cos2 (problem 3.4.1)

Percentage Accurate: 51.4% → 99.8%

Time: 9.5s

Alternatives: 5

Speedup: 1.0×

Specification

Math

\[\begin{array}{l} \\ \frac{1 - \cos x}{x \cdot x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (/ (- 1.0 (cos x)) (* x x)))

C

double code(double x) {
	return (1.0 - cos(x)) / (x * x);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (1.0d0 - cos(x)) / (x * x)
end function

Java

public static double code(double x) {
	return (1.0 - Math.cos(x)) / (x * x);
}

Python

def code(x):
	return (1.0 - math.cos(x)) / (x * x)

Julia

function code(x)
	return Float64(Float64(1.0 - cos(x)) / Float64(x * x))
end

MATLAB

function tmp = code(x)
	tmp = (1.0 - cos(x)) / (x * x);
end

Wolfram

code[x_] := N[(N[(1.0 - N[Cos[x], $MachinePrecision]), $MachinePrecision] / N[(x * x), $MachinePrecision]), $MachinePrecision]

Initial Program: 51.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{1 - \cos x}{x \cdot x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (/ (- 1.0 (cos x)) (* x x)))

C

double code(double x) {
	return (1.0 - cos(x)) / (x * x);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (1.0d0 - cos(x)) / (x * x)
end function

Java

public static double code(double x) {
	return (1.0 - Math.cos(x)) / (x * x);
}

Python

def code(x):
	return (1.0 - math.cos(x)) / (x * x)

Julia

function code(x)
	return Float64(Float64(1.0 - cos(x)) / Float64(x * x))
end

MATLAB

function tmp = code(x)
	tmp = (1.0 - cos(x)) / (x * x);
end

Wolfram

code[x_] := N[(N[(1.0 - N[Cos[x], $MachinePrecision]), $MachinePrecision] / N[(x * x), $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.8% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \frac{\tan \left(x \cdot 0.5\right)}{x} \cdot \frac{\sin x}{x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (* (/ (tan (* x 0.5)) x) (/ (sin x) x)))

C

double code(double x) {
	return (tan((x * 0.5)) / x) * (sin(x) / x);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (tan((x * 0.5d0)) / x) * (sin(x) / x)
end function

Java

public static double code(double x) {
	return (Math.tan((x * 0.5)) / x) * (Math.sin(x) / x);
}

Python

def code(x):
	return (math.tan((x * 0.5)) / x) * (math.sin(x) / x)

Julia

function code(x)
	return Float64(Float64(tan(Float64(x * 0.5)) / x) * Float64(sin(x) / x))
end

MATLAB

function tmp = code(x)
	tmp = (tan((x * 0.5)) / x) * (sin(x) / x);
end

Wolfram

code[x_] := N[(N[(N[Tan[N[(x * 0.5), $MachinePrecision]], $MachinePrecision] / x), $MachinePrecision] * N[(N[Sin[x], $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.3%

   \[\frac{1 - \cos x}{x \cdot x} \]
2. Step-by-step derivation

   1. flip-- 53.1%

      \[\leadsto \frac{\color{blue}{\frac{1 \cdot 1 - \cos x \cdot \cos x}{1 + \cos x}}}{x \cdot x} \]

   2. div-inv 53.0%

      \[\leadsto \frac{\color{blue}{\left(1 \cdot 1 - \cos x \cdot \cos x\right) \cdot \frac{1}{1 + \cos x}}}{x \cdot x} \]

   3. metadata-eval 53.0%

      \[\leadsto \frac{\left(\color{blue}{1} - \cos x \cdot \cos x\right) \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   4. 1-sub-cos 74.4%

      \[\leadsto \frac{\color{blue}{\left(\sin x \cdot \sin x\right)} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   5. pow2 74.4%

      \[\leadsto \frac{\color{blue}{{\sin x}^{2}} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

3. Applied egg-rr 74.4%

   \[\leadsto \frac{\color{blue}{{\sin x}^{2} \cdot \frac{1}{1 + \cos x}}}{x \cdot x} \]
4. Step-by-step derivation

   1. unpow2 74.4%

      \[\leadsto \frac{\color{blue}{\left(\sin x \cdot \sin x\right)} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   2. associate-*l* 74.4%

      \[\leadsto \frac{\color{blue}{\sin x \cdot \left(\sin x \cdot \frac{1}{1 + \cos x}\right)}}{x \cdot x} \]

   3. associate-*r/ 74.4%

      \[\leadsto \frac{\sin x \cdot \color{blue}{\frac{\sin x \cdot 1}{1 + \cos x}}}{x \cdot x} \]

   4. *-rgt-identity 74.4%

      \[\leadsto \frac{\sin x \cdot \frac{\color{blue}{\sin x}}{1 + \cos x}}{x \cdot x} \]

   5. hang-0p-tan 74.7%

      \[\leadsto \frac{\sin x \cdot \color{blue}{\tan \left(\frac{x}{2}\right)}}{x \cdot x} \]

5. Simplified 74.7%

   \[\leadsto \frac{\color{blue}{\sin x \cdot \tan \left(\frac{x}{2}\right)}}{x \cdot x} \]
6. Step-by-step derivation

   1. *-commutative 74.7%

      \[\leadsto \frac{\color{blue}{\tan \left(\frac{x}{2}\right) \cdot \sin x}}{x \cdot x} \]

   2. times-frac 99.8%

      \[\leadsto \color{blue}{\frac{\tan \left(\frac{x}{2}\right)}{x} \cdot \frac{\sin x}{x}} \]

   3. div-inv 99.8%

      \[\leadsto \frac{\tan \color{blue}{\left(x \cdot \frac{1}{2}\right)}}{x} \cdot \frac{\sin x}{x} \]

   4. metadata-eval 99.8%

      \[\leadsto \frac{\tan \left(x \cdot \color{blue}{0.5}\right)}{x} \cdot \frac{\sin x}{x} \]

7. Applied egg-rr 99.8%

   \[\leadsto \color{blue}{\frac{\tan \left(x \cdot 0.5\right)}{x} \cdot \frac{\sin x}{x}} \]

8. Final simplification 99.8%

   \[\leadsto \frac{\tan \left(x \cdot 0.5\right)}{x} \cdot \frac{\sin x}{x} \]

Alternative 2: 74.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 0.0045:\\ \;\;\;\;0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \cos x}{x \cdot x}\\ \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (if (<= x 0.0045)
   (+ 0.5 (* -0.041666666666666664 (* x x)))
   (/ (- 1.0 (cos x)) (* x x))))

C

double code(double x) {
	double tmp;
	if (x <= 0.0045) {
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	} else {
		tmp = (1.0 - cos(x)) / (x * x);
	}
	return tmp;
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 0.0045d0) then
        tmp = 0.5d0 + ((-0.041666666666666664d0) * (x * x))
    else
        tmp = (1.0d0 - cos(x)) / (x * x)
    end if
    code = tmp
end function

Java

public static double code(double x) {
	double tmp;
	if (x <= 0.0045) {
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	} else {
		tmp = (1.0 - Math.cos(x)) / (x * x);
	}
	return tmp;
}

Python

def code(x):
	tmp = 0
	if x <= 0.0045:
		tmp = 0.5 + (-0.041666666666666664 * (x * x))
	else:
		tmp = (1.0 - math.cos(x)) / (x * x)
	return tmp

Julia

function code(x)
	tmp = 0.0
	if (x <= 0.0045)
		tmp = Float64(0.5 + Float64(-0.041666666666666664 * Float64(x * x)));
	else
		tmp = Float64(Float64(1.0 - cos(x)) / Float64(x * x));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 0.0045)
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	else
		tmp = (1.0 - cos(x)) / (x * x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := If[LessEqual[x, 0.0045], N[(0.5 + N[(-0.041666666666666664 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 - N[Cos[x], $MachinePrecision]), $MachinePrecision] / N[(x * x), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 0.00449999999999999966

   1. Initial program 35.7%

      \[\frac{1 - \cos x}{x \cdot x} \]

   2. Taylor expanded in x around 0 66.3%

      \[\leadsto \color{blue}{0.5 + -0.041666666666666664 \cdot {x}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 66.3%

         \[\leadsto 0.5 + -0.041666666666666664 \cdot \color{blue}{\left(x \cdot x\right)} \]

   4. Simplified 66.3%

      \[\leadsto \color{blue}{0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)} \]

   if 0.00449999999999999966 < x

   1. Initial program 98.0%

      \[\frac{1 - \cos x}{x \cdot x} \]
3. Recombined 2 regimes into one program.

4. Final simplification 75.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.0045:\\ \;\;\;\;0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \cos x}{x \cdot x}\\ \end{array} \]

Alternative 3: 74.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq 0.0045:\\ \;\;\;\;0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1 - \cos x}{x}}{x}\\ \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (if (<= x 0.0045)
   (+ 0.5 (* -0.041666666666666664 (* x x)))
   (/ (/ (- 1.0 (cos x)) x) x)))

C

double code(double x) {
	double tmp;
	if (x <= 0.0045) {
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	} else {
		tmp = ((1.0 - cos(x)) / x) / x;
	}
	return tmp;
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 0.0045d0) then
        tmp = 0.5d0 + ((-0.041666666666666664d0) * (x * x))
    else
        tmp = ((1.0d0 - cos(x)) / x) / x
    end if
    code = tmp
end function

Java

public static double code(double x) {
	double tmp;
	if (x <= 0.0045) {
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	} else {
		tmp = ((1.0 - Math.cos(x)) / x) / x;
	}
	return tmp;
}

Python

def code(x):
	tmp = 0
	if x <= 0.0045:
		tmp = 0.5 + (-0.041666666666666664 * (x * x))
	else:
		tmp = ((1.0 - math.cos(x)) / x) / x
	return tmp

Julia

function code(x)
	tmp = 0.0
	if (x <= 0.0045)
		tmp = Float64(0.5 + Float64(-0.041666666666666664 * Float64(x * x)));
	else
		tmp = Float64(Float64(Float64(1.0 - cos(x)) / x) / x);
	end
	return tmp
end

MATLAB

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 0.0045)
		tmp = 0.5 + (-0.041666666666666664 * (x * x));
	else
		tmp = ((1.0 - cos(x)) / x) / x;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := If[LessEqual[x, 0.0045], N[(0.5 + N[(-0.041666666666666664 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(1.0 - N[Cos[x], $MachinePrecision]), $MachinePrecision] / x), $MachinePrecision] / x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 0.00449999999999999966

   1. Initial program 35.7%

      \[\frac{1 - \cos x}{x \cdot x} \]

   2. Taylor expanded in x around 0 66.3%

      \[\leadsto \color{blue}{0.5 + -0.041666666666666664 \cdot {x}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 66.3%

         \[\leadsto 0.5 + -0.041666666666666664 \cdot \color{blue}{\left(x \cdot x\right)} \]

   4. Simplified 66.3%

      \[\leadsto \color{blue}{0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)} \]

   if 0.00449999999999999966 < x

   1. Initial program 98.0%

      \[\frac{1 - \cos x}{x \cdot x} \]
   2. Step-by-step derivation

      1. associate-/r* 99.2%

         \[\leadsto \color{blue}{\frac{\frac{1 - \cos x}{x}}{x}} \]

      2. div-inv 99.2%

         \[\leadsto \color{blue}{\frac{1 - \cos x}{x} \cdot \frac{1}{x}} \]

   3. Applied egg-rr 99.2%

      \[\leadsto \color{blue}{\frac{1 - \cos x}{x} \cdot \frac{1}{x}} \]
   4. Step-by-step derivation

      1. un-div-inv 99.2%

         \[\leadsto \color{blue}{\frac{\frac{1 - \cos x}{x}}{x}} \]

   5. Applied egg-rr 99.2%

      \[\leadsto \color{blue}{\frac{\frac{1 - \cos x}{x}}{x}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 75.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.0045:\\ \;\;\;\;0.5 + -0.041666666666666664 \cdot \left(x \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1 - \cos x}{x}}{x}\\ \end{array} \]

Alternative 4: 52.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ 0.5 \cdot \frac{\sin x}{x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (* 0.5 (/ (sin x) x)))

C

double code(double x) {
	return 0.5 * (sin(x) / x);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = 0.5d0 * (sin(x) / x)
end function

Java

public static double code(double x) {
	return 0.5 * (Math.sin(x) / x);
}

Python

def code(x):
	return 0.5 * (math.sin(x) / x)

Julia

function code(x)
	return Float64(0.5 * Float64(sin(x) / x))
end

MATLAB

function tmp = code(x)
	tmp = 0.5 * (sin(x) / x);
end

Wolfram

code[x_] := N[(0.5 * N[(N[Sin[x], $MachinePrecision] / x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.3%

   \[\frac{1 - \cos x}{x \cdot x} \]
2. Step-by-step derivation

   1. flip-- 53.1%

      \[\leadsto \frac{\color{blue}{\frac{1 \cdot 1 - \cos x \cdot \cos x}{1 + \cos x}}}{x \cdot x} \]

   2. div-inv 53.0%

      \[\leadsto \frac{\color{blue}{\left(1 \cdot 1 - \cos x \cdot \cos x\right) \cdot \frac{1}{1 + \cos x}}}{x \cdot x} \]

   3. metadata-eval 53.0%

      \[\leadsto \frac{\left(\color{blue}{1} - \cos x \cdot \cos x\right) \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   4. 1-sub-cos 74.4%

      \[\leadsto \frac{\color{blue}{\left(\sin x \cdot \sin x\right)} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   5. pow2 74.4%

      \[\leadsto \frac{\color{blue}{{\sin x}^{2}} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

3. Applied egg-rr 74.4%

   \[\leadsto \frac{\color{blue}{{\sin x}^{2} \cdot \frac{1}{1 + \cos x}}}{x \cdot x} \]
4. Step-by-step derivation

   1. unpow2 74.4%

      \[\leadsto \frac{\color{blue}{\left(\sin x \cdot \sin x\right)} \cdot \frac{1}{1 + \cos x}}{x \cdot x} \]

   2. associate-*l* 74.4%

      \[\leadsto \frac{\color{blue}{\sin x \cdot \left(\sin x \cdot \frac{1}{1 + \cos x}\right)}}{x \cdot x} \]

   3. associate-*r/ 74.4%

      \[\leadsto \frac{\sin x \cdot \color{blue}{\frac{\sin x \cdot 1}{1 + \cos x}}}{x \cdot x} \]

   4. *-rgt-identity 74.4%

      \[\leadsto \frac{\sin x \cdot \frac{\color{blue}{\sin x}}{1 + \cos x}}{x \cdot x} \]

   5. hang-0p-tan 74.7%

      \[\leadsto \frac{\sin x \cdot \color{blue}{\tan \left(\frac{x}{2}\right)}}{x \cdot x} \]

5. Simplified 74.7%

   \[\leadsto \frac{\color{blue}{\sin x \cdot \tan \left(\frac{x}{2}\right)}}{x \cdot x} \]
6. Step-by-step derivation

   1. *-commutative 74.7%

      \[\leadsto \frac{\color{blue}{\tan \left(\frac{x}{2}\right) \cdot \sin x}}{x \cdot x} \]

   2. times-frac 99.8%

      \[\leadsto \color{blue}{\frac{\tan \left(\frac{x}{2}\right)}{x} \cdot \frac{\sin x}{x}} \]

   3. div-inv 99.8%

      \[\leadsto \frac{\tan \color{blue}{\left(x \cdot \frac{1}{2}\right)}}{x} \cdot \frac{\sin x}{x} \]

   4. metadata-eval 99.8%

      \[\leadsto \frac{\tan \left(x \cdot \color{blue}{0.5}\right)}{x} \cdot \frac{\sin x}{x} \]

7. Applied egg-rr 99.8%

   \[\leadsto \color{blue}{\frac{\tan \left(x \cdot 0.5\right)}{x} \cdot \frac{\sin x}{x}} \]

8. Taylor expanded in x around 0 50.3%

   \[\leadsto \color{blue}{0.5} \cdot \frac{\sin x}{x} \]

9. Final simplification 50.3%

   \[\leadsto 0.5 \cdot \frac{\sin x}{x} \]

Alternative 5: 51.1% accurate, 107.0× speedup

Math

\[\begin{array}{l} \\ 0.5 \end{array} \]

FPCore

(FPCore (x) :precision binary64 0.5)

C

double code(double x) {
	return 0.5;
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = 0.5d0
end function

Java

public static double code(double x) {
	return 0.5;
}

Python

def code(x):
	return 0.5

Julia

function code(x)
	return 0.5
end

MATLAB

function tmp = code(x)
	tmp = 0.5;
end

Wolfram

code[x_] := 0.5

Derivation

1. Initial program 53.3%

   \[\frac{1 - \cos x}{x \cdot x} \]

2. Taylor expanded in x around 0 49.2%

   \[\leadsto \color{blue}{0.5} \]

3. Final simplification 49.2%

   \[\leadsto 0.5 \]

Reproduce

herbie shell --seed 2023192 
(FPCore (x)
  :name "cos2 (problem 3.4.1)"
  :precision binary64
  (/ (- 1.0 (cos x)) (* x x)))