I have

f(t,k)=int(1/sqrt(1-k^2*sin(s)^2),s=0..phi)

so suppose I write

f(t,k)=F(phi,k)

where F(phi,k) is the solution of the integral.  Then

invF(f(t,k))=invF(F(phi,k))

Since I did a change of variable sin(theta/2)=k*sin(phi), the only way to have the answer wich is:

theta=2*arcsin(k*invF(f(t,k))

is to prove that invF(F(phi,k)) is equal to sin(phi)

Hope this will help.

mario.lemelin@cgocable.ca

First of all, sorry for my poor phrasing, english is not my first language.

The fact is that I have this:
f(t,k) =int(1/sqrt(1-k^2*sin(phi)^2),s=0..phi)
if you do
FunctionAdvisor( definition, InverseJacobiAM);
you see that this is exactly my integral and it's with no restriction on k and phi.

So theoretically, Maple should give me
f(t,k)=InverseJacobiAM(phi,k)

Now I do
JacobiAM(f(t,k),k)=JacobiAM(InverseJacobiAM(phi,k)
JacobiAM(f(t,k),k)=phi
but having made a change of variable earlier : sin(theta/2)=k*sin(phi)
The only way to have to solution (that I know) is that
JacobiAM(f(t,k),k) must be equal to sin(phi)
so I can have
theta=2*arcsin(k*JakobiAM(f(t,k),k))
And that's where I am stuck theoretically

Why Maple give EllipticF(sin(phi),k)? I don't know the inverse of this special fonction.
Note that in my book, JacobiAM is replace be sn. And if you do ?sn, you are redirected immediatelly at those JacobiAM and others.
So my problem is that last step.

mario.lemelin@cgocable.ca

First of all, sorry for my poor phrasing, english is not my first language.

The fact is that I have this:

f(t,k) =int(1/sqrt(1-k^2*sin(phi)^2),s=0..phi)

if you do

FunctionAdvisor( definition, InverseJacobiAM);

you see that this is exactly my integral and it's with no restriction on k and phi.  So theoretically, Maple should give me

f(t,k)=InverseJacobi(phi,k)

Now I do

JacobiAM(f(t,k),k)=JacobiAM(InverseJacobi(phi,k)

JacobiAM(f(t,k),k)=phi

but having made a change of variable earlier : sin(theta/2)=k*sin(phi)

The only way to have to solution (that I know) is that

JacobiAM(f(t,k),k) must be equal to sin(phi)

so I can have

theta=2*arcsin(k*JakobiAM(f(t,k),k))

And that's where I am stuck theoretically.  Why Maple give EllipticF(sin(phi),k)? I don't know the inverse of this special fonction.

Note that in my book, JacobiAM is replace be sn.  And if you do ?sn, you are redirected immediatelly at those JacobiAM and others.

So my problem is that last step.

mario.lemelin@cgocable.ca

Thanks for so far, but there is another step to make

> sol1 := int(1/sqrt(1-k^2*sin(s)^2), s = 0 .. phi) assuming phi > 0, phi < (1/2)*Pi;

                                    (1/2)                       
                     /            2\                            
                     \1 - sin(phi) /      EllipticF(sin(phi), k)
             sol1 := -------------------------------------------
                                      cos(phi)                  
> sol2 := simplify(sol1, symbolic);

                       sol2 := EllipticF(sin(phi), k)

But if you do ?sn, you will see  the following:

FunctionAdvisor( definition, InverseJacobiAM);

wich is exactly my integral and moreover, it is stated that there is no restriction on k and phi.

But there is a subtil difference.  It's suppose to be

InverseJacobiAM(phi,k) which Maple doesn't give????

so I can do

JacobiAM(InverseJacobiAM(phi,k))=phi  #but....

To guide you in where I am going with this is:  I have:

f(t,k)=u  where u = EllipticF(phi,k)

sn(f(t,k),k)=sn(u,k)=sin(phi) #this is the critical part

Now, having made a change of variable earlier : sin(theta/2)=k*sin(phi) I have

theta=2*arcsin(k*sn(f(t,k),k))

wich give me the only thing I know wich is the answer and it is exacty what I need meaning theta in fonction of the time and for a k fixed

Could you please help me for the last step?

Thanks

mario.lemelin@cgocable.ca

> de1 := diff(theta(t), t) = y(t);

                                 d                 
                         de1 := --- theta(t) = y(t)
                                 dt                
> de2 := diff(y(t), t) = -sin(theta(t));

                              d                       
                      de2 := --- y(t) = -sin(theta(t))
                              dt                      


> DEplot([de1, de2], [theta(t), y(t)], t = 0 .. 10, theta = -Pi .. Pi, y = -Pi .. Pi, [[theta(0) = 0, y(0) = 1]], arrows = small, animatecurves = true);

mario.lemelin@cgocable.ca

taking g/L =1

> with(DEtools);
> dfieldplot([diff(theta(t), t) = y(t), diff(y(t), t) = -sin(theta(t))], [theta, y], t = 0 .. 10, theta = -Pi .. Pi, y = -Pi .. Pi);

How can I plot this AND a particular initial conditions solution with display?

mario.lemelin@cgocable.ca

instead of:

solx := int(solv, t = 0 .. t)+C;

it's

solx := int(solv, t )+C;

mario.lemelin@cgocable.ca

If you look carefully at "solx", you will see that there is two terms that souldn't be there meaning:

m*vo/k  and -m^2*g/k

Just by looking at the integral, you can see that there will be only 3 terms.  Where those two additionnals terms come from?

mario.lemelin@cgocable.ca

mario.lemelin@cgocable.ca

I saw in previous thread the use of the option "inline", for example:

map[inline](....)

or

f:=proc(x) option inline statement end proc

When I read the help pages, since I speak french, it's really not clear to me what are the advantage of using this option.  Could someone have the patience to clear that up for me with an example.

Thanks in advance

mario.lemelin@cgocable.ca

There is a bug in the parser in document mode:

> L := [1, 2, 3, 4];
> p := (L::evaln) -> map2(cat,L,eval(L));#wich is suppose to work

Error, invalid parameters for inline function

it seems that Maple don't like the (L::evaln) so I try another way

> p := L -> map2(cat, L, eval(L));

                       p := L -> map2(cat, L, eval(L))
> p(L);

   [ || [1, 2, 3, 4] || 1,  || [1, 2, 3, 4] || 2,  || [1, 2, 3, 4] || 3, 

      || [1, 2, 3, 4] || 4]
> q := L -> map2(cat, evaln(L), eval(L))

                   q := L -> map2(cat, evaln(L), eval(L))
> q(L);

Error, (in q) illegal use of a formal parameter

I don't see another way but by doing a standard procedure.

mario.lemelin@cgocable.ca

>Digits:=20:

>L := [seq(evalf(k/Pi^4), k = 100 .. 120)];

>[seq(identify(L[i]), i = 1 .. nops(L))];

# Only 100, 105, 108, 112 and 120 is identify correctly

> map(ifactors, [100, 105, 108, 112, 120]);

[[1, [[2, 2], [5, 2]]], [1, [[3, 1], [5, 1], [7, 1]]], [1, [[2, 2], [3, 3]]],

  [1, [[2, 4], [7, 1]]], [1, [[2, 3], [3, 1], [5, 1]]]]

I don't know if you find the same thing.  I think that this should be report as a bug in the beta site.

mario.lemelin@cgocable.ca

> seq(cat(evaln(m), L[i]), i = 1 .. nops(L));

                                 m1, m5, m3

mario.lemelin@cgocable.ca

Thanks for your help and patience

mario.lemelin@cgocable.ca

I put

lhs(`if`(r::range,r,rhs(r))

and it's still doesn't work.

And how about the arrow  procedure for H?

mario.lemelin@cgocable.ca