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These are replies submitted by vv

The usual way of presenting this would be:

Write the projection as a procedure. E.g. the projection onto the unit sphere is
P:=(x,y,z) -> (x/sqrt(x^2+y^2+z^2), y/sqrt(x^2+y^2+z^2), z/sqrt(x^2+y^2+z^2));

Define the curve. E.g.
C:=t -> (cos(t),sin(t),t);

Plot the projection:
plots:-spacecurve( [P(C(t))],t=0..8*Pi);
plots:-animate(plots:-spacecurve, [[P(C(t))],t=0..a],a=0..8*Pi );

Then your posts will have a larger audience.







For rather small intervals you can check all the possible cases.

Once you have a plausible formula, you could try to compute the sum using Maple (with assume facility, maybe fixing some parameters). Maple is "competent" in computing this kind of sums.

But if you have derived correctly the formula, a comprehensive random test should be enough for your peace of mind.


Yes, it's odd that the rotation matrix for rotate(Plot, u,v,w)  is Rz(-w).Ry(-v).Rx(-u).

You have to write mathematically the problem.

For example, a similar problem (actually more interesting) would be:

Find x  for which the path (on the real line):
        b -->  x -->  d --> a --> x --> b --> x --> c  --> d --> x -->  b

has a minimal length.

write a function and plot it.



In this new version you should use:

Bdyprt1 := unapply('display'('point'([d[1][1], d[1][2], d[1][3]]), colour = red, symbolsize = 50, symbol = solidsphere), t):
Bdyprt1a := 'animate'('Bdyprt1', [t], t = 0 .. 8*Pi, frames = 100):

(but your approach does not seem very "natural").

Unfortunately the situation seems to be worse in Windows for 3d vector graphics:



I agree!  But you must admit that it is not normal that a syntax accepted by  line is rejected by its twin brother points.


The unpleasant fact is that absolutely similar procedures such as line and points have distinct evaluation rules which must be guessed by trial and error.


Yes, I know the theorem. It applies usually to extend (y0,z0) to (x0,y0,z0) and here the extension is not always possible, even in very simple examples. (I have used the notations from the question).

But I have not seen an impossible extension from (z0) to (x0,y0,z0). Have you?

Thank you for the answer.



point does some strange manipulations. Workaround:

ff:=unapply(display(POINTS([d[1][1], d[1][2], d[1][3]]), colour = red, symbolsize = 50, symbol = solidsphere), t);


@mehdi jafari 

The syntax problem is that (applied to your example):

fsolve({seq(eq||i,i=1..3)},{a1=-5..0,a2,a3});  #error
fsolve({seq(eq||i,i=1..3)},{a1=-5..0,a2=-infinity..infinity,a3=-infinity..infinity}); #ok

(Or set the ranges separately, after vars, as you did).


Have you read the answer+comments+example below?


You have two elementary mistakes:

1. You define xi := 2*x/a - 1, eta := ...
After that, how could you use xi, eta as new variables in Change?

2. (A more subtle one) You define the procedure U0 in terms of xi, eta (already defined).
This is not correct, you must use unapply in such cases.

@Markiyan Hirnyk 

And what was your "base" recommending  MatrixPolynomialAlgebra:-SmithForm instead of LinearAlgebra:-SmithForm
when they are the same?

@Markiyan Hirnyk 

Have you tried the example?
Please note that exactly the same text appears in the help page of LinearAlgebra[SmithForm].

P.S. You are free to use it for complex coefficients. I will not.

BTW, MatrixPolynomialAlgebra:-SmithForm  calls  LinearAlgebra:-SmithForm
so everything should be clear.

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