recursive function

[HighlightReel44]

I need to write a recursive function to compute the sum of the first n terms of the following series

1/3 - 2/5 + 3/7 - 4/9 + 5/11 - 6/13.....

Thanks ina dvance for the help

[whodgson]

Hmmm....If you arranged to have the numerators of each fraction in one array and the denominators of each in a second array you could access each element in each array (each having the same index [] value) and divide one into the other.
To decide on whether to add or subtract make the values with even indexes[] positive + and those with odd indexes [] negative -. the running sum+= could take care of calculating the series sum. Post some code to show us how you are progressing.

[HighlightReel44]

@whodgson
Any chance you can give me an example source code so i have something to follow along with

[jkmyoung]

That array method is too confusing and unnecessary!

so numerator = i
and denominator = 2i+1?

Have a parity flag.

loop on i -> n 
  if flag = positive
    add i/(2i+1)
    set flag = negative
  else // flag = negative
    subtract i/(2i+1)
    set flag = positive
end loop
 

Could we see how your source code differs so that we can help merge the algorithm in?

[HighlightReel44]

@jkmyoung
#include <iostream>
using namespace std;

int main()
{
double sum = 0;
for (int i = 1; i <= 97; i += 2)
sum += 1.0 * i / (i + 2);

cout << "sum is " << sum;

return 0;
}


This waht is tarted doin but i know its not right

[HighlightReel44]

@jkmyoung
I also formulated soemthing like this but this is only for a series with all positive values how do i change it to include the negatives


#include <iostream>
using namespace std;
double seriesFunc(int n){
if (n == 0) return 0;
else if (n == 1) return 1/3.0;
else return (n / (2 * n + 1)) + seriesFunc(n-1);}
int main()
{ cout << seriesFunc(6) << endl;
cout << seriesFunc(2) << endl;
return 0;}

[jkmyoung]

Try plugging in values:

sum += 1.0 * i / (i + 2);
> i = 1
sum += 1.0 * 1 / (1 + 2);
sum += 1 / 3;
> i = 2
sum += 1.0 * 2 / (2 + 2);
sum += 2 / 4

doesn't seem like what you want.
====
Recursion:
Notice when i is odd, you add. When i is even, you subtract.
i%2 == 0 -> i is even
So:

else if i is even  
   return  - i/(2i+1) plus sum of the rest
else 
   return  +i/(2i+1) plus sum of the rest

[HighlightReel44]

@jkmyoung
Thanks a lot this was very much appreciated