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procedure...

vv 12453
December 13 2022
0 0

You may use:

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

 

Impossible...

vv 12453
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 12453
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 12453
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 12453
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 12453
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 12453
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 12453
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 12453
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 12453
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 12453
October 03 2022
2 1

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

A bug...

vv 12453
September 28 2022
0 5
> 

restart;

 

It seems to be a bug.
If we alter a bit the bindings, the result is as expected
For x=0, y=0  the results should be the same, but they are not!

 

> 

use a=a+b in
  use b=a-b in
    a*b
  end use             #   ==>   a*(a-b)  
end use;              #   ==>   (a+b)*a              ##??
   

(a+b)^2

 (1)
> 

use a=a+b+x in
  use b=a-b+y in
    a*b                     
  end use             # ==>    a*(a-b+y)
end use;              # ==>  (a+b+x)*(a+b+x -b +y)   ##OK                

(a+b+x)*(a+x+y)

 (2)
> 

eval(%,[x=0,y=0]);

(a+b)*a

 (3)

 

Maybe this...

vv 12453
September 23 2022
1 1 
expr:=1+3*diff(y(x),x):
select( has, ["itself"=expr, op(expr)], diff(y(x),x));

solve...

vv 12453
September 22 2022
0 0

It is simpler to just use the obvious:
solve(Determinant(A),w);

sinh(x) / x...

vv 12453
September 14 2022
1 0

Mathematicaly, you need the inverse of the function

f : (0, oo) -> (1, oo),  f(x) = sinh(x) / x.

The inverse exists, let's denote it by F : (1, oo) --> (0, oo);
it ca be developed in power series but cannot be expressed in terms of known functions.

Your equation has a unique positive solution  C = mu*g*a/F(L/(mu*g*a))
iff  L/(mu*g*a) > 1  (provided that your constants are >0).

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