Average Error: 0.5 → 0.5
Time: 4.5s
Precision: 64
\[\log \left(1 + e^{x}\right) - x \cdot y\]
\[\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(1 + e^{x}\right)\right)\right) - x \cdot y\]
\log \left(1 + e^{x}\right) - x \cdot y
\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(1 + e^{x}\right)\right)\right) - x \cdot y
double code(double x, double y) {
	return (log((1.0 + exp(x))) - (x * y));
}
double code(double x, double y) {
	return (expm1(log1p(log((1.0 + exp(x))))) - (x * y));
}

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 expm1-log1p-u0.5

    \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(1 + e^{x}\right)\right)\right)} - x \cdot y\]
  4. Final simplification0.5

    \[\leadsto \mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(1 + e^{x}\right)\right)\right) - x \cdot y\]

Reproduce

herbie shell --seed 2020078 +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)))