2nthrt (problem 3.4.6)

Average Error: 33.0 → 1.3

Time: 27.0s

Precision: binary64

Cost: 52292

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \sqrt[3]{t_0}\\ t_2 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\\ \mathbf{if}\;\frac{1}{n} \leq -2:\\ \;\;\;\;\mathsf{fma}\left({t_1}^{2}, -t_1, t_2\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{{t_2}^{2} - {x}^{\left(\frac{1}{n} \cdot 2\right)}}{t_0 + t_2}\\ \end{array} \]

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 (pow x (/ 1.0 n))) (t_1 (cbrt t_0)) (t_2 (exp (/ (log1p x) n))))
   (if (<= (/ 1.0 n) -2.0)
     (fma (pow t_1 2.0) (- t_1) t_2)
     (if (<= (/ 1.0 n) 2e-8)
       (/ (log1p (/ 1.0 x)) n)
       (/ (- (pow t_2 2.0) (pow x (* (/ 1.0 n) 2.0))) (+ t_0 t_2))))))

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 = pow(x, (1.0 / n));
	double t_1 = cbrt(t_0);
	double t_2 = exp((log1p(x) / n));
	double tmp;
	if ((1.0 / n) <= -2.0) {
		tmp = fma(pow(t_1, 2.0), -t_1, t_2);
	} else if ((1.0 / n) <= 2e-8) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = (pow(t_2, 2.0) - pow(x, ((1.0 / n) * 2.0))) / (t_0 + t_2);
	}
	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 = x ^ Float64(1.0 / n)
	t_1 = cbrt(t_0)
	t_2 = exp(Float64(log1p(x) / n))
	tmp = 0.0
	if (Float64(1.0 / n) <= -2.0)
		tmp = fma((t_1 ^ 2.0), Float64(-t_1), t_2);
	elseif (Float64(1.0 / n) <= 2e-8)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(Float64((t_2 ^ 2.0) - (x ^ Float64(Float64(1.0 / n) * 2.0))) / Float64(t_0 + t_2));
	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[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[Power[t$95$0, 1/3], $MachinePrecision]}, Block[{t$95$2 = N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2.0], N[(N[Power[t$95$1, 2.0], $MachinePrecision] * (-t$95$1) + t$95$2), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-8], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[(N[(N[Power[t$95$2, 2.0], $MachinePrecision] - N[Power[x, N[(N[(1.0 / n), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(t$95$0 + t$95$2), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -2

   1. Initial program 0.1

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

   2. Applied egg-rr 0.1

      \[\leadsto \color{blue}{\mathsf{fma}\left({\left(\sqrt[3]{{x}^{\left(\frac{1}{n}\right)}}\right)}^{2}, -\sqrt[3]{{x}^{\left(\frac{1}{n}\right)}}, e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)} \]

   if -2 < (/.f64 1 n) < 2e-8

   1. Initial program 45.0

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

   2. Taylor expanded in n around inf 14.8

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

   3. Simplified 14.8

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
      Proof
      (/.f64 (-.f64 (log1p.f64 x) (log.f64 x)) n): 0 points increase in error, 0 points decrease in error
      (/.f64 (-.f64 (Rewrite<= log1p-def_binary64 (log.f64 (+.f64 1 x))) (log.f64 x)) n): 1 points increase in error, 0 points decrease in error
      (/.f64 (-.f64 (Rewrite<= +-rgt-identity_binary64 (+.f64 (log.f64 (+.f64 1 x)) 0)) (log.f64 x)) n): 0 points increase in error, 0 points decrease in error
      (/.f64 (-.f64 (+.f64 (log.f64 (+.f64 1 x)) (Rewrite<= metadata-eval (log.f64 1))) (log.f64 x)) n): 0 points increase in error, 0 points decrease in error
      (/.f64 (-.f64 (+.f64 (log.f64 (+.f64 1 x)) (Rewrite=> metadata-eval 0)) (log.f64 x)) n): 0 points increase in error, 0 points decrease in error
      (/.f64 (-.f64 (Rewrite=> +-rgt-identity_binary64 (log.f64 (+.f64 1 x))) (log.f64 x)) n): 0 points increase in error, 0 points decrease in error

   4. Applied egg-rr 14.7

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} \]

   5. Taylor expanded in x around 0 14.7

      \[\leadsto \frac{\log \color{blue}{\left(1 + \frac{1}{x}\right)}}{n} \]

   6. Applied egg-rr 1.4

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 2e-8 < (/.f64 1 n)

   1. Initial program 6.1

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

   2. Applied egg-rr 2.9

      \[\leadsto \color{blue}{\frac{{\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)}^{2} - {x}^{\left(2 \cdot \frac{1}{n}\right)}}{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} + {x}^{\left(\frac{1}{n}\right)}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 1.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2:\\ \;\;\;\;\mathsf{fma}\left({\left(\sqrt[3]{{x}^{\left(\frac{1}{n}\right)}}\right)}^{2}, -\sqrt[3]{{x}^{\left(\frac{1}{n}\right)}}, e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)}^{2} - {x}^{\left(\frac{1}{n} \cdot 2\right)}}{{x}^{\left(\frac{1}{n}\right)} + e^{\frac{\mathsf{log1p}\left(x\right)}{n}}}\\ \end{array} \]

Alternatives

Alternative 1
Error	1.3
Cost	46600

\[\begin{array}{l} t_0 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\\ t_1 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2:\\ \;\;\;\;e^{\frac{x}{n}} - t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{{t_0}^{2} - {x}^{\left(\frac{1}{n} \cdot 2\right)}}{t_1 + t_0}\\ \end{array} \]

Alternative 2
Error	1.3
Cost	20232

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \]

Alternative 3
Error	1.2
Cost	13576

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -8786.59698654033:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 505814.07699599635:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 4
Error	1.4
Cost	13508

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -0.0001:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-8}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 5
Error	1.5
Cost	7436

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -0.44:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 0:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 505814.07699599635:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 6
Error	1.6
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -0.44:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 0:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 505814.07699599635:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 7
Error	6.8
Cost	6980

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2.5:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 8
Error	16.1
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.034:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.35 \cdot 10^{+115}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{1}{x \cdot x} \cdot \left(\frac{0.3333333333333333}{x} + -0.5\right)}{n}\\ \mathbf{elif}\;x \leq 6.8 \cdot 10^{+163}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 2.55 \cdot 10^{+207}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 9
Error	33.4
Cost	840

\[\begin{array}{l} t_0 := \frac{1}{n \cdot \left(x + 0.5\right)}\\ \mathbf{if}\;n \leq -8786.59698654033:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -2.9 \cdot 10^{-281}:\\ \;\;\;\;\left(1 + \frac{1}{n \cdot x}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 10
Error	27.1
Cost	708

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2.5:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot \left(x + 0.5\right)}\\ \end{array} \]

Alternative 11
Error	38.3
Cost	448

\[\frac{1}{n \cdot \left(x + 0.5\right)} \]

Alternative 12
Error	40.5
Cost	320

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

Alternative 13
Error	61.0
Cost	192

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

Reproduce

herbie shell --seed 2022289 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))