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The question is not very clear. What mea...

vv 14257
November 26 2020
0 0

The question is not very clear. What means that f is continuous over an interval J? Must f be continuous at each point of J, or the restriction f|J must be continuous? How e removable discontinuity is it treated?

Note. Suppose that discont returns the empty set. We cannot conclude that f is continuous on R, because probably discont ignores the branches which define f (even if usually they are chosen such that f be continuous on R when possible).

 

@Assinat  A^+ is the transpose of ...

vv 14257
November 25 2020
0 0

@Assinat 

A^+ is the transpose of the matrix A. 

column-red.mw

Column reduction ?...

vv 14257
November 25 2020
0 0


 

> 

B:=Matrix(8, 8, [[1, -7, 0, 0, 0, 0, 0, 0], [-3, 21, 0, 0, 0, 0, 0, 0], [-1, 8, 1, -7, 0, 0, 0, 0], [2, -17, -3, 21, 0, 0, 0, 0], [1, -8, -1, 8, 1, -7, 0, 0], [-2, 9, 2, -17, -3, 21, 0, 0], [-1/3, 16/3, 1, -8, -1, 8, 1, -7], [4/3, -3, -2, 9, 2, -17, -3, 21]]);

Matrix(8, 8, {(1, 1) = 1, (1, 2) = -7, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (1, 6) = 0, (1, 7) = 0, (1, 8) = 0, (2, 1) = -3, (2, 2) = 21, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (2, 6) = 0, (2, 7) = 0, (2, 8) = 0, (3, 1) = -1, (3, 2) = 8, (3, 3) = 1, (3, 4) = -7, (3, 5) = 0, (3, 6) = 0, (3, 7) = 0, (3, 8) = 0, (4, 1) = 2, (4, 2) = -17, (4, 3) = -3, (4, 4) = 21, (4, 5) = 0, (4, 6) = 0, (4, 7) = 0, (4, 8) = 0, (5, 1) = 1, (5, 2) = -8, (5, 3) = -1, (5, 4) = 8, (5, 5) = 1, (5, 6) = -7, (5, 7) = 0, (5, 8) = 0, (6, 1) = -2, (6, 2) = 9, (6, 3) = 2, (6, 4) = -17, (6, 5) = -3, (6, 6) = 21, (6, 7) = 0, (6, 8) = 0, (7, 1) = -1/3, (7, 2) = 16/3, (7, 3) = 1, (7, 4) = -8, (7, 5) = -1, (7, 6) = 8, (7, 7) = 1, (7, 8) = -7, (8, 1) = 4/3, (8, 2) = -3, (8, 3) = -2, (8, 4) = 9, (8, 5) = 2, (8, 6) = -17, (8, 7) = -3, (8, 8) = 21})

 (1)
> 

K := LinearAlgebra:-NullSpace(B); k:=nops(K):

{Vector(8, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 49, (6) = 7, (7) = 0, (8) = 1}), Vector(8, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = -7, (6) = -1, (7) = 1, (8) = 0})}

 (2)

It seems that you want a column reduction:

> 

(LinearAlgebra:-ReducedRowEchelonForm(`<|>`(K[])^+))[[seq(k..1,-1)]]^+;

Matrix(8, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = 0, (3, 1) = 0, (3, 2) = 0, (4, 1) = 0, (4, 2) = 0, (5, 1) = 0, (5, 2) = 1, (6, 1) = 0, (6, 2) = 1/7, (7, 1) = 1, (7, 2) = 0, (8, 1) = 1/7, (8, 2) = 1/49})

 (3)
> 

 

 

Not quite....

vv 14257
November 24 2020
0 0

@Carl Love Not quite. E.g. for p=1 your result is complex.
 

homogeneous coordinates...

vv 14257
November 24 2020
0 0

@Earl  The proc uses practically "homogeneous coordinates". See Homogeneous coordinates - Wikipedia and the references therein. You may want to check directly that the image under f of three collinear points are also collinear. 

BF...

vv 14257
November 22 2020
0 0

@mmcdara You can use this version of BruteForce and then 5/121 works. It also includes your first proc and obtains other representations too, e.g. BF(13/12, 2) = 1/2+1/3+1/4

BF := proc(a, mindenom:=1)
  local A:=a, b:=mindenom-1, s:=NULL:
  while A > 0 do
    b := max(ceil(1/A),b+1):
    A := A-1/b:
    s := s, 1/b:
  end do:
  return s;
end proc:

 

@student_md  It's not difficul...

vv 14257
November 19 2020
0 0

@student_md 

It's not difficult to see yourself that  int(Psi(s), s=0..t) = P(1)*Psi(t)

implies Psi(t)=0. [it reduces to a simple linear ODE].

Psi = 0?...

vv 14257
November 19 2020
0 0

@student_md 

Let's take for simplicity the dimension N=1.
Then the first line in Eq (1) ==> Psi(t) = 0. So, your method does not seem to be useful.

+1...

vv 14257
November 18 2020
0 0

@pjv70 

It is hard without sources. If you just want the correct series, simply add 1, or use:

JT3:=convert(JacobiTheta3(z,q), Sum, dummy = n) assuming abs(q)<1:
series(JT3, z=0, 4);

          

symbolic?...

vv 14257
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 14257
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 14257
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 14257
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 14257
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 14257
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"? 

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