Min x
Hello everyone!I need help for something that is more a maths exercise than a programming.So i need to find the min x for this equation x+10^6=x.
what i have done
but this is not correct.
what i have done
|
|
but this is not correct.
Mathematically this is not a valid equation:
x + 10^6 = x indeed <=> 10^6 = 0
which is absurd!
x + 10^6 = x indeed <=> 10^6 = 0
which is absurd!
#include <stdio.h>
#include <math.h>
int main()
{
double e = 2;
while(e + pow(10,6) > e)
{
e *= e;
}
printf("%lf\n",e);
return 0;
}
#include <math.h>
int main()
{
double e = 2;
while(e + pow(10,6) > e)
{
e *= e;
}
printf("%lf\n",e);
return 0;
}
rich1 wrote:
but this condition is, obviously, always fulfilled and the loop is endless!!!
|
|
but this condition is, obviously, always fulfilled and the loop is endless!!!
Condor this not absurd equation.look this x+1=1
in your computer the zero number is not only 0.
Rich1 this what i have done but why this is the min x?
in your computer the zero number is not only 0.
Rich1 this what i have done but why this is the min x?
In your original post you wrote this "equation": x + 10^6 = x. Is this equivalent with 10^6 = 0 ? Yes, it is! And it's absurd! This I noticed previously.
Your second post is: x + 1 = 1 which is something else! Equation perfect valid!
Your second post is: x + 1 = 1 which is something else! Equation perfect valid!
Please condor type in google numerical analysis and computer epsilon.for those that know what i am telling
about please help
about please help
Topic archived. No new replies allowed.