For 0 <= x < 1 one can use a (inverse) transformation ]0 ... 1[ ---> IR like to be derived
from a probability density or arctan(x)/(Pi/2) or similar, it will depend a bit on the
problem and its (assumed) solution and if a good scaling is possible it would help. 

Then no side-relations are needed, just final transformations to the original task.

Digits:=14;

# the system
eq_hyrnik:={-16.75+(37.45*10^(-3))*A2+(961.07*10^(-3))*B2+(40.48*10^(-3))*C2+1.04*D2-1.00*B1=0,
 (2.27*10^(-6))*A2+(5.83*10^(-5))*B2-(2.46*10^(-6))*D2+(6.30*10^(-5))*C2-(6.59*10^(-5))*B1=0,
 (4.36*10^(-7))*A2-(4.72*10^(-7))*B2+(4.72*10^(-7))*C2-(5.10*10^(-7))*D2-(5.35*10^(-7))*A1+
 (5.35*10^(-7))*B1=0,B1-1.53*10^0=0,-(1.55*10^(-8))*a3^(4.00*10^0)+
 (3.33*10^(-1))*A3*a3^(3.00*10^0)+(5.00*10^(-1))*B3*a3^(2.00*10^0)+D3+1.67*10-
 (1.00*10^0)*B2-(1.00*10^0)*D2=0,-(6.21*10^(-8))*a3^(3.00*10^0)+A3*a3^(2.00*10^0)+B3*a3-
 (7.79*10^(-3))*A2+(7.79*10^(-3))*B2-(7.79*10^(-3))*C2-(7.79*10^(-3))*D2=0,-
 (1.86*10^(-7))*a3^(2.00*10^0)+(2.00*10^0)*A3*a3+B3-(1.21*10^(-4))*B2-
 (1.21*10^(-4))*C2=0,-(3.73*10^(-7))*a3+(2.00*10^0)*A3-(9.45*10^(-7))*A2+
 (9.45*10^(-7))*B2-(9.45*10^(-7))*C2+(9.45*10^(-7))*D2=0,-2.52*10+B2+D2=0,D3-1.
 *alfa=0,A3-(6.24*10^(-7))*P=0,(7.19*10^(-3))*A2-(7.78*10^(-3))*B2+(8.41*10^(-3))*C2+
 (7.78*10^(-3))*D2-(8.12*10^(-3))*A1+(8.12*10^(-3))*B1=0};
 
# writing exactly, depends on Digits
eq_hyrnik:=map(convert,eq_hyrnik,rational); 

# solve it
sol:=solve(eq_hyrnik,
   {A1, A2, A3, B1, B2, B3, C2, D2, D3, P, a3, alfa});
# and look at it numerical
sol_num:=evalf(%);

# check the solution numerically
eval(eq_hyrnik, sol_num): evalf(%);
 map(fnormal, %);
                                   -10                  -7
     {-0. = 0., 0. = 0., 0.14291 10    = 0., 0.130002 10   = 0.}

# and symbolical
eval(eq_hyrnik, sol): simplify(%);

                               {0 = 0}

Note: this is one solution, there is no reason why it is the only
one (or why there should be only a finite number, in general)

Digits:=14;

# the system
eq_hyrnik:={-16.75+(37.45*10^(-3))*A2+(961.07*10^(-3))*B2+(40.48*10^(-3))*C2+1.04*D2-1.00*B1=0,
 (2.27*10^(-6))*A2+(5.83*10^(-5))*B2-(2.46*10^(-6))*D2+(6.30*10^(-5))*C2-(6.59*10^(-5))*B1=0,
 (4.36*10^(-7))*A2-(4.72*10^(-7))*B2+(4.72*10^(-7))*C2-(5.10*10^(-7))*D2-(5.35*10^(-7))*A1+
 (5.35*10^(-7))*B1=0,B1-1.53*10^0=0,-(1.55*10^(-8))*a3^(4.00*10^0)+
 (3.33*10^(-1))*A3*a3^(3.00*10^0)+(5.00*10^(-1))*B3*a3^(2.00*10^0)+D3+1.67*10-
 (1.00*10^0)*B2-(1.00*10^0)*D2=0,-(6.21*10^(-8))*a3^(3.00*10^0)+A3*a3^(2.00*10^0)+B3*a3-
 (7.79*10^(-3))*A2+(7.79*10^(-3))*B2-(7.79*10^(-3))*C2-(7.79*10^(-3))*D2=0,-
 (1.86*10^(-7))*a3^(2.00*10^0)+(2.00*10^0)*A3*a3+B3-(1.21*10^(-4))*B2-
 (1.21*10^(-4))*C2=0,-(3.73*10^(-7))*a3+(2.00*10^0)*A3-(9.45*10^(-7))*A2+
 (9.45*10^(-7))*B2-(9.45*10^(-7))*C2+(9.45*10^(-7))*D2=0,-2.52*10+B2+D2=0,D3-1.
 *alfa=0,A3-(6.24*10^(-7))*P=0,(7.19*10^(-3))*A2-(7.78*10^(-3))*B2+(8.41*10^(-3))*C2+
 (7.78*10^(-3))*D2-(8.12*10^(-3))*A1+(8.12*10^(-3))*B1=0};
 
# writing exactly, depends on Digits
eq_hyrnik:=map(convert,eq_hyrnik,rational); 

# solve it
sol:=solve(eq_hyrnik,
   {A1, A2, A3, B1, B2, B3, C2, D2, D3, P, a3, alfa});
# and look at it numerical
sol_num:=evalf(%);

# check the solution numerically
eval(eq_hyrnik, sol_num): evalf(%);
 map(fnormal, %);
                                   -10                  -7
     {-0. = 0., 0. = 0., 0.14291 10    = 0., 0.130002 10   = 0.}

# and symbolical
eval(eq_hyrnik, sol): simplify(%);

                               {0 = 0}

Note: this is one solution, there is no reason why it is the only
one (or why there should be only a finite number, in general)

Thank you, that was quite instructive (though I will have it forgotten
if I will need it in future [sigh ...]) !

My reason to use log(cdf) for tails is that:

Look at plot(ln(erfc(x)), x=0 ..18) and now increase the upper bound to
something unreasonable, say 2000 instead of 18. It still will look quite
'linear' and Maple can solve erfc(x)= 1e-2000 after applying log (which
one can take either as solution or as initial guess for the actual task),
even with the usual 10 Digits.

Looking at asymptotics it is like to solve x^2 + ln(x) = some y, 
so its derivative is 2*x + 1/x ~ 2*x as x must be 'large'.

For other distributions with fatter/meager tails it should be similar. 

The other point (for me) is, that 1.0 - verySmall lets me switch to the
complementary, since it would be absorbed and viewed to be as 1.0, or in
words for that setting: instead of 'almost sure' numerical 'almost never'
is more convenient (as far as talking about extreme situations).

Thank you, that was quite instructive (though I will have it forgotten
if I will need it in future [sigh ...]) !

My reason to use log(cdf) for tails is that:

Look at plot(ln(erfc(x)), x=0 ..18) and now increase the upper bound to
something unreasonable, say 2000 instead of 18. It still will look quite
'linear' and Maple can solve erfc(x)= 1e-2000 after applying log (which
one can take either as solution or as initial guess for the actual task),
even with the usual 10 Digits.

Looking at asymptotics it is like to solve x^2 + ln(x) = some y, 
so its derivative is 2*x + 1/x ~ 2*x as x must be 'large'.

For other distributions with fatter/meager tails it should be similar. 

The other point (for me) is, that 1.0 - verySmall lets me switch to the
complementary, since it would be absorbed and viewed to be as 1.0, or in
words for that setting: instead of 'almost sure' numerical 'almost never'
is more convenient (as far as talking about extreme situations).

So we can not have Fat Freddy's cat looking at a Riemannian surface ?

So we can not have Fat Freddy's cat looking at a Riemannian surface ?

formatted printing uses a C-like syntax:

printf("x=%.2f", 12345.6789);
x=12345.68

a:=sprintf("x=%.2f", 12345.6789):
parse(%);
                             x = 12345.68

formatted printing uses a C-like syntax:

printf("x=%.2f", 12345.6789);
x=12345.68

a:=sprintf("x=%.2f", 12345.6789):
parse(%);
                             x = 12345.68

> f := Sum((-1)^j*binomial(t-j-1, 4-j)*binomial(g+t-4+j, j)*
> (Sum(factorial(g+2*t-8)*binomial(4-j, i)/(factorial(t-j-1)*factorial(t-3+j-i)*factorial(g-4+i)), 
> i = 0 .. 4-j)), j = 0 .. 4);

> combine(f);
> op(1,%): 
> b:=op(1,%);
> convert(%, GAMMA);
> a:=simplify(%) assuming (i::nonnegint, j::nonnegint);
> eval(%, g=30): 
> A:=eval(%, t=20);

    4373479246236240*GAMMA(47+j)/GAMMA(1+j)*(-1)^j/GAMMA(1+i)/GAMMA(5-j-i)/GAMMA(18+j-i)/GAMMA(27+i)

A is the expression over which you do your summation, it is always positive and never 0,
since there is no factor 1 / Gamma( neg integer)

> for j from 0 to 4 do
>   for i from 0 to 4 - j do
>     print(5-j-i, 18 +j+i); 
>   end do;
> end do; i:='i': j:='j':

The message may be caused by some 1 / GAMMA( negative integer ), which Maple does not take as 0.

For the result "0" in your example I would guess it is not correct

Most of us - at least that is true for me - learned by asking questions which I hoped would have been forgotten for ever. However I remind several of them, even before the www :-)

So: do not mind ...

Most of us - at least that is true for me - learned by asking questions which I hoped would have been forgotten for ever. However I remind several of them, even before the www :-)

So: do not mind ...

I do not have Maple 13, but it seems so ... while for my taste 'map' is clearer as command then the tilde '~'

I do not have Maple 13, but it seems so ... while for my taste 'map' is clearer as command then the tilde '~'