Jmat.Real.erfi, branch x greater than or equal to 5

Percentage Accurate: 100.0% → 100.0%

Time: 10.8s

Alternatives: 12

Speedup: 3.5×

• 99.4% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

\[x \geq 0.5\]

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{\left|x\right|}\\ t_1 := \left(t_0 \cdot t_0\right) \cdot t_0\\ t_2 := \left(t_1 \cdot t_0\right) \cdot t_0\\ \left(\frac{1}{\sqrt{\pi}} \cdot e^{\left|x\right| \cdot \left|x\right|}\right) \cdot \left(\left(\left(t_0 + \frac{1}{2} \cdot t_1\right) + \frac{3}{4} \cdot t_2\right) + \frac{15}{8} \cdot \left(\left(t_2 \cdot t_0\right) \cdot t_0\right)\right) \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (let* ((t_0 (/ 1.0 (fabs x)))
        (t_1 (* (* t_0 t_0) t_0))
        (t_2 (* (* t_1 t_0) t_0)))
   (*
    (* (/ 1.0 (sqrt PI)) (exp (* (fabs x) (fabs x))))
    (+
     (+ (+ t_0 (* (/ 1.0 2.0) t_1)) (* (/ 3.0 4.0) t_2))
     (* (/ 15.0 8.0) (* (* t_2 t_0) t_0))))))

C

double code(double x) {
	double t_0 = 1.0 / fabs(x);
	double t_1 = (t_0 * t_0) * t_0;
	double t_2 = (t_1 * t_0) * t_0;
	return ((1.0 / sqrt(((double) M_PI))) * exp((fabs(x) * fabs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
}

Java

public static double code(double x) {
	double t_0 = 1.0 / Math.abs(x);
	double t_1 = (t_0 * t_0) * t_0;
	double t_2 = (t_1 * t_0) * t_0;
	return ((1.0 / Math.sqrt(Math.PI)) * Math.exp((Math.abs(x) * Math.abs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
}

Python

def code(x):
	t_0 = 1.0 / math.fabs(x)
	t_1 = (t_0 * t_0) * t_0
	t_2 = (t_1 * t_0) * t_0
	return ((1.0 / math.sqrt(math.pi)) * math.exp((math.fabs(x) * math.fabs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)))

Julia

function code(x)
	t_0 = Float64(1.0 / abs(x))
	t_1 = Float64(Float64(t_0 * t_0) * t_0)
	t_2 = Float64(Float64(t_1 * t_0) * t_0)
	return Float64(Float64(Float64(1.0 / sqrt(pi)) * exp(Float64(abs(x) * abs(x)))) * Float64(Float64(Float64(t_0 + Float64(Float64(1.0 / 2.0) * t_1)) + Float64(Float64(3.0 / 4.0) * t_2)) + Float64(Float64(15.0 / 8.0) * Float64(Float64(t_2 * t_0) * t_0))))
end

MATLAB

function tmp = code(x)
	t_0 = 1.0 / abs(x);
	t_1 = (t_0 * t_0) * t_0;
	t_2 = (t_1 * t_0) * t_0;
	tmp = ((1.0 / sqrt(pi)) * exp((abs(x) * abs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
end

Wolfram

code[x_] := Block[{t$95$0 = N[(1.0 / N[Abs[x], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(N[(t$95$1 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]}, N[(N[(N[(1.0 / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[Exp[N[(N[Abs[x], $MachinePrecision] * N[Abs[x], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(N[(N[(t$95$0 + N[(N[(1.0 / 2.0), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision] + N[(N[(3.0 / 4.0), $MachinePrecision] * t$95$2), $MachinePrecision]), $MachinePrecision] + N[(N[(15.0 / 8.0), $MachinePrecision] * N[(N[(t$95$2 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Initial Program: 100.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{\left|x\right|}\\ t_1 := \left(t_0 \cdot t_0\right) \cdot t_0\\ t_2 := \left(t_1 \cdot t_0\right) \cdot t_0\\ \left(\frac{1}{\sqrt{\pi}} \cdot e^{\left|x\right| \cdot \left|x\right|}\right) \cdot \left(\left(\left(t_0 + \frac{1}{2} \cdot t_1\right) + \frac{3}{4} \cdot t_2\right) + \frac{15}{8} \cdot \left(\left(t_2 \cdot t_0\right) \cdot t_0\right)\right) \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (let* ((t_0 (/ 1.0 (fabs x)))
        (t_1 (* (* t_0 t_0) t_0))
        (t_2 (* (* t_1 t_0) t_0)))
   (*
    (* (/ 1.0 (sqrt PI)) (exp (* (fabs x) (fabs x))))
    (+
     (+ (+ t_0 (* (/ 1.0 2.0) t_1)) (* (/ 3.0 4.0) t_2))
     (* (/ 15.0 8.0) (* (* t_2 t_0) t_0))))))

C

double code(double x) {
	double t_0 = 1.0 / fabs(x);
	double t_1 = (t_0 * t_0) * t_0;
	double t_2 = (t_1 * t_0) * t_0;
	return ((1.0 / sqrt(((double) M_PI))) * exp((fabs(x) * fabs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
}

Java

public static double code(double x) {
	double t_0 = 1.0 / Math.abs(x);
	double t_1 = (t_0 * t_0) * t_0;
	double t_2 = (t_1 * t_0) * t_0;
	return ((1.0 / Math.sqrt(Math.PI)) * Math.exp((Math.abs(x) * Math.abs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
}

Python

def code(x):
	t_0 = 1.0 / math.fabs(x)
	t_1 = (t_0 * t_0) * t_0
	t_2 = (t_1 * t_0) * t_0
	return ((1.0 / math.sqrt(math.pi)) * math.exp((math.fabs(x) * math.fabs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)))

Julia

function code(x)
	t_0 = Float64(1.0 / abs(x))
	t_1 = Float64(Float64(t_0 * t_0) * t_0)
	t_2 = Float64(Float64(t_1 * t_0) * t_0)
	return Float64(Float64(Float64(1.0 / sqrt(pi)) * exp(Float64(abs(x) * abs(x)))) * Float64(Float64(Float64(t_0 + Float64(Float64(1.0 / 2.0) * t_1)) + Float64(Float64(3.0 / 4.0) * t_2)) + Float64(Float64(15.0 / 8.0) * Float64(Float64(t_2 * t_0) * t_0))))
end

MATLAB

function tmp = code(x)
	t_0 = 1.0 / abs(x);
	t_1 = (t_0 * t_0) * t_0;
	t_2 = (t_1 * t_0) * t_0;
	tmp = ((1.0 / sqrt(pi)) * exp((abs(x) * abs(x)))) * (((t_0 + ((1.0 / 2.0) * t_1)) + ((3.0 / 4.0) * t_2)) + ((15.0 / 8.0) * ((t_2 * t_0) * t_0)));
end

Wolfram

code[x_] := Block[{t$95$0 = N[(1.0 / N[Abs[x], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(N[(t$95$1 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]}, N[(N[(N[(1.0 / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[Exp[N[(N[Abs[x], $MachinePrecision] * N[Abs[x], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(N[(N[(t$95$0 + N[(N[(1.0 / 2.0), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision] + N[(N[(3.0 / 4.0), $MachinePrecision] * t$95$2), $MachinePrecision]), $MachinePrecision] + N[(N[(15.0 / 8.0), $MachinePrecision] * N[(N[(t$95$2 * t$95$0), $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Alternative 1: 100.0% accurate, 2.4× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} t_0 := \frac{1}{\left|x\right|}\\ \mathsf{fma}\left(1 + \frac{0.5}{x \cdot x}, t_0, \frac{{t_0}^{4}}{\left|x\right|} \cdot \left(0.75 + \frac{1.875}{x \cdot x}\right)\right) \cdot \frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (let* ((t_0 (/ 1.0 (fabs x))))
   (*
    (fma
     (+ 1.0 (/ 0.5 (* x x)))
     t_0
     (* (/ (pow t_0 4.0) (fabs x)) (+ 0.75 (/ 1.875 (* x x)))))
    (/ (pow (exp x) x) (sqrt PI)))))

C

x = abs(x);
double code(double x) {
	double t_0 = 1.0 / fabs(x);
	return fma((1.0 + (0.5 / (x * x))), t_0, ((pow(t_0, 4.0) / fabs(x)) * (0.75 + (1.875 / (x * x))))) * (pow(exp(x), x) / sqrt(((double) M_PI)));
}

Julia

x = abs(x)
function code(x)
	t_0 = Float64(1.0 / abs(x))
	return Float64(fma(Float64(1.0 + Float64(0.5 / Float64(x * x))), t_0, Float64(Float64((t_0 ^ 4.0) / abs(x)) * Float64(0.75 + Float64(1.875 / Float64(x * x))))) * Float64((exp(x) ^ x) / sqrt(pi)))
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := Block[{t$95$0 = N[(1.0 / N[Abs[x], $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(1.0 + N[(0.5 / N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * t$95$0 + N[(N[(N[Power[t$95$0, 4.0], $MachinePrecision] / N[Abs[x], $MachinePrecision]), $MachinePrecision] * N[(0.75 + N[(1.875 / N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[Exp[x], $MachinePrecision], x], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 100.0%

   \[\mathsf{fma}\left(1 + \frac{0.5}{x \cdot x}, \frac{1}{\left|x\right|}, \frac{{\left(\frac{1}{\left|x\right|}\right)}^{4}}{\left|x\right|} \cdot \left(0.75 + \frac{1.875}{x \cdot x}\right)\right) \cdot \frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \]

Alternative 2: 100.0% accurate, 3.5× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + {\left(\frac{1}{x}\right)}^{5} \cdot \left(0.75 + \frac{\frac{1.875}{x}}{x}\right)\right) \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (*
  (/ (pow (exp x) x) (sqrt PI))
  (+
   (/ (+ 1.0 (/ 0.5 (* x x))) (fabs x))
   (* (pow (/ 1.0 x) 5.0) (+ 0.75 (/ (/ 1.875 x) x))))))

C

x = abs(x);
double code(double x) {
	return (pow(exp(x), x) / sqrt(((double) M_PI))) * (((1.0 + (0.5 / (x * x))) / fabs(x)) + (pow((1.0 / x), 5.0) * (0.75 + ((1.875 / x) / x))));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return (Math.pow(Math.exp(x), x) / Math.sqrt(Math.PI)) * (((1.0 + (0.5 / (x * x))) / Math.abs(x)) + (Math.pow((1.0 / x), 5.0) * (0.75 + ((1.875 / x) / x))));
}

Python

x = abs(x)
def code(x):
	return (math.pow(math.exp(x), x) / math.sqrt(math.pi)) * (((1.0 + (0.5 / (x * x))) / math.fabs(x)) + (math.pow((1.0 / x), 5.0) * (0.75 + ((1.875 / x) / x))))

Julia

x = abs(x)
function code(x)
	return Float64(Float64((exp(x) ^ x) / sqrt(pi)) * Float64(Float64(Float64(1.0 + Float64(0.5 / Float64(x * x))) / abs(x)) + Float64((Float64(1.0 / x) ^ 5.0) * Float64(0.75 + Float64(Float64(1.875 / x) / x)))))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((exp(x) ^ x) / sqrt(pi)) * (((1.0 + (0.5 / (x * x))) / abs(x)) + (((1.0 / x) ^ 5.0) * (0.75 + ((1.875 / x) / x))));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[Power[N[Exp[x], $MachinePrecision], x], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[(N[(N[(1.0 + N[(0.5 / N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Abs[x], $MachinePrecision]), $MachinePrecision] + N[(N[Power[N[(1.0 / x), $MachinePrecision], 5.0], $MachinePrecision] * N[(0.75 + N[(N[(1.875 / x), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[\frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + {\left(\frac{1}{x}\right)}^{5} \cdot \left(0.75 + \frac{\frac{1.875}{x}}{x}\right)\right) \]

Alternative 3: 100.0% accurate, 3.5× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + \left(0.75 + \frac{\frac{1.875}{x}}{x}\right) \cdot {x}^{-5}\right) \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (*
  (/ (pow (exp x) x) (sqrt PI))
  (+
   (/ (+ 1.0 (/ 0.5 (* x x))) (fabs x))
   (* (+ 0.75 (/ (/ 1.875 x) x)) (pow x -5.0)))))

C

x = abs(x);
double code(double x) {
	return (pow(exp(x), x) / sqrt(((double) M_PI))) * (((1.0 + (0.5 / (x * x))) / fabs(x)) + ((0.75 + ((1.875 / x) / x)) * pow(x, -5.0)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return (Math.pow(Math.exp(x), x) / Math.sqrt(Math.PI)) * (((1.0 + (0.5 / (x * x))) / Math.abs(x)) + ((0.75 + ((1.875 / x) / x)) * Math.pow(x, -5.0)));
}

Python

x = abs(x)
def code(x):
	return (math.pow(math.exp(x), x) / math.sqrt(math.pi)) * (((1.0 + (0.5 / (x * x))) / math.fabs(x)) + ((0.75 + ((1.875 / x) / x)) * math.pow(x, -5.0)))

Julia

x = abs(x)
function code(x)
	return Float64(Float64((exp(x) ^ x) / sqrt(pi)) * Float64(Float64(Float64(1.0 + Float64(0.5 / Float64(x * x))) / abs(x)) + Float64(Float64(0.75 + Float64(Float64(1.875 / x) / x)) * (x ^ -5.0))))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((exp(x) ^ x) / sqrt(pi)) * (((1.0 + (0.5 / (x * x))) / abs(x)) + ((0.75 + ((1.875 / x) / x)) * (x ^ -5.0)));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[Power[N[Exp[x], $MachinePrecision], x], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[(N[(N[(1.0 + N[(0.5 / N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Abs[x], $MachinePrecision]), $MachinePrecision] + N[(N[(0.75 + N[(N[(1.875 / x), $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision] * N[Power[x, -5.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[\frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + \left(0.75 + \frac{\frac{1.875}{x}}{x}\right) \cdot {x}^{-5}\right) \]

Alternative 4: 99.6% accurate, 3.5× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + \frac{1.875}{{x}^{7}}\right) \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (*
  (/ (pow (exp x) x) (sqrt PI))
  (+ (/ (+ 1.0 (/ 0.5 (* x x))) (fabs x)) (/ 1.875 (pow x 7.0)))))

C

x = abs(x);
double code(double x) {
	return (pow(exp(x), x) / sqrt(((double) M_PI))) * (((1.0 + (0.5 / (x * x))) / fabs(x)) + (1.875 / pow(x, 7.0)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return (Math.pow(Math.exp(x), x) / Math.sqrt(Math.PI)) * (((1.0 + (0.5 / (x * x))) / Math.abs(x)) + (1.875 / Math.pow(x, 7.0)));
}

Python

x = abs(x)
def code(x):
	return (math.pow(math.exp(x), x) / math.sqrt(math.pi)) * (((1.0 + (0.5 / (x * x))) / math.fabs(x)) + (1.875 / math.pow(x, 7.0)))

Julia

x = abs(x)
function code(x)
	return Float64(Float64((exp(x) ^ x) / sqrt(pi)) * Float64(Float64(Float64(1.0 + Float64(0.5 / Float64(x * x))) / abs(x)) + Float64(1.875 / (x ^ 7.0))))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((exp(x) ^ x) / sqrt(pi)) * (((1.0 + (0.5 / (x * x))) / abs(x)) + (1.875 / (x ^ 7.0)));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[Power[N[Exp[x], $MachinePrecision], x], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[(N[(N[(1.0 + N[(0.5 / N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Abs[x], $MachinePrecision]), $MachinePrecision] + N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{{\left(e^{x}\right)}^{x}}{\sqrt{\pi}} \cdot \left(\frac{1 + \frac{0.5}{x \cdot x}}{\left|x\right|} + \frac{1.875}{{x}^{7}}\right) \]

Alternative 5: 99.6% accurate, 4.3× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot e^{x \cdot x - \log \left(\sqrt{\pi}\right)} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 1.875 (pow x 7.0))) (exp (- (* x x) (log (sqrt PI))))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (1.875 / pow(x, 7.0))) * exp(((x * x) - log(sqrt(((double) M_PI)))));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (1.875 / Math.pow(x, 7.0))) * Math.exp(((x * x) - Math.log(Math.sqrt(Math.PI))));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (1.875 / math.pow(x, 7.0))) * math.exp(((x * x) - math.log(math.sqrt(math.pi))))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(1.875 / (x ^ 7.0))) * exp(Float64(Float64(x * x) - log(sqrt(pi)))))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (1.875 / (x ^ 7.0))) * exp(((x * x) - log(sqrt(pi))));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Exp[N[(N[(x * x), $MachinePrecision] - N[Log[N[Sqrt[Pi], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot e^{x \cdot x - \log \left(\sqrt{\pi}\right)} \]

Alternative 6: 99.6% accurate, 5.3× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \frac{e^{x \cdot x}}{\sqrt{\pi}} \cdot \left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (/ (exp (* x x)) (sqrt PI)) (+ (/ 1.0 x) (/ 0.75 (pow x 5.0)))))

C

x = abs(x);
double code(double x) {
	return (exp((x * x)) / sqrt(((double) M_PI))) * ((1.0 / x) + (0.75 / pow(x, 5.0)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return (Math.exp((x * x)) / Math.sqrt(Math.PI)) * ((1.0 / x) + (0.75 / Math.pow(x, 5.0)));
}

Python

x = abs(x)
def code(x):
	return (math.exp((x * x)) / math.sqrt(math.pi)) * ((1.0 / x) + (0.75 / math.pow(x, 5.0)))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(exp(Float64(x * x)) / sqrt(pi)) * Float64(Float64(1.0 / x) + Float64(0.75 / (x ^ 5.0))))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = (exp((x * x)) / sqrt(pi)) * ((1.0 / x) + (0.75 / (x ^ 5.0)));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[Exp[N[(x * x), $MachinePrecision]], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 / x), $MachinePrecision] + N[(0.75 / N[Power[x, 5.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\frac{e^{x \cdot x}}{\sqrt{\pi}} \cdot \left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \]

Alternative 7: 99.6% accurate, 5.3× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{e^{x \cdot x}}{\sqrt{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 1.875 (pow x 7.0))) (/ (exp (* x x)) (sqrt PI))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (1.875 / pow(x, 7.0))) * (exp((x * x)) / sqrt(((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (1.875 / Math.pow(x, 7.0))) * (Math.exp((x * x)) / Math.sqrt(Math.PI));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (1.875 / math.pow(x, 7.0))) * (math.exp((x * x)) / math.sqrt(math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(1.875 / (x ^ 7.0))) * Float64(exp(Float64(x * x)) / sqrt(pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (1.875 / (x ^ 7.0))) * (exp((x * x)) / sqrt(pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Exp[N[(x * x), $MachinePrecision]], $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

   \[\left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{e^{x \cdot x}}{\sqrt{\pi}} \]

Alternative 8: 50.9% accurate, 6.9× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \cdot \frac{1 + x \cdot x}{\sqrt{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 0.75 (pow x 5.0))) (/ (+ 1.0 (* x x)) (sqrt PI))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (0.75 / pow(x, 5.0))) * ((1.0 + (x * x)) / sqrt(((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (0.75 / Math.pow(x, 5.0))) * ((1.0 + (x * x)) / Math.sqrt(Math.PI));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (0.75 / math.pow(x, 5.0))) * ((1.0 + (x * x)) / math.sqrt(math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(0.75 / (x ^ 5.0))) * Float64(Float64(1.0 + Float64(x * x)) / sqrt(pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (0.75 / (x ^ 5.0))) * ((1.0 + (x * x)) / sqrt(pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(0.75 / N[Power[x, 5.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 + N[(x * x), $MachinePrecision]), $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.6%

   \[\left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \cdot \frac{1 + x \cdot x}{\sqrt{\pi}} \]

Alternative 9: 50.9% accurate, 6.9× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{1 + x \cdot x}{\sqrt{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 1.875 (pow x 7.0))) (/ (+ 1.0 (* x x)) (sqrt PI))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (1.875 / pow(x, 7.0))) * ((1.0 + (x * x)) / sqrt(((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (1.875 / Math.pow(x, 7.0))) * ((1.0 + (x * x)) / Math.sqrt(Math.PI));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (1.875 / math.pow(x, 7.0))) * ((1.0 + (x * x)) / math.sqrt(math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(1.875 / (x ^ 7.0))) * Float64(Float64(1.0 + Float64(x * x)) / sqrt(pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (1.875 / (x ^ 7.0))) * ((1.0 + (x * x)) / sqrt(pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 + N[(x * x), $MachinePrecision]), $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 51.7%

   \[\left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{1 + x \cdot x}{\sqrt{\pi}} \]

Alternative 10: 2.3% accurate, 7.0× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \cdot \frac{1}{\sqrt{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 0.75 (pow x 5.0))) (/ 1.0 (sqrt PI))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (0.75 / pow(x, 5.0))) * (1.0 / sqrt(((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (0.75 / Math.pow(x, 5.0))) * (1.0 / Math.sqrt(Math.PI));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (0.75 / math.pow(x, 5.0))) * (1.0 / math.sqrt(math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(0.75 / (x ^ 5.0))) * Float64(1.0 / sqrt(pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (0.75 / (x ^ 5.0))) * (1.0 / sqrt(pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(0.75 / N[Power[x, 5.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(1.0 / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 2.3%

   \[\left(\frac{1}{x} + \frac{0.75}{{x}^{5}}\right) \cdot \frac{1}{\sqrt{\pi}} \]

Alternative 11: 2.3% accurate, 7.0× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{1}{\sqrt{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (* (+ (/ 1.0 x) (/ 1.875 (pow x 7.0))) (/ 1.0 (sqrt PI))))

C

x = abs(x);
double code(double x) {
	return ((1.0 / x) + (1.875 / pow(x, 7.0))) * (1.0 / sqrt(((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return ((1.0 / x) + (1.875 / Math.pow(x, 7.0))) * (1.0 / Math.sqrt(Math.PI));
}

Python

x = abs(x)
def code(x):
	return ((1.0 / x) + (1.875 / math.pow(x, 7.0))) * (1.0 / math.sqrt(math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(Float64(1.0 / x) + Float64(1.875 / (x ^ 7.0))) * Float64(1.0 / sqrt(pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = ((1.0 / x) + (1.875 / (x ^ 7.0))) * (1.0 / sqrt(pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(N[(1.0 / x), $MachinePrecision] + N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(1.0 / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 2.4%

   \[\left(\frac{1}{x} + \frac{1.875}{{x}^{7}}\right) \cdot \frac{1}{\sqrt{\pi}} \]

Alternative 12: 1.7% accurate, 7.1× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \frac{1.875}{{x}^{7}} \cdot \sqrt{\frac{1}{\pi}} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 (* (/ 1.875 (pow x 7.0)) (sqrt (/ 1.0 PI))))

C

x = abs(x);
double code(double x) {
	return (1.875 / pow(x, 7.0)) * sqrt((1.0 / ((double) M_PI)));
}

Java

x = Math.abs(x);
public static double code(double x) {
	return (1.875 / Math.pow(x, 7.0)) * Math.sqrt((1.0 / Math.PI));
}

Python

x = abs(x)
def code(x):
	return (1.875 / math.pow(x, 7.0)) * math.sqrt((1.0 / math.pi))

Julia

x = abs(x)
function code(x)
	return Float64(Float64(1.875 / (x ^ 7.0)) * sqrt(Float64(1.0 / pi)))
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = (1.875 / (x ^ 7.0)) * sqrt((1.0 / pi));
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := N[(N[(1.875 / N[Power[x, 7.0], $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(1.0 / Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 1.8%

   \[\frac{1.875}{{x}^{7}} \cdot \sqrt{\frac{1}{\pi}} \]

Reproduce

herbie shell --seed 2023297 
(FPCore (x)
  :name "Jmat.Real.erfi, branch x greater than or equal to 5"
  :precision binary64
  :pre (>= x 0.5)
  (* (* (/ 1.0 (sqrt PI)) (exp (* (fabs x) (fabs x)))) (+ (+ (+ (/ 1.0 (fabs x)) (* (/ 1.0 2.0) (* (* (/ 1.0 (fabs x)) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))))) (* (/ 3.0 4.0) (* (* (* (* (/ 1.0 (fabs x)) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))))) (* (/ 15.0 8.0) (* (* (* (* (* (* (/ 1.0 (fabs x)) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x))) (/ 1.0 (fabs x)))))))