2nthrt (problem 3.4.6)

Average Accuracy: 47.7% → 97.7%

Time: 23.0s

Precision: binary64

Cost: 39496

Math

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

↓

\[\begin{array}{l} t_0 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left({n}^{-1}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;\log \left(e^{t_0}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;{\left({t_0}^{3}\right)}^{0.3333333333333333}\\ \end{array} \]

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 (pow n -1.0)))))
   (if (<= (/ 1.0 n) -5e-7)
     (log (exp t_0))
     (if (<= (/ 1.0 n) 1e-18)
       (/ (log1p (/ 1.0 x)) n)
       (pow (pow t_0 3.0) 0.3333333333333333)))))

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, pow(n, -1.0));
	double tmp;
	if ((1.0 / n) <= -5e-7) {
		tmp = log(exp(t_0));
	} else if ((1.0 / n) <= 1e-18) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = pow(pow(t_0, 3.0), 0.3333333333333333);
	}
	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, Math.pow(n, -1.0));
	double tmp;
	if ((1.0 / n) <= -5e-7) {
		tmp = Math.log(Math.exp(t_0));
	} else if ((1.0 / n) <= 1e-18) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = Math.pow(Math.pow(t_0, 3.0), 0.3333333333333333);
	}
	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, math.pow(n, -1.0))
	tmp = 0
	if (1.0 / n) <= -5e-7:
		tmp = math.log(math.exp(t_0))
	elif (1.0 / n) <= 1e-18:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = math.pow(math.pow(t_0, 3.0), 0.3333333333333333)
	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 ^ (n ^ -1.0)))
	tmp = 0.0
	if (Float64(1.0 / n) <= -5e-7)
		tmp = log(exp(t_0));
	elseif (Float64(1.0 / n) <= 1e-18)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = (t_0 ^ 3.0) ^ 0.3333333333333333;
	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[Power[n, -1.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e-7], N[Log[N[Exp[t$95$0], $MachinePrecision]], $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 1e-18], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[Power[N[Power[t$95$0, 3.0], $MachinePrecision], 0.3333333333333333], $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -4.99999999999999977e-7

   1. Initial program 98.3%

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

   2. Applied egg-rr 98.0%

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

      [Start] 98.3	\[ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
      add-log-exp [=>] 98.0	\[ \color{blue}{\log \left(e^{{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}}\right)} \]
      pow-to-exp [=>] 98.0	\[ \log \left(e^{\color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)}}\right) \]
      un-div-inv [=>] 98.0	\[ \log \left(e^{e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)}}\right) \]
      +-commutative [=>] 98.0	\[ \log \left(e^{e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right) \]
      log1p-def [=>] 98.0	\[ \log \left(e^{e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right) \]
      inv-pow [=>] 98.0	\[ \log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left({n}^{-1}\right)}}}\right) \]

   if -4.99999999999999977e-7 < (/.f64 1 n) < 1.0000000000000001e-18

   1. Initial program 28.7%

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

   2. Taylor expanded in n around inf 78.0%

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

   3. Simplified 78.0%

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

      [Start] 78.0	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 78.0	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 78.2%

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

      [Start] 78.0	\[ \frac{\mathsf{log1p}\left(x\right) - \log x}{n} \]
      add-log-exp [=>] 78.0	\[ \frac{\color{blue}{\log \left(e^{\mathsf{log1p}\left(x\right) - \log x}\right)}}{n} \]
      exp-diff [=>] 78.0	\[ \frac{\log \color{blue}{\left(\frac{e^{\mathsf{log1p}\left(x\right)}}{e^{\log x}}\right)}}{n} \]
      log1p-udef [=>] 78.0	\[ \frac{\log \left(\frac{e^{\color{blue}{\log \left(1 + x\right)}}}{e^{\log x}}\right)}{n} \]
      add-exp-log [<=] 56.5	\[ \frac{\log \left(\frac{\color{blue}{1 + x}}{e^{\log x}}\right)}{n} \]
      +-commutative [=>] 56.5	\[ \frac{\log \left(\frac{\color{blue}{x + 1}}{e^{\log x}}\right)}{n} \]
      add-exp-log [<=] 78.2	\[ \frac{\log \left(\frac{x + 1}{\color{blue}{x}}\right)}{n} \]

   5. Applied egg-rr 78.2%

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

      [Start] 78.2	\[ \frac{\log \left(\frac{x + 1}{x}\right)}{n} \]
      log1p-expm1-u [=>] 78.2	\[ \frac{\color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(\frac{x + 1}{x}\right)\right)\right)}}{n} \]
      expm1-udef [=>] 78.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{e^{\log \left(\frac{x + 1}{x}\right)} - 1}\right)}{n} \]
      add-exp-log [<=] 78.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{x + 1}{x}} - 1\right)}{n} \]

   6. Simplified 99.2%

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

      [Start] 78.2	\[ \frac{\mathsf{log1p}\left(\frac{x + 1}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 78.2	\[ \frac{\mathsf{log1p}\left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x} - 1\right)}{n} \]
      associate-*l/ [<=] 75.1	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(x + 1\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 75.1	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 75.1	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 75.1	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 78.2	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      associate--l+ [=>] 99.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 99.2	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   7. Taylor expanded in n around 0 78.2%

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

   8. Simplified 99.2%

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

      [Start] 78.2	\[ \frac{\log \left(1 + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 99.2	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 1.0000000000000001e-18 < (/.f64 1 n)

   1. Initial program 83.5%

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

   2. Applied egg-rr 87.5%

      \[\leadsto \color{blue}{{\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left({n}^{-1}\right)}\right)}^{3}\right)}^{0.3333333333333333}} \]
      Proof

      [Start] 83.5	\[ {\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
      add-cbrt-cube [=>] 83.5	\[ \color{blue}{\sqrt[3]{\left(\left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right) \cdot \left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right)\right) \cdot \left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right)}} \]
      pow1/3 [=>] 83.5	\[ \color{blue}{{\left(\left(\left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right) \cdot \left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right)\right) \cdot \left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right)\right)}^{0.3333333333333333}} \]
      pow3 [=>] 83.5	\[ {\color{blue}{\left({\left({\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\right)}^{3}\right)}}^{0.3333333333333333} \]
      pow-to-exp [=>] 83.4	\[ {\left({\left(\color{blue}{e^{\log \left(x + 1\right) \cdot \frac{1}{n}}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{3}\right)}^{0.3333333333333333} \]
      un-div-inv [=>] 83.4	\[ {\left({\left(e^{\color{blue}{\frac{\log \left(x + 1\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{3}\right)}^{0.3333333333333333} \]
      +-commutative [=>] 83.4	\[ {\left({\left(e^{\frac{\log \color{blue}{\left(1 + x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{3}\right)}^{0.3333333333333333} \]
      log1p-def [=>] 87.5	\[ {\left({\left(e^{\frac{\color{blue}{\mathsf{log1p}\left(x\right)}}{n}} - {x}^{\left(\frac{1}{n}\right)}\right)}^{3}\right)}^{0.3333333333333333} \]
      inv-pow [=>] 87.5	\[ {\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\color{blue}{\left({n}^{-1}\right)}}\right)}^{3}\right)}^{0.3333333333333333} \]

3. Recombined 3 regimes into one program.

4. Final simplification 97.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;\log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left({n}^{-1}\right)}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;{\left({\left(e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left({n}^{-1}\right)}\right)}^{3}\right)}^{0.3333333333333333}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	97.7%
Cost	39108

\[\begin{array}{l} t_0 := e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;\log \left(e^{t_0 - {x}^{\left({n}^{-1}\right)}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;t_0 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Accuracy	97.8%
Cost	20232

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \]

Alternative 3
Accuracy	97.7%
Cost	14344

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7}:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n} + -0.5 \cdot \frac{x \cdot x}{n}} - t_0\\ \end{array} \]

Alternative 4
Accuracy	97.7%
Cost	13833

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{-7} \lor \neg \left(\frac{1}{n} \leq 10^{-18}\right):\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \end{array} \]

Alternative 5
Accuracy	97.2%
Cost	7560

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -50:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 10^{-18}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 6
Accuracy	97.7%
Cost	7180

\[\begin{array}{l} t_0 := \frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{if}\;n \leq -5.5:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -5 \cdot 10^{-308}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;n \leq 13200000:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 7
Accuracy	89.8%
Cost	6985

\[\begin{array}{l} \mathbf{if}\;n \leq -7.4 \lor \neg \left(n \leq 2 \cdot 10^{-307}\right):\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 8
Accuracy	74.7%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.6:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.7 \cdot 10^{+61}:\\ \;\;\;\;\frac{\left(\frac{1}{x} + \frac{1}{x} \cdot \frac{0.3333333333333333}{x \cdot x}\right) + \frac{-0.5}{x \cdot x}}{n}\\ \mathbf{elif}\;x \leq 3.2 \cdot 10^{+87}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 4.8 \cdot 10^{+175}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;x \leq 1.3 \cdot 10^{+207}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 3.6 \cdot 10^{+234}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 9
Accuracy	44.4%
Cost	840

\[\begin{array}{l} \mathbf{if}\;n \leq -1150000:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;n \leq -5 \cdot 10^{-247}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{n \cdot x}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \end{array} \]

Alternative 10
Accuracy	54.2%
Cost	712

\[\begin{array}{l} \mathbf{if}\;n \leq -5.5:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \mathbf{elif}\;n \leq 4.5 \cdot 10^{-101}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n} \cdot \frac{1}{x}\\ \end{array} \]

Alternative 11
Accuracy	36.0%
Cost	320

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

Alternative 12
Accuracy	36.6%
Cost	320

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

Alternative 13
Accuracy	4.7%
Cost	192

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

Reproduce

herbie shell --seed 2023138 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))