?

Average Error: 32.8 → 7.0
Time: 39.9s
Precision: binary64
Cost: 85828

?

\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
\[\begin{array}{l} t_0 := \log \left(1 + x\right)\\ \mathbf{if}\;x \leq 9500000:\\ \;\;\;\;\left(0.5 \cdot \frac{{t_0}^{2}}{{n}^{2}} + \left(\left(-\frac{\log x - t_0}{n}\right) - \frac{-0.16666666666666666 \cdot \left({t_0}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\left(0 - \left(-1 - \frac{\log x}{n}\right)\right) - 1}}{x \cdot n}\\ \end{array} \]
(FPCore (x n)
 :precision binary64
 (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))
(FPCore (x n)
 :precision binary64
 (let* ((t_0 (log (+ 1.0 x))))
   (if (<= x 9500000.0)
     (-
      (+
       (* 0.5 (/ (pow t_0 2.0) (pow n 2.0)))
       (-
        (- (/ (- (log x) t_0) n))
        (/
         (* -0.16666666666666666 (- (pow t_0 3.0) (pow (log x) 3.0)))
         (pow n 3.0))))
      (* 0.5 (/ (pow (log x) 2.0) (pow n 2.0))))
     (/ (exp (- (- 0.0 (- -1.0 (/ (log x) n))) 1.0)) (* x n)))))
double code(double x, double n) {
	return pow((x + 1.0), (1.0 / n)) - pow(x, (1.0 / n));
}
double code(double x, double n) {
	double t_0 = log((1.0 + x));
	double tmp;
	if (x <= 9500000.0) {
		tmp = ((0.5 * (pow(t_0, 2.0) / pow(n, 2.0))) + (-((log(x) - t_0) / n) - ((-0.16666666666666666 * (pow(t_0, 3.0) - pow(log(x), 3.0))) / pow(n, 3.0)))) - (0.5 * (pow(log(x), 2.0) / pow(n, 2.0)));
	} else {
		tmp = exp(((0.0 - (-1.0 - (log(x) / n))) - 1.0)) / (x * n);
	}
	return tmp;
}
real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    code = ((x + 1.0d0) ** (1.0d0 / n)) - (x ** (1.0d0 / n))
end function
real(8) function code(x, n)
    real(8), intent (in) :: x
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = log((1.0d0 + x))
    if (x <= 9500000.0d0) then
        tmp = ((0.5d0 * ((t_0 ** 2.0d0) / (n ** 2.0d0))) + (-((log(x) - t_0) / n) - (((-0.16666666666666666d0) * ((t_0 ** 3.0d0) - (log(x) ** 3.0d0))) / (n ** 3.0d0)))) - (0.5d0 * ((log(x) ** 2.0d0) / (n ** 2.0d0)))
    else
        tmp = exp(((0.0d0 - ((-1.0d0) - (log(x) / n))) - 1.0d0)) / (x * n)
    end if
    code = tmp
end function
public static double code(double x, double n) {
	return Math.pow((x + 1.0), (1.0 / n)) - Math.pow(x, (1.0 / n));
}
public static double code(double x, double n) {
	double t_0 = Math.log((1.0 + x));
	double tmp;
	if (x <= 9500000.0) {
		tmp = ((0.5 * (Math.pow(t_0, 2.0) / Math.pow(n, 2.0))) + (-((Math.log(x) - t_0) / n) - ((-0.16666666666666666 * (Math.pow(t_0, 3.0) - Math.pow(Math.log(x), 3.0))) / Math.pow(n, 3.0)))) - (0.5 * (Math.pow(Math.log(x), 2.0) / Math.pow(n, 2.0)));
	} else {
		tmp = Math.exp(((0.0 - (-1.0 - (Math.log(x) / n))) - 1.0)) / (x * n);
	}
	return tmp;
}
def code(x, n):
	return math.pow((x + 1.0), (1.0 / n)) - math.pow(x, (1.0 / n))
def code(x, n):
	t_0 = math.log((1.0 + x))
	tmp = 0
	if x <= 9500000.0:
		tmp = ((0.5 * (math.pow(t_0, 2.0) / math.pow(n, 2.0))) + (-((math.log(x) - t_0) / n) - ((-0.16666666666666666 * (math.pow(t_0, 3.0) - math.pow(math.log(x), 3.0))) / math.pow(n, 3.0)))) - (0.5 * (math.pow(math.log(x), 2.0) / math.pow(n, 2.0)))
	else:
		tmp = math.exp(((0.0 - (-1.0 - (math.log(x) / n))) - 1.0)) / (x * n)
	return tmp
function code(x, n)
	return Float64((Float64(x + 1.0) ^ Float64(1.0 / n)) - (x ^ Float64(1.0 / n)))
end
function code(x, n)
	t_0 = log(Float64(1.0 + x))
	tmp = 0.0
	if (x <= 9500000.0)
		tmp = Float64(Float64(Float64(0.5 * Float64((t_0 ^ 2.0) / (n ^ 2.0))) + Float64(Float64(-Float64(Float64(log(x) - t_0) / n)) - Float64(Float64(-0.16666666666666666 * Float64((t_0 ^ 3.0) - (log(x) ^ 3.0))) / (n ^ 3.0)))) - Float64(0.5 * Float64((log(x) ^ 2.0) / (n ^ 2.0))));
	else
		tmp = Float64(exp(Float64(Float64(0.0 - Float64(-1.0 - Float64(log(x) / n))) - 1.0)) / Float64(x * n));
	end
	return tmp
end
function tmp = code(x, n)
	tmp = ((x + 1.0) ^ (1.0 / n)) - (x ^ (1.0 / n));
end
function tmp_2 = code(x, n)
	t_0 = log((1.0 + x));
	tmp = 0.0;
	if (x <= 9500000.0)
		tmp = ((0.5 * ((t_0 ^ 2.0) / (n ^ 2.0))) + (-((log(x) - t_0) / n) - ((-0.16666666666666666 * ((t_0 ^ 3.0) - (log(x) ^ 3.0))) / (n ^ 3.0)))) - (0.5 * ((log(x) ^ 2.0) / (n ^ 2.0)));
	else
		tmp = exp(((0.0 - (-1.0 - (log(x) / n))) - 1.0)) / (x * n);
	end
	tmp_2 = tmp;
end
code[x_, n_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], N[(1.0 / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
code[x_, n_] := Block[{t$95$0 = N[Log[N[(1.0 + x), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[x, 9500000.0], N[(N[(N[(0.5 * N[(N[Power[t$95$0, 2.0], $MachinePrecision] / N[Power[n, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[((-N[(N[(N[Log[x], $MachinePrecision] - t$95$0), $MachinePrecision] / n), $MachinePrecision]) - N[(N[(-0.16666666666666666 * N[(N[Power[t$95$0, 3.0], $MachinePrecision] - N[Power[N[Log[x], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[n, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(0.5 * N[(N[Power[N[Log[x], $MachinePrecision], 2.0], $MachinePrecision] / N[Power[n, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Exp[N[(N[(0.0 - N[(-1.0 - N[(N[Log[x], $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 1.0), $MachinePrecision]], $MachinePrecision] / N[(x * n), $MachinePrecision]), $MachinePrecision]]]
{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)}
\begin{array}{l}
t_0 := \log \left(1 + x\right)\\
\mathbf{if}\;x \leq 9500000:\\
\;\;\;\;\left(0.5 \cdot \frac{{t_0}^{2}}{{n}^{2}} + \left(\left(-\frac{\log x - t_0}{n}\right) - \frac{-0.16666666666666666 \cdot \left({t_0}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\frac{e^{\left(0 - \left(-1 - \frac{\log x}{n}\right)\right) - 1}}{x \cdot n}\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes
  2. if x < 9.5e6

    1. Initial program 47.0

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in n around -inf 13.6

      \[\leadsto \color{blue}{\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(-1 \cdot \frac{-0.16666666666666666 \cdot {\log \left(1 + x\right)}^{3} - -0.16666666666666666 \cdot {\log x}^{3}}{{n}^{3}} + -1 \cdot \frac{-1 \cdot \log \left(1 + x\right) - -1 \cdot \log x}{n}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}} \]
    3. Simplified13.6

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

      [Start]13.6

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

    if 9.5e6 < x

    1. Initial program 20.1

      \[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
    2. Taylor expanded in x around inf 1.2

      \[\leadsto \color{blue}{\frac{e^{-1 \cdot \frac{\log \left(\frac{1}{x}\right)}{n}}}{n \cdot x}} \]
    3. Simplified1.2

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

      [Start]1.2

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

      rational.json-simplify-2 [=>]1.2

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

      rational.json-simplify-9 [=>]1.2

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

      rational.json-simplify-2 [=>]1.2

      \[ \frac{e^{-\frac{\log \left(\frac{1}{x}\right)}{n}}}{\color{blue}{x \cdot n}} \]
    4. Taylor expanded in x around 0 1.2

      \[\leadsto \frac{\color{blue}{e^{--1 \cdot \frac{\log x}{n}}}}{x \cdot n} \]
    5. Simplified1.2

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

      [Start]1.2

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

      rational.json-simplify-12 [=>]1.2

      \[ \frac{e^{\color{blue}{0 - -1 \cdot \frac{\log x}{n}}}}{x \cdot n} \]

      rational.json-simplify-2 [=>]1.2

      \[ \frac{e^{0 - \color{blue}{\frac{\log x}{n} \cdot -1}}}{x \cdot n} \]

      rational.json-simplify-9 [=>]1.2

      \[ \frac{e^{0 - \color{blue}{\left(-\frac{\log x}{n}\right)}}}{x \cdot n} \]

      rational.json-simplify-12 [=>]1.2

      \[ \frac{e^{0 - \color{blue}{\left(0 - \frac{\log x}{n}\right)}}}{x \cdot n} \]

      rational.json-simplify-44 [=>]1.2

      \[ \frac{e^{\color{blue}{\frac{\log x}{n} - \left(0 - 0\right)}}}{x \cdot n} \]

      metadata-eval [=>]1.2

      \[ \frac{e^{\frac{\log x}{n} - \color{blue}{0}}}{x \cdot n} \]

      rational.json-simplify-5 [=>]1.2

      \[ \frac{e^{\color{blue}{\frac{\log x}{n}}}}{x \cdot n} \]
    6. Applied egg-rr1.2

      \[\leadsto \frac{e^{\color{blue}{\left(0 - \left(-1 - \frac{\log x}{n}\right)\right) - 1}}}{x \cdot n} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification7.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 9500000:\\ \;\;\;\;\left(0.5 \cdot \frac{{\log \left(1 + x\right)}^{2}}{{n}^{2}} + \left(\left(-\frac{\log x - \log \left(1 + x\right)}{n}\right) - \frac{-0.16666666666666666 \cdot \left({\log \left(1 + x\right)}^{3} - {\log x}^{3}\right)}{{n}^{3}}\right)\right) - 0.5 \cdot \frac{{\log x}^{2}}{{n}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\left(0 - \left(-1 - \frac{\log x}{n}\right)\right) - 1}}{x \cdot n}\\ \end{array} \]

Alternatives

Alternative 1
Error7.2
Cost13764
\[\begin{array}{l} \mathbf{if}\;x \leq 6800000:\\ \;\;\;\;\frac{\log \left(x - -1\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\left(0 - \left(-1 - \frac{\log x}{n}\right)\right) - 1}}{x \cdot n}\\ \end{array} \]
Alternative 2
Error7.3
Cost13380
\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{x \cdot n}\\ \end{array} \]
Alternative 3
Error7.2
Cost13380
\[\begin{array}{l} \mathbf{if}\;x \leq 6800000:\\ \;\;\;\;\frac{\log \left(x - -1\right) - \log x}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\frac{\log x}{n}}}{x \cdot n}\\ \end{array} \]
Alternative 4
Error16.4
Cost7432
\[\begin{array}{l} \mathbf{if}\;x \leq 0.95:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 7.5 \cdot 10^{+230}:\\ \;\;\;\;\frac{\frac{1}{x} - 0.5 \cdot \frac{1}{{x}^{2}}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{{x}^{3} \cdot n}\\ \end{array} \]
Alternative 5
Error16.6
Cost7048
\[\begin{array}{l} \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 2.95 \cdot 10^{+231}:\\ \;\;\;\;\frac{\frac{1}{x}}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333}{{x}^{3} \cdot n}\\ \end{array} \]
Alternative 6
Error17.9
Cost6852
\[\begin{array}{l} t_0 := \frac{\frac{1}{x}}{n}\\ \mathbf{if}\;x \leq 1:\\ \;\;\;\;\frac{x - \log x}{n}\\ \mathbf{elif}\;x \leq 3.3 \cdot 10^{+231}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 8.2 \cdot 10^{+284}:\\ \;\;\;\;\left(0 - \left(-1 - \frac{1}{x \cdot n}\right)\right) - 1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 7
Error18.1
Cost6788
\[\begin{array}{l} t_0 := \frac{\frac{1}{x}}{n}\\ \mathbf{if}\;x \leq 0.55:\\ \;\;\;\;\frac{-\log x}{n}\\ \mathbf{elif}\;x \leq 3.4 \cdot 10^{+231}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 1.25 \cdot 10^{+287}:\\ \;\;\;\;\left(0 - \left(-1 - \frac{1}{x \cdot n}\right)\right) - 1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 8
Error36.2
Cost968
\[\begin{array}{l} t_0 := \frac{\frac{1}{x}}{n}\\ \mathbf{if}\;n \leq -870000:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq -3.5 \cdot 10^{-185}:\\ \;\;\;\;\left(0 - \left(-1 - \frac{1}{x \cdot n}\right)\right) - 1\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 9
Error40.7
Cost320
\[\frac{1}{x \cdot n} \]
Alternative 10
Error40.3
Cost320
\[\frac{\frac{1}{x}}{n} \]
Alternative 11
Error61.1
Cost192
\[\frac{x}{n} \]

Error

Reproduce?

herbie shell --seed 2023077 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))