one man

Alexey Ivanov

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9 years, 139 days
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f := exp(t)+exp(-t^2-2*t); 
g := exp(-t^2)+exp(-t^2-2*t); 
solve([f-g], t); 
plot([f, g], t = -3 .. 3, discont = true, color = [red, blue])


You have one equation with three variables, It has a periodic solution in the form of inclined "tubes". Solution for a single "tube" can be obtained numerically using the Draghilev method, or another numerical method. After that, you can select the desired range for each variable from the resulting set of solutions.
In Maple Pi, not pi.

restart; with(plots):
 f := cos(2*Pi*(x+y-2*z))+cos(2*Pi*(y+z-2*x))+cos(2*Pi*(z+x-2*y)); 
#solve({f, 0 <= x, x <= 1, y <= 1, z <= 1, x < y, y < z}, [x, y, z], explicit);
implicitplot3d(f, x = -.5 .. .5, y = -.5 .. .5, z = -.5 .. .5, axes = normal, numpoints = 5000, color = blue, transparency = .5, style = surface, scaling = constrained, axes = boxed)


I remember that similar questions have already arisen many times.
For example, there is such a technique 

a := `(3*Pi)/10`; print(a)


In this case, this method gives a parametric solution. In the text of the program, the Draghilev method is highlighted in green.
(If satisfied with numerical parameterization, the examples can be very complex. But as far as analytical parameterization is concerned, even in this case, the method can perform an example more complicated.)



  solve(r2,{ksi});  or  fsolve(r2); 

And if just combine "transparency" and "thickness"?
For example

restart: with(plots):
p1:=plot(f1(x),x=0..2,color=red,legend = f1(x)): 
p2:=plot(f2(x),x=0..2,color=blue,legend = f2(x)): 
p3:=plot(f3(x),x=0..2,color=green,thickness = 10, transparency = .7, legend = f3(x)): 


Just in case, search for "Draghilev method" (or Dragilev here: )
The method, in particular, finds continuous solutions of systems with free variables (underdetermined systems of equations).

You can do this:

 a := plot({seq((6*x-2*t)/x^2, t = 1 .. 3)}, x = -1 .. 5, y = -1 .. 6):
 b := plot(3/x, x = 0 .. 5, y = -1 .. 6, color = black, thickness = 3):
 plots[display](a, b)


Since childhood, I try to avoid strict zeros in coordinates (and generally avoid strictly identical values) due to formulas, because expressions may be nullified after substitution. It is better to shift the point o or d along the oX axis, for example:

point(o, 0.1e-11, 0.); 
point(A, 0., 1.); 
point(d, 0., 2.);
point(F, .8944271920, 1.4472135960); 
line(lOD, [o, d]); 
line(lAF, [A, F]); 
alpha := FindAngle(lOD, lAF);

alpha = 1.107148718

CompleteSquare(x^2+y^2-2*x-y-2 = 10, x);


EQ2 :=-1186578.220*R*k^2*wr-312683.0293*k^5-288960.9621*k^3*R:
 allvalues(solve([EQ1, EQ2], [k, wr]));


Looks like he's alive and well. Those who have lost hope of communicating with this person can find him, for example, here and here.


allvalues(solve({eq1, eq2}, {A, B}));

For calculating the kinematics of the manipulator, Maple of almost any version is quite suitable directly. If a system of nonlinear equations is used to describe the model of a manipulator, then it is very easy mathematically to fix any desired degree of freedom.
Perhaps you can be useful the messages, which can be found at this link:


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