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

Percentage Accurate: 100.0% → 99.6%

Time: 7.0s

Alternatives: 1

Speedup: N/A×

• 99.5% 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: 99.6% accurate, 4.3× speedup

Math

\[\begin{array}{l} \\ \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \frac{1.875}{{x}^{6}}\right) \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (* (/ (/ (exp (* x x)) (fabs x)) (sqrt PI)) (+ 1.0 (/ 1.875 (pow x 6.0)))))

C

double code(double x) {
	return ((exp((x * x)) / fabs(x)) / sqrt(((double) M_PI))) * (1.0 + (1.875 / pow(x, 6.0)));
}

Java

public static double code(double x) {
	return ((Math.exp((x * x)) / Math.abs(x)) / Math.sqrt(Math.PI)) * (1.0 + (1.875 / Math.pow(x, 6.0)));
}

Python

def code(x):
	return ((math.exp((x * x)) / math.fabs(x)) / math.sqrt(math.pi)) * (1.0 + (1.875 / math.pow(x, 6.0)))

Julia

function code(x)
	return Float64(Float64(Float64(exp(Float64(x * x)) / abs(x)) / sqrt(pi)) * Float64(1.0 + Float64(1.875 / (x ^ 6.0))))
end

MATLAB

function tmp = code(x)
	tmp = ((exp((x * x)) / abs(x)) / sqrt(pi)) * (1.0 + (1.875 / (x ^ 6.0)));
end

Wolfram

code[x_] := N[(N[(N[(N[Exp[N[(x * x), $MachinePrecision]], $MachinePrecision] / N[Abs[x], $MachinePrecision]), $MachinePrecision] / N[Sqrt[Pi], $MachinePrecision]), $MachinePrecision] * N[(1.0 + N[(1.875 / N[Power[x, 6.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[\left(\frac{1}{\sqrt{\pi}} \cdot e^{\left|x\right| \cdot \left|x\right|}\right) \cdot \left(\left(\left(\frac{1}{\left|x\right|} + \frac{1}{2} \cdot \left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right)\right) + \frac{3}{4} \cdot \left(\left(\left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right)\right) + \frac{15}{8} \cdot \left(\left(\left(\left(\left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right)\right) \]
2. Step-by-step derivation

   1. associate-+l+ 100.0%

      \[\leadsto \left(\frac{1}{\sqrt{\pi}} \cdot e^{\left|x\right| \cdot \left|x\right|}\right) \cdot \color{blue}{\left(\left(\frac{1}{\left|x\right|} + \frac{1}{2} \cdot \left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right)\right) + \left(\frac{3}{4} \cdot \left(\left(\left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) + \frac{15}{8} \cdot \left(\left(\left(\left(\left(\left(\frac{1}{\left|x\right|} \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right) \cdot \frac{1}{\left|x\right|}\right)\right)\right)} \]

3. Simplified 100.0%

   \[\leadsto \color{blue}{\frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \left(\frac{1.875}{{x}^{6}} + \frac{0.5 + \frac{0.75}{x \cdot x}}{x \cdot x}\right)\right)} \]

4. Taylor expanded in x around inf 100.0%

   \[\leadsto \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \left(\frac{1.875}{{x}^{6}} + \color{blue}{\frac{0.5}{{x}^{2}}}\right)\right) \]
5. Step-by-step derivation

   1. unpow2 100.0%

      \[\leadsto \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \left(\frac{1.875}{{x}^{6}} + \frac{0.5}{\color{blue}{x \cdot x}}\right)\right) \]

6. Simplified 100.0%

   \[\leadsto \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \left(\frac{1.875}{{x}^{6}} + \color{blue}{\frac{0.5}{x \cdot x}}\right)\right) \]

7. Taylor expanded in x around 0 100.0%

   \[\leadsto \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \color{blue}{\frac{1.875}{{x}^{6}}}\right) \]

8. Final simplification 100.0%

   \[\leadsto \frac{\frac{e^{x \cdot x}}{\left|x\right|}}{\sqrt{\pi}} \cdot \left(1 + \frac{1.875}{{x}^{6}}\right) \]

Reproduce

herbie shell --seed 2023187 
(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)))))))