How to find am approximate numerical value beyond x ~ 14

Otherwise it smells like Fourier stuff, no?

PS: the 'new' board still does not allow me to even read your post with Firefox.

Usually that is done though the IP address (of your provider) since it gives a geolocation which the server can use to send specific content.

Markting guys believe that is a good service for consumers and customers ... The usual switch for languages should be at the top, so one can see it easily.

Here is a simple info for your access - what country is displayed for you?

A somewhat messy way at first glance, but I think one can justify it and may should work in more complicated situations:

  sin(4*x)^2;   eval(%, x=arccos(c));   #MultiSeries:-   series(%, c=0, 9);

                   2       4        6       8      10                16 c  - 80 c  + 128 c  - 64 c  + O(c  )

Writing as polynomial and re-substitution allows Maple to to see, that this is the searched identity and valid (use 'combine' or 'simplify').

That is to say: "Is sin(4*arccos(c))^2 a polynomial in c? Find it."

  n:=13; # natural   k:=4;  # even   sin(n*x)^k;     eval(%, x=arccos(c)):   series(%,c=0, k*n+1): convert(%, polynom);

          <lengthy stuff, polynom of degree k*n=52, even powers>     subs(c=cos(x), %):   sin(n*x)^k - %: simplify(%);

                                 0

PS: sorry for the ugly formatation, but my usual settings with FF in the new board still do not work
and with IE it does not take "preformated" properly What a shame!

Find a sheet attached (even the upload is broken)

MP_sin_to_cos_power.mws

I am very rusty on curves, but enjoyed a partial recall :-)

f := 10000*y^5+10000*y^4-5109*y^3-595*y^2-10000*u^3+5109*u^2+595*u;

Plotting it shows, that over the Reals it has more than 1 connected component:

  myRange:= u=-20..20, y=-5...5;
  plots[implicitplot](f, myRange, gridrefine = 2, crossingrefine = 2);

That made doubt that one can 'write it down' by a parameter, since that should
produce 1 connected component (by continuity). But Complex restricted to Reals
often fails.

So look at the help for the command: "Such a parametrization exists if and only
if the genus is 0 and the curve is irreducible (which can be checked by AIrreduc)"

  AIrreduc(f); evala(%);
                                 true

So look at the 'genus':

  with(algcurves);
  genus(f,y,u);
                                  4

Done. I think that genus=0 is needed by the Theorem of Domain Invariance
(silently assuming that the parametrization is injective+continuous)

It would be worth to look deeper using 'differentials' of the package (relating
to the homology of the compact curve, giving by homogenization of f)

There is no "a" or "b" in your system (and it may be contratictory, for example dividing by 0)

As it is said at Wiki (by David Eppstein ?) that comes through the "momentum curve"
f: C^1 ---> C^3, x |---> (x, x^2, x^3) and viewing at this in projective coordinates
it is (x:1) |---> (x : x^2 : x^3 : 1).

To extendend it to projective space one has to 'homogenize' that polynomial map to
a homogenous one of the same, maximal degree, which then is 

P^1 |---> P^3,  (x:t) |---> (x*t^2 : x^2*t : x^3 : t^3)

Change ordering, if you feel the need.


The affine (parametrized) curve gives those points where Y = X^2 and Z = X^3, where
I use coordinates X,Y,Z in C^3. Thus it is the intersection of those 2 hyperplanes,
at least as set. But we can do better:

  with(Groebner);
  with(PolynomialIdeals);

  A := [X^2-Y, X^3-Z];   # set-theoretic vanishing ideal
  J := <A>;

  Basis(A, plex(X,Y,Z)); # find a 'basis' for the ideal

                   3    2         2            2
                 [Y  - Z , X Z - Y , X Y - Z, X  - Y]

Which does not look that nice, but just decompose into prime components

  PrimeDecomposition(J);
                             2       3
                          <-X  + Y, X  - Z>

So we see, that our initial approach was fine and it is a curve, i.e. dim = 1:

  HilbertDimension(A);
                                  1


Now for the projective situation. Here one has to 'homogenize' the ideal, since
one works in the algebra of homogenous polynomials. Maple can do that:

  Homogenize(J, 'W');
                           2          2
                  <-W Y + X , -X Z + Y , -W Z + X Y>


Using 'PrimeDecomposition' for that gives the same, so it is the intersection
of those 3 (homogenous) quadrics.

(*

   some lines of code

*)

I do not understand what you mean by 'for each branch', but I do it as follows:

g:= z -> 1/sqrt(4*z^2+4*z+3);

c:= t -> exp(2*Pi*I*t);
Int(g, z=c..``) = 'Int(g(c(t))*D(c)(t), t=0..1)';
`` = value(rhs(%));

                                1
                   /           /
                  |           |
                  |   g dz =  |   g(c(t)) D(c)(t) dt
                  |           |
                 /           /
                   c           0


                                  = 0

Some thoughts (and not caring much about domains of possible inversion) you also
may look into ?simplify/@ (while ?invfunc did not give me much):

invErf:= (erf@@(-1));

invErf(erf(x)); # automatic simplification 
                                  x

erf(invErf(y)); # NO automatic simplification 
simplify(%);    # may mess up complicated expressions
                                  y

Which menas that Maple treats left and right inverse differently.

This does not handle correct domains (use any fct instead). 


And the numerical case has do be done separately. With care - and fsolve may be
not the very way. If the fct becomes very flat/steep, then exp and log of the
task is better (and increasing Digits is a good idea). Providing some good guess
for a starting point will speed up the stuff.

w:=x+2+sin(x+2);
subs(x=u-2,w);

does what you want. Or use 'solve' in more complicated situations

w:=x+2+sin(x+2);
x+2=u; my:=isolate(%, x);
subs(my,w);
                              u + sin(u)

Just look up the help for convert / binary, it provides examples:

You can do it in your Maple or online (corrected my link)

Maple should be able to apply "Little Fermat" and its generalizations

In that case - modulo an integer - I would use f:= x -> exp(-x) and then

  # shift to 0 ... newUpper and back again
  lower:= -1;
  F:= x -> f(modp(x-lower,2)+lower); # periodic extension by period=2

  myRange:= -6 .. 4; 
  plot(F, myRange, scaling = constrained);

Limitations are seen by (Digits=15)

  myRange:= 10^6 .. 10^6 + 10; 
  plot(F, myRange);
  #plot('F'(x), x=myRange); # use quotation marks for that!

Have you tried to use https://en.wikipedia.org/wiki/Fibonacci_number ? You would need to check the conventions for starting (those are not unique). They are partly based on closed formulae for Fib(n) and also give criteria to recognize naturals as Fibonacci numbers (have not checked them). And of course lots of references.

NB: For the first 2 your question is not "well posed", so just treat it as special case

You have a linear system in 3 variables and your task has only "trivial"
solutions, because the according matrix is invertible:

  with(LinearAlgebra):

  sys:= convert({E1, E2, E3}, rational);
  var:= [c, d, e];
  A,x:=GenerateMatrix(sys,var );

  Determinant(A); evalf(%);