Average Error: 39.1 → 0.2
Time: 2.9s
Precision: binary64
Cost: 6913
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \leq 0.0001449948157992588:\\ \;\;\;\;x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]
\log \left(1 + x\right)
↓
\begin{array}{l}
\mathbf{if}\;x \leq 0.0001449948157992588:\\
\;\;\;\;x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)\\

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

\end{array}
(FPCore (x) :precision binary64 (log (+ 1.0 x)))
↓
(FPCore (x)
 :precision binary64
 (if (<= x 0.0001449948157992588)
   (* x (+ 1.0 (* x (+ -0.5 (* x 0.3333333333333333)))))
   (log (+ x 1.0))))
double code(double x) {
	return log(1.0 + x);
}
↓
double code(double x) {
	double tmp;
	if (x <= 0.0001449948157992588) {
		tmp = x * (1.0 + (x * (-0.5 + (x * 0.3333333333333333))));
	} else {
		tmp = log(x + 1.0);
	}
	return tmp;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Alternatives

Alternative 1
Error0.7
Cost6785
\[\begin{array}{l} \mathbf{if}\;x \leq 1.7105269244035937:\\ \;\;\;\;x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log x\\ \end{array}\]
Alternative 2
Error18.6
Cost1025
\[\begin{array}{l} \mathbf{if}\;x \leq 1.142643275337932:\\ \;\;\;\;x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array}\]
Alternative 3
Error18.7
Cost769
\[\begin{array}{l} \mathbf{if}\;x \leq 1.1071505472713281:\\ \;\;\;\;x - \left(x \cdot x\right) \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array}\]
Alternative 4
Error19.2
Cost385
\[\begin{array}{l} \mathbf{if}\;x \leq 1.0006723630715166:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array}\]
Alternative 5
Error59.4
Cost64
\[1\]

Error

Derivation

  1. Split input into 2 regimes

  2. if x < 1.44994815799258809e-4

    1. Initial program 59.0

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

    2. Taylor expanded around 0 0.2

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

    3. Simplified0.2

      \[\leadsto \color{blue}{x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)}\]

    4. Simplified0.2

      \[\leadsto \color{blue}{x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)}\]

    if 1.44994815799258809e-4 < x

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\log \left(x + 1\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.0001449948157992588:\\ \;\;\;\;x \cdot \left(1 + x \cdot \left(-0.5 + x \cdot 0.3333333333333333\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2021044 
(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)))