2nthrt (problem 3.4.6)

Average Error: 32.9 → 1.4

Time: 20.2s

Precision: binary64

Cost: 13764

Math

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

↓

\[\begin{array}{l} t_0 := {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{if}\;\frac{1}{n} \leq -3 \cdot 10^{-5}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - t_0\\ \end{array} \]

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) -3e-5)
     (- (pow (+ 1.0 x) (/ 1.0 n)) t_0)
     (if (<= (/ 1.0 n) 0.02)
       (/ (log1p (/ 1.0 x)) n)
       (- (+ 1.0 (/ x n)) 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) <= -3e-5) {
		tmp = pow((1.0 + x), (1.0 / n)) - t_0;
	} else if ((1.0 / n) <= 0.02) {
		tmp = log1p((1.0 / x)) / n;
	} else {
		tmp = (1.0 + (x / n)) - 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) <= -3e-5) {
		tmp = Math.pow((1.0 + x), (1.0 / n)) - t_0;
	} else if ((1.0 / n) <= 0.02) {
		tmp = Math.log1p((1.0 / x)) / n;
	} else {
		tmp = (1.0 + (x / n)) - 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) <= -3e-5:
		tmp = math.pow((1.0 + x), (1.0 / n)) - t_0
	elif (1.0 / n) <= 0.02:
		tmp = math.log1p((1.0 / x)) / n
	else:
		tmp = (1.0 + (x / n)) - 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) <= -3e-5)
		tmp = Float64((Float64(1.0 + x) ^ Float64(1.0 / n)) - t_0);
	elseif (Float64(1.0 / n) <= 0.02)
		tmp = Float64(log1p(Float64(1.0 / x)) / n);
	else
		tmp = Float64(Float64(1.0 + Float64(x / n)) - 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], -3e-5], 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], 0.02], N[(N[Log[1 + N[(1.0 / x), $MachinePrecision]], $MachinePrecision] / n), $MachinePrecision], N[(N[(1.0 + N[(x / n), $MachinePrecision]), $MachinePrecision] - t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 1 n) < -3.00000000000000008e-5

   1. Initial program 1.0

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

   if -3.00000000000000008e-5 < (/.f64 1 n) < 0.0200000000000000004

   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 15.4

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

   3. Simplified 15.4

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

      [Start] 15.4	\[ \frac{\log \left(1 + x\right) - \log x}{n} \]
      log1p-def [=>] 15.4	\[ \frac{\color{blue}{\mathsf{log1p}\left(x\right)} - \log x}{n} \]

   4. Applied egg-rr 15.3

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

   5. Applied egg-rr 15.3

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

   6. Simplified 1.4

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

      [Start] 15.3	\[ \frac{\mathsf{log1p}\left(\frac{x + 1}{x} - 1\right)}{n} \]
      *-lft-identity [<=] 15.3	\[ \frac{\mathsf{log1p}\left(\frac{\color{blue}{1 \cdot \left(x + 1\right)}}{x} - 1\right)}{n} \]
      associate-*l/ [<=] 17.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} \cdot \left(x + 1\right)} - 1\right)}{n} \]
      distribute-rgt-in [=>] 17.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(x \cdot \frac{1}{x} + 1 \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      +-commutative [=>] 17.2	\[ \frac{\mathsf{log1p}\left(\color{blue}{\left(1 \cdot \frac{1}{x} + x \cdot \frac{1}{x}\right)} - 1\right)}{n} \]
      *-lft-identity [=>] 17.2	\[ \frac{\mathsf{log1p}\left(\left(\color{blue}{\frac{1}{x}} + x \cdot \frac{1}{x}\right) - 1\right)}{n} \]
      rgt-mult-inverse [=>] 15.3	\[ \frac{\mathsf{log1p}\left(\left(\frac{1}{x} + \color{blue}{1}\right) - 1\right)}{n} \]
      associate--l+ [=>] 1.4	\[ \frac{\mathsf{log1p}\left(\color{blue}{\frac{1}{x} + \left(1 - 1\right)}\right)}{n} \]
      metadata-eval [=>] 1.4	\[ \frac{\mathsf{log1p}\left(\frac{1}{x} + \color{blue}{0}\right)}{n} \]

   if 0.0200000000000000004 < (/.f64 1 n)

   1. Initial program 3.8

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

   2. Taylor expanded in x around 0 2.6

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

4. Final simplification 1.4

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -3 \cdot 10^{-5}:\\ \;\;\;\;{\left(1 + x\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternatives

Alternative 1
Error	1.4
Cost	13508

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2 \cdot 10^{-5}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{n \cdot x}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 2
Error	12.5
Cost	7628

\[\begin{array}{l} t_0 := \frac{\log \left(\frac{1 + x}{x}\right)}{n}\\ \mathbf{if}\;\frac{1}{n} \leq 5 \cdot 10^{-99}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;\frac{1}{n} \leq 2 \cdot 10^{-54}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 3
Error	1.9
Cost	7560

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2000000000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{x}{n}\right) - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 4
Error	2.0
Cost	7304

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2000000000:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;\frac{1}{n} \leq 0.02:\\ \;\;\;\;\frac{\mathsf{log1p}\left(\frac{1}{x}\right)}{n}\\ \mathbf{else}:\\ \;\;\;\;1 - {x}^{\left(\frac{1}{n}\right)}\\ \end{array} \]

Alternative 5
Error	15.9
Cost	6980

\[\begin{array}{l} \mathbf{if}\;x \leq 0.85:\\ \;\;\;\;\frac{x}{n} - \frac{\log x}{n}\\ \mathbf{elif}\;x \leq 2.4 \cdot 10^{+86}:\\ \;\;\;\;\frac{\frac{1}{x} + \left(\frac{1}{x} \cdot \frac{0.3333333333333333}{x \cdot x} + \frac{-0.5}{x \cdot x}\right)}{n}\\ \mathbf{elif}\;x \leq 1.15 \cdot 10^{+137}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 2.7 \cdot 10^{+149}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 6
Error	15.9
Cost	6852

\[\begin{array}{l} \mathbf{if}\;x \leq 0.85:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 2.5 \cdot 10^{+86}:\\ \;\;\;\;\frac{\frac{1}{x} + \left(\frac{1}{x} \cdot \frac{0.3333333333333333}{x \cdot x} + \frac{-0.5}{x \cdot x}\right)}{n}\\ \mathbf{elif}\;x \leq 1.58 \cdot 10^{+137}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 10^{+153}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 7
Error	16.1
Cost	6788

\[\begin{array}{l} \mathbf{if}\;x \leq 0.6:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 1.75 \cdot 10^{+86}:\\ \;\;\;\;\frac{\frac{1}{x} + \left(\frac{1}{x} \cdot \frac{0.3333333333333333}{x \cdot x} + \frac{-0.5}{x \cdot x}\right)}{n}\\ \mathbf{elif}\;x \leq 1.3 \cdot 10^{+137}:\\ \;\;\;\;\frac{0}{n}\\ \mathbf{elif}\;x \leq 7.8 \cdot 10^{+152}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 8
Error	35.7
Cost	836

\[\begin{array}{l} \mathbf{if}\;\frac{1}{n} \leq -2000000000:\\ \;\;\;\;\left(1 + \frac{1}{n \cdot x}\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \end{array} \]

Alternative 9
Error	29.0
Cost	585

\[\begin{array}{l} \mathbf{if}\;n \leq -9.8 \lor \neg \left(n \leq 1.45 \cdot 10^{-60}\right):\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{n}\\ \end{array} \]

Alternative 10
Error	41.0
Cost	320

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

Alternative 11
Error	40.6
Cost	320

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

Alternative 12
Error	61.0
Cost	192

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

Reproduce

herbie shell --seed 2023017 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))