How do I do a summation series with in c++ and have it print out the terms?

#include %26lt;iostream.h%26gt;


using namespace std;


#include %26lt;math.h%26gt;





int main()


{





std::cout %26lt;%26lt; "The Zeta Function" %26lt;%26lt; std::endl;


int a, n, m, x, sum, total;


std::cout %26lt;%26lt; "Enter value for m, where m is the number of terms in the series that the series begins on" %26lt;%26lt; std::endl;


cin %26gt;%26gt; m;


std::cout %26lt;%26lt; "Enter value for n, where n is the number of terms in the series that the series ends on" %26lt;%26lt; std::endl;


cin %26gt;%26gt; n;





for (int x = m; x %26lt;= n; x++)


total = (1/(x^2));





std::cout %26lt;%26lt; "The sum of the series is " %26lt;%26lt; total %26lt;%26lt; "\n";


return 0;


}





The math function doesn't work and I want every term sumed up to that point to be printed then I want the total sum to be printed. The function is the Zeta Function

How do I do a summation series with in c++ and have it print out the terms?
Your code is very close but there are a few things wrong with it.





1. ^ is a bitwise operator XOR which can actually result in zero which means your code will try an do an illegal division by zero if you set m and n to be 2 and 3 for example.





You need use pow here. 1/x^2 = x^-2 so you can use


pow (x , -2).





2. All variables have been declared as integers which means any division 1/y = 0 for all y%26gt;1. So for all x^2 %26gt; 1 the division will lose all remainder and be zero.





You need to declare your variables as doubles for maximum precision.





double a, n, m, x, sum, total;





3. You are overwriting the value total each loop iteration and total is never initialised.





So you need to keep adding to total once it has been initialised to zero.





total = 0;


for (int x = m; x %26lt;= n; x++)


total += pow (x , -2);





4. User could type in values so m %26gt; n which means the loop terminates immediately leaving total as zero so we can swap these values if that's the case.





if( m%26gt; n) {


a=m;


m=n;


n=a;


}





Here is a corrected version of your code:





#include %26lt;iostream.h%26gt;


#include %26lt;math.h%26gt;


using namespace std;





int main()


{


std::cout %26lt;%26lt; "The Zeta Function" %26lt;%26lt; std::endl;


double a, n, m, x, sum, total;


std::cout %26lt;%26lt; "Enter value for m, where m is the number of terms in the series that the series begins on" %26lt;%26lt; std::endl;


cin %26gt;%26gt; m;


std::cout %26lt;%26lt; "Enter value for n, where n is the number of terms in the series that the series ends on" %26lt;%26lt; std::endl;


cin %26gt;%26gt; n;


if( m%26gt; n) {


a=m;


m=n;


n=a;


}


total = 0;


for (int x = m; x %26lt;= n; x++)


total += pow (x , -2);





std::cout %26lt;%26lt; "The sum of the series is " %26lt;%26lt; total %26lt;%26lt; "\n";


return 0;


}





Additional checks you might want are seeing if the user has input non-zero values and actual integers.
Reply:nstead of doing x^2 which is not supported, do x*x
Reply:^ is a bitwise XOR .. not the pow





you can use pow





double pow(double x, double y);


long double powl(long double x, long double y);


double pow10(int p);


long double pow10l(int p);


double pow10(int p);


long double pow10l(int p);