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A faster solution...

vv 14107
November 20 2018
2 1

Here is a faster solution.
Note that in Acer's solution a,b  must be a bit larger not to miss solutions.

> 

restart;
> 

P:=isolve(u^2+v^2 = 225):
> 

Q:=isolve(u^2+v^2 = 125):
> 

NS:=0:
for p in [P] do for q in [Q] do
  NS:=NS+1;
  s:=eval([u,v],p); t:=eval([u,v],q);
  x-a=s[1],y-b=s[2],x+1=t[1],y+3=t[2];
  S[NS]:= eval([a,b,x,y],solve({%}))
od od:
S:=sort(convert(S,list)):NS;

192

 (1)
> 

nr:=0:NT:=0:
for i to NS-1 do
if S[i][[1,2]] = S[i+1][[1,2]] then eq:=true; nr:=nr+1 else eq:=false fi;
if eq=false then
   if nr=1 then NT:=NT+1;  lprint(NT=S[i-1],S[i])fi;
   nr:=0
fi;
od:

1 = [-21, -13, -12, -1], [-21, -13, -6, -13]
2 = [-21, 7, -12, -5], [-21, 7, -6, 7]
3 = [-15, -5, -6, 7], [-15, -5, -3, -14]
4 = [-15, -1, -6, -13], [-15, -1, -3, 8]
5 = [-11, -23, -11, -8], [-11, -23, 1, -14]
6 = [-11, -13, -11, 2], [-11, -13, 4, -13]
7 = [-11, 7, -11, -8], [-11, 7, 4, 7]
8 = [-11, 17, -11, 2], [-11, 17, 1, 8]
9 = [-8, -4, 1, 8], [-8, -4, 4, -13]
10 = [-8, -2, 1, -14], [-8, -2, 4, 7]
11 = [-6, -8, -6, 7], [-6, -8, 9, -8]
12 = [-6, 2, -6, -13], [-6, 2, 9, 2]
13 = [-5, -5, 4, 7], [-5, -5, 10, -5]
14 = [-5, -1, 4, -13], [-5, -1, 10, -1]
15 = [-3, -17, -12, -5], [-3, -17, 9, -8]
16 = [-3, -7, -3, 8], [-3, -7, 9, 2]
17 = [-3, 1, -3, -14], [-3, 1, 9, -8]
18 = [-3, 11, -12, -1], [-3, 11, 9, 2]
19 = [-2, -10, -11, 2], [-2, -10, 10, -1]
20 = [-2, 4, -11, -8], [-2, 4, 10, -5]
21 = [0, -10, -12, -1], [0, -10, 9, 2]
22 = [0, 4, -12, -5], [0, 4, 9, -8]
23 = [1, -17, -11, -8], [1, -17, 10, -5]
24 = [1, -7, -11, 2], [1, -7, 1, 8]
25 = [1, 1, -11, -8], [1, 1, 1, -14]
26 = [1, 11, -11, 2], [1, 11, 10, -1]
27 = [3, -5, -12, -5], [3, -5, -6, 7]
28 = [3, -1, -12, -1], [3, -1, -6, -13]
29 = [4, -8, -11, -8], [4, -8, 4, 7]
30 = [4, 2, -11, 2], [4, 2, 4, -13]
31 = [6, -4, -6, -13], [6, -4, -3, 8]
32 = [6, -2, -6, 7], [6, -2, -3, -14]
33 = [9, -23, -3, -14], [9, -23, 9, -8]
34 = [9, -13, -6, -13], [9, -13, 9, 2]
35 = [9, 7, -6, 7], [9, 7, 9, -8]
36 = [9, 17, -3, 8], [9, 17, 9, 2]
37 = [13, -5, 1, -14], [13, -5, 4, 7]
38 = [13, -1, 1, 8], [13, -1, 4, -13]
39 = [19, -13, 4, -13], [19, -13, 10, -1]
40 = [19, 7, 4, 7], [19, 7, 10, -5]

  
> 

 


 

Download aisolve.mw

Inverse...

vv 14107
November 19 2018
0 1

The inverse function is elementary

W:=solve(Y=vs, w):
Y1:=fsolve(W=0):
Y2:=fsolve(W=10):
plot([W,Y, Y=Y2 .. Y1]);

parfrac...

vv 14107
November 18 2018
1 0

You don't have to compute this way the partial fractions. Simply use:

convert(Denom1, parfrac, s);


 

Basis...

vv 14107
November 18 2018
1 1

You must take the leading monomials of the Groebner basis for J, not of J itself.

LinearSolve...

vv 14107
November 15 2018
2 5

solve has problems with so many variables and almost crashes Maple (and Windows).
Replace solve with LinearSolve this way:

V:=[indets(Eqns1)[]]:
AB:=GenerateMatrix(Eqns1,V,augmented):
Vsol:=LinearSolve(AB):
V=~Vsol;


 

redefine...

vv 14107
November 14 2018
2 0

The simplest way seems to be:

impdiff := (f,y,x) -> -diff(f,x)/diff(f,y):
impdiff( (lhs-rhs)(eqn), x, y);

If you simplify the result the expansion appears again. But in Acer's solution too.

 

a*b...

vv 14107
November 12 2018
2 3

In your procedure EEA you cannot use for example a*b.
You must use ED[`*`](a, b)

 

Not a bug...

vv 14107
November 11 2018
1 0

KroneckerDelta   is a tensor supposed to be used in Physics, being related with the spacetime dimension.
Use piecewise(m=0,1)  instead; it is even shorter!

The unpleasant fact is that the integral...

vv 14107
November 11 2018
0 4

The unpleasant fact is that the integral is difficult to manage with IntegrationTools:-Parts because it is transformed in terms of WhittakerM.
We must do part of the work by hand to obtain the final result 

(1 - (b+1)^(-a))*GAMMA(a)/b.

 

Neutral operators...

vv 14107
November 10 2018
0 3

For programming the solution is to use neutral operators.

`&x` :=LinearAlgebra:-KroneckerProduct;
A := <a,b;c,d>;
B := <1,2;3,4>;
A &x B &x A;

 

 

2x...

vv 14107
November 09 2018
0 0

It works if you correct  2x  to  2*x.

Period...

vv 14107
November 09 2018
1 4

Let f be a superposition of two functions f1, f2 . i.e. f(x) = g(f1(x),f2(x)) .

If f1 has period T1 and f2 has period T2 then f will have generally as period the lcm of T1 and T2, provided that T1, T2 are commensurable (i.e. T1/T2 is rational).
In your example we must know Omega. E.g. for Omega=1 the function is not periodic.

CommonPeriod:=(T1::Not(0),T2::Not(0)) -> `if`(type(T2/T1,rational), abs(numer(T2/T1)*T1), infinity);

For example, the function    cos(4*x) + sin((3/2)*x)     will have the period
CommonPeriod( 2*Pi / 4,  2*Pi / (3/2));
      4*Pi

convert...

vv 14107
November 07 2018
0 1

A convert and a correct assumption are needed.

simplify(convert(exp(I*t)^q, exp)) assuming t > -Pi, t <= Pi, q > 0, q < 1;

 

 

lexdeg...

vv 14107
November 06 2018
0 2

Why don't you use the monomial order lexdeg([x,y,z], [u,v,w]) ; it is used exactly to eliminate x,y,z.

Your monomial order can be obtained:

t2:=plex(u,v,w);  t1:=wdeg([1,1,1,0,0,0],[x,y,z,u,v,w]);

t:=prod(t1, t2);  # your order

 

 

indets and satisfies...

vv 14107
November 05 2018
1 0 
sat:=a -> satisfies(u -> (op([0,0,0],u)=D and op([0,1],u)=a)):
difford:= (e,a) -> max(nops~(map2([op],[0,0,..],indets(e,sat(a))))):

# Example
expr:=(diff(a(x, y), x, y))*x*y+(diff(a(x, y), x, x, y))*x+diff(b(x, y), x, x)-sin(z):
expr:=convert(expr,D):
has(expr,a); # depends on a?
                              true
has(expr,c); # depends on c?
                             false
difford(expr,a);
                               3
difford(expr,b);
                               2

 

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