I would expect LCState is a variable that belongs the module containing the GetState proc. That would be one of the modules:  RandomTools:-LinearCongruence and RandomTools.

MapleQuestions_227381.mw

I do not have the shoot library. Where is it stored?

Perhaps this gives the correct values:

F := proc(f) Re(evalf(convert(f, Sum), 20)); end proc;
seq(sum(((binomial(2*k, k))/(4^k))^d, k = 1 .. infinity), d = [3, 4, 5, 6]);
F~([%]);

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[seq(convert(X, set), X = n__a)];
map(convert, [n__a], set);

or

[seq(convert(X, list), X = n__a)];
map(convert, [n__a], list);

The anser appears to be:
E__fv = -48.46001884+(6.283185307*I)*n, E__fv = -46.13690406+3.141592654*I+(6.283185307*I)*n;
So the first one gives
E__fv = -48.46001884;
as the one real solution.


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 frontend(collect, [eqq, [d+H, H, A, a], factor], [({Non})(identical(d+H)), {}]);

 subs(`d+H` = d+H, collect(simplify(eqq, [d+H-`d+H`], [d, `d+H`]), [`d+H`, H, A, a], factor));

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Edit:
I also managed to obtain this form:

https://www.mapleprimes.com/DocumentFiles/260082_Answer/Annotation_2019-06-19_230004.png

#1:
L := [[1,2,3],[7,8,9],[13,12,11]] ;
map2(op, 3, L);

F := proc (R0, M, K) local X, Y, beta; unapply([(M*X + R0)*cos(beta*K), (M*X + R0)*sin(beta*K), M*Y], X, Y, beta) end proc;
f := F(1, 1, (Pi/2)/0.5);
A := U[1,6](x, theta);
cylinderplot(A - 0.1, theta = 0..2*Pi, x = 0..0.5, grid = [50,50], transparency = 0.3, scaling = constrained);
plottools[transform](f)(%);

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I believe there is a Windows setting that controls the presentation of numbers. Could it be this control affects Maple too?

The reason for this is the sqrt command. Instead use ^(1/2).

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and you can obtain the following form of your system:

{p^3*c[2]+p^2*q*c[1]+p*q^2*c[2]+q^3*c[1]-p*c[4], -p^3*c[1]+p^2*q*c[2]-p*q^2*c[1]+q^3*c[2]-q*c[4]};

Quite straightforward from here.

What do you want to do exactly with this?


 

There are no automatic simplifications for this as the number of choices is simply too great. Every conversion you use will be a procedure written in Maple by you. Personally I like this form:

-beta*(2*lambda^2+(mu+2*lambda)*mu)*(2*lambda+nu)+((mu+2*lambda)*(2*lambda+nu)*(-1+c)-beta*mu)*mu^2;

To clarify, the above is the second formula given in the post:

((2*c-2)*lambda-beta+nu*(-1+c))*mu^3+(2*(lambda*(-1+c)-(1/2)*beta))*(2*lambda+nu)*mu^2-2*beta*lambda*(2*lambda+nu)*mu-2*lambda^2*beta*(2*lambda+nu);

Edit:

Here is a simplified version of the first formula:

(1/2)*(((mu^2+(2*lambda+mu)^2)*(2*lambda+nu)+2*mu^3)*beta-2*(2*lambda+nu)*mu^2*(2*lambda+mu)*(c-1))*sigma+(2*lambda+nu)*beta*mu^2*c*(2*lambda+mu);



The above is mmcdara's S0:

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So what you are doing is this:

V := f[0], f[2];

fn_U := unapply(U, V, a, nu);

Eq_U := [D[1], D[2]](fn_U)(V, b, 1/3);

solve(Eq_U, [V]);

You should copy and paste your definition of U to this thread.

I managed to reduce the problem statement into a boolean expression with parameter a using some manipulation and SolveTools:-SemiAlgebraic.

piecewise(a < -3,x <= 1 and 0 < x or x <= 0 or 1 < x and x < -a,-3 <= a and a < -1,x <= 1 and 0 < x or 1 < x and x < -a,-1 <= a and a <= 0,x <= 1 and 0 < x,0 < a and a < 1,a^(1/2) < x and x < 1,1 <= a,false);

Hope this suffices. Thumb if You like.