#C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

#C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

#C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

#C, 115 bytes, user4867444

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

#C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.