2nthrt (problem 3.4.6)

Average Error: 32.7 → 1.5

Time: 21.9s

Precision: binary64

Cost: 33028

Math

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

↓

\[\begin{array}{l} t_0 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{\frac{1}{\log \left(e^{t_0}\right)}}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

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 (- (exp (/ (log1p x) n)) (pow x (/ 1.0 n)))))
   (if (<= (/ 1.0 n) -1e-7)
     (/ 1.0 (/ 1.0 (log (exp t_0))))
     (if (<= (/ 1.0 n) 4e-16) (/ (log1p (/ 1.0 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 = exp((log1p(x) / n)) - pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e-7) {
		tmp = 1.0 / (1.0 / log(exp(t_0)));
	} else if ((1.0 / n) <= 4e-16) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = 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.exp((Math.log1p(x) / n)) - Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -1e-7) {
		tmp = 1.0 / (1.0 / Math.log(Math.exp(t_0)));
	} else if ((1.0 / n) <= 4e-16) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = 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.exp((math.log1p(x) / n)) - math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -1e-7:
		tmp = 1.0 / (1.0 / math.log(math.exp(t_0)))
	elif (1.0 / n) <= 4e-16:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = 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 = Float64(exp(Float64(log1p(x) / n)) - (x ^ Float64(1.0 / n)))
	tmp = 0.0
	if (Float64(1.0 / n) <= -1e-7)
		tmp = Float64(1.0 / Float64(1.0 / log(exp(t_0))));
	elseif (Float64(1.0 / n) <= 4e-16)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = 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[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -1e-7], N[(1.0 / N[(1.0 / N[Log[N[Exp[t$95$0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 4e-16], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], t$95$0]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -9.9999999999999995e-8

   1. Initial program 1.5

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

   2. Applied egg-rr 1.6

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

   3. Applied egg-rr 1.9

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

   if -9.9999999999999995e-8 < (/.f64 1 n) < 3.9999999999999999e-16

   1. Initial program 45.1

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

   2. Taylor expanded in n around inf 14.5

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

   3. Applied egg-rr 14.4

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

   4. Applied egg-rr 14.4

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

   5. Simplified 0.5

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

      [Start] 14.4	\[ \frac{\mathsf{log1p}\left(\frac{1 + x}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 14.4	\[ \frac{\mathsf{log1p}\left(\color{blue}{1 \cdot \frac{1 + x}{x}} - 1\right)}{n} \]
      associate-*r/ [=>] 14.4	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1 \cdot \left(1 + x\right)}{x}} - 1\right)}{n} \]
      associate-*l/ [<=] 16.6	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(1 + x\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 16.6	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 16.6	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 14.4	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      +-commutative [<=] 14.4	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 + \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 14.4	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(\frac{1}{x} + 1\right)} - 1\right)}{n} \]
      associate--l+ [=>] 0.5	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 0.5	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   6. Taylor expanded in n around 0 14.4

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

   7. Simplified 0.5

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

      [Start] 14.4	\[ \frac{\log \left(1 + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 0.5	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 3.9999999999999999e-16 < (/.f64 1 n)

   1. Initial program 9.2

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

   2. Taylor expanded in n around 0 9.2

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

   3. Simplified 6.5

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

      [Start] 9.2	\[ e^{\frac{\log \left(1 + x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)} \]
      log1p-def [=>] 6.5	\[ e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)} \]

3. Recombined 3 regimes into one program.

4. Final simplification 1.5

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{\frac{1}{\log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right)}}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternatives

Alternative 1
Error	1.4
Cost	26564

\[\begin{array}{l} t_0 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{\frac{1}{t_0 - {x}^{\left({n}^{-1}\right)}}}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;t_0 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Error	1.4
Cost	20232

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-7}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \]

Alternative 3
Error	1.6
Cost	13764

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{-7}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\left(1 + \frac{x}{n}\right) + 0.5 \cdot \left(\frac{x}{n} \cdot \frac{x}{n}\right)\right) - t_0\\ \end{array} \]

Alternative 4
Error	1.6
Cost	13508

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\left(1 + \frac{x}{n}\right) + 0.5 \cdot \left(\frac{x}{n} \cdot \frac{x}{n}\right)\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 5
Error	1.9
Cost	8200

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -40000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(\left(1 + \frac{x}{n}\right) + 0.5 \cdot \left(\frac{x}{n} \cdot \frac{x}{n}\right)\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 6
Error	1.9
Cost	7560

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -40000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 7
Error	2.0
Cost	7304

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -40000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-16}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 8
Error	7.3
Cost	6980

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -40000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 9
Error	19.3
Cost	6788

\[\begin{array}{l} t_0 := \frac{\frac{1}{n}}{x}\\ \mathbf{if}\;x \leq 0.7:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 4.8 \cdot 10^{+116}:\\ \;\;\;\;t_0 + \frac{-0.5}{n \cdot \left(x \cdot x\right)}\\ \mathbf{elif}\;x \leq 3.5 \cdot 10^{+140}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 10
Error	36.0
Cost	840

\[\begin{array}{l} \mathbf{if}\;n \leq -6.2:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;n \leq -1.75 \cdot 10^{-184}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{n \cdot x}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 11
Error	28.9
Cost	584

\[\begin{array}{l} \mathbf{if}\;n \leq -7:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;n \leq 5.8 \cdot 10^{-15}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 12
Error	40.7
Cost	320

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

Alternative 13
Error	40.3
Cost	320

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

Alternative 14
Error	61.0
Cost	192

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

Reproduce

herbie shell --seed 2023066 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))