Average Error: 0.6 → 0.6
Time: 9.4s
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 r146540 = 1.0;
        double r146541 = x;
        double r146542 = exp(r146541);
        double r146543 = r146540 + r146542;
        double r146544 = log(r146543);
        double r146545 = y;
        double r146546 = r146541 * r146545;
        double r146547 = r146544 - r146546;
        return r146547;
}

double f(double x, double y) {
        double r146548 = 1.0;
        double r146549 = x;
        double r146550 = exp(r146549);
        double r146551 = r146548 + r146550;
        double r146552 = log(r146551);
        double r146553 = y;
        double r146554 = r146549 * r146553;
        double r146555 = r146552 - r146554;
        return r146555;
}

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 
(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)))