Kahan's exp quotient

Average Error: 40.5 → 0.0

Time: 5.2s

Precision: binary64

Cost: 6592

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	40.5
Target	41.1
Herbie	0.0

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

Derivation

1. Initial program 40.5

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

2. Simplified 0.0

   \[\leadsto \color{blue}{\frac{\mathsf{expm1}\left(x\right)}{x}} \]
   Proof
   (/.f64 (expm1.f64 x) x): 0 points increase in error, 0 points decrease in error
   (/.f64 (Rewrite<= expm1-def_binary64 (-.f64 (exp.f64 x) 1)) x): 171 points increase in error, 1 points decrease in error

3. Final simplification 0.0

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

Alternatives

Alternative 1
Error	17.4
Cost	452

\[\begin{array}{l} \mathbf{if}\;x \leq -1.6:\\ \;\;\;\;\frac{-2}{x}\\ \mathbf{else}:\\ \;\;\;\;1 + x \cdot 0.5\\ \end{array} \]

Alternative 2
Error	17.4
Cost	448

\[\frac{1}{1 + x \cdot -0.5} \]

Alternative 3
Error	17.8
Cost	324

\[\begin{array}{l} \mathbf{if}\;x \leq -2:\\ \;\;\;\;\frac{-2}{x}\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 4
Error	20.6
Cost	64

\[1 \]

Reproduce

herbie shell --seed 2022296 
(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))