那么你将有泰勒级数，可以改写为更好的收敛

Well you would have the Taylor series, that can be rewritten for better convergence

要改变这个漂亮的平等纳入一个算法，你必须了解的收敛一系列工作：每学期越来越小。这种减少会发生足够快，以使总和为有限的值：LN（y）的

To transform this nice equality into an algorithm, you have to understand how a converging series work : each term is smaller and smaller. This decrease happens fast enough so that the total sum is a finite value : ln(y).

的实数好的属性，因为，你可以考虑序列收敛于LN（Y）：

Because of nice properties of the real numbers, you may consider the sequence converging to ln(y) :

• L（1）= 2/1 *（Y-1）/（Y + 1）
• L（2）= 2/1 *（γ-1）/（Y + 1）+ 2/3 *（（Y-1）/（Y + 1））^ 3
• L（3）= 2/1 *（γ-1）/（Y + 1）+ 2/3 *（（Y-1）/（Y + 1））^ 3 + 2/5 *（（ Y-1）/（Y + 1））^ 5

..等。

显然，算法计算该序列很容易：

Obviously, the algorithm to compute this sequence is easy :

x = (y-1)/(y+1);
z = x * x;
L = 0;
k = 0;
for(k=1; x > epsilon; k+=2)
{
    L += 2 * x / k;
    x *= z;
}

目前一些点，你的x将变得如此小，它无助到L的有趣位数了，而不是仅修改小得多位数。当这些修改开始是太微不足道了你的目的，你可能会停止。

At some point, your x will become so small that it will not contribute to the interesting digits of L anymore, instead only modifying the much smaller digits. When these modifications start to be too insignificant for your purposes, you may stop.

因此​​，如果你想达到一个precision 1E ^ -20，集小量是合理的小于，你是好去。

Thus if you want to achieve a precision 1e^-20, set epsilon to be reasonably smaller than that, and you're good to go.

不要忘了，如果你能在日志中因式分解。如果是例如LN（A²）完全平方= 2号法律公告（一） 事实上，该系列将收敛更快时（γ-1）/（Y + 1）较小，因此当y较小（或者说，越接近1，但应该是等价的，如果你打算使用整数）。

Don't forget to factorize within the log if you can. If it's a perfect square for example, ln(a²) = 2 ln(a) Indeed, the series will converge faster when (y-1)/(y+1) is smaller, thus when y is smaller (or rather, closer to 1, but that should be equivalent if you're planning on using integers).