# fast_tanh 疑問
> [2025-03-18 問答筆記](https://hackmd.io/EFs_Nfs6TOmIA5ldA_tZsQ?view#EricccTaiwan)

float fast_tanh(float v)
{
    /* Constants for polynomial approximation */
    const float c1 = 0.03138777f;
    const float c2 = 0.276281267f;
    const float log2_e = 1.442695022f; /* log2(e) */

    /* Scale input by log2(e) for exponentiation base 2 */
    v *= log2_e;

    /* Separate integer and fractional parts */
    int int_part = (int) v;
    float frac_part = v - int_part;

    /* Compute polynomial approximation terms */
    float frac_part_sq = frac_part * frac_part;

    float exp_frac_approx = log2_e + c2 * frac_part_sq;
    float exp_frac_linear = frac_part + (c1 * frac_part_sq * frac_part);

    float exp_approx = exp_frac_linear + exp_frac_approx;

    /* Efficiently compute scaling by 2^(int_part) */
    union {
        float f;    // holds the actual floating-point value
        uint32_t i; // used for manipulating the IEEE 754 bit representation
    } exp_scaled = {.f = exp_approx};
    exp_scaled.i += (uint32_t) (int_part << 23);

    float exp_neg_approx = exp_frac_linear - exp_frac_approx;

    /* Combine terms to approximate tanh(v) */
    return (exp_scaled.f + exp_neg_approx) / (exp_scaled.f - exp_neg_approx);
}

### Q1

這邊計算小數部份， lin 8 ~ 11

$$
2^{\text{frac_part}} \approx log_2e + 1\times\text{frac_part}+\text{c2}\times\text{frac_part}^2 + c1 \times \text{frac_part}^3
$$
但這跟實作 Minimax approximation algorithm 近似法去求 `c1` 和 `c2` 不是不一樣嘛
$$
2^{x} \approx 1 + \ln2\times x+\text{c2}\times x^2 + c1 \times x^3
$$
單論，0 和 1 次項的係數就差了 $\ln2$ ，`c1`和`c2`甚至可以共用，不太理解?
> 對的，就是差 $\ln2$ 倍數，用 jenkas 的係數所算出來的 $e^{0.7} = 2.905217$ 剛好與實際值 $e^{0.7}= 2.013752707$ 相差 $\ln2$ 倍數

### Q2

為何 `exp_neg_approx` 計算 ($e^{-x}$)，不需要做整數的縮放，只有計算小數部份而已 (line 30)?

### Q3
`exp_frac_approx` 和 `exp_frac_linear` 改成 `exp_frac_even_term` 和 `exp_frac_odd_term` 是否更符合泰勒展開的意思呢?

### Q4
`fast_tanh()` 也有使用到浮點數運算，所以與理想上的定點數操作，還有段改進空間?

### 處理 $e^{-x}$

#include <stdio.h>
#include <stdint.h>

float fast_tanh(float v)
{
    /* Constants for polynomial approximation */
    const float c1 = 0.03138777f;
    const float c2 = 0.276281267f;
    const float log2_e = 1.442695022f; /* log2(e) */

    /* Scale input by log2(e) for exponentiation base 2 */
    float v_pos = v * log2_e;
    float v_neg = -v_pos;

    // ------- exp(+v) -------
    int int_part_pos = (int)v_pos;
    float frac_part_pos = v_pos - int_part_pos;
    float frac_sq_pos = frac_part_pos * frac_part_pos;
    float even_pos = log2_e + c2 * frac_sq_pos;
    float odd_pos = frac_part_pos + c1 * frac_sq_pos * frac_part_pos;
    float approx_pos = even_pos + odd_pos;
    union {
        float f;
        uint32_t i;
    } exp_pos = {.f = approx_pos};
    exp_pos.i += (uint32_t)(int_part_pos << 23);

    // ------- exp(-v) -------
    int int_part_neg = (int)v_neg;
    float approx_neg = even_pos - odd_pos;
    union {
        float f;
        uint32_t i;
    } exp_neg = {.f = approx_neg};
    exp_neg.i += (uint32_t)(int_part_neg << 23);

    // ------- 結果 -------
    printf("exp(+v) = %.6f\n", exp_pos.f);
    printf("exp(-v) = %.6f\n", exp_neg.f);
    return (exp_pos.f - exp_neg.f) / (exp_pos.f + exp_neg.f);
}

int main()
{
    float v = 0.7f ;
    float result = fast_tanh(v);
    printf("hi\n");
    printf("fast_tanh(%.2f) = %.6f\n", v, result);
    return 0;
}

\begin{align*}
\frac{e^{2x}-1}{e^{-2x}+1}
&= \frac{1 + (\ln 2)(2x) + \frac{(\ln 2)^2}{2}(2x)^2 + \frac{(\ln 2)^3}{6}(2x)^3 - 1}
{1 + (\ln 2)(-2x) + \frac{(\ln 2)^2}{2}(-2x)^2 + \frac{(\ln 2)^3}{6}(-2x)^3 + 1} \\
&= \frac{(\ln 2)(2x) + 2 (\ln 2)^2 x^2 +\frac{4(\ln 2)^3}{3}x^3}{2 - (\ln2) (2x)+2(\ln2)^2x^2+\frac{4(\ln)^3}{3}x^3} \\
&= q
\end{align*}