Average Error: 0.5 → 0.5
Time: 13.8s
Precision: 64
\[\log \left(1 + e^{x}\right) - x \cdot y\]
↓
\[\log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \left(\log \left(e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1\right) + x \cdot y\right)\]
\log \left(1 + e^{x}\right) - x \cdot y
↓
\log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \left(\log \left(e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1\right) + x \cdot y\right)
double f(double x, double y) {
        double r127551 = 1.0;
        double r127552 = x;
        double r127553 = exp(r127552);
        double r127554 = r127551 + r127553;
        double r127555 = log(r127554);
        double r127556 = y;
        double r127557 = r127552 * r127556;
        double r127558 = r127555 - r127557;
        return r127558;
}
↓
double f(double x, double y) {
        double r127559 = 1.0;
        double r127560 = 3.0;
        double r127561 = pow(r127559, r127560);
        double r127562 = x;
        double r127563 = exp(r127562);
        double r127564 = pow(r127563, r127560);
        double r127565 = r127561 + r127564;
        double r127566 = log(r127565);
        double r127567 = r127563 - r127559;
        double r127568 = r127563 * r127567;
        double r127569 = r127559 * r127559;
        double r127570 = r127568 + r127569;
        double r127571 = log(r127570);
        double r127572 = y;
        double r127573 = r127562 * r127572;
        double r127574 = r127571 + r127573;
        double r127575 = r127566 - r127574;
        return r127575;
}

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
Target0.1
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. Using strategy rm

  3. Applied flip3-+0.5

    \[\leadsto \log \color{blue}{\left(\frac{{1}^{3} + {\left(e^{x}\right)}^{3}}{1 \cdot 1 + \left(e^{x} \cdot e^{x} - 1 \cdot e^{x}\right)}\right)} - x \cdot y\]

  4. Applied log-div0.5

    \[\leadsto \color{blue}{\left(\log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \log \left(1 \cdot 1 + \left(e^{x} \cdot e^{x} - 1 \cdot e^{x}\right)\right)\right)} - x \cdot y\]

  5. Applied associate--l-0.5

    \[\leadsto \color{blue}{\log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \left(\log \left(1 \cdot 1 + \left(e^{x} \cdot e^{x} - 1 \cdot e^{x}\right)\right) + x \cdot y\right)}\]

  6. Simplified0.5

    \[\leadsto \log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \color{blue}{\left(\log \left(e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1\right) + x \cdot y\right)}\]

  7. Final simplification0.5

    \[\leadsto \log \left({1}^{3} + {\left(e^{x}\right)}^{3}\right) - \left(\log \left(e^{x} \cdot \left(e^{x} - 1\right) + 1 \cdot 1\right) + x \cdot y\right)\]

Reproduce

herbie shell --seed 2019303 
(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)))