Average Error: 39.1 → 0.4
Time: 12.3s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000007547740210611664224416017532:\\ \;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x + 1 \le 1.000000000007547740210611664224416017532:\\
\;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\

\mathbf{else}:\\
\;\;\;\;\log \left(x + 1\right)\\

\end{array}
double f(double x) {
        double r3229038 = 1.0;
        double r3229039 = x;
        double r3229040 = r3229038 + r3229039;
        double r3229041 = log(r3229040);
        return r3229041;
}
↓
double f(double x) {
        double r3229042 = x;
        double r3229043 = 1.0;
        double r3229044 = r3229042 + r3229043;
        double r3229045 = 1.0000000000075477;
        bool r3229046 = r3229044 <= r3229045;
        double r3229047 = r3229043 * r3229042;
        double r3229048 = log(r3229043);
        double r3229049 = r3229042 / r3229043;
        double r3229050 = 0.5;
        double r3229051 = r3229050 * r3229049;
        double r3229052 = r3229049 * r3229051;
        double r3229053 = r3229048 - r3229052;
        double r3229054 = r3229047 + r3229053;
        double r3229055 = log(r3229044);
        double r3229056 = r3229046 ? r3229054 : r3229055;
        return r3229056;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.1
Target0.3
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000000000075477

    1. Initial program 59.4

      \[\log \left(1 + x\right)\]

    2. Taylor expanded around 0 0.3

      \[\leadsto \color{blue}{\left(\log 1 + 1 \cdot x\right) - \frac{1}{2} \cdot \frac{{x}^{2}}{{1}^{2}}}\]

    3. Simplified0.3

      \[\leadsto \color{blue}{\left(\log 1 - \frac{x}{1} \cdot \left(\frac{x}{1} \cdot \frac{1}{2}\right)\right) + 1 \cdot x}\]

    if 1.0000000000075477 < (+ 1.0 x)

    1. Initial program 0.5

      \[\log \left(1 + x\right)\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x + 1 \le 1.000000000007547740210611664224416017532:\\ \;\;\;\;1 \cdot x + \left(\log 1 - \frac{x}{1} \cdot \left(\frac{1}{2} \cdot \frac{x}{1}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019179 
(FPCore (x)
  :name "ln(1 + x)"

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))