递归code。在一系列的4/1 - 4/3 + 4/5 [英] Recursion code in series of 4/1 - 4/3 + 4/5

问题描述

我试图让一个C程序中的问题是：   X = 4/1 - 4/3 + 4/5 - 4/7 + 4/9 ....

I tried to make a C program in which the problem is: x = 4/1 - 4/3 + 4/5 - 4/7 + 4/9 ....

的停止条件是当x的值被计算在previous相互作用低于0.0001

The stop condition is when the value of "x" is calculated in the previous interaction is less than 0.0001

#include <iostream>

using namespace std;

float recursion( float n ){

    int f = 0 , i = 1;
    float calc = 4;

    f = i%2;

    if (calc == 0.0001)
        return 0;


        if ( f != 0 ){
            calc += recursion (4/n) - recursion (4/n+2);
            n += 2;
        }
            else
                calc += recursion (4/n) + recursion (4/n+2);
                n+=2;

            return calc;
}
int main(){

    float n , result;

    cout << "Enter the value of N" ;
    cin >> n;

    result = recursion (n);

    cout << "The value of N is ..." << result ;
}

我试图让会计师来改变符号+，以 - 或 - 到+。 我不知道为什么codito不工作。任何帮助是值得欢迎的。

I tried to get an accountant to change the sign "+" to "-" or "-" to "+". I do not know why codito not working. Any help is welcome

推荐答案

我不认为递归是必要的。这是很容易解决了迭代。 我看到了下面的模式。分母是1,3,5,7-奇数。这是我％2。 分数的符号被交替+和 - 之间 - 。使用迭代的模数2 -1或不乘以电流份额。那么+ =运算总和。

I don't think recursion is necessary. This is easily solved with iteration. I see the following pattern. The denominator is 1,3,5,7- odd numbers. That's i%2. The sign of the fraction is alternating between + and -. Use the modulus 2 of the iteration to multiply the current fraction by -1 or not. Then += to the running sum.

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