Continued Fractions Solving

Question

Here's what I thought of so far which works for 2 elements in the array. the elements in the array are the variables to be plugged in to the continued fraction.

double continuedFraction(int a[], int size)
{
    double fraction = a[0];

    for(int i = 1; i < size; i++)
    { 
        fraction += (double)1/(double)a[i];
    }

    return fraction;
}

btw I'm not using recursion I need to be able to get the continued fraction result.

I'm sure it shouldn't be too hard I really can't visualize this. I really appreciate anyone's guidance. — What, May 08 '14 at 21:58

score 0 · Accepted Answer · answered May 08 '14 at 22:16

For the first iteration, you get (a0 + 1/a1). In further iterations, your code keeps adding the inverse of given number which is (a0 + 1/a1) + (1/a2) + (1/a3) + ...

What you ideally need is an + ... 1/(a2 + 1/(a1 + 1/a0))...)

double continuedFraction(int a[], int size)
{
    double fraction = a[0];

    for(int i = 1; i < size; i++)
    { 
        fraction = a[i] + (1/fraction);
    }

    return fraction;
}

If you want the other way around, a0 + 1/(a1 + 1/(a2 + 1/a3 + ... ))...) you could run the loop from array_size-1 to 0

Thanks! you enlightened me now I'm thinking about it all the points to ya whenever I get the chance I love you — What, May 08 '14 at 22:27

Continued Fractions Solving

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