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sqrt(Pi)...

vv 11843
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 11843
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 11843
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 11843
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 11843
December 13 2022
0 0

You may use:

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

 

Impossible...

vv 11843
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 11843
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 11843
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 11843
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.

 

 

> 

 

 

op...

vv 11843
October 10 2022
1 2

convert(s, list)  simply returns [op(s)]  i.e. the list of the operands, see ?type,series.
It seems that you are mainly interested in Taylor series. In this case, you may use

s:=series(10/(1-x^2)^2, x, 10);
             s := 10+20*x^2+30*x^4+40*x^6+50*x^8+O(x^10)
PolynomialTools:-CoefficientList(convert(s,polynom),x);
            [10, 0, 20, 0, 30, 0, 40, 0, 50]


 

simplify...

vv 11843
October 09 2022
2 1

Definitely, simplify & co must be revised, as pointed out in this forum several times.
Even very simple 0 expressions are not recognized!

> 

f:= (exp(2*x)-1)/(exp(2*x)+1);

(exp(2*x)-1)/(exp(2*x)+1)

 (1)
> 

simplify(f - tanh(x));  # ?

((-tanh(x)+1)*exp(2*x)-tanh(x)-1)/(exp(2*x)+1)

 (2)
> 

is(f - tanh(x) = 0);

true

 (3)

 

You probably want...

vv 11843
October 05 2022
2 0

You probably want:

L[1, 0] := t + Pi*t/2 + (4*t^(3/2))/3:
L[2, 0] := t^(1/2) + Pi*t/4:
ans := int((L[1, 0] + L[2, 0])/(x - t)^(1/2), t = 0 .. x) assuming x>0;

                            

Simply solve...

vv 11843
October 05 2022
0 0 
solve([f, diff(f,x)], explicit); # x1=x2;  (f is the polynomial);
evalf(%);

The result contains the answers for all your questions.

T2...

vv 11843
October 04 2022
1 6

Remove or comment the first line    afa := 0.277 

J := Int(T2, [y=-Pi/6+afa..Pi/6, x=0..Pi/2], method = _CubaCuhre, epsilon = 1e-2):
afa:=0.277:
'afa'=afa, 'T2'=evalf( eval(T2, [y=afa, x=Pi/4]) );
#               afa = 0.277, T2 = 2.35384223415597*10^9

evalf(J);
#               1.83729944313523*10^9 + 0.*I

forget(evalf);
afa := 0.13;
evalf(J);
#                   afa := 0.13
#                   2.69229399550637*10^9 + 0.*I

forget(evalf);
afa := 0.33;
evalf(J);
#                          afa := 0.33
#                         1.54057599303060*10^9 + 0.*I

global...

vv 11843
October 03 2022
2 1

Just add global main_module;   (e.g. after local ... ;).

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