Average Error: 0.5 → 0.5
Time: 2.2s
Precision: 64
\[\log \left(1 + e^{x}\right) - x \cdot y\]
↓
\[\log \left(1 + e^{x}\right) - x \cdot y\]
\log \left(1 + e^{x}\right) - x \cdot y
↓
\log \left(1 + e^{x}\right) - x \cdot y
double f(double x, double y) {
        double r160852 = 1.0;
        double r160853 = x;
        double r160854 = exp(r160853);
        double r160855 = r160852 + r160854;
        double r160856 = log(r160855);
        double r160857 = y;
        double r160858 = r160853 * r160857;
        double r160859 = r160856 - r160858;
        return r160859;
}
↓
double f(double x, double y) {
        double r160860 = 1.0;
        double r160861 = x;
        double r160862 = exp(r160861);
        double r160863 = r160860 + r160862;
        double r160864 = log(r160863);
        double r160865 = y;
        double r160866 = r160861 * r160865;
        double r160867 = r160864 - r160866;
        return r160867;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.5
Target0.0
Herbie0.5
\[\begin{array}{l} \mathbf{if}\;x \le 0.0:\\ \;\;\;\;\log \left(1 + e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + e^{-x}\right) - \left(-x\right) \cdot \left(1 - y\right)\\ \end{array}\]

Derivation

  1. Initial program 0.5

    \[\log \left(1 + e^{x}\right) - x \cdot y\]

  2. Final simplification0.5

    \[\leadsto \log \left(1 + e^{x}\right) - x \cdot y\]

Reproduce

herbie shell --seed 2020100 
(FPCore (x y)
  :name "Logistic regression 2"
  :precision binary64

  :herbie-target
  (if (<= x 0.0) (- (log (+ 1 (exp x))) (* x y)) (- (log (+ 1 (exp (- x)))) (* (- x) (- 1 y))))

  (- (log (+ 1 (exp x))) (* x y)))