2nthrt (problem 3.4.6)

Average Error: 32.5 → 6.6

Time: 26.7s

Precision: binary64

Cost: 85380

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;x \leq 1.9:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) + \left(\frac{{\log x}^{2}}{n \cdot n} \cdot -0.5 - \frac{\mathsf{fma}\left(-1, \mathsf{log1p}\left(x\right), \log x\right)}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-e^{\frac{\log x}{n}}}{n}}{-x}\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (if (<= x 1.9)
   (+
    (fma
     0.5
     (/ (pow (log1p x) 2.0) (* n n))
     (/
      (* 0.16666666666666666 (- (pow (log1p x) 3.0) (pow (log x) 3.0)))
      (pow n 3.0)))
    (-
     (* (/ (pow (log x) 2.0) (* n n)) -0.5)
     (/ (fma -1.0 (log1p x) (log x)) n)))
   (/ (/ (- (exp (/ (log x) n))) n) (- x))))

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 tmp;
	if (x <= 1.9) {
		tmp = fma(0.5, (pow(log1p(x), 2.0) / (n * n)), ((0.16666666666666666 * (pow(log1p(x), 3.0) - pow(log(x), 3.0))) / pow(n, 3.0))) + (((pow(log(x), 2.0) / (n * n)) * -0.5) - (fma(-1.0, log1p(x), log(x)) / n));
	} else {
		tmp = (-exp((log(x) / n)) / n) / -x;
	}
	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)
	tmp = 0.0
	if (x <= 1.9)
		tmp = Float64(fma(0.5, Float64((log1p(x) ^ 2.0) / Float64(n * n)), Float64(Float64(0.16666666666666666 * Float64((log1p(x) ^ 3.0) - (log(x) ^ 3.0))) / (n ^ 3.0))) + Float64(Float64(Float64((log(x) ^ 2.0) / Float64(n * n)) * -0.5) - Float64(fma(-1.0, log1p(x), log(x)) / n)));
	else
		tmp = Float64(Float64(Float64(-exp(Float64(log(x) / n))) / n) / Float64(-x));
	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_] := If[LessEqual[x, 1.9], N[(N[(0.5 * N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] / N[(n * n), $MachinePrecision]), $MachinePrecision] + N[(N[(0.16666666666666666 * N[(N[Power[N[Log[1 + x], $MachinePrecision], 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[n, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / N[(n * n), $MachinePrecision]), $MachinePrecision] * -0.5), $MachinePrecision] - N[(N[(-1.0 * N[Log[1 + x], $MachinePrecision] + N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[((-N[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision] / (-x)), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 1.8999999999999999

   1. Initial program 47.0

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

   2. Taylor expanded in n around -inf 13.6

      \[\leadsto \color{blue}{\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}} \]

   3. Simplified 13.6

      \[\leadsto \color{blue}{\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) - \left(\frac{\mathsf{fma}\left(-1, \mathsf{log1p}\left(x\right), \log x\right)}{n} - \frac{{\log x}^{2}}{n \cdot n} \cdot -0.5\right)} \]
      Proof

      [Start] 13.6	\[ \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}} \]

   if 1.8999999999999999 < x

   1. Initial program 20.9

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

   2. Applied egg-rr 20.9

      \[\leadsto \color{blue}{\left({\left(x + 1\right)}^{\left(\frac{0.5}{n}\right)} - {x}^{\left(\frac{0.5}{n}\right)}\right) \cdot \left({\left(x + 1\right)}^{\left(\frac{0.5}{n}\right)} + {x}^{\left(\frac{0.5}{n}\right)}\right)} \]

   3. Taylor expanded in x around inf 1.8

      \[\leadsto \color{blue}{\frac{{\left(e^{-0.5 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}\right)}^{2}}{n \cdot x}} \]

   4. Applied egg-rr 1.0

      \[\leadsto \color{blue}{-\frac{\frac{e^{\frac{-\log x}{n} \cdot -1}}{n}}{-x}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 6.6

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.9:\\ \;\;\;\;\mathsf{fma}\left(0.5, \frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n \cdot n}, \frac{0.16666666666666666 \cdot \left({\left(\mathsf{log1p}\left(x\right)\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right) + \left(\frac{{\log x}^{2}}{n \cdot n} \cdot -0.5 - \frac{\mathsf{fma}\left(-1, \mathsf{log1p}\left(x\right), \log x\right)}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-e^{\frac{\log x}{n}}}{n}}{-x}\\ \end{array} \]

Alternatives

Alternative 1
Error	6.5
Cost	26820

\[\begin{array}{l} \mathbf{if}\;x \leq 24000000:\\ \;\;\;\;\frac{0.5 \cdot \left(\mathsf{log1p}\left(x\right) - \log x\right)}{n} \cdot \left(2 \cdot e^{-0.5 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-e^{\frac{\log x}{n}}}{n}}{-x}\\ \end{array} \]

Alternative 2
Error	6.5
Cost	20676

\[\begin{array}{l} \mathbf{if}\;x \leq 56000000:\\ \;\;\;\;\frac{0.5 \cdot \log \left(\frac{x + 1}{x}\right)}{n} \cdot \left({\left(x + 1\right)}^{\left(\frac{0.5}{n}\right)} + {x}^{\left(\frac{0.5}{n}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-e^{\frac{\log x}{n}}}{n}}{-x}\\ \end{array} \]

Alternative 3
Error	6.7
Cost	13508

\[\begin{array}{l} \mathbf{if}\;x \leq 1.9:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-e^{\frac{\log x}{n}}}{n}}{-x}\\ \end{array} \]

Alternative 4
Error	7.2
Cost	13380

\[\begin{array}{l} \mathbf{if}\;x \leq 1.9:\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{x \cdot n}\\ \end{array} \]

Alternative 5
Error	15.3
Cost	7689

\[\begin{array}{l} \mathbf{if}\;x \leq 12200 \lor \neg \left(x \leq 9 \cdot 10^{+185}\right):\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x} + \left(\frac{0.3333333333333333}{{x}^{3}} + \frac{-0.5}{x \cdot x}\right)}{n}\\ \end{array} \]

Alternative 6
Error	15.5
Cost	7113

\[\begin{array}{l} \mathbf{if}\;x \leq 1.9 \lor \neg \left(x \leq 6 \cdot 10^{+185}\right):\\ \;\;\;\;\frac{\log \left(\frac{x + 1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{\frac{-0.5}{x}}{x}}{n}\\ \end{array} \]

Alternative 7
Error	16.9
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 0.96:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 2.05 \cdot 10^{+198}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{\frac{-0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{x \cdot n}\right)\\ \end{array} \]

Alternative 8
Error	17.1
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.7:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.7 \cdot 10^{+198}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{\frac{-0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;-1 + \left(1 + \frac{1}{x \cdot n}\right)\\ \end{array} \]

Alternative 9
Error	34.6
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -20:\\ \;\;\;\;-1 + \left(1 + \frac{1}{x \cdot n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{n}}{x}\\ \end{array} \]

Alternative 10
Error	40.2
Cost	320

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

Alternative 11
Error	39.7
Cost	320

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

Reproduce

herbie shell --seed 2023057 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))