2nthrt (problem 3.4.6)

Average Error: 51.31% → 1.82%

Time: 19.5s

Precision: binary64

Cost: 20232

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{t_0}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \]

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))))
   (if (<= (/ 1.0 n) -4e-9)
     (* (/ 1.0 n) (/ t_0 x))
     (if (<= (/ 1.0 n) 4e-12)
       (/ (log1p (/ 1.0 x)) n)
       (- (exp (/ (log1p x) n)) t_0)))))

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 tmp;
	if ((1.0 / n) <= -4e-9) {
		tmp = (1.0 / n) * (t_0 / x);
	} else if ((1.0 / n) <= 4e-12) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	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.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -4e-9) {
		tmp = (1.0 / n) * (t_0 / x);
	} else if ((1.0 / n) <= 4e-12) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
	}
	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.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -4e-9:
		tmp = (1.0 / n) * (t_0 / x)
	elif (1.0 / n) <= 4e-12:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = math.exp((math.log1p(x) / n)) - t_0
	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)
	tmp = 0.0
	if (Float64(1.0 / n) <= -4e-9)
		tmp = Float64(Float64(1.0 / n) * Float64(t_0 / x));
	elseif (Float64(1.0 / n) <= 4e-12)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	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]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -4e-9], N[(N[(1.0 / n), $MachinePrecision] * N[(t$95$0 / x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 4e-12], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -4.00000000000000025e-9

   1. Initial program 2.66

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

   2. Taylor expanded in x around inf 2.76

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

   3. Simplified 2.76

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
      Proof

      [Start] 2.76	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      mul-1-neg [=>] 2.76	\[ \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]
      log-rec [=>] 2.76	\[ \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]
      mul-1-neg [<=] 2.76	\[ \frac{e^{-\frac{\color{blue}{-1 \cdot \log x}}{n}}}{n \cdot x} \]
      distribute-neg-frac [=>] 2.76	\[ \frac{e^{\color{blue}{\frac{--1 \cdot \log x}{n}}}}{n \cdot x} \]
      mul-1-neg [=>] 2.76	\[ \frac{e^{\frac{-\color{blue}{\left(-\log x\right)}}{n}}}{n \cdot x} \]
      remove-double-neg [=>] 2.76	\[ \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]
      *-commutative [=>] 2.76	\[ \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Applied egg-rr 2.76

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

   if -4.00000000000000025e-9 < (/.f64 1 n) < 3.99999999999999992e-12

   1. Initial program 70.31

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

   2. Taylor expanded in n around inf 22.87

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

   3. Applied egg-rr 22.72

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

   4. Applied egg-rr 22.71

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

   5. Simplified 0.83

      \[\leadsto \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x} + 0\right)}}{n} \]
      Proof

      [Start] 22.71	\[ \frac{\mathsf{log1p}\left(\frac{1 + x}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 22.71	\[ \frac{\mathsf{log1p}\left(\color{blue}{1 \cdot \frac{1 + x}{x}} - 1\right)}{n} \]
      associate-*r/ [=>] 22.71	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1 \cdot \left(1 + x\right)}{x}} - 1\right)}{n} \]
      associate-*l/ [<=] 26.19	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(1 + x\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 26.19	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 26.19	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 22.72	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      +-commutative [<=] 22.72	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 + \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 22.72	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(\frac{1}{x} + 1\right)} - 1\right)}{n} \]
      associate--l+ [=>] 0.83	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 0.83	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   6. Taylor expanded in n around 0 22.72

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

   7. Simplified 0.83

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

      [Start] 22.72	\[ \frac{\log \left(1 + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 0.83	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 3.99999999999999992e-12 < (/.f64 1 n)

   1. Initial program 11.64

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

   2. Taylor expanded in n around 0 11.67

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

   3. Simplified 6.72

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

      [Start] 11.67	\[ e^{\frac{\log \left(1 + x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      log1p-def [=>] 6.72	\[ e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]

3. Recombined 3 regimes into one program.

4. Final simplification 1.82

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{{x}^{\left(\frac{1}{n}\right)}}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternatives

Alternative 1
Error	2.19%
Cost	7560

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{t_0}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \end{array} \]

Alternative 2
Error	2.94%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 3
Error	2.39%
Cost	7304

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]

Alternative 4
Error	2.39%
Cost	7304

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{t_0}{x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-12}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]

Alternative 5
Error	11.24%
Cost	6980

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5000000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 6
Error	24.39%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.68:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.15 \cdot 10^{+160}:\\ \;\;\;\;\frac{\frac{1}{n}}{x} - \frac{0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 7
Error	54.51%
Cost	841

\[\begin{array}{l} \mathbf{if}\;n \leq -6.8 \lor \neg \left(n \leq -1 \cdot 10^{-282}\right):\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{1}{n \cdot x}\right) + -1\\ \end{array} \]

Alternative 8
Error	45.14%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -18 \lor \neg \left(n \leq 2.75 \cdot 10^{-36}\right):\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 9
Error	63.27%
Cost	320

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

Alternative 10
Error	62.49%
Cost	320

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

Reproduce

herbie shell --seed 2023125 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))