expax (section 3.5)

Percentage Accurate: 65.9% → 100.0%

Time: 4.7s

Precision: binary64

Cost: 6592

• Unsound rule application detected in e-graph. Results may not be sound.

• 24.9% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[e^{a \cdot x} - 1 \]

↓

\[\mathsf{expm1}\left(a \cdot x\right) \]

FPCore

(FPCore (a x) :precision binary64 (- (exp (* a x)) 1.0))

↓

(FPCore (a x) :precision binary64 (expm1 (* a x)))

C

double code(double a, double x) {
	return exp((a * x)) - 1.0;
}

↓

double code(double a, double x) {
	return expm1((a * x));
}

Java

public static double code(double a, double x) {
	return Math.exp((a * x)) - 1.0;
}

↓

public static double code(double a, double x) {
	return Math.expm1((a * x));
}

Python

def code(a, x):
	return math.exp((a * x)) - 1.0

↓

def code(a, x):
	return math.expm1((a * x))

Julia

function code(a, x)
	return Float64(exp(Float64(a * x)) - 1.0)
end

↓

function code(a, x)
	return expm1(Float64(a * x))
end

Wolfram

code[a_, x_] := N[(N[Exp[N[(a * x), $MachinePrecision]], $MachinePrecision] - 1.0), $MachinePrecision]

↓

code[a_, x_] := N[(Exp[N[(a * x), $MachinePrecision]] - 1), $MachinePrecision]

Target

Original	65.9%
Target	99.8%
Herbie	100.0%

\[\begin{array}{l} \mathbf{if}\;\left|a \cdot x\right| < 0.1:\\ \;\;\;\;\left(a \cdot x\right) \cdot \left(1 + \left(\frac{a \cdot x}{2} + \frac{{\left(a \cdot x\right)}^{2}}{6}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;e^{a \cdot x} - 1\\ \end{array} \]

Derivation

1. Initial program 62.3%

   \[e^{a \cdot x} - 1 \]

2. Simplified 100.0%

   \[\leadsto \color{blue}{\mathsf{expm1}\left(a \cdot x\right)} \]
   Step-by-step derivation

   [Start] 62.3%	\[ e^{a \cdot x} - 1 \]
   expm1-def [=>] 100.0%	\[ \color{blue}{\mathsf{expm1}\left(a \cdot x\right)} \]

3. Final simplification 100.0%

   \[\leadsto \mathsf{expm1}\left(a \cdot x\right) \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	6592

\[\mathsf{expm1}\left(a \cdot x\right) \]

Alternative 2
Accuracy	67.8%
Cost	964

\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq 10^{-13}:\\ \;\;\;\;a \cdot x\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(x + a \cdot \left(x \cdot \left(x \cdot 0.5\right)\right)\right)\\ \end{array} \]

Alternative 3
Accuracy	67.7%
Cost	836

\[\begin{array}{l} \mathbf{if}\;a \cdot x \leq 2 \cdot 10^{+51}:\\ \;\;\;\;a \cdot x\\ \mathbf{else}:\\ \;\;\;\;0.5 \cdot \left(\left(x \cdot x\right) \cdot \left(a \cdot a\right)\right)\\ \end{array} \]

Alternative 4
Accuracy	57.3%
Cost	192

\[a \cdot x \]

Reproduce

herbie shell --seed 2023271 
(FPCore (a x)
  :name "expax (section 3.5)"
  :precision binary64

  :herbie-target
  (if (< (fabs (* a x)) 0.1) (* (* a x) (+ 1.0 (+ (/ (* a x) 2.0) (/ (pow (* a x) 2.0) 6.0)))) (- (exp (* a x)) 1.0))

  (- (exp (* a x)) 1.0))