## 622 Reputation

14 years, 262 days

## Modules....

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?

## Try this....

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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## Like this:...

```[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);
```

## One real solution....

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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## There are two ways:...

``` 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:

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

## Consider this code:...

```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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## Windows....

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

## Change the sqrt....

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

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## Introduce substitution...

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?

## Not automated....

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:

 S0 := 4*beta*c*lambda^2*mu^2+2*beta*c*lambda*mu^3+2*beta*c*lambda*mu^2*nu+beta*c*mu^3*nu-4*c*lambda^2*mu^2*sigma-2*c*lambda*mu^3*sigma-2*c*lambda*mu^2*nu*sigma-c*mu^3*nu*sigma+4*beta*lambda^3*sigma+4*beta*lambda^2*mu*sigma+2*beta*lambda^2*nu*sigma+2*beta*lambda*mu^2*sigma+2*beta*lambda*mu*nu*sigma+beta*mu^3*sigma+beta*mu^2*nu*sigma+4*lambda^2*mu^2*sigma+2*lambda*mu^3*sigma+2*lambda*mu^2*nu*sigma+mu^3*nu*sigma

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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.

## A solution....

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);`

$\left\{\begin{array}{cc}{x}\le {1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{0}<{x}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}or\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}\le {0}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}or\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{1}<{x}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}<-{a}& {a}<-{3}\\ {x}\le {1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{0}<{x}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}or\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{1}<{x}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}<-{a}& -{3}\le {a}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{a}<-{1}\\ {x}\le {1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{0}<{x}& -{1}\le {a}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{a}\le {0}\\ \sqrt{{a}}<{x}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{x}<{1}& {0}<{a}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}and\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{a}<{1}\\ {\mathrm{false}}& {1}\le {a}\end{array}$

Hope this suffices. Thumb if You like.

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