2nthrt (problem 3.4.6)

Percentage Accurate: 53.2% → 86.0%

Time: 21.2s

Alternatives: 17

Speedup: 1.9×

• could not determine a ground truth

  1. x = -2
  2. n = 4.94066e-324

Specification

Math

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

FPCore

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

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

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

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

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

Initial Program: 53.2% accurate, 1.0× speedup

Math

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

FPCore

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

C

double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}

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

Java

public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}

Python

def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))

Julia

function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end

MATLAB

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

Alternative 1: 86.0% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n} + 0.5 \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{{n}^{2}} - \frac{{\log x}^{2}}{{n}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-14)
       (+
        (/ (- (log1p x) (log x)) n)
        (*
         0.5
         (-
          (/ (pow (log1p x) 2.0) (pow n 2.0))
          (/ (pow (log x) 2.0) (pow n 2.0)))))
       (- (exp (/ (log1p x) n)) t_0)))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = ((log1p(x) - log(x)) / n) + (0.5 * ((pow(log1p(x), 2.0) / pow(n, 2.0)) - (pow(log(x), 2.0) / pow(n, 2.0))));
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = ((Math.log1p(x) - Math.log(x)) / n) + (0.5 * ((Math.pow(Math.log1p(x), 2.0) / Math.pow(n, 2.0)) - (Math.pow(Math.log(x), 2.0) / Math.pow(n, 2.0))));
	} else {
		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-14:
		tmp = ((math.log1p(x) - math.log(x)) / n) + (0.5 * ((math.pow(math.log1p(x), 2.0) / math.pow(n, 2.0)) - (math.pow(math.log(x), 2.0) / math.pow(n, 2.0))))
	else:
		tmp = math.exp((math.log1p(x) / n)) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-14)
		tmp = Float64(Float64(Float64(log1p(x) - log(x)) / n) + Float64(0.5 * Float64(Float64((log1p(x) ^ 2.0) / (n ^ 2.0)) - Float64((log(x) ^ 2.0) / (n ^ 2.0)))));
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[(N[(N[(N[Log[1 + x], $MachinePrecision] - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision] + N[(0.5 * N[(N[(N[Power[N[Log[1 + x], $MachinePrecision], 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], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2e-14

   1. Initial program 38.2%

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

   2. Taylor expanded in n around inf 87.5%

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

      1. associate--l+ 78.6%

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

      2. +-commutative 78.6%

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

      3. associate--r+ 87.5%

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

      4. div-sub 87.6%

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

      5. remove-double-neg 87.6%

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

      6. mul-1-neg 87.6%

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

      7. distribute-lft-out-- 87.6%

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

      8. mul-1-neg 87.6%

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

      9. associate-*r/ 87.6%

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

   4. Simplified 87.6%

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

   if 2e-14 < (/.f64 1 n)

   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 0 45.1%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
   3. Step-by-step derivation

      1. log1p-def 95.1%

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

   4. Simplified 95.1%

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

4. Final simplification 92.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{\mathsf{log1p}\left(x\right) - \log x}{n} + 0.5 \cdot \left(\frac{{\left(\mathsf{log1p}\left(x\right)\right)}^{2}}{{n}^{2}} - \frac{{\log x}^{2}}{{n}^{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2: 86.1% accurate, 0.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-14)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (- (exp (/ (log1p x) n)) t_0)))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -log((x / (1.0 + x))) / n;
	} else {
		tmp = exp((log1p(x) / n)) - t_0;
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else {
		tmp = Math.exp((Math.log1p(x) / n)) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-14:
		tmp = -math.log((x / (1.0 + x))) / n
	else:
		tmp = math.exp((math.log1p(x) / n)) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-14)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	else
		tmp = Float64(exp(Float64(log1p(x) / n)) - t_0);
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2e-14

   1. Initial program 38.2%

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

   2. Taylor expanded in n around inf 87.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 87.3%

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

   4. Simplified 87.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 87.3%

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

      2. diff-log 87.3%

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

   6. Applied egg-rr 87.3%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 87.3%

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

      2. log-rec 87.4%

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

   8. Applied egg-rr 87.4%

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

   if 2e-14 < (/.f64 1 n)

   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 0 45.1%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
   3. Step-by-step derivation

      1. log1p-def 95.1%

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

   4. Simplified 95.1%

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

4. Final simplification 92.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 3: 82.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 6 \cdot 10^{+109}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\frac{1}{n}}{x}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-14)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (if (<= (/ 1.0 n) 6e+109)
         (- (+ 1.0 (/ x n)) t_0)
         (log1p (expm1 (/ (/ 1.0 n) x))))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 6e+109) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = log1p(expm1(((1.0 / n) / x)));
	}
	return tmp;
}

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 6e+109) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = Math.log1p(Math.expm1(((1.0 / n) / x)));
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-14:
		tmp = -math.log((x / (1.0 + x))) / n
	elif (1.0 / n) <= 6e+109:
		tmp = (1.0 + (x / n)) - t_0
	else:
		tmp = math.log1p(math.expm1(((1.0 / n) / x)))
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-14)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	elseif (Float64(1.0 / n) <= 6e+109)
		tmp = Float64(Float64(1.0 + Float64(x / n)) - t_0);
	else
		tmp = log1p(expm1(Float64(Float64(1.0 / n) / x)));
	end
	return tmp
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 6e+109], N[(N[(1.0 + N[(x / n), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision], N[Log[1 + N[(Exp[N[(N[(1.0 / n), $MachinePrecision] / x), $MachinePrecision]] - 1), $MachinePrecision]], $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2e-14

   1. Initial program 38.2%

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

   2. Taylor expanded in n around inf 87.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 87.3%

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

   4. Simplified 87.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 87.3%

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

      2. diff-log 87.3%

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

   6. Applied egg-rr 87.3%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 87.3%

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

      2. log-rec 87.4%

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

   8. Applied egg-rr 87.4%

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

   if 2e-14 < (/.f64 1 n) < 6.00000000000000031e109

   1. Initial program 72.3%

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

   2. Taylor expanded in x around 0 74.8%

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

   if 6.00000000000000031e109 < (/.f64 1 n)

   1. Initial program 24.0%

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

   2. Taylor expanded in n around inf 6.4%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 6.4%

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

   4. Simplified 6.4%

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

   5. Taylor expanded in x around inf 44.4%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 44.4%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
   8. Step-by-step derivation

      1. log1p-expm1-u 78.0%

         \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{1}{x \cdot n}\right)\right)} \]

      2. *-commutative 78.0%

         \[\leadsto \mathsf{log1p}\left(\mathsf{expm1}\left(\frac{1}{\color{blue}{n \cdot x}}\right)\right) \]

      3. associate-/r* 78.0%

         \[\leadsto \mathsf{log1p}\left(\mathsf{expm1}\left(\color{blue}{\frac{\frac{1}{n}}{x}}\right)\right) \]

   9. Applied egg-rr 78.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\frac{1}{n}}{x}\right)\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 89.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 6 \cdot 10^{+109}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\frac{1}{n}}{x}\right)\right)\\ \end{array} \]

Alternative 4: 81.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 6 \cdot 10^{+109}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\sqrt{{\left(n \cdot x\right)}^{-2}}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-14)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (if (<= (/ 1.0 n) 6e+109)
         (- (+ 1.0 (/ x n)) t_0)
         (sqrt (pow (* n x) -2.0)))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 6e+109) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = sqrt(pow((n * x), -2.0));
	}
	return tmp;
}

Fortran

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 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 2d-14) then
        tmp = -log((x / (1.0d0 + x))) / n
    else if ((1.0d0 / n) <= 6d+109) then
        tmp = (1.0d0 + (x / n)) - t_0
    else
        tmp = sqrt(((n * x) ** (-2.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 6e+109) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = Math.sqrt(Math.pow((n * x), -2.0));
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-14:
		tmp = -math.log((x / (1.0 + x))) / n
	elif (1.0 / n) <= 6e+109:
		tmp = (1.0 + (x / n)) - t_0
	else:
		tmp = math.sqrt(math.pow((n * x), -2.0))
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-14)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	elseif (Float64(1.0 / n) <= 6e+109)
		tmp = Float64(Float64(1.0 + Float64(x / n)) - t_0);
	else
		tmp = sqrt((Float64(n * x) ^ -2.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-58)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 2e-14)
		tmp = -log((x / (1.0 + x))) / n;
	elseif ((1.0 / n) <= 6e+109)
		tmp = (1.0 + (x / n)) - t_0;
	else
		tmp = sqrt(((n * x) ^ -2.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 6e+109], N[(N[(1.0 + N[(x / n), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision], N[Sqrt[N[Power[N[(n * x), $MachinePrecision], -2.0], $MachinePrecision]], $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2e-14

   1. Initial program 38.2%

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

   2. Taylor expanded in n around inf 87.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 87.3%

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

   4. Simplified 87.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 87.3%

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

      2. diff-log 87.3%

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

   6. Applied egg-rr 87.3%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 87.3%

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

      2. log-rec 87.4%

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

   8. Applied egg-rr 87.4%

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

   if 2e-14 < (/.f64 1 n) < 6.00000000000000031e109

   1. Initial program 72.3%

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

   2. Taylor expanded in x around 0 74.8%

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

   if 6.00000000000000031e109 < (/.f64 1 n)

   1. Initial program 24.0%

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

   2. Taylor expanded in n around inf 6.4%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 6.4%

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

   4. Simplified 6.4%

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

   5. Taylor expanded in x around inf 44.4%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 44.4%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
   8. Step-by-step derivation

      1. add-sqr-sqrt 44.4%

         \[\leadsto \color{blue}{\sqrt{\frac{1}{x \cdot n}} \cdot \sqrt{\frac{1}{x \cdot n}}} \]

      2. sqrt-unprod 73.6%

         \[\leadsto \color{blue}{\sqrt{\frac{1}{x \cdot n} \cdot \frac{1}{x \cdot n}}} \]

      3. inv-pow 73.6%

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

      4. inv-pow 73.6%

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

      5. pow-prod-up 73.6%

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

      6. metadata-eval 73.6%

         \[\leadsto \sqrt{{\left(x \cdot n\right)}^{\color{blue}{-2}}} \]

   9. Applied egg-rr 73.6%

      \[\leadsto \color{blue}{\sqrt{{\left(x \cdot n\right)}^{-2}}} \]
3. Recombined 4 regimes into one program.

4. Final simplification 89.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 6 \cdot 10^{+109}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{{\left(n \cdot x\right)}^{-2}}\\ \end{array} \]

Alternative 5: 86.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - t_0\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-5)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (- (exp (/ x n)) t_0)))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -log((x / (1.0 + x))) / n;
	} else {
		tmp = exp((x / n)) - t_0;
	}
	return tmp;
}

Fortran

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 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 2d-5) then
        tmp = -log((x / (1.0d0 + x))) / n
    else
        tmp = exp((x / n)) - t_0
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else {
		tmp = Math.exp((x / n)) - t_0;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-5:
		tmp = -math.log((x / (1.0 + x))) / n
	else:
		tmp = math.exp((x / n)) - t_0
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-5)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	else
		tmp = Float64(exp(Float64(x / n)) - t_0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-58)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 2e-5)
		tmp = -log((x / (1.0 + x))) / n;
	else
		tmp = exp((x / n)) - t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-5], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], N[(N[Exp[N[(x / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2.00000000000000016e-5

   1. Initial program 37.6%

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

   2. Taylor expanded in n around inf 85.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 85.8%

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

   4. Simplified 85.8%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 85.8%

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

      2. diff-log 85.9%

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

   6. Applied egg-rr 85.9%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 85.9%

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

      2. log-rec 86.0%

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

   8. Applied egg-rr 86.0%

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

   if 2.00000000000000016e-5 < (/.f64 1 n)

   1. Initial program 47.3%

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

   2. Taylor expanded in n around 0 47.3%

      \[\leadsto \color{blue}{e^{\frac{\log \left(1 + x\right)}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
   3. Step-by-step derivation

      1. log1p-def 100.0%

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

   4. Simplified 100.0%

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

   5. Taylor expanded in x around 0 100.0%

      \[\leadsto e^{\color{blue}{\frac{x}{n}}} - {x}^{\left(\frac{1}{n}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 92.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;e^{\frac{x}{n}} - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 6: 66.8% accurate, 1.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ t_2 := \frac{1}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+201}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+140}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+66}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-11}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_2\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (- 1.0 (pow x (/ 1.0 n))))
        (t_1 (/ (log (/ (+ 1.0 x) x)) n))
        (t_2 (/ 1.0 (* n x))))
   (if (<= (/ 1.0 n) -5e+201)
     t_1
     (if (<= (/ 1.0 n) -5e+140)
       t_0
       (if (<= (/ 1.0 n) -5e+66)
         (/ 0.0 n)
         (if (<= (/ 1.0 n) -2e-11)
           t_0
           (if (<= (/ 1.0 n) -2e-58)
             t_2
             (if (<= (/ 1.0 n) 2e-5)
               t_1
               (if (<= (/ 1.0 n) 5e+190) t_0 t_2)))))))))

C

double code(double x, double n) {
	double t_0 = 1.0 - pow(x, (1.0 / n));
	double t_1 = log(((1.0 + x) / x)) / n;
	double t_2 = 1.0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -5e+201) {
		tmp = t_1;
	} else if ((1.0 / n) <= -5e+140) {
		tmp = t_0;
	} else if ((1.0 / n) <= -5e+66) {
		tmp = 0.0 / n;
	} else if ((1.0 / n) <= -2e-11) {
		tmp = t_0;
	} else if ((1.0 / n) <= -2e-58) {
		tmp = t_2;
	} else if ((1.0 / n) <= 2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = t_0;
	} else {
		tmp = t_2;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: t_2
    real(8) :: tmp
    t_0 = 1.0d0 - (x ** (1.0d0 / n))
    t_1 = log(((1.0d0 + x) / x)) / n
    t_2 = 1.0d0 / (n * x)
    if ((1.0d0 / n) <= (-5d+201)) then
        tmp = t_1
    else if ((1.0d0 / n) <= (-5d+140)) then
        tmp = t_0
    else if ((1.0d0 / n) <= (-5d+66)) then
        tmp = 0.0d0 / n
    else if ((1.0d0 / n) <= (-2d-11)) then
        tmp = t_0
    else if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_2
    else if ((1.0d0 / n) <= 2d-5) then
        tmp = t_1
    else if ((1.0d0 / n) <= 5d+190) then
        tmp = t_0
    else
        tmp = t_2
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = 1.0 - Math.pow(x, (1.0 / n));
	double t_1 = Math.log(((1.0 + x) / x)) / n;
	double t_2 = 1.0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -5e+201) {
		tmp = t_1;
	} else if ((1.0 / n) <= -5e+140) {
		tmp = t_0;
	} else if ((1.0 / n) <= -5e+66) {
		tmp = 0.0 / n;
	} else if ((1.0 / n) <= -2e-11) {
		tmp = t_0;
	} else if ((1.0 / n) <= -2e-58) {
		tmp = t_2;
	} else if ((1.0 / n) <= 2e-5) {
		tmp = t_1;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = t_0;
	} else {
		tmp = t_2;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = 1.0 - math.pow(x, (1.0 / n))
	t_1 = math.log(((1.0 + x) / x)) / n
	t_2 = 1.0 / (n * x)
	tmp = 0
	if (1.0 / n) <= -5e+201:
		tmp = t_1
	elif (1.0 / n) <= -5e+140:
		tmp = t_0
	elif (1.0 / n) <= -5e+66:
		tmp = 0.0 / n
	elif (1.0 / n) <= -2e-11:
		tmp = t_0
	elif (1.0 / n) <= -2e-58:
		tmp = t_2
	elif (1.0 / n) <= 2e-5:
		tmp = t_1
	elif (1.0 / n) <= 5e+190:
		tmp = t_0
	else:
		tmp = t_2
	return tmp

Julia

function code(x, n)
	t_0 = Float64(1.0 - (x ^ Float64(1.0 / n)))
	t_1 = Float64(log(Float64(Float64(1.0 + x) / x)) / n)
	t_2 = Float64(1.0 / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -5e+201)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= -5e+140)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= -5e+66)
		tmp = Float64(0.0 / n);
	elseif (Float64(1.0 / n) <= -2e-11)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= -2e-58)
		tmp = t_2;
	elseif (Float64(1.0 / n) <= 2e-5)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 5e+190)
		tmp = t_0;
	else
		tmp = t_2;
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = 1.0 - (x ^ (1.0 / n));
	t_1 = log(((1.0 + x) / x)) / n;
	t_2 = 1.0 / (n * x);
	tmp = 0.0;
	if ((1.0 / n) <= -5e+201)
		tmp = t_1;
	elseif ((1.0 / n) <= -5e+140)
		tmp = t_0;
	elseif ((1.0 / n) <= -5e+66)
		tmp = 0.0 / n;
	elseif ((1.0 / n) <= -2e-11)
		tmp = t_0;
	elseif ((1.0 / n) <= -2e-58)
		tmp = t_2;
	elseif ((1.0 / n) <= 2e-5)
		tmp = t_1;
	elseif ((1.0 / n) <= 5e+190)
		tmp = t_0;
	else
		tmp = t_2;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision]}, Block[{t$95$2 = N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+201], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+140], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+66], N[(0.0 / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-11], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], t$95$2, If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-5], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e+190], t$95$0, t$95$2]]]]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 1 n) < -4.9999999999999995e201 or -2.0000000000000001e-58 < (/.f64 1 n) < 2.00000000000000016e-5

   1. Initial program 47.1%

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

   2. Taylor expanded in n around inf 82.6%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 82.6%

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

   4. Simplified 82.6%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 82.6%

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

      2. diff-log 82.7%

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

   6. Applied egg-rr 82.7%

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

   if -4.9999999999999995e201 < (/.f64 1 n) < -5.00000000000000008e140 or -4.99999999999999991e66 < (/.f64 1 n) < -1.99999999999999988e-11 or 2.00000000000000016e-5 < (/.f64 1 n) < 5.00000000000000036e190

   1. Initial program 85.1%

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

   2. Taylor expanded in x around 0 66.5%

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

   if -5.00000000000000008e140 < (/.f64 1 n) < -4.99999999999999991e66

   1. Initial program 100.0%

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

   2. Taylor expanded in n around inf 63.0%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 63.0%

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

   4. Simplified 63.0%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 63.0%

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

      2. diff-log 63.0%

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

   6. Applied egg-rr 63.0%

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

   7. Taylor expanded in x around inf 69.7%

      \[\leadsto \frac{\log \color{blue}{1}}{n} \]

   if -1.99999999999999988e-11 < (/.f64 1 n) < -2.0000000000000001e-58 or 5.00000000000000036e190 < (/.f64 1 n)

   1. Initial program 4.3%

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

   2. Taylor expanded in n around inf 12.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 12.8%

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

   4. Simplified 12.8%

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

   5. Taylor expanded in x around inf 90.3%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 90.3%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 90.3%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
3. Recombined 4 regimes into one program.

4. Final simplification 77.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+140}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+66}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-11}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \]

Alternative 7: 66.8% accurate, 1.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ t_1 := \frac{1}{n \cdot x}\\ \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+140}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+66}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-11}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (- 1.0 (pow x (/ 1.0 n)))) (t_1 (/ 1.0 (* n x))))
   (if (<= (/ 1.0 n) -5e+201)
     (/ (log (/ (+ 1.0 x) x)) n)
     (if (<= (/ 1.0 n) -5e+140)
       t_0
       (if (<= (/ 1.0 n) -5e+66)
         (/ 0.0 n)
         (if (<= (/ 1.0 n) -2e-11)
           t_0
           (if (<= (/ 1.0 n) -2e-58)
             t_1
             (if (<= (/ 1.0 n) 2e-5)
               (/ (- (log (/ x (+ 1.0 x)))) n)
               (if (<= (/ 1.0 n) 5e+190) t_0 t_1)))))))))

C

double code(double x, double n) {
	double t_0 = 1.0 - pow(x, (1.0 / n));
	double t_1 = 1.0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -5e+201) {
		tmp = log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= -5e+140) {
		tmp = t_0;
	} else if ((1.0 / n) <= -5e+66) {
		tmp = 0.0 / n;
	} else if ((1.0 / n) <= -2e-11) {
		tmp = t_0;
	} else if ((1.0 / n) <= -2e-58) {
		tmp = t_1;
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = 1.0d0 - (x ** (1.0d0 / n))
    t_1 = 1.0d0 / (n * x)
    if ((1.0d0 / n) <= (-5d+201)) then
        tmp = log(((1.0d0 + x) / x)) / n
    else if ((1.0d0 / n) <= (-5d+140)) then
        tmp = t_0
    else if ((1.0d0 / n) <= (-5d+66)) then
        tmp = 0.0d0 / n
    else if ((1.0d0 / n) <= (-2d-11)) then
        tmp = t_0
    else if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_1
    else if ((1.0d0 / n) <= 2d-5) then
        tmp = -log((x / (1.0d0 + x))) / n
    else if ((1.0d0 / n) <= 5d+190) then
        tmp = t_0
    else
        tmp = t_1
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = 1.0 - Math.pow(x, (1.0 / n));
	double t_1 = 1.0 / (n * x);
	double tmp;
	if ((1.0 / n) <= -5e+201) {
		tmp = Math.log(((1.0 + x) / x)) / n;
	} else if ((1.0 / n) <= -5e+140) {
		tmp = t_0;
	} else if ((1.0 / n) <= -5e+66) {
		tmp = 0.0 / n;
	} else if ((1.0 / n) <= -2e-11) {
		tmp = t_0;
	} else if ((1.0 / n) <= -2e-58) {
		tmp = t_1;
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = 1.0 - math.pow(x, (1.0 / n))
	t_1 = 1.0 / (n * x)
	tmp = 0
	if (1.0 / n) <= -5e+201:
		tmp = math.log(((1.0 + x) / x)) / n
	elif (1.0 / n) <= -5e+140:
		tmp = t_0
	elif (1.0 / n) <= -5e+66:
		tmp = 0.0 / n
	elif (1.0 / n) <= -2e-11:
		tmp = t_0
	elif (1.0 / n) <= -2e-58:
		tmp = t_1
	elif (1.0 / n) <= 2e-5:
		tmp = -math.log((x / (1.0 + x))) / n
	elif (1.0 / n) <= 5e+190:
		tmp = t_0
	else:
		tmp = t_1
	return tmp

Julia

function code(x, n)
	t_0 = Float64(1.0 - (x ^ Float64(1.0 / n)))
	t_1 = Float64(1.0 / Float64(n * x))
	tmp = 0.0
	if (Float64(1.0 / n) <= -5e+201)
		tmp = Float64(log(Float64(Float64(1.0 + x) / x)) / n);
	elseif (Float64(1.0 / n) <= -5e+140)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= -5e+66)
		tmp = Float64(0.0 / n);
	elseif (Float64(1.0 / n) <= -2e-11)
		tmp = t_0;
	elseif (Float64(1.0 / n) <= -2e-58)
		tmp = t_1;
	elseif (Float64(1.0 / n) <= 2e-5)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	elseif (Float64(1.0 / n) <= 5e+190)
		tmp = t_0;
	else
		tmp = t_1;
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = 1.0 - (x ^ (1.0 / n));
	t_1 = 1.0 / (n * x);
	tmp = 0.0;
	if ((1.0 / n) <= -5e+201)
		tmp = log(((1.0 + x) / x)) / n;
	elseif ((1.0 / n) <= -5e+140)
		tmp = t_0;
	elseif ((1.0 / n) <= -5e+66)
		tmp = 0.0 / n;
	elseif ((1.0 / n) <= -2e-11)
		tmp = t_0;
	elseif ((1.0 / n) <= -2e-58)
		tmp = t_1;
	elseif ((1.0 / n) <= 2e-5)
		tmp = -log((x / (1.0 + x))) / n;
	elseif ((1.0 / n) <= 5e+190)
		tmp = t_0;
	else
		tmp = t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+201], N[(N[Log[N[(N[(1.0 + x), $MachinePrecision] / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+140], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], -5e+66], N[(0.0 / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-11], t$95$0, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], t$95$1, If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-5], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e+190], t$95$0, t$95$1]]]]]]]]]

Derivation

1. Split input into 5 regimes

2. if (/.f64 1 n) < -4.9999999999999995e201

   1. Initial program 100.0%

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

   2. Taylor expanded in n around inf 64.6%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 64.6%

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

   4. Simplified 64.6%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 64.6%

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

      2. diff-log 64.6%

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

   6. Applied egg-rr 64.6%

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

   if -4.9999999999999995e201 < (/.f64 1 n) < -5.00000000000000008e140 or -4.99999999999999991e66 < (/.f64 1 n) < -1.99999999999999988e-11 or 2.00000000000000016e-5 < (/.f64 1 n) < 5.00000000000000036e190

   1. Initial program 85.1%

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

   2. Taylor expanded in x around 0 66.5%

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

   if -5.00000000000000008e140 < (/.f64 1 n) < -4.99999999999999991e66

   1. Initial program 100.0%

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

   2. Taylor expanded in n around inf 63.0%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 63.0%

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

   4. Simplified 63.0%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 63.0%

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

      2. diff-log 63.0%

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

   6. Applied egg-rr 63.0%

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

   7. Taylor expanded in x around inf 69.7%

      \[\leadsto \frac{\log \color{blue}{1}}{n} \]

   if -1.99999999999999988e-11 < (/.f64 1 n) < -2.0000000000000001e-58 or 5.00000000000000036e190 < (/.f64 1 n)

   1. Initial program 4.3%

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

   2. Taylor expanded in n around inf 12.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 12.8%

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

   4. Simplified 12.8%

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

   5. Taylor expanded in x around inf 90.3%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 90.3%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 90.3%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2.00000000000000016e-5

   1. Initial program 37.6%

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

   2. Taylor expanded in n around inf 85.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 85.8%

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

   4. Simplified 85.8%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 85.8%

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

      2. diff-log 85.9%

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

   6. Applied egg-rr 85.9%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 85.9%

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

      2. log-rec 86.0%

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

   8. Applied egg-rr 86.0%

      \[\leadsto \frac{\color{blue}{-\log \left(\frac{x}{1 + x}\right)}}{n} \]
3. Recombined 5 regimes into one program.

4. Final simplification 77.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -5 \cdot 10^{+201}:\\ \;\;\;\;\frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+140}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -5 \cdot 10^{+66}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-11}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \]

Alternative 8: 81.4% accurate, 1.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-14)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (if (<= (/ 1.0 n) 5e+190) (- (+ 1.0 (/ x n)) t_0) (/ 1.0 (* n x)))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = 1.0 / (n * x);
	}
	return tmp;
}

Fortran

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 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 2d-14) then
        tmp = -log((x / (1.0d0 + x))) / n
    else if ((1.0d0 / n) <= 5d+190) then
        tmp = (1.0d0 + (x / n)) - t_0
    else
        tmp = 1.0d0 / (n * x)
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-14) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = (1.0 + (x / n)) - t_0;
	} else {
		tmp = 1.0 / (n * x);
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-14:
		tmp = -math.log((x / (1.0 + x))) / n
	elif (1.0 / n) <= 5e+190:
		tmp = (1.0 + (x / n)) - t_0
	else:
		tmp = 1.0 / (n * x)
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-14)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	elseif (Float64(1.0 / n) <= 5e+190)
		tmp = Float64(Float64(1.0 + Float64(x / n)) - t_0);
	else
		tmp = Float64(1.0 / Float64(n * x));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-58)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 2e-14)
		tmp = -log((x / (1.0 + x))) / n;
	elseif ((1.0 / n) <= 5e+190)
		tmp = (1.0 + (x / n)) - t_0;
	else
		tmp = 1.0 / (n * x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-14], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e+190], N[(N[(1.0 + N[(x / n), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision], N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2e-14

   1. Initial program 38.2%

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

   2. Taylor expanded in n around inf 87.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 87.3%

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

   4. Simplified 87.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 87.3%

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

      2. diff-log 87.3%

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

   6. Applied egg-rr 87.3%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 87.3%

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

      2. log-rec 87.4%

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

   8. Applied egg-rr 87.4%

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

   if 2e-14 < (/.f64 1 n) < 5.00000000000000036e190

   1. Initial program 59.5%

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

   2. Taylor expanded in x around 0 59.3%

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

   if 5.00000000000000036e190 < (/.f64 1 n)

   1. Initial program 3.1%

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

   2. Taylor expanded in n around inf 8.6%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 8.6%

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

   4. Simplified 8.6%

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

   5. Taylor expanded in x around inf 91.7%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 91.7%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 91.7%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
3. Recombined 4 regimes into one program.

4. Final simplification 88.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \]

Alternative 9: 56.0% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{-\log x}{n}\\ t_1 := 1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;x \leq 5.7 \cdot 10^{-216}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 10^{-193}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 7 \cdot 10^{-181}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 2.1 \cdot 10^{-87}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 4.6 \cdot 10^{-49}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 31500:\\ \;\;\;\;t_1\\ \mathbf{elif}\;x \leq 1.45 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (- (log x)) n)) (t_1 (- 1.0 (pow x (/ 1.0 n)))))
   (if (<= x 5.7e-216)
     t_0
     (if (<= x 1e-193)
       t_1
       (if (<= x 7e-181)
         t_0
         (if (<= x 2.1e-87)
           t_1
           (if (<= x 4.6e-49)
             t_0
             (if (<= x 31500.0)
               t_1
               (if (<= x 1.45e+100) (/ (/ 1.0 x) n) (/ 0.0 n))))))))))

C

double code(double x, double n) {
	double t_0 = -log(x) / n;
	double t_1 = 1.0 - pow(x, (1.0 / n));
	double tmp;
	if (x <= 5.7e-216) {
		tmp = t_0;
	} else if (x <= 1e-193) {
		tmp = t_1;
	} else if (x <= 7e-181) {
		tmp = t_0;
	} else if (x <= 2.1e-87) {
		tmp = t_1;
	} else if (x <= 4.6e-49) {
		tmp = t_0;
	} else if (x <= 31500.0) {
		tmp = t_1;
	} else if (x <= 1.45e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = -log(x) / n
    t_1 = 1.0d0 - (x ** (1.0d0 / n))
    if (x <= 5.7d-216) then
        tmp = t_0
    else if (x <= 1d-193) then
        tmp = t_1
    else if (x <= 7d-181) then
        tmp = t_0
    else if (x <= 2.1d-87) then
        tmp = t_1
    else if (x <= 4.6d-49) then
        tmp = t_0
    else if (x <= 31500.0d0) then
        tmp = t_1
    else if (x <= 1.45d+100) then
        tmp = (1.0d0 / x) / n
    else
        tmp = 0.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = -Math.log(x) / n;
	double t_1 = 1.0 - Math.pow(x, (1.0 / n));
	double tmp;
	if (x <= 5.7e-216) {
		tmp = t_0;
	} else if (x <= 1e-193) {
		tmp = t_1;
	} else if (x <= 7e-181) {
		tmp = t_0;
	} else if (x <= 2.1e-87) {
		tmp = t_1;
	} else if (x <= 4.6e-49) {
		tmp = t_0;
	} else if (x <= 31500.0) {
		tmp = t_1;
	} else if (x <= 1.45e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = -math.log(x) / n
	t_1 = 1.0 - math.pow(x, (1.0 / n))
	tmp = 0
	if x <= 5.7e-216:
		tmp = t_0
	elif x <= 1e-193:
		tmp = t_1
	elif x <= 7e-181:
		tmp = t_0
	elif x <= 2.1e-87:
		tmp = t_1
	elif x <= 4.6e-49:
		tmp = t_0
	elif x <= 31500.0:
		tmp = t_1
	elif x <= 1.45e+100:
		tmp = (1.0 / x) / n
	else:
		tmp = 0.0 / n
	return tmp

Julia

function code(x, n)
	t_0 = Float64(Float64(-log(x)) / n)
	t_1 = Float64(1.0 - (x ^ Float64(1.0 / n)))
	tmp = 0.0
	if (x <= 5.7e-216)
		tmp = t_0;
	elseif (x <= 1e-193)
		tmp = t_1;
	elseif (x <= 7e-181)
		tmp = t_0;
	elseif (x <= 2.1e-87)
		tmp = t_1;
	elseif (x <= 4.6e-49)
		tmp = t_0;
	elseif (x <= 31500.0)
		tmp = t_1;
	elseif (x <= 1.45e+100)
		tmp = Float64(Float64(1.0 / x) / n);
	else
		tmp = Float64(0.0 / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = -log(x) / n;
	t_1 = 1.0 - (x ^ (1.0 / n));
	tmp = 0.0;
	if (x <= 5.7e-216)
		tmp = t_0;
	elseif (x <= 1e-193)
		tmp = t_1;
	elseif (x <= 7e-181)
		tmp = t_0;
	elseif (x <= 2.1e-87)
		tmp = t_1;
	elseif (x <= 4.6e-49)
		tmp = t_0;
	elseif (x <= 31500.0)
		tmp = t_1;
	elseif (x <= 1.45e+100)
		tmp = (1.0 / x) / n;
	else
		tmp = 0.0 / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[((-N[Log[x], $MachinePrecision]) / n), $MachinePrecision]}, Block[{t$95$1 = N[(1.0 - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[x, 5.7e-216], t$95$0, If[LessEqual[x, 1e-193], t$95$1, If[LessEqual[x, 7e-181], t$95$0, If[LessEqual[x, 2.1e-87], t$95$1, If[LessEqual[x, 4.6e-49], t$95$0, If[LessEqual[x, 31500.0], t$95$1, If[LessEqual[x, 1.45e+100], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(0.0 / n), $MachinePrecision]]]]]]]]]]

Derivation

1. Split input into 4 regimes

2. if x < 5.70000000000000004e-216 or 1e-193 < x < 6.99999999999999993e-181 or 2.10000000000000007e-87 < x < 4.5999999999999998e-49

   1. Initial program 28.3%

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

   2. Taylor expanded in x around 0 28.3%

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

   3. Taylor expanded in n around inf 65.6%

      \[\leadsto \color{blue}{-1 \cdot \frac{\log x}{n}} \]
   4. Step-by-step derivation

      1. neg-mul-1 65.6%

         \[\leadsto \color{blue}{-\frac{\log x}{n}} \]

      2. distribute-neg-frac 65.6%

         \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   5. Simplified 65.6%

      \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   if 5.70000000000000004e-216 < x < 1e-193 or 6.99999999999999993e-181 < x < 2.10000000000000007e-87 or 4.5999999999999998e-49 < x < 31500

   1. Initial program 63.5%

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

   2. Taylor expanded in x around 0 58.7%

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

   if 31500 < x < 1.45e100

   1. Initial program 37.3%

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

   2. Taylor expanded in n around inf 40.4%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 40.4%

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

   4. Simplified 40.4%

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

   5. Taylor expanded in x around inf 75.2%

      \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

   if 1.45e100 < x

   1. Initial program 86.3%

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

   2. Taylor expanded in n around inf 86.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 86.3%

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

   4. Simplified 86.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 86.3%

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

      2. diff-log 86.3%

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

   6. Applied egg-rr 86.3%

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

   7. Taylor expanded in x around inf 86.3%

      \[\leadsto \frac{\log \color{blue}{1}}{n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 70.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 5.7 \cdot 10^{-216}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 10^{-193}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 7 \cdot 10^{-181}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.1 \cdot 10^{-87}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 4.6 \cdot 10^{-49}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 31500:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;x \leq 1.45 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 10: 81.2% accurate, 1.8× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{t_0}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;1 - t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -2e-58)
     (/ t_0 (* n x))
     (if (<= (/ 1.0 n) 2e-5)
       (/ (- (log (/ x (+ 1.0 x)))) n)
       (if (<= (/ 1.0 n) 5e+190) (- 1.0 t_0) (/ 1.0 (* n x)))))))

C

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = 1.0 - t_0;
	} else {
		tmp = 1.0 / (n * x);
	}
	return tmp;
}

Fortran

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 = x ** (1.0d0 / n)
    if ((1.0d0 / n) <= (-2d-58)) then
        tmp = t_0 / (n * x)
    else if ((1.0d0 / n) <= 2d-5) then
        tmp = -log((x / (1.0d0 + x))) / n
    else if ((1.0d0 / n) <= 5d+190) then
        tmp = 1.0d0 - t_0
    else
        tmp = 1.0d0 / (n * x)
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -2e-58) {
		tmp = t_0 / (n * x);
	} else if ((1.0 / n) <= 2e-5) {
		tmp = -Math.log((x / (1.0 + x))) / n;
	} else if ((1.0 / n) <= 5e+190) {
		tmp = 1.0 - t_0;
	} else {
		tmp = 1.0 / (n * x);
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -2e-58:
		tmp = t_0 / (n * x)
	elif (1.0 / n) <= 2e-5:
		tmp = -math.log((x / (1.0 + x))) / n
	elif (1.0 / n) <= 5e+190:
		tmp = 1.0 - t_0
	else:
		tmp = 1.0 / (n * x)
	return tmp

Julia

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -2e-58)
		tmp = Float64(t_0 / Float64(n * x));
	elseif (Float64(1.0 / n) <= 2e-5)
		tmp = Float64(Float64(-log(Float64(x / Float64(1.0 + x)))) / n);
	elseif (Float64(1.0 / n) <= 5e+190)
		tmp = Float64(1.0 - t_0);
	else
		tmp = Float64(1.0 / Float64(n * x));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = x ^ (1.0 / n);
	tmp = 0.0;
	if ((1.0 / n) <= -2e-58)
		tmp = t_0 / (n * x);
	elseif ((1.0 / n) <= 2e-5)
		tmp = -log((x / (1.0 + x))) / n;
	elseif ((1.0 / n) <= 5e+190)
		tmp = 1.0 - t_0;
	else
		tmp = 1.0 / (n * x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -2e-58], N[(t$95$0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 2e-5], N[((-N[Log[N[(x / N[(1.0 + x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 5e+190], N[(1.0 - t$95$0), $MachinePrecision], N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 1 n) < -2.0000000000000001e-58

   1. Initial program 86.6%

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

   2. Taylor expanded in x around inf 97.7%

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
   3. Step-by-step derivation

      1. mul-1-neg 97.7%

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

      2. log-rec 97.7%

         \[\leadsto \frac{e^{-\frac{\color{blue}{-\log x}}{n}}}{n \cdot x} \]

      3. mul-1-neg 97.7%

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

      4. distribute-neg-frac 97.7%

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

      5. mul-1-neg 97.7%

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

      6. remove-double-neg 97.7%

         \[\leadsto \frac{e^{\frac{\color{blue}{\log x}}{n}}}{n \cdot x} \]

      7. *-commutative 97.7%

         \[\leadsto \frac{e^{\frac{\log x}{n}}}{\color{blue}{x \cdot n}} \]

   4. Simplified 97.7%

      \[\leadsto \color{blue}{\frac{e^{\frac{\log x}{n}}}{x \cdot n}} \]
   5. Step-by-step derivation

      1. div-inv 97.7%

         \[\leadsto \color{blue}{e^{\frac{\log x}{n}} \cdot \frac{1}{x \cdot n}} \]

      2. div-inv 97.7%

         \[\leadsto e^{\color{blue}{\log x \cdot \frac{1}{n}}} \cdot \frac{1}{x \cdot n} \]

      3. pow-to-exp 97.7%

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

   6. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{{x}^{\left(\frac{1}{n}\right)} \cdot \frac{1}{x \cdot n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.7%

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

      2. *-rgt-identity 97.7%

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

   8. Simplified 97.7%

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

   if -2.0000000000000001e-58 < (/.f64 1 n) < 2.00000000000000016e-5

   1. Initial program 37.6%

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

   2. Taylor expanded in n around inf 85.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 85.8%

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

   4. Simplified 85.8%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 85.8%

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

      2. diff-log 85.9%

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

   6. Applied egg-rr 85.9%

      \[\leadsto \frac{\color{blue}{\log \left(\frac{1 + x}{x}\right)}}{n} \]
   7. Step-by-step derivation

      1. clear-num 85.9%

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

      2. log-rec 86.0%

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

   8. Applied egg-rr 86.0%

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

   if 2.00000000000000016e-5 < (/.f64 1 n) < 5.00000000000000036e190

   1. Initial program 63.7%

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

   2. Taylor expanded in x around 0 60.1%

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

   if 5.00000000000000036e190 < (/.f64 1 n)

   1. Initial program 3.1%

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

   2. Taylor expanded in n around inf 8.6%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 8.6%

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

   4. Simplified 8.6%

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

   5. Taylor expanded in x around inf 91.7%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 91.7%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 91.7%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
3. Recombined 4 regimes into one program.

4. Final simplification 88.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-58}:\\ \;\;\;\;\frac{{x}^{\left(\frac{1}{n}\right)}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-5}:\\ \;\;\;\;\frac{-\log \left(\frac{x}{1 + x}\right)}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 5 \cdot 10^{+190}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \end{array} \]

Alternative 11: 59.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 5.5 \cdot 10^{-166}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.8 \cdot 10^{-140}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;x \leq 1.8 \cdot 10^{-9}:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 5.3 \cdot 10^{+99}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (if (<= x 5.5e-166)
   (/ (- (log x)) n)
   (if (<= x 3.8e-140)
     (/ 1.0 (* n x))
     (if (<= x 1.8e-9)
       (/ (- x (log x)) n)
       (if (<= x 5.3e+99) (/ (/ 1.0 x) n) (/ 0.0 n))))))

C

double code(double x, double n) {
	double tmp;
	if (x <= 5.5e-166) {
		tmp = -log(x) / n;
	} else if (x <= 3.8e-140) {
		tmp = 1.0 / (n * x);
	} else if (x <= 1.8e-9) {
		tmp = (x - log(x)) / n;
	} else if (x <= 5.3e+99) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (x <= 5.5d-166) then
        tmp = -log(x) / n
    else if (x <= 3.8d-140) then
        tmp = 1.0d0 / (n * x)
    else if (x <= 1.8d-9) then
        tmp = (x - log(x)) / n
    else if (x <= 5.3d+99) then
        tmp = (1.0d0 / x) / n
    else
        tmp = 0.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double tmp;
	if (x <= 5.5e-166) {
		tmp = -Math.log(x) / n;
	} else if (x <= 3.8e-140) {
		tmp = 1.0 / (n * x);
	} else if (x <= 1.8e-9) {
		tmp = (x - Math.log(x)) / n;
	} else if (x <= 5.3e+99) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Python

def code(x, n):
	tmp = 0
	if x <= 5.5e-166:
		tmp = -math.log(x) / n
	elif x <= 3.8e-140:
		tmp = 1.0 / (n * x)
	elif x <= 1.8e-9:
		tmp = (x - math.log(x)) / n
	elif x <= 5.3e+99:
		tmp = (1.0 / x) / n
	else:
		tmp = 0.0 / n
	return tmp

Julia

function code(x, n)
	tmp = 0.0
	if (x <= 5.5e-166)
		tmp = Float64(Float64(-log(x)) / n);
	elseif (x <= 3.8e-140)
		tmp = Float64(1.0 / Float64(n * x));
	elseif (x <= 1.8e-9)
		tmp = Float64(Float64(x - log(x)) / n);
	elseif (x <= 5.3e+99)
		tmp = Float64(Float64(1.0 / x) / n);
	else
		tmp = Float64(0.0 / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (x <= 5.5e-166)
		tmp = -log(x) / n;
	elseif (x <= 3.8e-140)
		tmp = 1.0 / (n * x);
	elseif (x <= 1.8e-9)
		tmp = (x - log(x)) / n;
	elseif (x <= 5.3e+99)
		tmp = (1.0 / x) / n;
	else
		tmp = 0.0 / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := If[LessEqual[x, 5.5e-166], N[((-N[Log[x], $MachinePrecision]) / n), $MachinePrecision], If[LessEqual[x, 3.8e-140], N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[x, 1.8e-9], N[(N[(x - N[Log[x], $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision], If[LessEqual[x, 5.3e+99], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(0.0 / n), $MachinePrecision]]]]]

Derivation

1. Split input into 5 regimes

2. if x < 5.4999999999999997e-166

   1. Initial program 41.9%

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

   2. Taylor expanded in x around 0 41.9%

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

   3. Taylor expanded in n around inf 57.7%

      \[\leadsto \color{blue}{-1 \cdot \frac{\log x}{n}} \]
   4. Step-by-step derivation

      1. neg-mul-1 57.7%

         \[\leadsto \color{blue}{-\frac{\log x}{n}} \]

      2. distribute-neg-frac 57.7%

         \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   5. Simplified 57.7%

      \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   if 5.4999999999999997e-166 < x < 3.79999999999999998e-140

   1. Initial program 66.1%

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

   2. Taylor expanded in n around inf 13.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 13.1%

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

   4. Simplified 13.1%

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

   5. Taylor expanded in x around inf 56.2%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 56.2%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 56.2%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]

   if 3.79999999999999998e-140 < x < 1.8e-9

   1. Initial program 45.3%

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

   2. Taylor expanded in n around inf 44.0%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 44.0%

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

   4. Simplified 44.0%

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

   5. Taylor expanded in x around 0 44.0%

      \[\leadsto \frac{\color{blue}{x + -1 \cdot \log x}}{n} \]
   6. Step-by-step derivation

      1. neg-mul-1 44.0%

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

      2. unsub-neg 44.0%

         \[\leadsto \frac{\color{blue}{x - \log x}}{n} \]

   7. Simplified 44.0%

      \[\leadsto \frac{\color{blue}{x - \log x}}{n} \]

   if 1.8e-9 < x < 5.30000000000000034e99

   1. Initial program 46.8%

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

   2. Taylor expanded in n around inf 34.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 34.8%

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

   4. Simplified 34.8%

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

   5. Taylor expanded in x around inf 64.3%

      \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

   if 5.30000000000000034e99 < x

   1. Initial program 86.3%

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

   2. Taylor expanded in n around inf 86.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 86.3%

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

   4. Simplified 86.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 86.3%

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

      2. diff-log 86.3%

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

   6. Applied egg-rr 86.3%

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

   7. Taylor expanded in x around inf 86.3%

      \[\leadsto \frac{\log \color{blue}{1}}{n} \]
3. Recombined 5 regimes into one program.

4. Final simplification 63.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 5.5 \cdot 10^{-166}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.8 \cdot 10^{-140}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;x \leq 1.8 \cdot 10^{-9}:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 5.3 \cdot 10^{+99}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 12: 59.5% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{-\log x}{n}\\ \mathbf{if}\;x \leq 3.8 \cdot 10^{-167}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 3.8 \cdot 10^{-140}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;x \leq 1.8 \cdot 10^{-9}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 2.2 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (/ (- (log x)) n)))
   (if (<= x 3.8e-167)
     t_0
     (if (<= x 3.8e-140)
       (/ 1.0 (* n x))
       (if (<= x 1.8e-9)
         t_0
         (if (<= x 2.2e+100) (/ (/ 1.0 x) n) (/ 0.0 n)))))))

C

double code(double x, double n) {
	double t_0 = -log(x) / n;
	double tmp;
	if (x <= 3.8e-167) {
		tmp = t_0;
	} else if (x <= 3.8e-140) {
		tmp = 1.0 / (n * x);
	} else if (x <= 1.8e-9) {
		tmp = t_0;
	} else if (x <= 2.2e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Fortran

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(x) / n
    if (x <= 3.8d-167) then
        tmp = t_0
    else if (x <= 3.8d-140) then
        tmp = 1.0d0 / (n * x)
    else if (x <= 1.8d-9) then
        tmp = t_0
    else if (x <= 2.2d+100) then
        tmp = (1.0d0 / x) / n
    else
        tmp = 0.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double t_0 = -Math.log(x) / n;
	double tmp;
	if (x <= 3.8e-167) {
		tmp = t_0;
	} else if (x <= 3.8e-140) {
		tmp = 1.0 / (n * x);
	} else if (x <= 1.8e-9) {
		tmp = t_0;
	} else if (x <= 2.2e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Python

def code(x, n):
	t_0 = -math.log(x) / n
	tmp = 0
	if x <= 3.8e-167:
		tmp = t_0
	elif x <= 3.8e-140:
		tmp = 1.0 / (n * x)
	elif x <= 1.8e-9:
		tmp = t_0
	elif x <= 2.2e+100:
		tmp = (1.0 / x) / n
	else:
		tmp = 0.0 / n
	return tmp

Julia

function code(x, n)
	t_0 = Float64(Float64(-log(x)) / n)
	tmp = 0.0
	if (x <= 3.8e-167)
		tmp = t_0;
	elseif (x <= 3.8e-140)
		tmp = Float64(1.0 / Float64(n * x));
	elseif (x <= 1.8e-9)
		tmp = t_0;
	elseif (x <= 2.2e+100)
		tmp = Float64(Float64(1.0 / x) / n);
	else
		tmp = Float64(0.0 / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	t_0 = -log(x) / n;
	tmp = 0.0;
	if (x <= 3.8e-167)
		tmp = t_0;
	elseif (x <= 3.8e-140)
		tmp = 1.0 / (n * x);
	elseif (x <= 1.8e-9)
		tmp = t_0;
	elseif (x <= 2.2e+100)
		tmp = (1.0 / x) / n;
	else
		tmp = 0.0 / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := Block[{t$95$0 = N[((-N[Log[x], $MachinePrecision]) / n), $MachinePrecision]}, If[LessEqual[x, 3.8e-167], t$95$0, If[LessEqual[x, 3.8e-140], N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision], If[LessEqual[x, 1.8e-9], t$95$0, If[LessEqual[x, 2.2e+100], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(0.0 / n), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if x < 3.79999999999999967e-167 or 3.79999999999999998e-140 < x < 1.8e-9

   1. Initial program 43.7%

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

   2. Taylor expanded in x around 0 43.7%

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

   3. Taylor expanded in n around inf 50.3%

      \[\leadsto \color{blue}{-1 \cdot \frac{\log x}{n}} \]
   4. Step-by-step derivation

      1. neg-mul-1 50.3%

         \[\leadsto \color{blue}{-\frac{\log x}{n}} \]

      2. distribute-neg-frac 50.3%

         \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   5. Simplified 50.3%

      \[\leadsto \color{blue}{\frac{-\log x}{n}} \]

   if 3.79999999999999967e-167 < x < 3.79999999999999998e-140

   1. Initial program 66.1%

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

   2. Taylor expanded in n around inf 13.1%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 13.1%

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

   4. Simplified 13.1%

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

   5. Taylor expanded in x around inf 56.2%

      \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
   6. Step-by-step derivation

      1. *-commutative 56.2%

         \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

   7. Simplified 56.2%

      \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]

   if 1.8e-9 < x < 2.2000000000000001e100

   1. Initial program 46.8%

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

   2. Taylor expanded in n around inf 34.8%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 34.8%

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

   4. Simplified 34.8%

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

   5. Taylor expanded in x around inf 64.3%

      \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

   if 2.2000000000000001e100 < x

   1. Initial program 86.3%

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

   2. Taylor expanded in n around inf 86.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 86.3%

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

   4. Simplified 86.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 86.3%

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

      2. diff-log 86.3%

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

   6. Applied egg-rr 86.3%

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

   7. Taylor expanded in x around inf 86.3%

      \[\leadsto \frac{\log \color{blue}{1}}{n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 63.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 3.8 \cdot 10^{-167}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.8 \cdot 10^{-140}:\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{elif}\;x \leq 1.8 \cdot 10^{-9}:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 2.2 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 13: 44.8% accurate, 29.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 9 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (if (<= x 9e+100) (/ (/ 1.0 x) n) (/ 0.0 n)))

C

double code(double x, double n) {
	double tmp;
	if (x <= 9e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: tmp
    if (x <= 9d+100) then
        tmp = (1.0d0 / x) / n
    else
        tmp = 0.0d0 / n
    end if
    code = tmp
end function

Java

public static double code(double x, double n) {
	double tmp;
	if (x <= 9e+100) {
		tmp = (1.0 / x) / n;
	} else {
		tmp = 0.0 / n;
	}
	return tmp;
}

Python

def code(x, n):
	tmp = 0
	if x <= 9e+100:
		tmp = (1.0 / x) / n
	else:
		tmp = 0.0 / n
	return tmp

Julia

function code(x, n)
	tmp = 0.0
	if (x <= 9e+100)
		tmp = Float64(Float64(1.0 / x) / n);
	else
		tmp = Float64(0.0 / n);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x, n)
	tmp = 0.0;
	if (x <= 9e+100)
		tmp = (1.0 / x) / n;
	else
		tmp = 0.0 / n;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, n_] := If[LessEqual[x, 9e+100], N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision], N[(0.0 / n), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 9.00000000000000073e100

   1. Initial program 45.9%

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

   2. Taylor expanded in n around inf 44.7%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 44.7%

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

   4. Simplified 44.7%

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

   5. Taylor expanded in x around inf 31.9%

      \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

   if 9.00000000000000073e100 < x

   1. Initial program 86.3%

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

   2. Taylor expanded in n around inf 86.3%

      \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
   3. Step-by-step derivation

      1. log1p-def 86.3%

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

   4. Simplified 86.3%

      \[\leadsto \color{blue}{\frac{\mathsf{log1p}\left(x\right) - \log x}{n}} \]
   5. Step-by-step derivation

      1. log1p-udef 86.3%

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

      2. diff-log 86.3%

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

   6. Applied egg-rr 86.3%

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

   7. Taylor expanded in x around inf 86.3%

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

4. Final simplification 48.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 9 \cdot 10^{+100}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 14: 40.3% accurate, 42.2× speedup

Math

\[\begin{array}{l} \\ \frac{1}{n \cdot x} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ 1.0 (* n x)))

C

double code(double x, double n) {
	return 1.0 / (n * x);
}

Fortran

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

Java

public static double code(double x, double n) {
	return 1.0 / (n * x);
}

Python

def code(x, n):
	return 1.0 / (n * x)

Julia

function code(x, n)
	return Float64(1.0 / Float64(n * x))
end

MATLAB

function tmp = code(x, n)
	tmp = 1.0 / (n * x);
end

Wolfram

code[x_, n_] := N[(1.0 / N[(n * x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 58.5%

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

2. Taylor expanded in n around inf 57.7%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
3. Step-by-step derivation

   1. log1p-def 57.7%

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

4. Simplified 57.7%

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

5. Taylor expanded in x around inf 39.9%

   \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
6. Step-by-step derivation

   1. *-commutative 39.9%

      \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

7. Simplified 39.9%

   \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]

8. Final simplification 39.9%

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

Alternative 15: 40.7% accurate, 42.2× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{1}{n}}{x} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ (/ 1.0 n) x))

C

double code(double x, double n) {
	return (1.0 / n) / x;
}

Fortran

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

Java

public static double code(double x, double n) {
	return (1.0 / n) / x;
}

Python

def code(x, n):
	return (1.0 / n) / x

Julia

function code(x, n)
	return Float64(Float64(1.0 / n) / x)
end

MATLAB

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

Wolfram

code[x_, n_] := N[(N[(1.0 / n), $MachinePrecision] / x), $MachinePrecision]

Derivation

1. Initial program 58.5%

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

2. Taylor expanded in n around inf 57.7%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
3. Step-by-step derivation

   1. log1p-def 57.7%

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

4. Simplified 57.7%

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

5. Taylor expanded in x around inf 39.9%

   \[\leadsto \color{blue}{\frac{1}{n \cdot x}} \]
6. Step-by-step derivation

   1. *-commutative 39.9%

      \[\leadsto \frac{1}{\color{blue}{x \cdot n}} \]

7. Simplified 39.9%

   \[\leadsto \color{blue}{\frac{1}{x \cdot n}} \]
8. Step-by-step derivation

   1. log1p-expm1-u 52.0%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{1}{x \cdot n}\right)\right)} \]

   2. *-commutative 52.0%

      \[\leadsto \mathsf{log1p}\left(\mathsf{expm1}\left(\frac{1}{\color{blue}{n \cdot x}}\right)\right) \]

   3. associate-/r* 52.5%

      \[\leadsto \mathsf{log1p}\left(\mathsf{expm1}\left(\color{blue}{\frac{\frac{1}{n}}{x}}\right)\right) \]

9. Applied egg-rr 52.5%

   \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\frac{\frac{1}{n}}{x}\right)\right)} \]
10. Step-by-step derivation

    1. log1p-expm1-u 40.4%

       \[\leadsto \color{blue}{\frac{\frac{1}{n}}{x}} \]

11. Applied egg-rr 40.4%

    \[\leadsto \color{blue}{\frac{\frac{1}{n}}{x}} \]

12. Final simplification 40.4%

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

Alternative 16: 40.7% accurate, 42.2× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{1}{x}}{n} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ (/ 1.0 x) n))

C

double code(double x, double n) {
	return (1.0 / x) / n;
}

Fortran

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

Java

public static double code(double x, double n) {
	return (1.0 / x) / n;
}

Python

def code(x, n):
	return (1.0 / x) / n

Julia

function code(x, n)
	return Float64(Float64(1.0 / x) / n)
end

MATLAB

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

Wolfram

code[x_, n_] := N[(N[(1.0 / x), $MachinePrecision] / n), $MachinePrecision]

Derivation

1. Initial program 58.5%

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

2. Taylor expanded in n around inf 57.7%

   \[\leadsto \color{blue}{\frac{\log \left(1 + x\right) - \log x}{n}} \]
3. Step-by-step derivation

   1. log1p-def 57.7%

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

4. Simplified 57.7%

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

5. Taylor expanded in x around inf 40.4%

   \[\leadsto \frac{\color{blue}{\frac{1}{x}}}{n} \]

6. Final simplification 40.4%

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

Alternative 17: 4.5% accurate, 70.3× speedup

Math

\[\begin{array}{l} \\ \frac{x}{n} \end{array} \]

FPCore

(FPCore (x n) :precision binary64 (/ x n))

C

double code(double x, double n) {
	return x / n;
}

Fortran

real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = x / n
end function

Java

public static double code(double x, double n) {
	return x / n;
}

Python

def code(x, n):
	return x / n

Julia

function code(x, n)
	return Float64(x / n)
end

MATLAB

function tmp = code(x, n)
	tmp = x / n;
end

Wolfram

code[x_, n_] := N[(x / n), $MachinePrecision]

Derivation

1. Initial program 58.5%

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

2. Taylor expanded in x around 0 33.9%

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

3. Taylor expanded in x around inf 4.3%

   \[\leadsto \color{blue}{\frac{x}{n}} \]

4. Final simplification 4.3%

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

Reproduce

herbie shell --seed 2023320 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))