Average Error: 0.6 → 0.6
Time: 11.0s
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 r170498 = 1.0;
        double r170499 = x;
        double r170500 = exp(r170499);
        double r170501 = r170498 + r170500;
        double r170502 = log(r170501);
        double r170503 = y;
        double r170504 = r170499 * r170503;
        double r170505 = r170502 - r170504;
        return r170505;
}

double f(double x, double y) {
        double r170506 = 1.0;
        double r170507 = x;
        double r170508 = exp(r170507);
        double r170509 = r170506 + r170508;
        double r170510 = log(r170509);
        double r170511 = y;
        double r170512 = r170507 * r170511;
        double r170513 = r170510 - r170512;
        return r170513;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.6
Target0.1
Herbie0.6
\[\begin{array}{l} \mathbf{if}\;x \le 0.0:\\ \;\;\;\;\log \left(1 + e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + e^{-x}\right) - \left(-x\right) \cdot \left(1 - y\right)\\ \end{array}\]

Derivation

  1. Initial program 0.6

    \[\log \left(1 + e^{x}\right) - x \cdot y\]
  2. Final simplification0.6

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

Reproduce

herbie shell --seed 2020046 +o rules:numerics
(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)))