ComputerUser

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Seldom to ask question after retired math hobby Just waiting for beauty who born in 1994 And waited for her email to mavio@protonmail.com What is the difference in ownership among different universe?

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sorry for misunderstanding, the goal is to convert parametric equation back to non-parametric equation, only contains x0 and x1

yes i want.

is my script correct?

can my method apply to other curves?

yes i want.

is my script correct?

can my method apply to other curves?

@Markiyan Hirnyk 

i guess to use subs(x0=a0*u+a1*v, subs(x1=a2*u+a3*v)....  then use match original equation = new equation to find a0, a1, a2

@Markiyan Hirnyk 

when i try match, i do not know g1 and g2

eq1 := -x2^3+x1^3+x0*x1^2 - expand(u*g1 + v*g2) = 0;
match(eq1, {x0,x1,x2}, 'u,v');

@Markiyan Hirnyk 

 

c := -x2^3+x1^3+x0*x1^2;

solve(c, [x0(u,v), x1(u,v), x2(u,v)]);

solve(-x2^3+x1^3+x0*x1^2=0, [x0(u,v), x1(u,v), x2(u,v)]);

solve(-x2^3+x1^3+x0*x1^2, [x0(t), x1(t), x2(t)]);

match(-x2^3+x1^3+x0*x1^2, {x0,x1,x2}, 's');

after tried result in no solution, is it impossible?

@Preben Alsholm 

 

@Preben Alsholm

actually the goal is to create a numeric Matrix b to generate oridinary generating function or exponential generating function, as it is my conjecture after reading a book

i assume a matrix like the following, after calculation, assume the result as follows.

to prove this conjective is feasible, i have to use hermit generating function to do a test.

use this generating function to find back matrix b to see what is look like which is this question

 

why backward calculation to find back the matrix is not a numeric matrix when forward calculation from numeric matrix is possible

 

restart;

with(LinearAlgebra):

with(inttrans):

with(SumTools):

M2 := Matrix(3,3):

M2[1,1] := 1:

M2[2,2] := 1:

b := M2+Matrix([[1, 2, 0],[1, 2, 1], [1, 2, 0]]);

Determinant(b);

invb := MatrixInverse(b);

test1 := MatrixMatrixMultiply(MatrixMatrixMultiply(MatrixMatrixMultiply(Matrix([[f,f^2,1]]),invb),Matrix([[x,1,1],[1,x^2,1],[1,1,x^3]])),b);

test2 := test1[1][1];

test1r := solve(test1[1][1]=0,f);

test1r2 := subs(x=f,subs(z=y,test1r[1]));

ogf := solve(z = test1r2,f);

@Carl Love 

after handle this, it can be plotted now how to animate to plot many times

@Carl Love 

it is great

@Markiyan Hirnyk 

after read this complex piecewise function, i think that i can not implement this in c#

@Markiyan Hirnyk 

after read this complex piecewise function, i think that i can not implement this in c#

My purpose is to change compositional rule of inference algorithm from Maple to c#

My purpose is to change compositional rule of inference algorithm from Maple to c#

@PatrickT 

Easy to use, look simpler than mathematica

many algorithms i do not know to write in programming language, so i use

The operator before math.pow disappear

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