2nthrt (problem 3.4.6)

Average Error: 32.7 → 2.0

Time: 1.7min

Precision: binary64

Cost: 13380

Math

\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

↓

\[\begin{array}{l} t_0 := \frac{\log x}{n}\\ \mathbf{if}\;x \leq 0.55:\\ \;\;\;\;-\mathsf{expm1}\left(t_0\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{t_0}}{x \cdot n}\\ \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))

↓

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (log x) n)))
   (if (<= x 0.55) (- (expm1 t_0)) (/ (exp t_0) (* x n)))))

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

↓

double code(double x, double n) {
	double t_0 = log(x) / n;
	double tmp;
	if (x <= 0.55) {
		tmp = -expm1(t_0);
	} else {
		tmp = exp(t_0) / (x * n);
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

↓

public static double code(double x, double n) {
	double t_0 = Math.log(x) / n;
	double tmp;
	if (x <= 0.55) {
		tmp = -Math.expm1(t_0);
	} else {
		tmp = Math.exp(t_0) / (x * n);
	}
	return tmp;
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

↓

def code(x, n):
	t_0 = math.log(x) / n
	tmp = 0
	if x <= 0.55:
		tmp = -math.expm1(t_0)
	else:
		tmp = math.exp(t_0) / (x * n)
	return tmp

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

↓

function code(x, n)
	t_0 = Float64(log(x) / n)
	tmp = 0.0
	if (x <= 0.55)
		tmp = Float64(-expm1(t_0));
	else
		tmp = Float64(exp(t_0) / Float64(x * n));
	end
	return tmp
end

Wolfram

code[x_, n_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

↓

code[x_, n_] := Block[{t$95$0 = N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]}, If[LessEqual[x, 0.55], (-N[(Exp[t$95$0] - 1), $MachinePrecision]), N[(N[Exp[t$95$0], $MachinePrecision] / N[(x * n), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if x < 0.55000000000000004

   1. Initial program 46.5

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   2. Taylor expanded in x around 0 46.6

      \[\leadsto \color{blue}{1 - e^{\frac{\log x}{n}}} \]

   3. Simplified 2.0

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

   if 0.55000000000000004 < x

   1. Initial program 21.2

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]

   2. Taylor expanded in x around inf 1.9

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]

   3. Simplified 1.9

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
      Proof
3. Recombined 2 regimes into one program.

Alternatives

Alternative 1
Error	11.0
Cost	13384

\[\begin{array}{l} \mathbf{if}\;x \leq 190:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{elif}\;x \leq 2.4 \cdot 10^{+66}:\\ \;\;\;\;\frac{\frac{1}{x} + -0.5 \cdot {\left(\frac{-1}{x}\right)}^{2}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n}\\ \end{array} \]

Alternative 2
Error	11.3
Cost	13188

\[\begin{array}{l} \mathbf{if}\;x \leq 190:\\ \;\;\;\;-\mathsf{expm1}\left(\frac{\log x}{n}\right)\\ \mathbf{elif}\;x \leq 5 \cdot 10^{+201}:\\ \;\;\;\;\frac{\frac{1}{x} + -0.5 \cdot {\left(\frac{-1}{x}\right)}^{2}}{n}\\ \mathbf{else}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \end{array} \]

Alternative 3
Error	17.1
Cost	7432

\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x}{n} + -1 \cdot \frac{\log x}{n}\\ \mathbf{elif}\;x \leq 7.7 \cdot 10^{+201}:\\ \;\;\;\;\frac{\frac{1}{x} + -0.5 \cdot {\left(\frac{-1}{x}\right)}^{2}}{n}\\ \mathbf{else}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \end{array} \]

Alternative 4
Error	17.3
Cost	7108

\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x}{n} + -1 \cdot \frac{\log x}{n}\\ \mathbf{elif}\;x \leq 2.5 \cdot 10^{+202}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \end{array} \]

Alternative 5
Error	17.3
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 8.9 \cdot 10^{+201}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \end{array} \]

Alternative 6
Error	17.5
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 1.5:\\ \;\;\;\;\frac{\log x}{-n}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{+200}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \end{array} \]

Alternative 7
Error	35.0
Cost	840

\[\begin{array}{l} \mathbf{if}\;n \leq -860000000:\\ \;\;\;\;\frac{-1}{n} \cdot \frac{-1}{x}\\ \mathbf{elif}\;n \leq -1.16 \cdot 10^{-259}:\\ \;\;\;\;6 - \left(6 + \frac{-1}{x \cdot n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 8
Error	40.5
Cost	320

\[\frac{1}{n \cdot x} \]

Alternative 9
Error	40.0
Cost	320

\[\frac{\frac{1}{x}}{n} \]

Alternative 10
Error	61.0
Cost	192

\[\frac{x}{n} \]

Reproduce

herbie shell --seed 2023033 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))