expq2 (section 3.11)

?

Percentage Accurate: 27.9% → 79.9%
Time: 4.2s
Precision: binary64
Cost: 19524

?

\[\frac{e^{x}}{e^{x} - 1} \]
↓
\[\begin{array}{l} \mathbf{if}\;e^{x} \leq 0.99999:\\ \;\;\;\;\frac{e^{x}}{\mathsf{expm1}\left(x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x} + 0.5\\ \end{array} \]
(FPCore (x) :precision binary64 (/ (exp x) (- (exp x) 1.0)))
↓
(FPCore (x)
 :precision binary64
 (if (<= (exp x) 0.99999) (/ (exp x) (expm1 x)) (+ (/ 1.0 x) 0.5)))
double code(double x) {
	return exp(x) / (exp(x) - 1.0);
}
↓
double code(double x) {
	double tmp;
	if (exp(x) <= 0.99999) {
		tmp = exp(x) / expm1(x);
	} else {
		tmp = (1.0 / x) + 0.5;
	}
	return tmp;
}

public static double code(double x) {
	return Math.exp(x) / (Math.exp(x) - 1.0);
}
↓
public static double code(double x) {
	double tmp;
	if (Math.exp(x) <= 0.99999) {
		tmp = Math.exp(x) / Math.expm1(x);
	} else {
		tmp = (1.0 / x) + 0.5;
	}
	return tmp;
}

def code(x):
	return math.exp(x) / (math.exp(x) - 1.0)
↓
def code(x):
	tmp = 0
	if math.exp(x) <= 0.99999:
		tmp = math.exp(x) / math.expm1(x)
	else:
		tmp = (1.0 / x) + 0.5
	return tmp

function code(x)
	return Float64(exp(x) / Float64(exp(x) - 1.0))
end
↓
function code(x)
	tmp = 0.0
	if (exp(x) <= 0.99999)
		tmp = Float64(exp(x) / expm1(x));
	else
		tmp = Float64(Float64(1.0 / x) + 0.5);
	end
	return tmp
end

code[x_] := N[(N[Exp[x], $MachinePrecision] / N[(N[Exp[x], $MachinePrecision] - 1.0), $MachinePrecision]), $MachinePrecision]
↓
code[x_] := If[LessEqual[N[Exp[x], $MachinePrecision], 0.99999], N[(N[Exp[x], $MachinePrecision] / N[(Exp[x] - 1), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 / x), $MachinePrecision] + 0.5), $MachinePrecision]]

\frac{e^{x}}{e^{x} - 1}
↓
\begin{array}{l}
\mathbf{if}\;e^{x} \leq 0.99999:\\
\;\;\;\;\frac{e^{x}}{\mathsf{expm1}\left(x\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{x} + 0.5\\


\end{array}

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 3 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Bogosity?

Bogosity

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original27.9%
Target52.2%
Herbie79.9%
\[\frac{1}{1 - e^{-x}} \]

Derivation?

  1. Split input into 2 regimes

  2. if (exp.f64 x) < 0.999990000000000046

    1. Initial program 99.6%

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

    2. Simplified100.0%

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

      [Start]99.6%

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

      expm1-def [=>]100.0%

      \[ \frac{e^{x}}{\color{blue}{\mathsf{expm1}\left(x\right)}} \]

    if 0.999990000000000046 < (exp.f64 x)

    1. Initial program 2.7%

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

    2. Simplified67.0%

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

      [Start]2.7%

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

      expm1-def [=>]67.0%

      \[ \frac{e^{x}}{\color{blue}{\mathsf{expm1}\left(x\right)}} \]

    3. Taylor expanded in x around 0 73.2%

      \[\leadsto \color{blue}{0.5 + \frac{1}{x}} \]

    4. Simplified73.2%

      \[\leadsto \color{blue}{\frac{1}{x} + 0.5} \]
      Step-by-step derivation

      [Start]73.2%

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

      +-commutative [=>]73.2%

      \[ \color{blue}{\frac{1}{x} + 0.5} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification79.7%

    \[\leadsto \begin{array}{l} \mathbf{if}\;e^{x} \leq 0.99999:\\ \;\;\;\;\frac{e^{x}}{\mathsf{expm1}\left(x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x} + 0.5\\ \end{array} \]

Alternatives

Alternative 1
Accuracy79.9%
Cost19524
\[\begin{array}{l} \mathbf{if}\;e^{x} \leq 0.99999:\\ \;\;\;\;\frac{e^{x}}{\mathsf{expm1}\left(x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{x} + 0.5\\ \end{array} \]
Alternative 2
Accuracy55.7%
Cost320
\[\frac{1}{x} + 0.5 \]
Alternative 3
Accuracy52.5%
Cost192
\[\frac{1}{x} \]

Reproduce?

herbie shell --seed 2023269 
(FPCore (x)
  :name "expq2 (section 3.11)"
  :precision binary64

  :herbie-target
  (/ 1.0 (- 1.0 (exp (- x))))

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