Kahan's exp quotient

Average Accuracy: 52.9% → 100.0%

Time: 2.1s

Precision: binary64

Cost: 6592

• 24.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

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

↓

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

FPCore

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

↓

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

C

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

↓

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

Python

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

↓

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

Julia

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

↓

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

Wolfram

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

↓

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

Target

Original	52.9%
Target	52.3%
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 52.7%

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

2. Simplified 100.0%

   \[\leadsto \color{blue}{\frac{\mathsf{expm1}\left(x\right)}{x}} \]
   Proof

   [Start] 52.7	\[ \frac{e^{x} - 1}{x} \]
   expm1-def [=>] 100.0	\[ \frac{\color{blue}{\mathsf{expm1}\left(x\right)}}{x} \]

3. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	51.3%
Cost	64

\[1 \]

Reproduce

herbie shell --seed 2023160 
(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))