@C1Ron  If you study the properties of the ellipse and its equation  first time,  then I think that for you would be much more useful to solve this simple example by hand, not with Maple. Of course, if you know how to solve these examples manually, with the help of Maple you can check your solution.

@John Fredsted  I do not understand the exact meaning of your statement " But if they are variables themselves, then the -> construction cannot work".  unapply  and  -> construction are the different commands. As OP problem they are equivalent. But in some cases we have to use  unapply command, while in other cases the   -> construction.

Here are two examples.

Example 1. We want for each value of the parameter a  to solve the inequality  sqrt(x-a) < x :

F:=unapply(solve(sqrt(x-a)<x, x), a);

G:=a->solve(sqrt(x-a)<x, x);

F(-1);

G(-1);

Example 2. We want for each natural  n  to solve the recurrent equation  u(n+1)=3+2*u(n), u(1)=1 :

f:=n->rsolve({u(n+1)=3+2*u(n), u(1)=1}, u(n));

g:=unapply(rsolve({u(n+1)=3+2*u(n), u(1)=1}, u(n)), n);

f(2);

g(2);

We see that in the first example  unapply  command fails, and in the second example  -> construction fails.

@John Fredsted  I think that OP wants to define the argument as a list. If no then the simpliest variant is an arrow-function:

test := (x,y,z)->[y,y*z-x,-15*x*y-x*z-x]:

test(a,b,c);

test(1,1,1);

@Preben Alsholm  Very witty and surprising method! Vote up.

@Carl Love  Your methods sometimes provide a more complete factorization (perhaps undesirable in some cases). Suppose we want to collect only  relatively  a^4*b :

S:=a^4*b*c*x^2+a^4*b*c*y+a^4*b*c+z;

subs(t=a^4*b, collect(algsubs(a^4*b=t, S), t));

subs(t=a^4*b, collect(simplify(S, {a^4*b=t}), t));

thaw(collect(simplify(S, {a^4*b= freeze(a^4*b)}), t));

@emendes 

interface(rtablesize=infinity):

aux:=[-x1*x2+u0,x1*x1-2*x2,u1,u2,u3,u4,u5,u6,u7,u8,u9,u10];

f:=Vector[row](aux);

@taro  Christian's code gives a more complete factorization. But the same can be made more simple and understandable code:

e:= g^((2*(-sigma+k+1))/(-1+sigma))-tau^2:

combine(factor(expand(e)));

or

normal(combine(factor(expand(e))));

without the factor  -1  ahead.

@Carl Love   Thanks. Take a look at my second method.

@Carl Love  I think the gold circle is the set of points of tangency of black circles.

@shani2775  Use the code below

restart;

u[0] := (4/3)*c^2*cos((1/4)*x)^2; alpha := 2;

iteration := 3;

for k from 0 while k <= iteration do

u[s] := eval(u[k], t = xi); u[k+1] := simplify(u[k]-(int(diff(u[s], [`$`(xi, alpha)])+diff(u[s]*u[s], x)+diff(u[s]*u[s], x, x, x), xi = 0 .. t)))

end do;

@Carl Love   From the original within Maple we can see

f := unapply((1), x, y)

@Markiyan Hirnyk  The code of  LSSF  was written and works in 1D input, but does not work in 2D. Replace  op(Ind minus {Ind[i]})=~0  by  op(Ind minus {Ind[i]}=~0)  in the code of the procedure. Now there no any problems.

@Carl Love  You wrote "As far as I recall, a true return from is is trustworthy."  However, any Maple's bug can be rewritten in terms of  is .

Example:

is(int(1/(x^4+4), x=0..1) < 0.2);

evalf(Int(1/(x^4+4),x=0..1));

                                                              true

                                                      0.2389834593

@Earl  You wrote "But, why are your values for DthetaZero neither the under/over values I found nor the exact value which Carl Love found above?"  But this value  

1.034385*Pi;

                                                 3.249616317

is also not exact (approximate only) and the estimate is true  

               DthetaZero1=3.2496<3.249616317<3.2497=DthetaZero2

@Preben Alsholm  Thank you for the detailed analysis of the issue.