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symbolic?...

vv 13805
November 14 2020
0 0

@Carl Love Maple gives only truncated series, not the full ones. It is easy to write the recurrence equations for the coefficients, but rsolve will not be able to solve the system.

symmetric...

vv 13805
November 14 2020
0 0

@emendes 

It's not clear for me how are you going to use the solutions with the symmetric ones removed. If the system is not symmetric, it will be impossible to know whether (x=a,y=b)  implies (x=b,y=a) too. Also, if the system is symmetric, it is better to treat it as such, returning only (x+y, x*y).

Comment...

vv 13805
November 04 2020
0 0

@Kitonum Yes, Mathematica's answer is much better, but my comment was about your remarks which were both incorrect.

Not quite...

vv 13805
November 03 2020
0 0

@Kitonum 

Not quite (for minimize/maximize):

> 

minimize(x^2+a*x+1, x=-infinity..infinity) assuming a::real;

1-(1/4)*a^2

 (1)
   
> 

minimize(x^2+a*x+1, x=-1..1) assuming a>2;

2-a

 (2)
> 

minimize(x^2+a*x+1, x=-1..1) assuming a<2,a>-2;

1-(1/4)*a^2

 (3)
> 

minimize(x^2+a*x+1, x=-1..1) assuming a<-2;

2+a

 (4)
> 

 

 

Edit. The result of Optimization:-Minimize(f, x=0..6) is not a bug, x=2 is a local minimum.
For a global minimum use:
Optimization:-Minimize(f, x=0..6, method=branchandbound);
        [-2., [x = 6.]]

 

 

@MapleEnthusiast  Th...

vv 13805
November 03 2020
0 0

@MapleEnthusiast 

The number of epsilon's depends on which pi[k,j]  are eliminated:

  

> 

restart;
> 

System := R[i] = sum(pi[i, j]*(-epsilon*sum(pi[k, j]*((R[i] - R[k]) + (-delta[i] + delta[k])), k = 1 .. J) + R[j]), j = 1 .. J):
> 

eq1 := simplify(expand(subs(J = 3, i = 1, System))):
> 

eq2 := simplify(expand(subs(J = 3, i = 2, System))):
> 

eq3 := simplify(expand(subs(J = 3, i = 3, System))):
> 

J := {$ (1 .. 3)}:
> 

Con := {seq(add(pi[i, j], i = J) = 1, j = J)}:
> 

SCon:=solve(Con);

{pi[1, 1] = 1-pi[2, 1]-pi[3, 1], pi[1, 2] = 1-pi[2, 2]-pi[3, 2], pi[1, 3] = 1-pi[2, 3]-pi[3, 3], pi[2, 1] = pi[2, 1], pi[2, 2] = pi[2, 2], pi[2, 3] = pi[2, 3], pi[3, 1] = pi[3, 1], pi[3, 2] = pi[3, 2], pi[3, 3] = pi[3, 3]}

 (1)
> 

A := simplify(solve(simplify({eq1, eq2, eq3}, Con), {R[1], R[2]})):
> 

diff(subs(R[3] = 0, delta[1] = 0, delta[3] = 0, numer(rhs(A[1]))), delta[2]);
> 

numboccur(%, epsilon);
indets(%%);

-epsilon*pi[1, 1]*pi[2, 1]*pi[2, 2]+epsilon*pi[1, 2]*pi[2, 1]^2+epsilon*pi[1, 2]*pi[2, 3]^2-epsilon*pi[1, 3]*pi[2, 2]*pi[2, 3]+epsilon*pi[1, 1]*pi[2, 1]-epsilon*pi[1, 2]*pi[2, 1]-epsilon*pi[1, 2]*pi[2, 3]+epsilon*pi[1, 3]*pi[2, 3]

  

8

  

{epsilon, pi[1, 1], pi[1, 2], pi[1, 3], pi[2, 1], pi[2, 2], pi[2, 3]}

 (2)
> 

B := simplify(solve(simplify({eq1, eq2, eq3}, Con), {R[1], R[3]})):
> 

DB:=diff(subs(R[2] = 0, delta[1] = 0, delta[2] = 0, numer(rhs(B[1]))), delta[3]);
> 

numboccur(%, epsilon);
indets(%%);

-epsilon*pi[1, 1]^2*pi[2, 3]+epsilon*pi[1, 1]*pi[1, 3]*pi[2, 1]-epsilon*pi[1, 1]*pi[2, 1]*pi[2, 3]-epsilon*pi[1, 2]^2*pi[2, 3]+epsilon*pi[1, 2]*pi[1, 3]*pi[2, 2]-epsilon*pi[1, 2]*pi[2, 2]*pi[2, 3]+epsilon*pi[1, 3]*pi[2, 1]^2+epsilon*pi[1, 3]*pi[2, 2]^2+epsilon*pi[1, 1]*pi[2, 3]+epsilon*pi[1, 2]*pi[2, 3]-epsilon*pi[1, 3]*pi[2, 1]-epsilon*pi[1, 3]*pi[2, 2]

  

12

  

{epsilon, pi[1, 1], pi[1, 2], pi[1, 3], pi[2, 1], pi[2, 2], pi[2, 3]}

 (3)
> 

expand(eval(DB, SCon));
numboccur(%, epsilon);
indets(%%);

epsilon*pi[2, 1]*pi[3, 1]*pi[3, 3]+epsilon*pi[2, 2]*pi[3, 2]*pi[3, 3]-epsilon*pi[2, 3]*pi[3, 1]^2-epsilon*pi[2, 3]*pi[3, 2]^2-epsilon*pi[2, 1]*pi[3, 1]-epsilon*pi[2, 2]*pi[3, 2]+epsilon*pi[2, 3]*pi[3, 1]+epsilon*pi[2, 3]*pi[3, 2]

  

8

  

{epsilon, pi[2, 1], pi[2, 2], pi[2, 3], pi[3, 1], pi[3, 2], pi[3, 3]}

 (4)
> 

 

Download puzzle4-vv.mw

math...

vv 13805
October 28 2020
0 0

@MapleEnthusiast I mean a mathematical reason, not heuristics. Note also that you have simplified wrt size.

(1+e)^3 = 1 + 3e + 3e^2 + e^3: e appears once in LHS and 3 times in RHS. Is this "puzzling"? 

A,B...

vv 13805
October 28 2020
0 0

@MapleEnthusiast So, the solutions for A and B are different (and you also took only the numerators!). There is no reason to obtain similar results. If you construct a simpler system (for yourself!) you will be able to check by hand and understand & see the problem.

puzzle?...

vv 13805
October 28 2020
0 0

@MapleEnthusiast Why should they be the same? Note that the R[1,1] could be different in A and B.
If you want to understand exactly what happens, take a smaller system with 2 or 3 parameters and don't use (so many) indices.

op(0,expr)...

vv 13805
October 28 2020
0 0

@Carl Love Actually, for an expr having a type other than rtable,table, list, Grid:-Map returns 
op(0,expr)(result);
It could be indeed an oversight. Or maybe they want to avoid a (possibly huge) set in the answer.

Rank...

vv 13805
October 20 2020
0 0

@acer Yes, I know, but it is anyway unstable; mathematically Rank : R^(n^2) --> N is discontinuous.

So, you are asking for more than the men...

vv 13805
October 15 2020
0 0

So, you are asking for more than the mentioned NP-hard problem!

0...

vv 13805
October 15 2020
0 0

@MapleEnthusiast The matrix has many symbolic parameters, so the pseudo inverse will be huge. But what do you want to do with it? For a system Ax=b, the pseudo inverse A' is generally used to compute the least squares solution A'b (as Carl said). Your system is homogeneous (b=0), so, A'b=0.

pseudo inverse always exists...

vv 13805
October 14 2020
0 0

@MapleEnthusiast The More-Penrose inverse exists for any matrix (even non square or 0). But I don't see how could it be useful here.
As in my example, the matrix of your system is singular if the relations are considered. So, it does not make sense to solve it before replacing the dependent parameters. 

example...

vv 13805
October 14 2020
0 0

@MapleEnthusiast 

 

Take a simple similar situation.

You have the system

> 

sys := {a*x + b*y + c*(a-b) = 0, x+y+c+d = 0};

{a*x+b*y+c*(a-b) = 0, x+y+c+d = 0}

 (1)

and the parameters satisfy the relations

> 

rels:=[a-b=0, c+d=0];

[a-b = 0, c+d = 0]

 (2)
> 

solve(sys,{x,y});

{x = -(a*c-2*b*c-b*d)/(a-b), y = -(a*d+b*c)/(a-b)}

 (3)
> 

simplify(%, rels);

Error, (in simplify/siderels:-Recurse) numeric exception: division by zero

  
> 

sys1:=simplify(sys, rels); # now the system is homogeneous (exactly like yours)

{a*(x+y) = 0, x+y = 0}

 (4)
> 

solve(sys1, {x,y});

{x = -y, y = y}

 (5)

Using a pseudo inverse cannot help here.

 

elsewhere?...

vv 13805
October 14 2020
0 0

@nm Now I can't find this info either. Probably I read about it elsewhere.

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