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Digits...

vv 13877
February 28 2023
1 0

To work correctly, MatrixPower needs a much higher precision, e.g. take Digits=40 in your example.

BTW, a matrix C has a square root if det(C) <>0.  <0,1; 0,0> has not a square root.

Logical equivalence...

vv 13877
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 13877
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 13877
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 13877
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 13877
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 13877
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 13877
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 13877
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 13877
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 13877
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 13877
December 13 2022
0 0

You may use:

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

 

Impossible...

vv 13877
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 13877
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 13877
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.

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