Average Error: 39.2 → 0.0
Time: 3.5s
Precision: 64
\[\log \left(1 + x\right)\]
↓
\[\log_* (1 + x)\]
\log \left(1 + x\right)
↓
\log_* (1 + x)
double f(double x) {
        double r1902483 = 1.0;
        double r1902484 = x;
        double r1902485 = r1902483 + r1902484;
        double r1902486 = log(r1902485);
        return r1902486;
}
↓
double f(double x) {
        double r1902487 = x;
        double r1902488 = log1p(r1902487);
        return r1902488;
}

Error

Bits error versus x

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Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

Derivation

  1. Initial program 39.2

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

  2. Simplified0.0

    \[\leadsto \color{blue}{\log_* (1 + x)}\]

  3. Final simplification0.0

    \[\leadsto \log_* (1 + x)\]

Reproduce

herbie shell --seed 2019104 +o rules:numerics
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))