2nthrt (problem 3.4.6)

Average Error: 32.2 → 11.9

Time: 17.2s

Precision: binary64

Math

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

↓

\[\begin{array}{l} t_0 := e^{\frac{\log x}{n}}\\ t_1 := \mathsf{fma}\left(\frac{t_0}{x \cdot x}, \frac{0.5}{n \cdot n} + \frac{-0.5}{n}, \frac{t_0}{n \cdot x}\right)\\ t_2 := \sqrt{{x}^{\left(\frac{1}{n}\right)}}\\ t_3 := \mathsf{fma}\left(-t_2, t_2, e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)\\ \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+158}:\\ \;\;\;\;t_3\\ \mathbf{elif}\;\frac{1}{n} \leq -1.9 \cdot 10^{-23}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-88}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n} + \frac{0.5}{n} \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n} - \frac{{\log x}^{2}}{n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0002:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;t_3\\ \end{array} \]

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 (/ (log x) n)))
        (t_1
         (fma (/ t_0 (* x x)) (+ (/ 0.5 (* n n)) (/ -0.5 n)) (/ t_0 (* n x))))
        (t_2 (sqrt (pow x (/ 1.0 n))))
        (t_3 (fma (- t_2) t_2 (exp (/ (log1p x) n)))))
   (if (<= (/ 1.0 n) -1e+158)
     t_3
     (if (<= (/ 1.0 n) -1.9e-23)
       t_1
       (if (<= (/ 1.0 n) 4e-88)
         (+
          (/ (log (/ (+ 1.0 x) x)) n)
          (* (/ 0.5 n) (- (/ (pow (log1p x) 2.0) n) (/ (pow (log x) 2.0) n))))
         (if (<= (/ 1.0 n) 0.0002) t_1 t_3))))))

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((log(x) / n));
	double t_1 = fma((t_0 / (x * x)), ((0.5 / (n * n)) + (-0.5 / n)), (t_0 / (n * x)));
	double t_2 = sqrt(pow(x, (1.0 / n)));
	double t_3 = fma(-t_2, t_2, exp((log1p(x) / n)));
	double tmp;
	if ((1.0 / n) <= -1e+158) {
		tmp = t_3;
	} else if ((1.0 / n) <= -1.9e-23) {
		tmp = t_1;
	} else if ((1.0 / n) <= 4e-88) {
		tmp = (log(((1.0 + x) / x)) / n) + ((0.5 / n) * ((pow(log1p(x), 2.0) / n) - (pow(log(x), 2.0) / n)));
	} else if ((1.0 / n) <= 0.0002) {
		tmp = t_1;
	} else {
		tmp = t_3;
	}
	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 = exp(Float64(log(x) / n))
	t_1 = fma(Float64(t_0 / Float64(x * x)), Float64(Float64(0.5 / Float64(n * n)) + Float64(-0.5 / n)), Float64(t_0 / Float64(n * x)))
	t_2 = sqrt((x ^ Float64(1.0 / n)))
	t_3 = fma(Float64(-t_2), t_2, exp(Float64(log1p(x) / n)))
	tmp = 0.0
	if (Float64(1.0 / n) <= -1e+158)
		tmp = t_3;
	elseif (Float64(1.0 / n) <= -1.9e-23)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 4e-88)
		tmp = Float64(Float64(log(Float64(Float64(1.0 + x) / x)) / n) + Float64(Float64(0.5 / n) * Float64(Float64((log1p(x) ^ 2.0) / n) - Float64((log(x) ^ 2.0) / n))));
	elseif (Float64(1.0 / n) <= 0.0002)
		tmp = t_1;
	else
		tmp = t_3;
	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[Exp[N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 / N[(x * x), $MachinePrecision]), $MachinePrecision] * N[(N[(0.5 / N[(n * n), $MachinePrecision]), $MachinePrecision] + N[(-0.5 / n), $MachinePrecision]), $MachinePrecision] + N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[Sqrt[N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]}, Block[{t$95$3 = N[((-t$95$2) * t$95$2 + N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -1e+158], t$95$3, If[LessEqual[N[(1.0 / n), $MachinePrecision], -1.9e-23], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 4e-88], N[(N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision] + N[(N[(0.5 / n), $MachinePrecision] * N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 2.0], $MachinePrecision] / n), $MachinePrecision] - N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 0.0002], t$95$1, t$95$3]]]]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -9.99999999999999953e157 or 2.0000000000000001e-4 < (/.f64 1 n)

   1. Initial program 3.0

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

   2. Applied egg-rr 0.9

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

   if -9.99999999999999953e157 < (/.f64 1 n) < -1.90000000000000006e-23 or 3.99999999999999974e-88 < (/.f64 1 n) < 2.0000000000000001e-4

   1. Initial program 28.5

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

   2. Taylor expanded in x around inf 20.2

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

   3. Simplified 20.2

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{\frac{\log x}{n}}}{x \cdot x}, \frac{0.5}{n \cdot n} + \frac{-0.5}{n}, \frac{e^{\frac{\log x}{n}}}{x \cdot n}\right)} \]

   if -1.90000000000000006e-23 < (/.f64 1 n) < 3.99999999999999974e-88

   1. Initial program 42.7

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

   2. Taylor expanded in n around inf 12.3

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

   3. Simplified 12.3

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n} + \frac{0.5}{n} \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n} - \frac{{\log x}^{2}}{n}\right)} \]

   4. Applied egg-rr 12.2

      \[\leadsto \frac{\color{blue}{\log \left(\frac{x + 1}{x}\right)}}{n} + \frac{0.5}{n} \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n} - \frac{{\log x}^{2}}{n}\right) \]
3. Recombined 3 regimes into one program.

4. Final simplification 11.9

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+158}:\\ \;\;\;\;\mathsf{fma}\left(-\sqrt{{x}^{\left(\frac{1}{n}\right)}}, \sqrt{{x}^{\left(\frac{1}{n}\right)}}, e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq -1.9 \cdot 10^{-23}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{\frac{\log x}{n}}}{x \cdot x}, \frac{0.5}{n \cdot n} + \frac{-0.5}{n}, \frac{e^{\frac{\log x}{n}}}{n \cdot x}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-88}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n} + \frac{0.5}{n} \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{n} - \frac{{\log x}^{2}}{n}\right)\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0002:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{\frac{\log x}{n}}}{x \cdot x}, \frac{0.5}{n \cdot n} + \frac{-0.5}{n}, \frac{e^{\frac{\log x}{n}}}{n \cdot x}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(-\sqrt{{x}^{\left(\frac{1}{n}\right)}}, \sqrt{{x}^{\left(\frac{1}{n}\right)}}, e^{\frac{\mathsf{log1p}\left(x\right)}{n}}\right)\\ \end{array} \]

Reproduce

herbie shell --seed 2022210 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))