Average Error: 39.0 → 0.3
Time: 13.2s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000104635906827965641241462435573339:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.000000104635906827965641241462435573339:\\
\;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\

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

\end{array}
double f(double x) {
        double r108227 = 1.0;
        double r108228 = x;
        double r108229 = r108227 + r108228;
        double r108230 = log(r108229);
        return r108230;
}
↓
double f(double x) {
        double r108231 = 1.0;
        double r108232 = x;
        double r108233 = r108231 + r108232;
        double r108234 = 1.0000001046359068;
        bool r108235 = r108233 <= r108234;
        double r108236 = r108232 * r108231;
        double r108237 = -0.5;
        double r108238 = r108231 * r108231;
        double r108239 = r108238 / r108232;
        double r108240 = r108237 / r108239;
        double r108241 = r108232 * r108240;
        double r108242 = r108236 + r108241;
        double r108243 = log(r108231);
        double r108244 = r108242 + r108243;
        double r108245 = log(r108233);
        double r108246 = r108235 ? r108244 : r108245;
        return r108246;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original39.0
Target0.2
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000001046359068

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.3

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

    3. Simplified0.3

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

    5. Applied distribute-lft-in0.3

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

    if 1.0000001046359068 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000104635906827965641241462435573339:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019351 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

  :herbie-target
  (if (== (+ 1 x) 1) x (/ (* x (log (+ 1 x))) (- (+ 1 x) 1)))

  (log (+ 1 x)))