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9 years, 72 days
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Logical equivalence...

vv 13118
February 25 2023
1 3

The equivalence is logical equivalence indeed. For the simple expression q := (x[1] &or x[2]) &and x[3],  the only nonidentical permutation is f defined by f(x1)=x2, f(x2)=x1, f(x3)=x3 [so, f(not x1) = not x2 etc].

As the help file (and  Christian Wolinski) says, the expression must be in CNF form. It is strange that Normalize (recommended in the help) does not work and we must use Canonicalize:

with(Logic);
q := (x[1] &or x[2]) &and x[3];
qn := Normalize(q, form=CNF);                       # wrong!
qn := Canonicalize(q, {x[1],x[2],x[3]}, form=CNF);  # ok
G, L := SymmetryGroup(qn, output = [group, expressions]);
#g1, g2 := Generators(G)[];
GroupOrder(G);  # 2,  ok

 

O...

vv 13118
February 18 2023
2 0

Here `O` is simply a local variable lost from some procedure and landed at top level, due to a bug.

Invariance of domain...

vv 13118
February 16 2023
0 0

According to a well known theorem, the Invariance of domain, an interval such as (380,780) cannot be homeomorphic to  (0,1)^3, so, a correspondence between WaveLength and (r,g,b) will be always problematic.

eliminate...

vv 13118
February 04 2023
2 1
> 

Pn := [ (5*X + 12*Y)*(-384 + 48*X + 55*Y)/(225*X^2 + 225*Y^2 - 1800*X - 602*Y),
         2*(12*X - 5*Y)*(-384 + 48*X + 55*Y)/(225*X^2 + 225*Y^2 - 1800*X - 602*Y)]:
> 

(x,y) =~ Pn[]

x = (5*X+12*Y)*(-384+48*X+55*Y)/(225*X^2+225*Y^2-1800*X-602*Y), y = 2*(12*X-5*Y)*(-384+48*X+55*Y)/(225*X^2+225*Y^2-1800*X-602*Y)

 (1)
> 

eliminate( {%, Y= 2+3*X}, {X,Y})

[{X = (4/3)*(38274*x+38914*y-126451)/(30102*x-25053*y-131066), Y = -(213300*x+105550*y-767936)/(-30102*x+25053*y+131066)}, {738*x^2+2133*x*y+1240*y^2-2952*x-8173*y}]

 (2)
> 

conic:=%[-1][];

738*x^2+2133*x*y+1240*y^2-2952*x-8173*y

 (3)
> 

plots:-implicitplot(conic, x=-30..30,y=-30..30);

 
> 

 


Download conic.mw

By hand...

vv 13118
January 28 2023
3 5
> 

B:=<0,-2,1;-1,0,0;-2,0,0>; A:=B+1;

Matrix(3, 3, {(1, 1) = 0, (1, 2) = -2, (1, 3) = 1, (2, 1) = -1, (2, 2) = 0, (2, 3) = 0, (3, 1) = -2, (3, 2) = 0, (3, 3) = 0})

  

Matrix(%id = 18446745955451478974)

 (1)
> 

# By hand:
> 

B^2,B^3

Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = -1, (3, 1) = 0, (3, 2) = 4, (3, 3) = -2}), Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

 (2)
> 

1 + n*B + binomial(n,2)*B^2:  'A'^n = expand~(%)

A^n = (Matrix(3, 3, {(1, 1) = 1, (1, 2) = -2*n, (1, 3) = n, (2, 1) = -n, (2, 2) = n^2-n+1, (2, 3) = -(1/2)*n^2+(1/2)*n, (3, 1) = -2*n, (3, 2) = 2*n^2-2*n, (3, 3) = -n^2+n+1}))

 (3)
   
> 

# By Maple:
> 

LinearAlgebra:-MatrixPower(A, n);

Matrix(3, 3, {(1, 1) = 1, (1, 2) = -2*n, (1, 3) = n, (2, 1) = -n, (2, 2) = n^2-n+1, (2, 3) = -(1/2)*n^2+(1/2)*n, (3, 1) = -2*n, (3, 2) = 2*n^2-2*n, (3, 3) = -n^2+n+1})

 (4)

 

Answer...

vv 13118
January 20 2023
3 4
> 

S:=Sum(Sum( a^2*b^2/sinh(Pi*(a+b))*(-1)^(a+b),  a=1..infinity), b=1..infinity);
# The double series converges very fast!

Sum(Sum(a^2*b^2*(-1)^(a+b)/sinh(Pi*(a+b)), a = 1 .. infinity), b = 1 .. infinity)

 (1)
> 

evalf[30](S);

0.265258238486492226281472938954e-2

 (2)
> 

S=identify(%);

Sum(Sum(a^2*b^2*(-1)^(a+b)/sinh(Pi*(a+b)), a = 1 .. infinity), b = 1 .. infinity) = (1/120)/Pi

 (3)
> 

# The equality holds but the proof is to be found!


 

sqrt(Pi)...

vv 13118
January 18 2023
2 0
> 

str:="(1/(2^
       j) ((k*Gamma[5 + 2 j] Gamma[
          1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,
           7/2 + j + l/2}, -1])/
       Gamma[6 + 2 j +
         l] + ((k + m) Gamma[7 + 2 j] Gamma[
          1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,
           9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -
       l) Gamma[5 + 2 j] Gamma[
     1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,
         3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +
      1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1,
         7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1]))";

"(1/(2^ j) ((k*Gamma[5 + 2 j] Gamma[ 1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1])/ Gamma[6 + 2 j + l] + ((k + m) Gamma[7 + 2 j] Gamma[ 1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j - l) Gamma[5 + 2 j] Gamma[ 1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j, 3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] + 1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1, 7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1]))"

 (1)
> 

ex:=convert(str, FromMma);

(k*GAMMA(5+2*j)*GAMMA(1+l)*hypergeom([1, 5/2+j, 3+j], [3+j+(1/2)*l, 7/2+j+(1/2)*l], -1)/GAMMA(6+2*j+l)+(k+m)*GAMMA(7+2*j)*GAMMA(1+l)*hypergeom([1, 7/2+j, 4+j], [4+j+(1/2)*l, 9/2+j+(1/2)*l], -1)/GAMMA(8+2*j+l))/(2^j*2^(-5-3*j-l)*GAMMA(5+2*j)*GAMMA(1+l)*(k*hypergeom([1, 5/2+j, 3+j], [3+j+(1/2)*l, 7/2+j+(1/2)*l], -1)/(GAMMA(3+j+(1/2)*l)*GAMMA(7/2+j+(1/2)*l))+(1/2)*(3+j)*(5+2*j)*(k+m)*hypergeom([1, 7/2+j, 4+j], [4+j+(1/2)*l, 9/2+j+(1/2)*l], -1)/(GAMMA(4+j+(1/2)*l)*GAMMA(9/2+j+(1/2)*l))))

 (2)
> 

simplify(ex);

Pi^(1/2)

 (3)

Download sqrtPi.mw

outline only...

vv 13118
January 16 2023
2 1

For a surface with color=red, the color of the interior of the polygons is red and the color of their borders is black.

But in your case, the surface is actually a curve (it does not depend on c), so the area of each polygon is zero! Hence, the color reduces to black 

typesetting...

vv 13118
January 09 2023
0 2

You are using interface(prettyprint=0). To get rid of those Typesetting:-mprintslash, use at the beginning:

interface(typesetting=standard);

extrema...

vv 13118
December 30 2022
4 1

Note first that the maxima/minima can be found directly using the command  extrema:

 

> 

F:=x^2-2*x*y+2*y^2-12;

x^2-2*x*y+2*y^2-12

 (1)
> 

extrema(x, {F}, {x,y}, 'Sx');  # min x   and  max x

{-2*6^(1/2), 2*6^(1/2)}

 (2)
> 

Sx;

{{x = -2*6^(1/2), y = -6^(1/2)}, {x = 2*6^(1/2), y = 6^(1/2)}}

 (3)
> 

extrema(y, {F}, {x,y}, 'Sy');  # min y   and  max y

{-2*3^(1/2), 2*3^(1/2)}

 (4)
> 

Sy;

{{x = -2*3^(1/2), y = -2*3^(1/2)}, {x = 2*3^(1/2), y = 2*3^(1/2)}}

 (5)

 

 

Now, if you want the math details and use the implicit function theory,

then dy/dx equals indeed   (x-y)/(x-2*y);  you can use

 

 

> 

dydx:=implicitdiff(F,y,x)

(-x+y)/(-x+2*y)

 (6)

Then you have to solve the system  {F=0,  dydx=0}.  

> 

solve({F=0,  dydx=0}, explicit);

{x = 2*3^(1/2), y = 2*3^(1/2)}, {x = -2*3^(1/2), y = -2*3^(1/2)}

 (7)

Similarly for y (i.e. dx/dy=0).

 

You may use (9)  in your worksheet only to check that the solutions correspond to min/max rather than inflexion points.

 

procedure...

vv 13118
December 13 2022
0 0

You may use:

Matrix(2,2, (i,j)-> undefined);

 

Impossible...

vv 13118
December 01 2022
2 0

An irreducible polynomial f with integer coefficients cannot have both sqrt(2) and sqrt(3) as roots!
(Because f would be divisible by the minimal poynomials).

Continuous arctan via dsolve...

vv 13118
October 17 2022
1 2
> 

restart;
> 

X:=cos(3*t); Y:=sin(t);

cos(3*t)

  

sin(t)

 (1)
> 

plot([X,Y, t=-0..2*Pi]);

 

 

> 

ARG_RADIUS:=proc(X,Y,t)
  local a,r,sol;
  sol := dsolve([
    diff(r(t)*cos(a(t)),t)=diff(X,t),  diff(r(t)*sin(a(t)),t)=diff(Y,t),
    r(0)=eval(sqrt(X^2+Y^2),t=0), a(0)=arctan(eval(Y,t=0),eval(X,t=0))
    ],  numeric, output=operator);
  eval((a,r)[],sol)
end:

> 

A,R := ARG_RADIUS(X,Y,t):
> 

plot(arctan(Y,X), t=0..2*Pi, discont);

 
> 

plot(A, 0..2*Pi);  # continuous arctan

 
> 

plot([R(t)*cos(A(t)), R(t)*sin(A(t)), t=0..2*Pi]); # Lissajous (check)

 

 

Download ARG-cont.mw

pdsolve...

vv 13118
October 13 2022
0 0

It seems you want to solve the PDE  u=0. Then:

pdsolve(u);

            f = _F1(x)*_F2(y)*_F3(z)*_F4(t), [_F1(x), _F2(y), _F3(z), _F4(t), ` are arbitrary functions.`]

Unfortunately this is not the general solution. For example   f = _F1(x) + _F4(t)   is a solution too.

Cannot be verified by Maple...

vv 13118
October 13 2022
4 1

Cannot be verified by Maple!

 

We must first define correctly the problem.

The equation contains two antiderivatives, so both contain an additive constant.

The only possibility is to consider that we accept some pair of constants
(for arbitrary constants, the equation cannot have solutions).
But in this case, Maple cannot verify a solution because it would imply to choose a convenient pair of constants.


Let's take a simple example (which actually dsolve refuses to solve because it does not contain a derivative).

 

int(y(x), x) + int(y(x)^2, x) - 1 = 0.

 

Applying diff ==>  y(x) + y(x)^2 = 0 ==>  y(x)=0  and  y(x)=-1.

y(x) = -1  ==>  -x + C1  +  x + C2 - 1 = 0, so we must have  C1+C2=1.

The same for y(x)=0. So, there are two solutions, but the integration constants have to be chosen carefully.

 

 

Other examples.

 

> 

restart;
> 

eq1:=int(diff(y(x),x)*x,x)+int(y(x),x) = 0;

int((diff(y(x), x))*x, x)+int(y(x), x) = 0

 (1)
> 

sol1:=dsolve(eq1);

y(x) = _C1/x

 (2)
> 

eq2:=int(diff(y(x),x)*x,x)+int(y(x),x) +1 = 0;

int((diff(y(x), x))*x, x)+int(y(x), x)+1 = 0

 (3)
> 

dsolve(eq2);
> 

odetest(sol1, eq1)

0

 (4)

So, no solution for eq2, but y(x) = _C1/x  is a solution  if we choose the constants of integration satisfying C1+C2 = -1.

eq1 works just because the int computed by Maple have (by chance) convenient constants.

 

 

> 

 

 

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