Kahan's exp quotient

Average Accuracy: 52.0% → 100.0%

Time: 2.5s

Precision: binary64

Cost: 19392

• 23.8% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\frac{e^{x} - 1}{x} \]

↓

\[\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\mathsf{expm1}\left(x\right)}{x}\right)\right) \]

FPCore

(FPCore (x) :precision binary64 (/ (- (exp x) 1.0) x))

↓

(FPCore (x) :precision binary64 (expm1 (log1p (/ (expm1 x) x))))

C

double code(double x) {
	return (exp(x) - 1.0) / x;
}

↓

double code(double x) {
	return expm1(log1p((expm1(x) / x)));
}

Java

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

↓

public static double code(double x) {
	return Math.expm1(Math.log1p((Math.expm1(x) / x)));
}

Python

def code(x):
	return (math.exp(x) - 1.0) / x

↓

def code(x):
	return math.expm1(math.log1p((math.expm1(x) / x)))

Julia

function code(x)
	return Float64(Float64(exp(x) - 1.0) / x)
end

↓

function code(x)
	return expm1(log1p(Float64(expm1(x) / x)))
end

Wolfram

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

↓

code[x_] := N[(Exp[N[Log[1 + N[(N[(Exp[x] - 1), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]

Target

Original	52.0%
Target	51.5%
Herbie	100.0%

\[\begin{array}{l} \mathbf{if}\;x < 1 \land x > -1:\\ \;\;\;\;\frac{e^{x} - 1}{\log \left(e^{x}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{x} - 1}{x}\\ \end{array} \]

Derivation

1. Initial program 50.0%

   \[\frac{e^{x} - 1}{x} \]

2. Simplified 100.0%

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

   [Start] 50.0	\[ \frac{e^{x} - 1}{x} \]
   expm1-def [=>] 100.0	\[ \frac{\color{blue}{\mathsf{expm1}\left(x\right)}}{x} \]

3. Applied egg-rr 100.0%

   \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\mathsf{expm1}\left(x\right)}{x}\right)\right)} \]
   Step-by-step derivation

   [Start] 100.0	\[ \frac{\mathsf{expm1}\left(x\right)}{x} \]
   expm1-log1p-u [=>] 100.0	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\mathsf{expm1}\left(x\right)}{x}\right)\right)} \]

4. Final simplification 100.0%

   \[\leadsto \mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\mathsf{expm1}\left(x\right)}{x}\right)\right) \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	6592

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

Alternative 2
Accuracy	52.5%
Cost	320

\[x \cdot 0.5 + 1 \]

Alternative 3
Accuracy	52.3%
Cost	64

\[1 \]

Reproduce

herbie shell --seed 2023164 
(FPCore (x)
  :name "Kahan's exp quotient"
  :precision binary64

  :herbie-target
  (if (and (< x 1.0) (> x -1.0)) (/ (- (exp x) 1.0) (log (exp x))) (/ (- (exp x) 1.0) x))

  (/ (- (exp x) 1.0) x))