Average Error: 39.3 → 0.2
Time: 4.6s
Precision: binary64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.0000001684850568:\\ \;\;\;\;\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + 0.5 \cdot \frac{{x}^{2}}{{1}^{2}}\right)\right) + 1 \cdot \left(x - \left({x}^{2} + \frac{{x}^{3}}{{1}^{2}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \le 1.0000001684850568:\\
\;\;\;\;\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + 0.5 \cdot \frac{{x}^{2}}{{1}^{2}}\right)\right) + 1 \cdot \left(x - \left({x}^{2} + \frac{{x}^{3}}{{1}^{2}}\right)\right)\\

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

\end{array}
double code(double x) {
	return ((double) log(((double) (1.0 + x))));
}
↓
double code(double x) {
	double VAR;
	if ((((double) (1.0 + x)) <= 1.0000001684850568)) {
		VAR = ((double) (((double) (((double) (0.3333333333333333 * ((double) (((double) pow(x, 3.0)) / ((double) pow(1.0, 3.0)))))) + ((double) (((double) (1.0 * ((double) pow(x, 3.0)))) + ((double) (0.5 * ((double) (((double) pow(x, 2.0)) / ((double) pow(1.0, 2.0)))))))))) + ((double) (1.0 * ((double) (x - ((double) (((double) pow(x, 2.0)) + ((double) (((double) pow(x, 3.0)) / ((double) pow(1.0, 2.0))))))))))));
	} else {
		VAR = ((double) log(((double) (1.0 + x))));
	}
	return VAR;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Split input into 2 regimes

  2. if (+ 1.0 x) < 1.0000001684850568

    1. Initial program 59.2

      \[\log \left(1 + x\right)\]
    2. Using strategy rm

    3. Applied flip3-+59.2

      \[\leadsto \log \color{blue}{\left(\frac{{1}^{3} + {x}^{3}}{1 \cdot 1 + \left(x \cdot x - 1 \cdot x\right)}\right)}\]

    4. Applied log-div59.2

      \[\leadsto \color{blue}{\log \left({1}^{3} + {x}^{3}\right) - \log \left(1 \cdot 1 + \left(x \cdot x - 1 \cdot x\right)\right)}\]

    5. Taylor expanded around 0 0.2

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + \left(0.5 \cdot \frac{{x}^{2}}{{1}^{2}} + 1 \cdot x\right)\right)\right) - \left(1 \cdot {x}^{2} + 1 \cdot \frac{{x}^{3}}{{1}^{2}}\right)}\]

    6. Simplified0.2

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + 0.5 \cdot \frac{{x}^{2}}{{1}^{2}}\right)\right) + 1 \cdot \left(x - \left({x}^{2} + \frac{{x}^{3}}{{1}^{2}}\right)\right)}\]

    if 1.0000001684850568 < (+ 1.0 x)

    1. Initial program 0.2

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.0000001684850568:\\ \;\;\;\;\left(0.333333333333333315 \cdot \frac{{x}^{3}}{{1}^{3}} + \left(1 \cdot {x}^{3} + 0.5 \cdot \frac{{x}^{2}}{{1}^{2}}\right)\right) + 1 \cdot \left(x - \left({x}^{2} + \frac{{x}^{3}}{{1}^{2}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

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

  (log (+ 1.0 x)))