2nthrt (problem 3.4.6)

Average Error: 33.2 → 6.7

Time: 39.7s

Precision: binary64

Cost: 85700

Math

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

↓

\[\begin{array}{l} t_0 := \log \left(1 + x\right)\\ \mathbf{if}\;x \leq 8.5:\\ \;\;\;\;\frac{t_0 - \log x}{n} + \left(\left(-\left({t_0}^{3} - {\log x}^{3}\right) \cdot \frac{-0.16666666666666666}{{n}^{3}}\right) + 0.5 \cdot \left(\frac{{t_0}^{2}}{{n}^{2}} - \frac{{\log x}^{2}}{{n}^{2}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log \left(\frac{1}{x}\right)}{-n}}}{x \cdot n}\\ \end{array} \]

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 (log (+ 1.0 x))))
   (if (<= x 8.5)
     (+
      (/ (- t_0 (log x)) n)
      (+
       (-
        (*
         (- (pow t_0 3.0) (pow (log x) 3.0))
         (/ -0.16666666666666666 (pow n 3.0))))
       (*
        0.5
        (- (/ (pow t_0 2.0) (pow n 2.0)) (/ (pow (log x) 2.0) (pow n 2.0))))))
     (/ (exp (/ (log (/ 1.0 x)) (- n))) (* x n)))))

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 = log((1.0 + x));
	double tmp;
	if (x <= 8.5) {
		tmp = ((t_0 - log(x)) / n) + (-((pow(t_0, 3.0) - pow(log(x), 3.0)) * (-0.16666666666666666 / pow(n, 3.0))) + (0.5 * ((pow(t_0, 2.0) / pow(n, 2.0)) - (pow(log(x), 2.0) / pow(n, 2.0)))));
	} else {
		tmp = exp((log((1.0 / x)) / -n)) / (x * n);
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((x + 1.0d0) ** (1.0d0 / n)) - (x ** (1.0d0 / n))
end function

↓

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = log((1.0d0 + x))
    if (x <= 8.5d0) then
        tmp = ((t_0 - log(x)) / n) + (-(((t_0 ** 3.0d0) - (log(x) ** 3.0d0)) * ((-0.16666666666666666d0) / (n ** 3.0d0))) + (0.5d0 * (((t_0 ** 2.0d0) / (n ** 2.0d0)) - ((log(x) ** 2.0d0) / (n ** 2.0d0)))))
    else
        tmp = exp((log((1.0d0 / x)) / -n)) / (x * n)
    end if
    code = tmp
end function

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.log((1.0 + x));
	double tmp;
	if (x <= 8.5) {
		tmp = ((t_0 - Math.log(x)) / n) + (-((Math.pow(t_0, 3.0) - Math.pow(Math.log(x), 3.0)) * (-0.16666666666666666 / Math.pow(n, 3.0))) + (0.5 * ((Math.pow(t_0, 2.0) / Math.pow(n, 2.0)) - (Math.pow(Math.log(x), 2.0) / Math.pow(n, 2.0)))));
	} else {
		tmp = Math.exp((Math.log((1.0 / x)) / -n)) / (x * n);
	}
	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.log((1.0 + x))
	tmp = 0
	if x <= 8.5:
		tmp = ((t_0 - math.log(x)) / n) + (-((math.pow(t_0, 3.0) - math.pow(math.log(x), 3.0)) * (-0.16666666666666666 / math.pow(n, 3.0))) + (0.5 * ((math.pow(t_0, 2.0) / math.pow(n, 2.0)) - (math.pow(math.log(x), 2.0) / math.pow(n, 2.0)))))
	else:
		tmp = math.exp((math.log((1.0 / x)) / -n)) / (x * n)
	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 = log(Float64(1.0 + x))
	tmp = 0.0
	if (x <= 8.5)
		tmp = Float64(Float64(Float64(t_0 - log(x)) / n) + Float64(Float64(-Float64(Float64((t_0 ^ 3.0) - (log(x) ^ 3.0)) * Float64(-0.16666666666666666 / (n ^ 3.0)))) + Float64(0.5 * Float64(Float64((t_0 ^ 2.0) / (n ^ 2.0)) - Float64((log(x) ^ 2.0) / (n ^ 2.0))))));
	else
		tmp = Float64(exp(Float64(log(Float64(1.0 / x)) / Float64(-n))) / Float64(x * n));
	end
	return tmp
end

MATLAB

function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end

↓

function tmp_2 = code(x, n)
	t_0 = log((1.0 + x));
	tmp = 0.0;
	if (x <= 8.5)
		tmp = ((t_0 - log(x)) / n) + (-(((t_0 ^ 3.0) - (log(x) ^ 3.0)) * (-0.16666666666666666 / (n ^ 3.0))) + (0.5 * (((t_0 ^ 2.0) / (n ^ 2.0)) - ((log(x) ^ 2.0) / (n ^ 2.0)))));
	else
		tmp = exp((log((1.0 / x)) / -n)) / (x * n);
	end
	tmp_2 = 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[Log[N[(1.0 + x), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[x, 8.5], N[(N[(N[(t$95$0 - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] + N[((-N[(N[(N[Power[t$95$0, 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision] * N[(-0.16666666666666666 / N[Power[n, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]) + N[(0.5 * N[(N[(N[Power[t$95$0, 2.0], $MachinePrecision] / N[Power[n, 2.0], $MachinePrecision]), $MachinePrecision] - N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / N[Power[n, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Exp[N[(N[Log[N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / (-n)), $MachinePrecision]], $MachinePrecision] / N[(x * n), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if x < 8.5

   1. Initial program 47.7

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

   2. Taylor expanded in n around -inf 12.7

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

   3. Simplified 12.7

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

      [Start] 12.7	\[ \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}} \]
      rational.json-simplify-48 [=>] 12.7	\[ \color{blue}{\left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right) + \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right)} \]
      rational.json-simplify-1 [=>] 12.7	\[ \color{blue}{\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right) + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)} \]
      rational.json-simplify-1 [=>] 12.7	\[ \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right) + \color{blue}{\left(-1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n} + -1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}}\right)} \]
      rational.json-simplify-41 [=>] 12.7	\[ \color{blue}{-1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + \left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\right)\right)} \]

   if 8.5 < x

   1. Initial program 20.7

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

   2. Taylor expanded in x around inf 1.6

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

   3. Simplified 1.6

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

      [Start] 1.6	\[ \frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x} \]
      rational.json-simplify-2 [=>] 1.6	\[ \frac{e^{\color{blue}{\frac{\log \left(\frac{1}{x}\right)}{n} \cdot -1}}}{n \cdot x} \]
      rational.json-simplify-9 [=>] 1.6	\[ \frac{e^{\color{blue}{-\frac{\log \left(\frac{1}{x}\right)}{n}}}}{n \cdot x} \]
      rational.json-simplify-10 [=>] 1.6	\[ \frac{e^{\color{blue}{\frac{\frac{\log \left(\frac{1}{x}\right)}{n}}{-1}}}}{n \cdot x} \]
      rational.json-simplify-47 [=>] 1.6	\[ \frac{e^{\color{blue}{\frac{\log \left(\frac{1}{x}\right)}{n \cdot -1}}}}{n \cdot x} \]
      rational.json-simplify-9 [=>] 1.6	\[ \frac{e^{\frac{\log \left(\frac{1}{x}\right)}{\color{blue}{-n}}}}{n \cdot x} \]
      rational.json-simplify-2 [=>] 1.6	\[ \frac{e^{\frac{\log \left(\frac{1}{x}\right)}{-n}}}{\color{blue}{x \cdot n}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 6.7

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 8.5:\\ \;\;\;\;\frac{\log \left(1 + x\right) - \log x}{n} + \left(\left(-\left({\log \left(1 + x\right)}^{3} - {\log x}^{3}\right) \cdot \frac{-0.16666666666666666}{{n}^{3}}\right) + 0.5 \cdot \left(\frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} - \frac{{\log x}^{2}}{{n}^{2}}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log \left(\frac{1}{x}\right)}{-n}}}{x \cdot n}\\ \end{array} \]

Alternatives

Alternative 1
Error	12.5
Cost	14028

\[\begin{array}{l} t_0 := \frac{\log \left(x - -1\right) - \log x}{n}\\ \mathbf{if}\;\frac{1}{n} \leq 5 \cdot 10^{-93}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 4 \cdot 10^{-50}:\\ \;\;\;\;\frac{1}{n \cdot \left(x + 0.5\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0001:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - e^{\frac{\log x}{n}}\\ \end{array} \]

Alternative 2
Error	6.9
Cost	13572

\[\begin{array}{l} \mathbf{if}\;x \leq 8.5:\\ \;\;\;\;\frac{\log \left(x - -1\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log \left(\frac{1}{x}\right)}{-n}}}{x \cdot n}\\ \end{array} \]

Alternative 3
Error	13.6
Cost	7044

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

Alternative 4
Error	13.6
Cost	6852

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

Alternative 5
Error	13.8
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.55:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{\left(-x\right) \cdot \frac{-n}{x}}{x \cdot n}}{n}\\ \end{array} \]

Alternative 6
Error	29.2
Cost	1416

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+216}:\\ \;\;\;\;\frac{\frac{\left(-x\right) \cdot \frac{-n}{x}}{x \cdot n}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -20:\\ \;\;\;\;\left(-x\right) \cdot \left(\left(1 - \frac{\frac{\frac{1}{x}}{x}}{n}\right) + -1\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot \left(x + 0.5\right)}\\ \end{array} \]

Alternative 7
Error	31.4
Cost	1096

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+140}:\\ \;\;\;\;\frac{-x}{-n \cdot \left(x \cdot x\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -20:\\ \;\;\;\;-1 + \left(1 - \frac{\frac{-1}{x}}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot \left(x + 0.5\right)}\\ \end{array} \]

Alternative 8
Error	30.3
Cost	1096

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -1 \cdot 10^{+140}:\\ \;\;\;\;\frac{\frac{\left(-x\right) \cdot \frac{-n}{x}}{x \cdot n}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -20:\\ \;\;\;\;-1 + \left(1 - \frac{\frac{-1}{x}}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot \left(x + 0.5\right)}\\ \end{array} \]

Alternative 9
Error	36.8
Cost	840

\[\begin{array}{l} t_0 := \frac{1}{n \cdot \left(x + 0.5\right)}\\ \mathbf{if}\;n \leq -9.5:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -1.05 \cdot 10^{-140}:\\ \;\;\;\;\frac{1}{x \cdot x} \cdot \frac{x}{n}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 10
Error	33.5
Cost	840

\[\begin{array}{l} t_0 := \frac{1}{n \cdot \left(x + 0.5\right)}\\ \mathbf{if}\;n \leq -7.5:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -1.4 \cdot 10^{-217}:\\ \;\;\;\;-1 + \left(1 - \frac{\frac{-1}{x}}{n}\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 11
Error	35.5
Cost	712

\[\begin{array}{l} t_0 := \frac{1}{n \cdot \left(x + 0.5\right)}\\ \mathbf{if}\;n \leq -4.5:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -2 \cdot 10^{-215}:\\ \;\;\;\;\frac{x}{x \cdot \left(x \cdot n\right)}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

Alternative 12
Error	38.1
Cost	448

\[\frac{1}{n \cdot \left(x + 0.5\right)} \]

Alternative 13
Error	40.7
Cost	320

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

Alternative 14
Error	40.2
Cost	320

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

Reproduce

herbie shell --seed 2023075 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))