Average Error: 32.6 → 32.5

Time: 20.8s

Precision: binary64

Cost: 32512

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

↓

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

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

↓

(FPCore (x n)
 :precision binary64
 (log (exp (- (exp (/ (log1p x) n)) (pow x (/ 1.0 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) {
	return log(exp((exp((log1p(x) / n)) - pow(x, (1.0 / n)))));
}

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) {
	return Math.log(Math.exp((Math.exp((Math.log1p(x) / n)) - Math.pow(x, (1.0 / n)))));
}

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

↓

def code(x, n):
	return math.log(math.exp((math.exp((math.log1p(x) / n)) - math.pow(x, (1.0 / n)))))

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

↓

function code(x, n)
	return log(exp(Float64(exp(Float64(log1p(x) / n)) - (x ^ Float64(1.0 / n)))))
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_] := N[Log[N[Exp[N[(N[Exp[N[(N[Log[1 + x], $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] - N[Power[x, N[(1.0 / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]

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

↓

\log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right)

Try it out

In
Out

Enter valid numbers for all inputs

Derivation

Initial program 32.6
\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} - {x}^{\left(\frac{1}{n}\right)} \]
Applied egg-rr32.5
\[\leadsto \color{blue}{\log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right)} \]
Final simplification32.5
\[\leadsto \log \left(e^{e^{\frac{\mathsf{log1p}\left(x\right)}{n}} - {x}^{\left(\frac{1}{n}\right)}}\right) \]

Alternatives

Alternative 1
Error	32.3
Cost	19712

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

Alternative 2
Error	32.7
Cost	13760

\[{\left(x + 1\right)}^{\left(\frac{1}{n}\right)} + 3 \cdot \left({x}^{\left(\frac{1}{n}\right)} \cdot -0.3333333333333333\right) \]

Alternative 3
Error	44.9
Cost	13504

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

Alternative 4
Error	32.6
Cost	13504

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

Reproduce

herbie shell --seed 2022334 
(FPCore (x n)
  :name "2nthrt (problem 3.4.6)"
  :precision binary64
  (- (pow (+ x 1.0) (/ 1.0 n)) (pow x (/ 1.0 n))))