ln(1 + x)

Average Error: 39.9 → 0.3

Time: 10.7s

Precision: 64

• Falling back on racket egraph because egg-herbie package not installed

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \le 1.000000464934527766303062890074215829372:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

C

double f(double x) {
        double r84888 = 1.0;
        double r84889 = x;
        double r84890 = r84888 + r84889;
        double r84891 = log(r84890);
        return r84891;
}

↓

double f(double x) {
        double r84892 = 1.0;
        double r84893 = x;
        double r84894 = r84892 + r84893;
        double r84895 = 1.0000004649345278;
        bool r84896 = r84894 <= r84895;
        double r84897 = r84893 * r84892;
        double r84898 = -0.5;
        double r84899 = r84892 * r84892;
        double r84900 = r84899 / r84893;
        double r84901 = r84898 / r84900;
        double r84902 = r84893 * r84901;
        double r84903 = r84897 + r84902;
        double r84904 = log(r84892);
        double r84905 = r84903 + r84904;
        double r84906 = log(r84894);
        double r84907 = r84896 ? r84905 : r84906;
        return r84907;
}

Error

Bits error versus x

Target

Original	39.9
Target	0.2
Herbie	0.3

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

Derivation

1. Split input into 2 regimes

2. if (+ 1.0 x) < 1.0000004649345278

   1. Initial program 59.2

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

   2. Taylor expanded around 0 0.3

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

   3. Simplified 0.3

      \[\leadsto \color{blue}{x \cdot \left(1 + \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1}\]
   4. Using strategy rm

   5. Applied distribute-lft-in 0.3

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

   if 1.0000004649345278 < (+ 1.0 x)

   1. Initial program 0.2

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

4. Final simplification 0.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \le 1.000000464934527766303062890074215829372:\\ \;\;\;\;\left(x \cdot 1 + x \cdot \frac{\frac{-1}{2}}{\frac{1 \cdot 1}{x}}\right) + \log 1\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019350 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

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

  (log (+ 1 x)))