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typo(syntax)...

vv 13822
August 01 2019
0 0

A:-B:-foo();           

anames,unames...

vv 13822
July 31 2019
2 0

In ?type,protected it is recommended:

select(type, {unames(), anames(anything)}, protected);

Executing this command several times, the number of names increases.
Probably the simple evaluation of some of these names generates other names which are not "seen" initially.

 

title...

vv 13822
July 29 2019
1 0

labels = ["x", ""], title=cat("y", " "$220)

approx & exact...

vv 13822
July 27 2019
2 0

restart;
a := 1/(i*sqrt(i+1)+(i+1)*sqrt(i)):
evalf(Sum(a, i=1..infinity));  # approx
                          1.000000000
 

b := (expand@simplify@expand@rationalize)(a):
sum(b, i=1..infinity); #exact
                               1

 

answer...

vv 13822
July 27 2019
0 7

I posted an answer to your deleted question.
I think that the moderator who deleted it was wrong because:
1. It was answered.
2. The content was distinct from the previous question. Not to mention that the old thread is difficult to follow (and loads slowly) being very long.

So, here is the answer again.

> 

restart;
> 

Digits:=15;

15

 (1)
> 

F[1, 1] := (r, theta) -> A[1, 1] + A[1, 2]*cos(theta) + A[2, 1]*(2*r - 1) + A[2, 2]*(2*r - 1)*cos(theta);
L[1, 1] := (r, theta, phi) -> ((2.803059644*10^12*cos(theta)^2*r^2 + 7.474825716*10^11*r^3*cos(theta)^3 + 7.474825716*10^10*r^4*cos(theta)^4 + 4.671766072*10^12*r*cos(theta) + 2.919853795*10^12)*diff(F[1, 1](r, theta), theta, theta) + (1.584823993*10^13*cos(theta)*r^3 + 9.508943959*10^12*cos(theta)^2*r^4 + 2.535718389*10^12*r^5*cos(theta)^3 + 2.535718389*10^11*r^6*cos(theta)^4 + 9.905149957*10^12*r^2)*diff(F[1, 1](r, theta), r, r) + (1.981029991*10^13*r^2*cos(theta) + 1.426341594*10^13*r^3*cos(theta)^2 + 4.437507181*10^12*r^4*cos(theta)^3 + 5.071436778*10^11*r^5*cos(theta)^4 + 9.905149957*10^12*r)*diff(F[1, 1](r, theta), r) + ((-1)*1.167941518*10^12*sin(theta)*r + (-1)*5.606119287*10^11*sin(theta)*r^3*cos(theta)^2 + (-1)*7.474825716*10^10*sin(theta)*r^4*cos(theta)^3 + (-1)*1.401529822*10^12*sin(theta)*r^2*cos(theta))*diff(F[1, 1](r, theta), theta) + ((-1)*5.071436778*10^11*r^4*cos(theta)^4 + (-1)*3.803577584*10^12*r^3*cos(theta)^3 + ((-1)*1.109376795*10^13*r^2 + (-1)*7.474825716*10^10*r^4)*cos(theta)^2 + ((-1)*1.584823993*10^13*r + (-1)*3.737412858*10^11*r^3)*cos(theta) + (-1)*9.905149957*10^12 + (-1)*4.671766072*10^11*r^2)*F[1, 1](r, theta))/((39.0625 + 62.5*r*cos(theta) + 37.5*cos(theta)^2*r^2 + 10.*r^3*cos(theta)^3 + r^4*cos(theta)^4)*r^2):

proc (r, theta) options operator, arrow; A[1, 1]+A[1, 2]*cos(theta)+A[2, 1]*(2*r-1)+A[2, 2]*(2*r-1)*cos(theta) end proc

 (2)
> 

for p to 1 do  for s to 1  do
f := L[p, s](r, theta, phi)*F[p, 1](r, theta):
Aij:=[indets(F[p,s](r,theta),indexed)[]];
g1:=expand(f, Aij, distributed);
g2:=coeffs(g1, Aij, 'T'):
C:=int~([g2], theta = 0 .. 2*Pi, r = 0.5 .. 1, epsilon=1e-8, numeric):
kk:=add(C .~ T);
kkk := evalf( int(F[p, 1](r, theta)^2, theta = 0 .. 2*Pi, r = 0.5 .. 1) );
k[p, s] := kk/kkk;
print([p, s] = %);
od od;

[1, 1] = (-1716301349235.42*A[1, 1]^2+745939588799.615*A[1, 1]*A[2, 1]+513374393378.777*A[2, 1]^2+161631455946.711*A[1, 1]*A[1, 2]+427274927473.220*A[2, 1]*A[1, 2]+135421805163.233*A[2, 2]*A[1, 2]+427274927473.220*A[1, 1]*A[2, 2]+406304482715.359*A[2, 1]*A[2, 2]-1120768744270.57*A[1, 2]^2+178063137194.727*A[2, 2]^2)/(1.04719755119660*A[2, 1]^2+.523598775598299*A[2, 2]^2+3.14159265358979*A[1, 1]*A[2, 1]+1.57079632679490*A[2, 2]*A[1, 2]+3.14159265358979*A[1, 1]^2+1.57079632679490*A[1, 2]^2)

 (3)

spherical...

vv 13822
July 26 2019
3 1 
f:=exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3));
F:=eval(f, [x=r*sin(v)*cos(u), y=r*sin(v)*sin(u), z=r*cos(v)]);
int(F*r^2*sin(v), r=0..2, u=0..2*Pi, v=0..Pi);
evalf(%);

4*(1/24 - exp(-64)/24)*Pi

0.5235987756

To obtain a numerical answer you must change the variables to integrate in a cube.
Of course the spherical one is the best here because the new integrand is continuous.
Or, use piecewise:

JJ:=Int(exp(-(sqrt(4*x^2+4*y^2+4*z^2)^3))*piecewise(x^2+y^2+z^2<4,1,0), x=-2..2,y=-2..2,z=-2..2):
evalf(JJ);

0.5235987756

coeffs...

vv 13822
July 26 2019
0 0
> 

restart;
> 

Digits:=15;

15

 (1)
> 

F[1, 1] := (r, theta) -> A[1, 1] + A[1, 2]*cos(theta) + A[2, 1]*(2*r - 1) + A[2, 2]*(2*r - 1)*cos(theta);
L[1, 1] := (r, theta, phi) -> ((2.803059644*10^12*cos(theta)^2*r^2 + 7.474825716*10^11*r^3*cos(theta)^3 + 7.474825716*10^10*r^4*cos(theta)^4 + 4.671766072*10^12*r*cos(theta) + 2.919853795*10^12)*diff(F[1, 1](r, theta), theta, theta) + (1.584823993*10^13*cos(theta)*r^3 + 9.508943959*10^12*cos(theta)^2*r^4 + 2.535718389*10^12*r^5*cos(theta)^3 + 2.535718389*10^11*r^6*cos(theta)^4 + 9.905149957*10^12*r^2)*diff(F[1, 1](r, theta), r, r) + (1.981029991*10^13*r^2*cos(theta) + 1.426341594*10^13*r^3*cos(theta)^2 + 4.437507181*10^12*r^4*cos(theta)^3 + 5.071436778*10^11*r^5*cos(theta)^4 + 9.905149957*10^12*r)*diff(F[1, 1](r, theta), r) + ((-1)*1.167941518*10^12*sin(theta)*r + (-1)*5.606119287*10^11*sin(theta)*r^3*cos(theta)^2 + (-1)*7.474825716*10^10*sin(theta)*r^4*cos(theta)^3 + (-1)*1.401529822*10^12*sin(theta)*r^2*cos(theta))*diff(F[1, 1](r, theta), theta) + ((-1)*5.071436778*10^11*r^4*cos(theta)^4 + (-1)*3.803577584*10^12*r^3*cos(theta)^3 + ((-1)*1.109376795*10^13*r^2 + (-1)*7.474825716*10^10*r^4)*cos(theta)^2 + ((-1)*1.584823993*10^13*r + (-1)*3.737412858*10^11*r^3)*cos(theta) + (-1)*9.905149957*10^12 + (-1)*4.671766072*10^11*r^2)*F[1, 1](r, theta))/((39.0625 + 62.5*r*cos(theta) + 37.5*cos(theta)^2*r^2 + 10.*r^3*cos(theta)^3 + r^4*cos(theta)^4)*r^2):

proc (r, theta) options operator, arrow; A[1, 1]+A[1, 2]*cos(theta)+A[2, 1]*(2*r-1)+A[2, 2]*(2*r-1)*cos(theta) end proc

 (2)
> 

for p to 1 do  for s to 1  do
f := L[p, s](r, theta, phi)*F[p, 1](r, theta):
Aij:=[indets(F[p,s](r,theta),indexed)[]];
g1:=expand(f, Aij, distributed);
g2:=coeffs(g1, Aij, 'T'):
C:=int~([g2], theta = 0 .. 2*Pi, r = 0.5 .. 1, epsilon=1e-8, numeric):
kk:=add(C .~ T);
kkk := evalf( int(F[p, 1](r, theta)^2, theta = 0 .. 2*Pi, r = 0.5 .. 1) );
k[p, s] := kk/kkk;
print([p, s] = %);
od od;

[1, 1] = (-1716301349235.42*A[1, 1]^2+745939588799.615*A[1, 1]*A[2, 1]+513374393378.777*A[2, 1]^2+161631455946.711*A[1, 1]*A[1, 2]+427274927473.220*A[2, 1]*A[1, 2]+135421805163.233*A[2, 2]*A[1, 2]+427274927473.220*A[1, 1]*A[2, 2]+406304482715.359*A[2, 1]*A[2, 2]-1120768744270.57*A[1, 2]^2+178063137194.727*A[2, 2]^2)/(1.04719755119660*A[2, 1]^2+.523598775598299*A[2, 2]^2+3.14159265358979*A[1, 1]*A[2, 1]+1.57079632679490*A[2, 2]*A[1, 2]+3.14159265358979*A[1, 1]^2+1.57079632679490*A[1, 2]^2)

 (3)

convertions fail...

vv 13822
July 26 2019
1 0

For a solution given implicitely, as in this case, odetest may fail.

Actually the implicit solution is

> 

Int( 1/(y^3 + 1)^(2/3), y) + Int( 1/(x^3 + 1)^(2/3), x) + c = 0;

Int(1/(y^3+1)^(2/3), y)+Int(1/(x^3+1)^(2/3), x)+c = 0

 (1)

The problem is that the integrals are computed in terms of hypergeom, and Maple is not able to further manipulate.

> 

f:=1/(x^3 + 1)^(2/3);

1/(x^3+1)^(2/3)

 (2)
> 

int(f, x)

x*hypergeom([1/3, 2/3], [4/3], -x^3)

 (3)
> 

Z:=diff(% ,x) - f; # Z should be 0

hypergeom([1/3, 2/3], [4/3], -x^3)-(1/2)*x^3*hypergeom([4/3, 5/3], [7/3], -x^3)-1/(x^3+1)^(2/3)

 (4)
> 

is(Z=0) assuming x>0, x<1;

FAIL

 (5)
> 

# It seems that any usual convertion fails here.

An approximation...

vv 13822
July 24 2019
1 3

Salut Viorel, bine ai venit!

If you are interested in an approximate solution based on series, here it is:

restart;
ode:= diff(g(r),r$2)- r/R*g(r)=0;
ic:=g(2*R)=0, D[1](g)(0)=R;
dsolve([ode,ic],g(r)) assuming R>0;
G:=rhs(%);
Order:=6;
simplify(series(G, r=0));
convert(%, polynom):
g1:=convert(series(%, R=0), polynom);  
R:=1/3;  #############
dsol:=dsolve([ode,ic],g(r), numeric);
plots:-odeplot(dsol, r=0..2*R);
plot(g1, r=0..2*R);

So, for small R and r, an approximation is

allsolutions...

vv 13822
July 22 2019
1 1

RealDomains is not very reliable but it works in this case (Maple 2019).

eq:=(-2*cos(x)^2+2*sin(x+(1/4)*Pi)^2-1)/sqrt(-x^2+4*x) = 0:
RealDomain:-solve({-x^2+4*x > 0, eq}, {x}, allsolutions);
RealDomain:-solve(eq, {x}, allsolutions);

 

compact...

vv 13822
July 21 2019
2 0

If you want a compact solution:

9*x-(1/3)*Pi = (-1)^k*(7*x-(1/3)*Pi) + k*Pi:
solve(%,x); # k in Z

COMMAND...

vv 13822
July 20 2019
0 1

I think that in a package where almost everything is a module, such a COMMAND could exist only if it is provided by the package itself.
But we can use:

X := ''RandomVariable''(Normal(mu, sigma)):
Y := eval(X, mu=1);

 

Maybe...

vv 13822
July 19 2019
1 1

something like:

plots:-pointplot([x1,y1], color=[seq(COLOR(HUE,(t+1)/2),t=z1)]);

 

residue...

vv 13822
July 18 2019
4 4

It is not that difficult but unfortunately Maple 2019 gives 0 for it.

It can be computed using residues. This must be done by hand because Maple cannot compute residues for essential singularities.

But it can compute the resulting sum, and the result is   2*Pi*BesselJ(m, b)*exp(I/2*m*Pi)

F2_...

vv 13822
July 14 2019
0 5

 

> 

F1 := (x, y) -> 9/8*y + 9/8 - 21/8*y*exp(-2*x) - 21/8*exp(-x);

proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x) end proc

 (1)
> 

S1 := y -> fsolve(F1(x, y), x);

proc (y) options operator, arrow; fsolve(F1(x, y), x) end proc

 (2)
> 

S1(0.2);

.7548332751

 (3)
> 

g := (U,y) -> U + S1(2*y*exp(-U));

proc (U, y) options operator, arrow; U+S1(2*y*exp(-U)) end proc

 (4)
> 

g(0.6,0.2);

1.347716146

 (5)
> 

fsolve('g'(U,0.3) = 0.2, U);

-.4051840484

 (6)
> 

U1 := (x,y) -> fsolve('g(U,y) = x', U);

proc (x, y) options operator, arrow; fsolve('g(U, y) = x', U) end proc

 (7)
> 

U1(0.4,0.6);

-.1591368141

 (8)
> 

F2 := (x, y) -> 9/8*y + 9/8 - 21/8*y*exp(-2*x) - 21/8*exp(-x) + 3/4*int(F1(x - z, 2*y*exp(-z)), z = 0 .. U1(x,y));

proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(3/4)*(int(F1(x-z, 2*y*exp(-z)), z = 0 .. U1(x, y))) end proc

 (9)
> 

F2(0.2,0.3);

-1.008550780

 (10)
> 

####################################################################
> 

X:=[seq(k*0.1, k=0..10)];
Y:=[seq(k*0.1 ,k=0..10)];

[0., .1, .2, .3, .4, .5, .6, .7, .8, .9, 1.0]

  

[0., .1, .2, .3, .4, .5, .6, .7, .8, .9, 1.0]

 (11)
> 

F1_:= Interpolation:-SplineInterpolation( [X, Y],  [seq([seq(F1(x,y),y=Y)],x=X)] );

Interpolation:-SplineInterpolation([_rtable[18446744074399121150], _rtable[18446744074399121270]], _rtable[18446744074399113214], verify = false)

 (12)
> 

plot3d([F1(x,y), '1+F1_(x,y)'], x=0..1, y=0..1);

 
> 

S1_:=Interpolation:-SplineInterpolation( X,  [seq(fsolve(F1_(x, y), x), y=Y)] );

Interpolation:-SplineInterpolation([_rtable[18446744074449580390]], Vector[row](11, {(1) = .8471962164, (2) = .7961144432, (3) = .7548618542000001, (4) = .7211061446, (5) = .6931443837, (6) = .6697004904, (7) = .6498166028, (8) = .6327736053999999, (9) = .6180256179, (10) = .6051531019, (11) = .593829713}), verify = false)

 (13)
> 

S1_(1/2);

HFloat(0.6697004904)

 (14)
> 

solu:=(x,y)->fsolve(U + 'S1_'(2*y*exp(-U))=x, U);

proc (x, y) options operator, arrow; fsolve(U+('S1_')(2*y*exp(-U)) = x, U) end proc

 (15)
> 

solu(0.3,0.4);

-.2865364247

 (16)
> 

U1_:= Interpolation:-SplineInterpolation( [X, Y],  [seq([seq(solu(x,y),y=Y)],x=X)] );

Interpolation:-SplineInterpolation([_rtable[18446744074439092814], _rtable[18446744074439092934]], _rtable[18446744074439093054], verify = false)

 (17)
> 

U1(0.1,0.2); U1_(0.1,0.2);

-.5354543130

  

HFloat(-0.5354485486)

 (18)
> 

F2_:=proc(x,y)
  local z, u:=evalf(U1_(x,y));
  evalf(9/8*y + 9/8 - 21/8*y*exp(-2*x) - 21/8*exp(-x)) + 3/4*int(''F1_''(x - z, 2*y*exp(-z)), z = 0 .. u, numeric)
end:
 
> 

F2(0.1,0.2), F2_(0.1,0.2);

-1.139973599, HFloat(-1.1400076769496776)

 (19)
> 

 


 

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