1).
That error displays by using 'Change' without having loaded IntegrationTools
(which I do in my ini file).

After changing coordinates by cos(t) = x the trigonometric integral becomes an
integral K0 over -1 .. +1, integrand = rationalFunction(x,z) / sqrt(1-x^2).

That rational function of degree = (2,3) in x (z is a parameter) is written as
partial fraction, for which Maple uses Sum(terms/(x-a), a=RootOf(somePolynom)).

It would display that using the notation _Z, which I change to w, to have a variable
and my notation is Pol:=somePolynom, it is of degree 3 in w (and it would use
_alpha, which I changed to alpha)

Note that Maple does it in a generic way: for z=1/2 that would be 'false', one
only needs degree=2 in that case.

I have gone that way, because 1/(x-a)/sqrt(1-x^2) has an integral over -1 .. +1
that only depends on the location of a: as long as a is not in that *closed*
interval ( a < -1 or 1 < a ) it works out fine. Moreover it becomes handy, if
a is real.

Since a will be one of the roots I want to find that range, see 2).


2).
For that polynomial I look at the discriminant to find the interval(s), where the
roots w.r.t variable w are Reals, determined by the condition 0 < discriminant. 

That range is given by 0 < -64*z^3*(27*z^3-4) and Maple says, it is an interval
from 0 to somewhat beyond 1/2.


3).
The next step is to see, that those 3 real roots do not fall into -1 .. +1,
which are the next steps in that section, which ends by sketching the shape
of the polynomial Pol (a cubic in w).

1).
That error displays by using 'Change' without having loaded IntegrationTools
(which I do in my ini file).

After changing coordinates by cos(t) = x the trigonometric integral becomes an
integral K0 over -1 .. +1, integrand = rationalFunction(x,z) / sqrt(1-x^2).

That rational function of degree = (2,3) in x (z is a parameter) is written as
partial fraction, for which Maple uses Sum(terms/(x-a), a=RootOf(somePolynom)).

It would display that using the notation _Z, which I change to w, to have a variable
and my notation is Pol:=somePolynom, it is of degree 3 in w (and it would use
_alpha, which I changed to alpha)

Note that Maple does it in a generic way: for z=1/2 that would be 'false', one
only needs degree=2 in that case.

I have gone that way, because 1/(x-a)/sqrt(1-x^2) has an integral over -1 .. +1
that only depends on the location of a: as long as a is not in that *closed*
interval ( a < -1 or 1 < a ) it works out fine. Moreover it becomes handy, if
a is real.

Since a will be one of the roots I want to find that range, see 2).


2).
For that polynomial I look at the discriminant to find the interval(s), where the
roots w.r.t variable w are Reals, determined by the condition 0 < discriminant. 

That range is given by 0 < -64*z^3*(27*z^3-4) and Maple says, it is an interval
from 0 to somewhat beyond 1/2.


3).
The next step is to see, that those 3 real roots do not fall into -1 .. +1,
which are the next steps in that section, which ends by sketching the shape
of the polynomial Pol (a cubic in w).

I have some stuff in my ini-file and think the following is
missing if executing the sheet directly (sorry for that):

with(SolveTools):
with(IntegrationTools):
completesquare:=student[completesquare]:

Would you mind to try again (I do not have Maple 13 and
M12 has problems for the limit of roots w1,w2)?

Will comment on the 'Roots' later - I guess you mean the
final step(s) for the branched curve?

I have some stuff in my ini-file and think the following is
missing if executing the sheet directly (sorry for that):

with(SolveTools):
with(IntegrationTools):
completesquare:=student[completesquare]:

Would you mind to try again (I do not have Maple 13 and
M12 has problems for the limit of roots w1,w2)?

Will comment on the 'Roots' later - I guess you mean the
final step(s) for the branched curve?

Those questions make be pondering:

Is it about how to use Maple? Then these are weird tasks ...

Is it about understanding Geometry? Then Maple is the false way ...

As an example for the last my view would be: plane rotates, fixed
in A and B. Let it fall onto the sphere - from above or below.

Which is what Kintonum says. I guess.

Or otherwise said: what is the benefit of such homework?

Since I do not answer to private mails directly:

If it works for you in abscence of a6, then just choose it,
for example a6 = 0.

And as you are very new to Maple: do not mind, it needs
some time - you have managed it so far very fine, really!

Since I do not answer to private mails directly:

If it works for you in abscence of a6, then just choose it,
for example a6 = 0.

And as you are very new to Maple: do not mind, it needs
some time - you have managed it so far very fine, really!

1/m = epsilon does not matter, You still say 1/m = epsilon = 0.

1/m = epsilon does not matter, You still say 1/m = epsilon = 0.

think I have it, will work it out at the weekend

Essentially one can reduce to the following situation:

z = -1/2*(-1+2*y)^2/(1-2*y+4*y^2)/(y-1);
plots:-implicitplot(%, z= 0 .. 0.6, y=-4 ..1, 
  gridrefine = 2, crossingrefine = 2);

The location of zero of the middle 'branch' of that curve is the the jump.

think I have it, will work it out at the weekend

Essentially one can reduce to the following situation:

z = -1/2*(-1+2*y)^2/(1-2*y+4*y^2)/(y-1);
plots:-implicitplot(%, z= 0 .. 0.6, y=-4 ..1, 
  gridrefine = 2, crossingrefine = 2);

The location of zero of the middle 'branch' of that curve is the the jump.

I think that close to z=1/2 (but not equal to that) one can write the integral
as summing over the 3 roots of 8*z^2*w^3-4*z*w^2+(-2*z-6*z^2)*w+1+2*z+2*z^2,
roots w.r.t. variable w for the term

  1/2*signum(alpha)*Pi*(-1+z*alpha+2*z*alpha^2-z)/
    (-1+alpha^2)^(1/2)/z/(-1-4*alpha+12*z*alpha^2-3*z);

                                                      2
            signum(alpha) Pi (-1 + z alpha + 2 z alpha  - z)
      1/2 -----------------------------------------------------
                     2 1/2                             2
          (-1 + alpha )    z (-1 - 4 alpha + 12 z alpha  - 3 z)

and all the roots are real, two are above +1, one is below -1 (that is some
experimental reduction close to z=1/2).

The general summand would be Sigma:=piecewise(And(-1 < alpha,alpha < 1),undefined,
1/2*I*signum(0,Im((1-alpha^2)^(1/2)/(alpha+1)),-1)*Pi*(-1+z*alpha+2*z*alpha^2-z)/
(1-alpha^2)^(1/2)/z/(-1-4*alpha+12*z*alpha^2-3*z))

and the solution (z <> 1/2) is F:=Sum(Sigma,rho), you may inspect it by plotting
plot(F, z=1/4 .. 3/4) versus plotting your task.

Here rho:= alpha = RootOf(8*z^2*_Z^3-4*z*_Z^2+(-2*z-6*z^2)*_Z+1+2*z+2*z^2) are
the roots of the above mentioned polynomial.

The above is the special situation close to z=1/2.

Maple finds the solution of your task after cos(t) = x and simplifying, then
evaluating - but it will use the roots of a polynomial of degree = 6 and some
logarithm * rational on them, the above is degree = 3 and the complicated for
signs by the signum * rational.

The reduction follows by using partial fractions of the task after cos(t) = x
(now z<>1/2) up to the factor 1/sqrt(1-x^2).

Hope it helps - but now I have to give up, again.

I think that close to z=1/2 (but not equal to that) one can write the integral
as summing over the 3 roots of 8*z^2*w^3-4*z*w^2+(-2*z-6*z^2)*w+1+2*z+2*z^2,
roots w.r.t. variable w for the term

  1/2*signum(alpha)*Pi*(-1+z*alpha+2*z*alpha^2-z)/
    (-1+alpha^2)^(1/2)/z/(-1-4*alpha+12*z*alpha^2-3*z);

                                                      2
            signum(alpha) Pi (-1 + z alpha + 2 z alpha  - z)
      1/2 -----------------------------------------------------
                     2 1/2                             2
          (-1 + alpha )    z (-1 - 4 alpha + 12 z alpha  - 3 z)

and all the roots are real, two are above +1, one is below -1 (that is some
experimental reduction close to z=1/2).

The general summand would be Sigma:=piecewise(And(-1 < alpha,alpha < 1),undefined,
1/2*I*signum(0,Im((1-alpha^2)^(1/2)/(alpha+1)),-1)*Pi*(-1+z*alpha+2*z*alpha^2-z)/
(1-alpha^2)^(1/2)/z/(-1-4*alpha+12*z*alpha^2-3*z))

and the solution (z <> 1/2) is F:=Sum(Sigma,rho), you may inspect it by plotting
plot(F, z=1/4 .. 3/4) versus plotting your task.

Here rho:= alpha = RootOf(8*z^2*_Z^3-4*z*_Z^2+(-2*z-6*z^2)*_Z+1+2*z+2*z^2) are
the roots of the above mentioned polynomial.

The above is the special situation close to z=1/2.

Maple finds the solution of your task after cos(t) = x and simplifying, then
evaluating - but it will use the roots of a polynomial of degree = 6 and some
logarithm * rational on them, the above is degree = 3 and the complicated for
signs by the signum * rational.

The reduction follows by using partial fractions of the task after cos(t) = x
(now z<>1/2) up to the factor 1/sqrt(1-x^2).

Hope it helps - but now I have to give up, again.

I would agree, that it would be better no have some 'norming'.

However as soon as floats are used in arithemtics operations
the trailing places will be cut off (here Digits equals 10):

  10.0000000000000000000001;
  %*1.0;

                             10.00000000

I would agree, that it would be better no have some 'norming'.

However as soon as floats are used in arithemtics operations
the trailing places will be cut off (here Digits equals 10):

  10.0000000000000000000001;
  %*1.0;

                             10.00000000