Average Error: 38.8 → 0.1
Time: 5.5s
Precision: binary64
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.0000335079842024:\\ \;\;\;\;x + \left(\left(x \cdot x\right) \cdot \left(x \cdot 0.3333333333333333 - 0.5\right) - 0.25 \cdot {x}^{4}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;1 + x \leq 1.0000335079842024:\\
\;\;\;\;x + \left(\left(x \cdot x\right) \cdot \left(x \cdot 0.3333333333333333 - 0.5\right) - 0.25 \cdot {x}^{4}\right)\\

\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) 1.0000335079842024)
   (+ x (- (* (* x x) (- (* x 0.3333333333333333) 0.5)) (* 0.25 (pow x 4.0))))
   (log (+ 1.0 x))))
double code(double x) {
	return log(1.0 + x);
}
↓
double code(double x) {
	double tmp;
	if ((1.0 + x) <= 1.0000335079842024) {
		tmp = x + (((x * x) * ((x * 0.3333333333333333) - 0.5)) - (0.25 * pow(x, 4.0)));
	} 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

Original38.8
Target0.2
Herbie0.1
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (+.f64 1 x) < 1.0000335079842024

    1. Initial program 59.0

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

    2. Taylor expanded around 0 0.1

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

    3. Simplified0.1

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

    if 1.0000335079842024 < (+.f64 1 x)

    1. Initial program 0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \leq 1.0000335079842024:\\ \;\;\;\;x + \left(\left(x \cdot x\right) \cdot \left(x \cdot 0.3333333333333333 - 0.5\right) - 0.25 \cdot {x}^{4}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2021032 
(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)))