to that question does is generalize the code to arbitrary functions. I guess that's what @Saurav is referring to.