Average Error: 0.5 → 0.5
Time: 4.6s
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 r194876 = 1.0;
        double r194877 = x;
        double r194878 = exp(r194877);
        double r194879 = r194876 + r194878;
        double r194880 = log(r194879);
        double r194881 = y;
        double r194882 = r194877 * r194881;
        double r194883 = r194880 - r194882;
        return r194883;
}
↓
double f(double x, double y) {
        double r194884 = 1.0;
        double r194885 = x;
        double r194886 = exp(r194885);
        double r194887 = r194884 + r194886;
        double r194888 = log(r194887);
        double r194889 = y;
        double r194890 = r194885 * r194889;
        double r194891 = r194888 - r194890;
        return r194891;
}

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 2019362 
(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)))