Average Error: 38.9 → 0.3
Time: 3.8s
Precision: binary64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.000000000011955:\\ \;\;\;\;\left(x + \left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right)\right) \cdot \left(x \cdot x\right)\right) + \left(x \cdot x\right) \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \leq 1.000000000011955:\\
\;\;\;\;\left(x + \left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right)\right) \cdot \left(x \cdot x\right)\right) + \left(x \cdot x\right) \cdot -0.5\\

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

\end{array}
(FPCore (x) :precision binary64 (log (+ 1.0 x)))
↓
(FPCore (x)
 :precision binary64
 (if (<= (+ 1.0 x) 1.000000000011955)
   (+
    (+ x (* (* x (+ 0.3333333333333333 (* x -0.25))) (* x x)))
    (* (* x x) -0.5))
   (log (+ 1.0 x))))
double code(double x) {
	return log(1.0 + x);
}
↓
double code(double x) {
	double tmp;
	if ((1.0 + x) <= 1.000000000011955) {
		tmp = (x + ((x * (0.3333333333333333 + (x * -0.25))) * (x * x))) + ((x * x) * -0.5);
	} else {
		tmp = log(1.0 + x);
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.9
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 (+.f64 1 x) < 1.0000000000119551

    1. Initial program 59.2

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

    2. Taylor expanded around 0 0.2

      \[\leadsto \color{blue}{\left(x + 0.3333333333333333 \cdot {x}^{3}\right) - \left(0.5 \cdot {x}^{2} + 0.25 \cdot {x}^{4}\right)}\]

    3. Simplified0.2

      \[\leadsto \color{blue}{x + \left(x \cdot x\right) \cdot \left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right) + -0.5\right)}\]
    4. Using strategy rm

    5. Applied distribute-rgt-in_binary64_10510.2

      \[\leadsto x + \color{blue}{\left(\left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right)\right) \cdot \left(x \cdot x\right) + -0.5 \cdot \left(x \cdot x\right)\right)}\]

    6. Applied associate-+r+_binary64_10330.2

      \[\leadsto \color{blue}{\left(x + \left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right)\right) \cdot \left(x \cdot x\right)\right) + -0.5 \cdot \left(x \cdot x\right)}\]

    if 1.0000000000119551 < (+.f64 1 x)

    1. Initial program 0.6

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \leq 1.000000000011955:\\ \;\;\;\;\left(x + \left(x \cdot \left(0.3333333333333333 + x \cdot -0.25\right)\right) \cdot \left(x \cdot x\right)\right) + \left(x \cdot x\right) \cdot -0.5\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2021045 
(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)))