2nthrt (problem 3.4.6)

Average Error: 50.59% → 2.59%

Time: 20.2s

Precision: binary64

Cost: 13764

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0002:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - t_0\\ \end{array} \]

FPCore

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

↓

(FPCore (x n)
 :precision binary64
 (let* ((t_0 (pow x (/ 1.0 n))))
   (if (<= (/ 1.0 n) -4e-9)
     (- (pow (+ 1.0 x) (/ 1.0 n)) t_0)
     (if (<= (/ 1.0 n) 0.0002) (/ (log1p (/ 1.0 x)) n) (- 1.0 t_0)))))

C

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

↓

double code(double x, double n) {
	double t_0 = pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -4e-9) {
		tmp = pow((1.0 + x), (1.0 / n)) - t_0;
	} else if ((1.0 / n) <= 0.0002) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = 1.0 - t_0;
	}
	return tmp;
}

Java

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

↓

public static double code(double x, double n) {
	double t_0 = Math.pow(x, (1.0 / n));
	double tmp;
	if ((1.0 / n) <= -4e-9) {
		tmp = Math.pow((1.0 + x), (1.0 / n)) - t_0;
	} else if ((1.0 / n) <= 0.0002) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = 1.0 - t_0;
	}
	return tmp;
}

Python

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

↓

def code(x, n):
	t_0 = math.pow(x, (1.0 / n))
	tmp = 0
	if (1.0 / n) <= -4e-9:
		tmp = math.pow((1.0 + x), (1.0 / n)) - t_0
	elif (1.0 / n) <= 0.0002:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = 1.0 - t_0
	return tmp

Julia

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

↓

function code(x, n)
	t_0 = x ^ Float64(1.0 / n)
	tmp = 0.0
	if (Float64(1.0 / n) <= -4e-9)
		tmp = Float64((Float64(1.0 + x) ^ Float64(1.0 / n)) - t_0);
	elseif (Float64(1.0 / n) <= 0.0002)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(1.0 - t_0);
	end
	return tmp
end

Wolfram

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

↓

code[x_, n_] := Block[{t$95$0 = N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[(1.0 / n), $MachinePrecision], -4e-9], N[(N[Power[N[(1.0 + x), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - t$95$0), $MachinePrecision], If[LessEqual[N[(1.0 / n), $MachinePrecision], 0.0002], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[(1.0 - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -4.00000000000000025e-9

   1. Initial program 2.4

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

   if -4.00000000000000025e-9 < (/.f64 1 n) < 2.0000000000000001e-4

   1. Initial program 69.45

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

   2. Taylor expanded in n around inf 23.85

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

   3. Simplified 23.84

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

      [Start] 23.85	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 23.84	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 23.64

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

   5. Applied egg-rr 23.64

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

   6. Simplified 1.62

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

      [Start] 23.64	\[ \frac{\mathsf{log1p}\left(\frac{x + 1}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 23.64	\[ \frac{\mathsf{log1p}\left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x} - 1\right)}{n} \]
      associate-*l/ [<=] 27.5	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(x + 1\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 27.5	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 27.5	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 27.5	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 23.64	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      associate--l+ [=>] 1.62	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 1.62	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   7. Taylor expanded in n around 0 23.64

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

   8. Simplified 1.62

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

      [Start] 23.64	\[ \frac{\log \left(1 + \frac{1}{x}\right)}{n} \]
      log1p-def [=>] 1.62	\[ \frac{\color{blue}{\mathsf{log1p}\left(\frac{1}{x}\right)}}{n} \]

   if 2.0000000000000001e-4 < (/.f64 1 n)

   1. Initial program 9.8

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

   2. Taylor expanded in x around 0 10.08

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

   3. Taylor expanded in x around inf 10.08

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

   4. Simplified 10.08

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

      [Start] 10.08	\[ 1 - e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}} \]
      log-rec [=>] 10.08	\[ 1 - e^{-1 \cdot \frac{\color{blue}{-\log x}}{n}} \]
      distribute-frac-neg [=>] 10.08	\[ 1 - e^{-1 \cdot \color{blue}{\left(-\frac{\log x}{n}\right)}} \]
      mul-1-neg [<=] 10.08	\[ 1 - e^{-1 \cdot \color{blue}{\left(-1 \cdot \frac{\log x}{n}\right)}} \]
      associate-*r* [=>] 10.08	\[ 1 - e^{\color{blue}{\left(-1 \cdot -1\right) \cdot \frac{\log x}{n}}} \]
      metadata-eval [=>] 10.08	\[ 1 - e^{\color{blue}{1} \cdot \frac{\log x}{n}} \]
      *-commutative [=>] 10.08	\[ 1 - e^{\color{blue}{\frac{\log x}{n} \cdot 1}} \]
      associate-*l/ [=>] 10.08	\[ 1 - e^{\color{blue}{\frac{\log x \cdot 1}{n}}} \]
      associate-*r/ [<=] 10.08	\[ 1 - e^{\color{blue}{\log x \cdot \frac{1}{n}}} \]
      exp-to-pow [=>] 10.08	\[ 1 - \color{blue}{{x}^{\left(\frac{1}{n}\right)}} \]

3. Recombined 3 regimes into one program.

4. Final simplification 2.59

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -4 \cdot 10^{-9}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0002:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternatives

Alternative 1
Error	3.19%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -500000:\\ \;\;\;\;0\\ \mathbf{elif}\;\frac{1}{n} \leq 0.0002:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Error	11.08%
Cost	6985

\[\begin{array}{l} \mathbf{if}\;n \leq -7.6 \lor \neg \left(n \leq -2 \cdot 10^{-306}\right):\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 3
Error	24.69%
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.68:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 1.85 \cdot 10^{+145}:\\ \;\;\;\;\frac{\frac{1}{x} + \frac{\frac{-0.5}{x}}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 4
Error	44.83%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -7.8 \lor \neg \left(n \leq 2.4 \cdot 10^{-82}\right):\\ \;\;\;\;\frac{1}{n \cdot x}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 5
Error	44.14%
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -18.5 \lor \neg \left(n \leq 1.75 \cdot 10^{-77}\right):\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 6
Error	60.3%
Cost	64

\[0 \]

Reproduce

herbie shell --seed 2023103 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))