What I understand is that your coefficients are numerical but your $x (as you wrote it) is just a symbol in an equation and not a variable in a computer program.

So, provided I understood correctly, what you are looking for is a package capable of symbolic calculations.

There is one in perl already:

use Math::Calculus::NewtonRaphson;

use strict;
use warnings;

my $exp = Math::Calculus::NewtonRaphson->new();
$exp->addVariable('x'); # our variable in the equations
$exp->setExpression('x+3*x-x^2');
if( $exp->getError() ){ print "got error, ".$exp->getError()."\n"; exi
+t(1) }
my $result = $exp->newtonRaphson('x', 0);
print "result: ".$result."\n";
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Which gets fatal errors unfortunately.

Do you have any more options?

Yes you do. At least two.

1) Use perl to produce the sum equation in symbolic form (e.g. 1/(10*x)) and then use the symbolic solver in package R.

library(rSymPy)
x <- Var("x")
sympy("solve(3+3*x-x*x, x)")
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The answer in this simple example is:

[1] "[3/2 - 21**(1/2)/2, 3/2 + 21**(1/2)/2]"
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But you still need perl in order to produce the final equation as a string containing numbers and the letter 'x' (which is your symbolic variable you want to solve for). In the following code I have omitted the $c==0 or $c==1 logic.

use strict;
use warnings;
my $i;
my @q = map { rand() } (1..1000); # your 1000 'q' coefficients as just
+ random numbers here
my @c = map { rand() } (1..1000); # your 1000 'c' coefficients as just
+ random numbers here


my ($a1, $a2, $a3);
my $sum_equ = "";

for ($i=1; $i<1000; $i++) {
      $a1 = (1-$c[$i])*(1-2*$q[$i]);
      $a2 = -2*$q[$i];
      $a3 = $c[$i]*(2*$q[$i] -1);
      # concatenate the equation of this time with the total equation:
      $sum_equ = $sum_equ . "($a1/($a2 * x + x + $q[$i]) + $a3/(2*q[$i
+]*x-x+1)) + ";
}
$sum_equ =~ s/\+\s*$//; # remove last +
print "library(rSymPy)\n\
x <- Var('x')\n\
sympy('solve($sum_equ, x)')\n
";
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Save the above code to a file called 'equ.pl' and then from the command-line do:

perl equ.pl | R
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It will probably give you the solution crash your computer (because the equation is huge).

2) Your second option (and your only option if your computer crashed in option #1) requires a bit more thinking because you need to handcraft your equations before concatenating in order to do some simplifications of the form 3*x + x => 4*x. How are you going to achieve that? You need to start from your basis equations and accumulate the coefficients of 'x' 'x^2' (if any) in a single variable etc..