I am trying to code the summation of the following series:

I can't figure out how to alternate the sign from "-" to " " to "-" and so on....

Here is my code:

#include<stdio.h>

int main()
{
    int i=1,n;
    float sum,num,den;
    den=2;
    sum=0;

    printf("Enter number of terms for summation: ");
    scanf("%d",&n);

    for(i=1;i<=n;i  )
    {
        num=i*(i 1);
        den=den*(i 2);
        sum =num/den;
    }
    printf("%f",sum);
}

CodePudding user response：

Start a variable at 1.

int coeff =  1;

then:

sum  = coeff * num/den;
coeff *= -1;  // Toggles -1 to  1 and back to -1 each loop.

When coeff is 1, then the expression is 1 * -1, which results in -1.

When coeff is -1, then the expression is -1 * -1, which results in 1.

the variable keeps toggling back-n-forth between 1 and -1.

CodePudding user response：

Let’s discuss the methods proposed in comments and other answers in descending order of nominal burden. (By “nominal burden,” I meant what the code literally says to calculate, disregarding potential compiler optimization.)

• sum =pow(-1,i)*num/den. This calls pow, which is a complicated routine. This solution involves a function call and branching through the various special cases that pow must handle.
• if ((i & 1) == 0) num = -num;. This involves testing, comparing, branching, negating, and assignment. For hardware design reasons, branching may impair performance.
• int coeff = 1;, sum = coeff * num/den;, and coeff *= -1;. Okay, now we are on track. We have added just one variable and two multiplications. The latter does not even have to be a multiplication; it could be just coeff = -coeff;.

Having learned from those, can we solve it any more simply? Yes, we make a simple change, from:

den=den*(i 2);

to:

den=-den*(i 2);

Then den will flip between positive and negative each iteration, with just a single negation. However, we do need to start den with a negative value; den = -2;.

But then, consider this:

den = den*(-2-i);

Although this involves more characters changed in the source code, it might not be any more work. i 2 may be calculated by taking 2 and adding i, whereas -2-i may be calculated by taking −2 and subtracting i, the same amount of work.

Thus, you can make your program work for alternating signs without any more nominal work than the unchanging sign version.

CodePudding user response：

int main()
{
    int i=1,n, sign = 1;
    float sum,num,den;
    den=2;
    sum=0;

    printf("Enter number of terms for summation: ");
    scanf("%d",&n);

    for(i=1;i<=n;i  )
    {
        num=i*(i 1);
        den=den*(i 2);
        sum =sign * num/den;
        sign *= -1;
    }
    printf("%f",sum);
}

CodePudding user response：

With x = i % 2 alternating between 0 and 1, you have two equations in two variables:

 1 = A * 0   B
-1 = A * 1   B

From which you can quickly infer:

B = 1
A = -2

So what you can do is:

    const int A = -2;
    const int B = 1;
    ...
    for (i = 0; i < n; i  )
    {
        int x = i % 2;
        int y = A * x   B;
        ...
        sum =y*num/den;
    }