# 求 log 近似值
## 使用 $\tanh^{-1}x$ 的泰勒展開來解釋
觀察 [kxo 作業說明](https://hackmd.io/@sysprog/linux2025-kxo/%2F%40sysprog%2Flinux2025-kxo-a#Fixed-point-log) 對 `fixed_log` 的實作：

Q23_8 fixed_log(int input)
{
    if (!input || input == (1U << Q))
        return 0;

    int64_t y = input << Q;
    y = ((y - (1U << Q)) << (Q)) / (y + (1U << Q));
    int64_t ans = 0U;
    for (unsigned i = 1; i < 20; i += 2) {
        int64_t z = (1U << Q);
        for (int j = 0; j < i; j++) {
            z *= y;
            z >>= Q;
        }
        z <<= Q;
        z /= (i << Q);

        ans += z;
    }
    ans <<= 1;
    return (Q23_8) ans;
}

以下以再不考慮定點數的觀點來看這個實作的數學部份：
它是在求 $\displaystyle 2\sum_{i=0}^{\infty}\frac{1}{2i+1}(\frac{y-1}{y+1})^{2i+1}$
會發現，$\displaystyle \tanh^{-1}x = \sum_{i=0}^{\infty}\frac{1}{2i+1}x^{2i+1}$，然後我們把上式帶入：

$\displaystyle 2\sum_{i=0}^{\infty}\frac{1}{2i+1}(\frac{y-1}{y+1})^{2i+1} = 2\tanh^{-1}(\frac{y-1}{y+1})$

$\displaystyle \text{Let } x = \frac{y-1}{y+1},\ k=2\tanh^{-1}x$

$\displaystyle \Rightarrow \tanh\frac{k}{2}=x$

$\displaystyle \Rightarrow x=\frac{e^{k}-1}{e^k+1}$

$\displaystyle \Rightarrow e^kx+x=e^k-1$

$\displaystyle \Rightarrow e^k=\frac{1+x}{1-x} = \frac{1+\frac{y-1}{y+1}}{1-\frac{y-1}{y+1}}=y$

$\displaystyle k=\ln y$

所以，實際上這個函數是使用 $\tanh^{-1}$ 來估計 $\ln$

## 使用 $\ln$ 的泰勒展開式來解釋
$\displaystyle 2\sum_{i=0}^{\infty}\frac{1}{2i+1}(\frac{y-1}{y+1})^{2i+1}$

$\displaystyle \sum_{i=1}^{\infty}\frac{1}{i}(\frac{y-1}{y+1})^i - \sum_{i=1}^{\infty}\frac{1}{i}(-\frac{y-1}{y+1})^i$

$\displaystyle = \ln(1+\frac{y-1}{y+1}) - \ln(1-\frac{y-1}{y+1})$

$\displaystyle = \ln(2y) - \ln2$

$\displaystyle = \ln y$