Average Error: 39.3 → 0.2
Time: 12.8s
Precision: 64
Internal Precision: 1344
\[\log \left(1 + x\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le 8.747595575536827 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

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 x < 8.747595575536827e-05

    1. Initial program 58.8

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

    2. Initial simplification58.8

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

    3. Taylor expanded around 0 0.3

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

    4. Simplified0.3

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

    if 8.747595575536827e-05 < x

    1. Initial program 0.1

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

    2. Initial simplification0.1

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le 8.747595575536827 \cdot 10^{-05}:\\ \;\;\;\;x + \left(x \cdot x\right) \cdot \left(\frac{1}{3} \cdot x - \frac{1}{2}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + 1\right)\\ \end{array}\]

Runtime

Time bar (total: 12.8s)Debug logProfile

herbie shell --seed 2018252 
(FPCore (x)
  :name "ln(1 + x)"

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

  (log (+ 1 x)))