Average Error: 0.5 → 0.5
Time: 14.5s
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 r138098 = 1.0;
        double r138099 = x;
        double r138100 = exp(r138099);
        double r138101 = r138098 + r138100;
        double r138102 = log(r138101);
        double r138103 = y;
        double r138104 = r138099 * r138103;
        double r138105 = r138102 - r138104;
        return r138105;
}
↓
double f(double x, double y) {
        double r138106 = 1.0;
        double r138107 = x;
        double r138108 = exp(r138107);
        double r138109 = r138106 + r138108;
        double r138110 = log(r138109);
        double r138111 = y;
        double r138112 = r138107 * r138111;
        double r138113 = r138110 - r138112;
        return r138113;
}

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 2019306 
(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)))