one man

Alexey Ivanov

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5 years, 108 days

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@fereydoon_shekofte  Sorry, I know English worse than I know Maple.

A small addition to the text, and we have a tangent cube.
One face of the cube belongs to the tangent plane.

Thank you, Yuri Nikolaevich, r: = sqrt (add (d [i] ^ 2, i = 1..nops (d))) is already available to my understanding.

r: = sqrt (`+` (`~` [`^`] (d, 2) [])); Instead of r: = sqrt (add (`~` [`^`] (d, 2))); ?
(I'll never understand this.)
Many thanks, now everything works.


I do not know how to use the "proc" function. I wanted, at least formally, to master this function also for another task.
But your  text is reproduced with an error. (My max available version is Maple 17)
Look, please, what's wrong:

> K := NPar([f1, f2], X0, 9.6, 100);

Error, (in NPar) invalid input: add expects 2 arguments, but received 1



@vv  I understand what feelings arise for professionals when they see my texts. Yes, I met Maple, but I did not learn to use it. I borrowed pieces of text, looked help. There was only a desire that the idea be confirmed and the algorithm will work.
But I will try to behave more civilized.
Thank you for specific assistance and constant attention.

@Ronan  Thank you for your interest to this topic.
Want to draw your attention that for my part there is practically no reaction to the replicas of a certain person (I will not mention his name). For myself, I have long concluded that it is useless to communicate with him for many reasons, or rather, even harmful to one's own reputation and health.

@fereydoon_shekofte  Thank you for your kind words.
When I had a job, I worked on developing algorithms, but I was not able to implement many ideas. A few years ago I met Maple, and Maple inspired me to return to those tasks.
I would be happy if this was useful for someone.

Do you understand that the graphs depict the equations of the first derivatives of f (\ epsilon, \ phi)?
You do not show me the f (\ epsilon, \ phi), and I myself integrated the partial derivatives and looked at the plot of
the f (\ epsilon, \ phi) with help of plot3d. And you yourself can see that the minimum of f (\ epsilon, \ phi) in this definition area is an infinite flat set.


Sorry, I still do not see your original function f (\ epsilon, \ phi).
As for your last equations, I took them from fsolve, and denoted x and y.
On an implicit graph x, it is clear that \ epsilon is practically 0, and the graph of y is empty.

I think you still have these options:
1) to look for min ((f (\ epsilon, \ phi)) ^ 2). It's working with the function itself without derivatives.
2) scale the variables "\ epsilon" and "\ phi" so that the range "\ phi" is much wider for fsolve.

@kuwait1 Ah, here's the thing. And where is your f (\ epsilon, \ phi)?

By the way, it may well be that your min has an infinite number of solutions.

@kuwait1  You really have only one equation: either x or y. There are many ways to solve one equation with several variables. First try the easiest way: set the desired values of the epsilon and  then solve one equation for another variable using fsolve.

The graphs show that your equations are equivalent:
with(plots, implicitplot):
implicitplot(x, `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = red);
implicitplot(y, `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = blue);
implicitplot([x, y], `ε` = -5 .. 5, phi = -5 .. 5, numpoints = 20000, color = [red,blue]);



Additional curling of a Möbius strip (rolling without slipping).

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