Do you mean 4*arctan (your equ) or 1*arctan (your plot)

fsolve ---> numerical

solve or RootOf + allvalues ---> 'analytical' (if possible, here it does not help)

In your case look up ?fsolve, "After finding a solution, to find additional solutions, use the avoid option in subsequent fsolve calls"

Even if you need one, no explicit solution even in the case of *specific
parameters*, as the example shows (up to RootOf).

Staring at the system this is 1 function and you do a 'cyclic permutation'
x -> y -> z -> x and a -> b -> c -> a etc to generate your system (while
d and M are fixed). So it seems you are computing some invariant under a
permutation.

And instead of fct = 0 you can try to write it as Fct = 1, because of exp,
using 2*a instead of a (etc) this becomes (x=0, x=1 do not solve your sys):

  eq1:= x/(1-x) = exp((c*d*z+a*(1-d)*y+1)*M) 

resp    x/(1-x) = mu^((c*d*z+a*(1-d)*y+1))  with mu = exp(1)^M 

while the other equations are similar (just do permute).


For the example {M=1,a=1,b=1,c=1,d=1} this looks 'simple',

  {x/(1-x) = exp(z+1), y/(1-y) = exp(x+1), z/(1-z) = exp(y+1)}

But even then it is not found explicitely (well, almost and in term of
the function (multivalued) 'RootOf', but d=-1 already lets it fail).


Probably you need some other (geometric ?) reasoning to find the common
zeros for such a system of functions = rational + transcendent.

And: it is not clear at all, whether that there will be a unique one.


Edited: what is the reason for the task, where does it come from?

Just put brackets around the sequence, [someSequence] is a list.

Similar {someSequence} is a set (and duplicates are eliminated),
convert(someSet, list) then makes it a list (but you can not rely
on ordering doing it that way).

Having XP (on a quite moderate machine) and older Maple I can not reproduce
your problem (but killed that for large N) - but what sholud it be good for?

More or less you have numbers between +-1 and plot them as horzontal lines,
for large N that will give almost a black box.

So: what do you want & expect?

It is not that I have the very idea, but these are sums of powers (and symmetric functions,
though I do not see how the related theory leads to a solution quickly).

May be that it is worth to ponder about.

I am not aware that Maple has such an ability, just inserting blanks is the way, unfortunately.

Using Maple 12:

t:=1/GAMMA(-j+2)/GAMMA(j+1);
T1:=Sum(%, j=0..1);
simplify(%);
                                  0

T2:=value(T1);
                               T2 := 2

# Reason (?)

simplify(t);
% assuming j::nonnegint;

                           sin(Pi (j + 1))
                           ---------------
                            j (j - 1) Pi

                                  0
# but

seq(t, j=0..1);
                                 1, 1

I mostly use the classical interface and through the menu you can
choose the option, to show 'invisible' characters). The same should
be possible in the standard interface.

But having M12 I always use pdfCreator and that works, it also does
a much better job, if working with MS Office 2007 vs the MS solution
for pdf (especially for graphics & rendering)

So you need to find A1 and A5 ... for that I assume, that the A_i do not
contain the symbol x (else your problem may not be clear).

Then 'expand(exp2)' shows, that A5 is almost the quadric term in x, the
same for exp1.

Thus you can find them through the Taylor polynom:

  -coeftayl(exp1, x=0, 2)/p;
                                  A1

  -coeftayl(exp2, x=0, 2)/p;
                                  A5

Likewise you can use '-coeff(exp1, x, 2)/p', compare the help pages.

exp1:=A1*x*(k-p*x)+A2*x+A3*p+A4;
exp2:=A5*x*(k-p*x);
exp3:=exp1/exp2;

'exp3 - A1/A5'; 
theRest:=simplify(%);

                                A2 x + A3 p + A4
                     theRest := ----------------
                                 A5 x (k - p x)

'exp3 = A1/A5 + theRest';
simplify(%); 
is(%);

                                 true

The integrand is pdfNormal * cdfLogistic (where mean and variance in the Normal is parametrized
and the Logistic is the most simple version; by change of variables you also can put the parameters
to the Logistic and have a usual Normal).

So it is an expectation value and may be you find it if combining a search for both the distributions
(have not look up Gradsteyn, whether that kind of integrals is given there)

This is a bit too vague.

Why not just posting the example?

If it is longer code then it is better to upload a worksheet (use the green arrow in the menu bar at the editor for answering)

Using Maple gives a result (but that is not a proof ...)

restart: interface(version); Digits:=trunc(evalhf(Digits));

A:= x -> cos(100*(x^4-x));

Int(A(x), x = 0 .. infinity); expand(%);  
a1:=value(%);

Int(A(x), x = -infinity .. 0); Change(%, x=-xi, xi); expand(%); 
a2:=value(%);

a:=collect(a1+a2, hypergeom);
evalf(a);
                                -1000.

However a cross check with increased precision give me doubts, so:

Now this is a sum of hypergeometrics and even if one trusts Maple,
that those hypergeometrics are computed correctly, one should check
for possible cancellations through adding terms

S:= [op(a)];
evalf(S): evalf[3](%);

                   16           16          17           17
          [0.980 10  , -0.222 10  , 0.129 10  , -0.205 10  ]

evalf[Digits+17](a);

                          -0.09036215081945

Then the product would depend heavily on hardware as well, not only on Operating System and the former is likely to develop and spread even faster

That would give a hard task for a producer ...

For me it is not clear, that it will run on (arbitrary) notebooks or on a PC, which I may buy in 2 - 3 years.

So I would not want to pay for such a bundeled solution

with(LinearAlgebra);

[seq( cat(eq,j), j=1..8)]; sys:=eval(%):
vars:=[A1[n], A2[n], C1[n], C2[n], B2[n], A[k,i], D2[n], A[0,i]]; # hm ...

(A, b) := GenerateMatrix( sys, vars ); # see the help for that

Where I am never sure, whether indexed variables are a good idea in Maple