2cbrt (problem 3.3.4)

Percentage Accurate: 53.3% → 99.6%
Time: 14.8s
Alternatives: 14
Speedup: 1.0×

Specification

?
\[\begin{array}{l} \\ \sqrt[3]{x + 1} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ x 1.0)) (cbrt x)))
double code(double x) {
	return cbrt((x + 1.0)) - cbrt(x);
}
public static double code(double x) {
	return Math.cbrt((x + 1.0)) - Math.cbrt(x);
}
function code(x)
	return Float64(cbrt(Float64(x + 1.0)) - cbrt(x))
end
code[x_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\sqrt[3]{x + 1} - \sqrt[3]{x}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 14 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 53.3% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \sqrt[3]{x + 1} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ x 1.0)) (cbrt x)))
double code(double x) {
	return cbrt((x + 1.0)) - cbrt(x);
}
public static double code(double x) {
	return Math.cbrt((x + 1.0)) - Math.cbrt(x);
}
function code(x)
	return Float64(cbrt(Float64(x + 1.0)) - cbrt(x))
end
code[x_] := N[(N[Power[N[(x + 1.0), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\sqrt[3]{x + 1} - \sqrt[3]{x}
\end{array}

Alternative 1: 99.6% accurate, 0.2× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{1 + x}\\ \frac{1}{\frac{t_0}{\frac{1}{t_0}} + \sqrt[3]{x} \cdot \frac{x + \left(1 + x\right)}{\mathsf{fma}\left(\sqrt[3]{x}, \sqrt[3]{x} - t_0, {t_0}^{2}\right)}} \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ 1.0 x))))
   (/
    1.0
    (+
     (/ t_0 (/ 1.0 t_0))
     (*
      (cbrt x)
      (/ (+ x (+ 1.0 x)) (fma (cbrt x) (- (cbrt x) t_0) (pow t_0 2.0))))))))
double code(double x) {
	double t_0 = cbrt((1.0 + x));
	return 1.0 / ((t_0 / (1.0 / t_0)) + (cbrt(x) * ((x + (1.0 + x)) / fma(cbrt(x), (cbrt(x) - t_0), pow(t_0, 2.0)))));
}
function code(x)
	t_0 = cbrt(Float64(1.0 + x))
	return Float64(1.0 / Float64(Float64(t_0 / Float64(1.0 / t_0)) + Float64(cbrt(x) * Float64(Float64(x + Float64(1.0 + x)) / fma(cbrt(x), Float64(cbrt(x) - t_0), (t_0 ^ 2.0))))))
end
code[x_] := Block[{t$95$0 = N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]}, N[(1.0 / N[(N[(t$95$0 / N[(1.0 / t$95$0), $MachinePrecision]), $MachinePrecision] + N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[(x + N[(1.0 + x), $MachinePrecision]), $MachinePrecision] / N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] - t$95$0), $MachinePrecision] + N[Power[t$95$0, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{1 + x}\\
\frac{1}{\frac{t_0}{\frac{1}{t_0}} + \sqrt[3]{x} \cdot \frac{x + \left(1 + x\right)}{\mathsf{fma}\left(\sqrt[3]{x}, \sqrt[3]{x} - t_0, {t_0}^{2}\right)}}
\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 2: 75.5% accurate, 0.3× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{1 + x}\\ t_1 := t_0 - \sqrt[3]{x}\\ \mathbf{if}\;t_1 \leq 2 \cdot 10^{-6}:\\ \;\;\;\;\frac{1}{\sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right) + \sqrt[3]{{x}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;e^{\left(3 \cdot \log t_1\right) \cdot 0.3333333333333333}\\ \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ 1.0 x))) (t_1 (- t_0 (cbrt x))))
   (if (<= t_1 2e-6)
     (/ 1.0 (+ (* (cbrt x) (+ (cbrt x) t_0)) (cbrt (pow x 2.0))))
     (exp (* (* 3.0 (log t_1)) 0.3333333333333333)))))
double code(double x) {
	double t_0 = cbrt((1.0 + x));
	double t_1 = t_0 - cbrt(x);
	double tmp;
	if (t_1 <= 2e-6) {
		tmp = 1.0 / ((cbrt(x) * (cbrt(x) + t_0)) + cbrt(pow(x, 2.0)));
	} else {
		tmp = exp(((3.0 * log(t_1)) * 0.3333333333333333));
	}
	return tmp;
}
public static double code(double x) {
	double t_0 = Math.cbrt((1.0 + x));
	double t_1 = t_0 - Math.cbrt(x);
	double tmp;
	if (t_1 <= 2e-6) {
		tmp = 1.0 / ((Math.cbrt(x) * (Math.cbrt(x) + t_0)) + Math.cbrt(Math.pow(x, 2.0)));
	} else {
		tmp = Math.exp(((3.0 * Math.log(t_1)) * 0.3333333333333333));
	}
	return tmp;
}
function code(x)
	t_0 = cbrt(Float64(1.0 + x))
	t_1 = Float64(t_0 - cbrt(x))
	tmp = 0.0
	if (t_1 <= 2e-6)
		tmp = Float64(1.0 / Float64(Float64(cbrt(x) * Float64(cbrt(x) + t_0)) + cbrt((x ^ 2.0))));
	else
		tmp = exp(Float64(Float64(3.0 * log(t_1)) * 0.3333333333333333));
	end
	return tmp
end
code[x_] := Block[{t$95$0 = N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 2e-6], N[(1.0 / N[(N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + t$95$0), $MachinePrecision]), $MachinePrecision] + N[Power[N[Power[x, 2.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Exp[N[(N[(3.0 * N[Log[t$95$1], $MachinePrecision]), $MachinePrecision] * 0.3333333333333333), $MachinePrecision]], $MachinePrecision]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{1 + x}\\
t_1 := t_0 - \sqrt[3]{x}\\
\mathbf{if}\;t_1 \leq 2 \cdot 10^{-6}:\\
\;\;\;\;\frac{1}{\sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right) + \sqrt[3]{{x}^{2}}}\\

\mathbf{else}:\\
\;\;\;\;e^{\left(3 \cdot \log t_1\right) \cdot 0.3333333333333333}\\


\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 3: 99.2% accurate, 0.3× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{1 + x}\\ \frac{1}{\mathsf{fma}\left(\sqrt[3]{x}, \sqrt[3]{x} + t_0, {t_0}^{2}\right)} \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ 1.0 x))))
   (/ 1.0 (fma (cbrt x) (+ (cbrt x) t_0) (pow t_0 2.0)))))
double code(double x) {
	double t_0 = cbrt((1.0 + x));
	return 1.0 / fma(cbrt(x), (cbrt(x) + t_0), pow(t_0, 2.0));
}
function code(x)
	t_0 = cbrt(Float64(1.0 + x))
	return Float64(1.0 / fma(cbrt(x), Float64(cbrt(x) + t_0), (t_0 ^ 2.0)))
end
code[x_] := Block[{t$95$0 = N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]}, N[(1.0 / N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + t$95$0), $MachinePrecision] + N[Power[t$95$0, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{1 + x}\\
\frac{1}{\mathsf{fma}\left(\sqrt[3]{x}, \sqrt[3]{x} + t_0, {t_0}^{2}\right)}
\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 4: 99.2% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{1 + x}\\ \frac{1}{\frac{t_0}{\frac{1}{t_0}} + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right)} \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ 1.0 x))))
   (/ 1.0 (+ (/ t_0 (/ 1.0 t_0)) (* (cbrt x) (+ (cbrt x) t_0))))))
double code(double x) {
	double t_0 = cbrt((1.0 + x));
	return 1.0 / ((t_0 / (1.0 / t_0)) + (cbrt(x) * (cbrt(x) + t_0)));
}
public static double code(double x) {
	double t_0 = Math.cbrt((1.0 + x));
	return 1.0 / ((t_0 / (1.0 / t_0)) + (Math.cbrt(x) * (Math.cbrt(x) + t_0)));
}
function code(x)
	t_0 = cbrt(Float64(1.0 + x))
	return Float64(1.0 / Float64(Float64(t_0 / Float64(1.0 / t_0)) + Float64(cbrt(x) * Float64(cbrt(x) + t_0))))
end
code[x_] := Block[{t$95$0 = N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]}, N[(1.0 / N[(N[(t$95$0 / N[(1.0 / t$95$0), $MachinePrecision]), $MachinePrecision] + N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{1 + x}\\
\frac{1}{\frac{t_0}{\frac{1}{t_0}} + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right)}
\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 5: 85.5% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)\\ \mathbf{if}\;x \leq -1:\\ \;\;\;\;\frac{1}{t_0 + \sqrt[3]{{x}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{t_0 + e^{0.6666666666666666 \cdot \mathsf{log1p}\left(x\right)}}\\ \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (* (cbrt x) (+ (cbrt x) (cbrt (+ 1.0 x))))))
   (if (<= x -1.0)
     (/ 1.0 (+ t_0 (cbrt (pow x 2.0))))
     (/ 1.0 (+ t_0 (exp (* 0.6666666666666666 (log1p x))))))))
double code(double x) {
	double t_0 = cbrt(x) * (cbrt(x) + cbrt((1.0 + x)));
	double tmp;
	if (x <= -1.0) {
		tmp = 1.0 / (t_0 + cbrt(pow(x, 2.0)));
	} else {
		tmp = 1.0 / (t_0 + exp((0.6666666666666666 * log1p(x))));
	}
	return tmp;
}
public static double code(double x) {
	double t_0 = Math.cbrt(x) * (Math.cbrt(x) + Math.cbrt((1.0 + x)));
	double tmp;
	if (x <= -1.0) {
		tmp = 1.0 / (t_0 + Math.cbrt(Math.pow(x, 2.0)));
	} else {
		tmp = 1.0 / (t_0 + Math.exp((0.6666666666666666 * Math.log1p(x))));
	}
	return tmp;
}
function code(x)
	t_0 = Float64(cbrt(x) * Float64(cbrt(x) + cbrt(Float64(1.0 + x))))
	tmp = 0.0
	if (x <= -1.0)
		tmp = Float64(1.0 / Float64(t_0 + cbrt((x ^ 2.0))));
	else
		tmp = Float64(1.0 / Float64(t_0 + exp(Float64(0.6666666666666666 * log1p(x)))));
	end
	return tmp
end
code[x_] := Block[{t$95$0 = N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[x, -1.0], N[(1.0 / N[(t$95$0 + N[Power[N[Power[x, 2.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(1.0 / N[(t$95$0 + N[Exp[N[(0.6666666666666666 * N[Log[1 + x], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)\\
\mathbf{if}\;x \leq -1:\\
\;\;\;\;\frac{1}{t_0 + \sqrt[3]{{x}^{2}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0 + e^{0.6666666666666666 \cdot \mathsf{log1p}\left(x\right)}}\\


\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 6: 85.4% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{x} + \sqrt[3]{1 + x}\\ \mathbf{if}\;x \leq -1:\\ \;\;\;\;\frac{1}{\sqrt[3]{x} \cdot t_0 + \sqrt[3]{{x}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\mathsf{fma}\left(\sqrt[3]{x}, t_0, {\left(1 + x\right)}^{0.6666666666666666}\right)}\\ \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (+ (cbrt x) (cbrt (+ 1.0 x)))))
   (if (<= x -1.0)
     (/ 1.0 (+ (* (cbrt x) t_0) (cbrt (pow x 2.0))))
     (/ 1.0 (fma (cbrt x) t_0 (pow (+ 1.0 x) 0.6666666666666666))))))
double code(double x) {
	double t_0 = cbrt(x) + cbrt((1.0 + x));
	double tmp;
	if (x <= -1.0) {
		tmp = 1.0 / ((cbrt(x) * t_0) + cbrt(pow(x, 2.0)));
	} else {
		tmp = 1.0 / fma(cbrt(x), t_0, pow((1.0 + x), 0.6666666666666666));
	}
	return tmp;
}
function code(x)
	t_0 = Float64(cbrt(x) + cbrt(Float64(1.0 + x)))
	tmp = 0.0
	if (x <= -1.0)
		tmp = Float64(1.0 / Float64(Float64(cbrt(x) * t_0) + cbrt((x ^ 2.0))));
	else
		tmp = Float64(1.0 / fma(cbrt(x), t_0, (Float64(1.0 + x) ^ 0.6666666666666666)));
	end
	return tmp
end
code[x_] := Block[{t$95$0 = N[(N[Power[x, 1/3], $MachinePrecision] + N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[x, -1.0], N[(1.0 / N[(N[(N[Power[x, 1/3], $MachinePrecision] * t$95$0), $MachinePrecision] + N[Power[N[Power[x, 2.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(1.0 / N[(N[Power[x, 1/3], $MachinePrecision] * t$95$0 + N[Power[N[(1.0 + x), $MachinePrecision], 0.6666666666666666], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{x} + \sqrt[3]{1 + x}\\
\mathbf{if}\;x \leq -1:\\
\;\;\;\;\frac{1}{\sqrt[3]{x} \cdot t_0 + \sqrt[3]{{x}^{2}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{\mathsf{fma}\left(\sqrt[3]{x}, t_0, {\left(1 + x\right)}^{0.6666666666666666}\right)}\\


\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 7: 99.2% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt[3]{1 + x}\\ \frac{1}{{t_0}^{2} + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right)} \end{array} \end{array} \]
(FPCore (x)
 :precision binary64
 (let* ((t_0 (cbrt (+ 1.0 x))))
   (/ 1.0 (+ (pow t_0 2.0) (* (cbrt x) (+ (cbrt x) t_0))))))
double code(double x) {
	double t_0 = cbrt((1.0 + x));
	return 1.0 / (pow(t_0, 2.0) + (cbrt(x) * (cbrt(x) + t_0)));
}
public static double code(double x) {
	double t_0 = Math.cbrt((1.0 + x));
	return 1.0 / (Math.pow(t_0, 2.0) + (Math.cbrt(x) * (Math.cbrt(x) + t_0)));
}
function code(x)
	t_0 = cbrt(Float64(1.0 + x))
	return Float64(1.0 / Float64((t_0 ^ 2.0) + Float64(cbrt(x) * Float64(cbrt(x) + t_0))))
end
code[x_] := Block[{t$95$0 = N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]}, N[(1.0 / N[(N[Power[t$95$0, 2.0], $MachinePrecision] + N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \sqrt[3]{1 + x}\\
\frac{1}{{t_0}^{2} + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + t_0\right)}
\end{array}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 8: 59.0% accurate, 0.4× speedup?

\[\begin{array}{l} \\ e^{-\mathsf{log1p}\left(\sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)\right)} \end{array} \]
(FPCore (x)
 :precision binary64
 (exp (- (log1p (* (cbrt x) (+ (cbrt x) (cbrt (+ 1.0 x))))))))
double code(double x) {
	return exp(-log1p((cbrt(x) * (cbrt(x) + cbrt((1.0 + x))))));
}
public static double code(double x) {
	return Math.exp(-Math.log1p((Math.cbrt(x) * (Math.cbrt(x) + Math.cbrt((1.0 + x))))));
}
function code(x)
	return exp(Float64(-log1p(Float64(cbrt(x) * Float64(cbrt(x) + cbrt(Float64(1.0 + x)))))))
end
code[x_] := N[Exp[(-N[Log[1 + N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])], $MachinePrecision]
\begin{array}{l}

\\
e^{-\mathsf{log1p}\left(\sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)\right)}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 9: 58.9% accurate, 0.7× speedup?

\[\begin{array}{l} \\ \frac{1}{1 + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)} \end{array} \]
(FPCore (x)
 :precision binary64
 (/ 1.0 (+ 1.0 (* (cbrt x) (+ (cbrt x) (cbrt (+ 1.0 x)))))))
double code(double x) {
	return 1.0 / (1.0 + (cbrt(x) * (cbrt(x) + cbrt((1.0 + x)))));
}
public static double code(double x) {
	return 1.0 / (1.0 + (Math.cbrt(x) * (Math.cbrt(x) + Math.cbrt((1.0 + x)))));
}
function code(x)
	return Float64(1.0 / Float64(1.0 + Float64(cbrt(x) * Float64(cbrt(x) + cbrt(Float64(1.0 + x))))))
end
code[x_] := N[(1.0 / N[(1.0 + N[(N[Power[x, 1/3], $MachinePrecision] * N[(N[Power[x, 1/3], $MachinePrecision] + N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\frac{1}{1 + \sqrt[3]{x} \cdot \left(\sqrt[3]{x} + \sqrt[3]{1 + x}\right)}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 10: 53.4% accurate, 0.7× speedup?

\[\begin{array}{l} \\ \left|\frac{1}{\frac{1}{\sqrt[3]{1 + x}}} - \sqrt[3]{x}\right| \end{array} \]
(FPCore (x)
 :precision binary64
 (fabs (- (/ 1.0 (/ 1.0 (cbrt (+ 1.0 x)))) (cbrt x))))
double code(double x) {
	return fabs(((1.0 / (1.0 / cbrt((1.0 + x)))) - cbrt(x)));
}
public static double code(double x) {
	return Math.abs(((1.0 / (1.0 / Math.cbrt((1.0 + x)))) - Math.cbrt(x)));
}
function code(x)
	return abs(Float64(Float64(1.0 / Float64(1.0 / cbrt(Float64(1.0 + x)))) - cbrt(x)))
end
code[x_] := N[Abs[N[(N[(1.0 / N[(1.0 / N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\left|\frac{1}{\frac{1}{\sqrt[3]{1 + x}}} - \sqrt[3]{x}\right|
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 11: 53.3% accurate, 0.7× speedup?

\[\begin{array}{l} \\ \left|\sqrt[3]{x} - \sqrt[3]{1 + x}\right| \end{array} \]
(FPCore (x) :precision binary64 (fabs (- (cbrt x) (cbrt (+ 1.0 x)))))
double code(double x) {
	return fabs((cbrt(x) - cbrt((1.0 + x))));
}
public static double code(double x) {
	return Math.abs((Math.cbrt(x) - Math.cbrt((1.0 + x))));
}
function code(x)
	return abs(Float64(cbrt(x) - cbrt(Float64(1.0 + x))))
end
code[x_] := N[Abs[N[(N[Power[x, 1/3], $MachinePrecision] - N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\left|\sqrt[3]{x} - \sqrt[3]{1 + x}\right|
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 12: 53.3% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \sqrt[3]{1 + x} - \sqrt[3]{x} \end{array} \]
(FPCore (x) :precision binary64 (- (cbrt (+ 1.0 x)) (cbrt x)))
double code(double x) {
	return cbrt((1.0 + x)) - cbrt(x);
}
public static double code(double x) {
	return Math.cbrt((1.0 + x)) - Math.cbrt(x);
}
function code(x)
	return Float64(cbrt(Float64(1.0 + x)) - cbrt(x))
end
code[x_] := N[(N[Power[N[(1.0 + x), $MachinePrecision], 1/3], $MachinePrecision] - N[Power[x, 1/3], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\sqrt[3]{1 + x} - \sqrt[3]{x}
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 13: 3.6% accurate, 205.0× speedup?

\[\begin{array}{l} \\ 0 \end{array} \]
(FPCore (x) :precision binary64 0.0)
double code(double x) {
	return 0.0;
}
real(8) function code(x)
    real(8), intent (in) :: x
    code = 0.0d0
end function
public static double code(double x) {
	return 0.0;
}
def code(x):
	return 0.0
function code(x)
	return 0.0
end
function tmp = code(x)
	tmp = 0.0;
end
code[x_] := 0.0
\begin{array}{l}

\\
0
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Alternative 14: 49.9% accurate, 205.0× speedup?

\[\begin{array}{l} \\ 1 \end{array} \]
(FPCore (x) :precision binary64 1.0)
double code(double x) {
	return 1.0;
}
real(8) function code(x)
    real(8), intent (in) :: x
    code = 1.0d0
end function
public static double code(double x) {
	return 1.0;
}
def code(x):
	return 1.0
function code(x)
	return 1.0
end
function tmp = code(x)
	tmp = 1.0;
end
code[x_] := 1.0
\begin{array}{l}

\\
1
\end{array}
Derivation
    &prev;&pcontext;&pcontext2;&ctx;
  1. Add Preprocessing

Reproduce

?
herbie shell --seed 2023343 
(FPCore (x)
  :name "2cbrt (problem 3.3.4)"
  :precision binary64
  (- (cbrt (+ x 1.0)) (cbrt x)))