Quadratic roots, wide range

Percentage Accurate: 17.7% → 97.7%

Time: 15.8s

Alternatives: 5

Speedup: 29.0×

Specification

\[\left(\left(4.930380657631324 \cdot 10^{-32} < a \land a < 2.028240960365167 \cdot 10^{+31}\right) \land \left(4.930380657631324 \cdot 10^{-32} < b \land b < 2.028240960365167 \cdot 10^{+31}\right)\right) \land \left(4.930380657631324 \cdot 10^{-32} < c \land c < 2.028240960365167 \cdot 10^{+31}\right)\]

Math

\[\begin{array}{l} \\ \frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (/ (+ (- b) (sqrt (- (* b b) (* (* 4.0 a) c)))) (* 2.0 a)))

C

double code(double a, double b, double c) {
	return (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}

Fortran

real(8) function code(a, b, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    code = (-b + sqrt(((b * b) - ((4.0d0 * a) * c)))) / (2.0d0 * a)
end function

Java

public static double code(double a, double b, double c) {
	return (-b + Math.sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}

Python

def code(a, b, c):
	return (-b + math.sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a)

Julia

function code(a, b, c)
	return Float64(Float64(Float64(-b) + sqrt(Float64(Float64(b * b) - Float64(Float64(4.0 * a) * c)))) / Float64(2.0 * a))
end

MATLAB

function tmp = code(a, b, c)
	tmp = (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
end

Wolfram

code[a_, b_, c_] := N[(N[((-b) + N[Sqrt[N[(N[(b * b), $MachinePrecision] - N[(N[(4.0 * a), $MachinePrecision] * c), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]

Initial Program: 17.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (/ (+ (- b) (sqrt (- (* b b) (* (* 4.0 a) c)))) (* 2.0 a)))

C

double code(double a, double b, double c) {
	return (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}

Fortran

real(8) function code(a, b, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    code = (-b + sqrt(((b * b) - ((4.0d0 * a) * c)))) / (2.0d0 * a)
end function

Java

public static double code(double a, double b, double c) {
	return (-b + Math.sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}

Python

def code(a, b, c):
	return (-b + math.sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a)

Julia

function code(a, b, c)
	return Float64(Float64(Float64(-b) + sqrt(Float64(Float64(b * b) - Float64(Float64(4.0 * a) * c)))) / Float64(2.0 * a))
end

MATLAB

function tmp = code(a, b, c)
	tmp = (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
end

Wolfram

code[a_, b_, c_] := N[(N[((-b) + N[Sqrt[N[(N[(b * b), $MachinePrecision] - N[(N[(4.0 * a), $MachinePrecision] * c), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]

Alternative 1: 97.7% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\frac{-5 \cdot \left({a}^{3} \cdot {c}^{4}\right)}{{b}^{6}}}{b}\right) - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (-
  (-
   (fma
    -2.0
    (* (/ (pow c 3.0) (pow b 5.0)) (* a a))
    (/ (/ (* -5.0 (* (pow a 3.0) (pow c 4.0))) (pow b 6.0)) b))
   (/ c b))
  (* a (/ c (/ (pow b 3.0) c)))))

C

double code(double a, double b, double c) {
	return (fma(-2.0, ((pow(c, 3.0) / pow(b, 5.0)) * (a * a)), (((-5.0 * (pow(a, 3.0) * pow(c, 4.0))) / pow(b, 6.0)) / b)) - (c / b)) - (a * (c / (pow(b, 3.0) / c)));
}

Julia

function code(a, b, c)
	return Float64(Float64(fma(-2.0, Float64(Float64((c ^ 3.0) / (b ^ 5.0)) * Float64(a * a)), Float64(Float64(Float64(-5.0 * Float64((a ^ 3.0) * (c ^ 4.0))) / (b ^ 6.0)) / b)) - Float64(c / b)) - Float64(a * Float64(c / Float64((b ^ 3.0) / c))))
end

Wolfram

code[a_, b_, c_] := N[(N[(N[(-2.0 * N[(N[(N[Power[c, 3.0], $MachinePrecision] / N[Power[b, 5.0], $MachinePrecision]), $MachinePrecision] * N[(a * a), $MachinePrecision]), $MachinePrecision] + N[(N[(N[(-5.0 * N[(N[Power[a, 3.0], $MachinePrecision] * N[Power[c, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[b, 6.0], $MachinePrecision]), $MachinePrecision] / b), $MachinePrecision]), $MachinePrecision] - N[(c / b), $MachinePrecision]), $MachinePrecision] - N[(a * N[(c / N[(N[Power[b, 3.0], $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.3%

   \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]

2. Taylor expanded in a around 0 98.0%

   \[\leadsto \color{blue}{-1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + \left(-1 \cdot \frac{c}{b} + \left(-0.25 \cdot \frac{{a}^{3} \cdot \left(16 \cdot \frac{{c}^{4}}{{b}^{6}} + {\left(-2 \cdot \frac{{c}^{2}}{{b}^{3}}\right)}^{2}\right)}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right)} \]

4. Simplified 98.0%

   \[\leadsto \color{blue}{\left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\left(-0.25 \cdot {a}^{3}\right) \cdot \mathsf{fma}\left(16, \frac{{c}^{4}}{{b}^{6}}, 4 \cdot \frac{{c}^{4}}{{b}^{6}}\right)}{b}\right) - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a} \]

5. Taylor expanded in c around 0 98.0%

   \[\leadsto \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\color{blue}{-5 \cdot \frac{{c}^{4} \cdot {a}^{3}}{{b}^{6}}}}{b}\right) - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
6. Step-by-step derivation

   1. associate-*r/ 98.0%

      \[\leadsto \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\color{blue}{\frac{-5 \cdot \left({c}^{4} \cdot {a}^{3}\right)}{{b}^{6}}}}{b}\right) - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

   2. *-commutative 98.0%

      \[\leadsto \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\frac{-5 \cdot \color{blue}{\left({a}^{3} \cdot {c}^{4}\right)}}{{b}^{6}}}{b}\right) - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

7. Simplified 98.0%

   \[\leadsto \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\color{blue}{\frac{-5 \cdot \left({a}^{3} \cdot {c}^{4}\right)}{{b}^{6}}}}{b}\right) - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

8. Final simplification 98.0%

   \[\leadsto \left(\mathsf{fma}\left(-2, \frac{{c}^{3}}{{b}^{5}} \cdot \left(a \cdot a\right), \frac{\frac{-5 \cdot \left({a}^{3} \cdot {c}^{4}\right)}{{b}^{6}}}{b}\right) - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \]

Alternative 2: 97.7% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \mathsf{fma}\left(-0.25, \frac{{\left(c \cdot a\right)}^{4}}{a} \cdot \frac{20}{{b}^{7}}, \frac{-2 \cdot \left(a \cdot \left({c}^{3} \cdot a\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{b \cdot \left(b \cdot b\right)}{c}} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (-
  (fma
   -0.25
   (* (/ (pow (* c a) 4.0) a) (/ 20.0 (pow b 7.0)))
   (- (/ (* -2.0 (* a (* (pow c 3.0) a))) (pow b 5.0)) (/ c b)))
  (* a (/ c (/ (* b (* b b)) c)))))

C

double code(double a, double b, double c) {
	return fma(-0.25, ((pow((c * a), 4.0) / a) * (20.0 / pow(b, 7.0))), (((-2.0 * (a * (pow(c, 3.0) * a))) / pow(b, 5.0)) - (c / b))) - (a * (c / ((b * (b * b)) / c)));
}

Julia

function code(a, b, c)
	return Float64(fma(-0.25, Float64(Float64((Float64(c * a) ^ 4.0) / a) * Float64(20.0 / (b ^ 7.0))), Float64(Float64(Float64(-2.0 * Float64(a * Float64((c ^ 3.0) * a))) / (b ^ 5.0)) - Float64(c / b))) - Float64(a * Float64(c / Float64(Float64(b * Float64(b * b)) / c))))
end

Wolfram

code[a_, b_, c_] := N[(N[(-0.25 * N[(N[(N[Power[N[(c * a), $MachinePrecision], 4.0], $MachinePrecision] / a), $MachinePrecision] * N[(20.0 / N[Power[b, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[(-2.0 * N[(a * N[(N[Power[c, 3.0], $MachinePrecision] * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[b, 5.0], $MachinePrecision]), $MachinePrecision] - N[(c / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(a * N[(c / N[(N[(b * N[(b * b), $MachinePrecision]), $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.3%

   \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]

2. Taylor expanded in b around inf 98.0%

   \[\leadsto \color{blue}{-1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + \left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right)} \]

4. Simplified 98.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(-0.25, \frac{\mathsf{fma}\left(16, {c}^{4} \cdot {a}^{4}, 4 \cdot \left({c}^{4} \cdot {a}^{4}\right)\right)}{a \cdot {b}^{7}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a} \]

5. Taylor expanded in c around 0 98.0%

   \[\leadsto \mathsf{fma}\left(-0.25, \color{blue}{\frac{{c}^{4} \cdot \left(4 \cdot {a}^{4} + 16 \cdot {a}^{4}\right)}{a \cdot {b}^{7}}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
6. Step-by-step derivation

   1. distribute-rgt-out 98.0%

      \[\leadsto \mathsf{fma}\left(-0.25, \frac{{c}^{4} \cdot \color{blue}{\left({a}^{4} \cdot \left(4 + 16\right)\right)}}{a \cdot {b}^{7}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

   2. associate-*r* 98.0%

      \[\leadsto \mathsf{fma}\left(-0.25, \frac{\color{blue}{\left({c}^{4} \cdot {a}^{4}\right) \cdot \left(4 + 16\right)}}{a \cdot {b}^{7}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

   3. times-frac 98.0%

      \[\leadsto \mathsf{fma}\left(-0.25, \color{blue}{\frac{{c}^{4} \cdot {a}^{4}}{a} \cdot \frac{4 + 16}{{b}^{7}}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

7. Simplified 98.0%

   \[\leadsto \mathsf{fma}\left(-0.25, \color{blue}{\frac{{\left(c \cdot a\right)}^{4}}{a} \cdot \frac{20}{{b}^{7}}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
8. Step-by-step derivation

   1. unpow3 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \frac{c \cdot a}{\frac{\color{blue}{\left(b \cdot b\right) \cdot b}}{c}} \]

9. Applied egg-rr 98.0%

   \[\leadsto \mathsf{fma}\left(-0.25, \frac{{\left(c \cdot a\right)}^{4}}{a} \cdot \frac{20}{{b}^{7}}, \frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{\color{blue}{\left(b \cdot b\right) \cdot b}}{c}} \cdot a \]

10. Final simplification 98.0%

    \[\leadsto \mathsf{fma}\left(-0.25, \frac{{\left(c \cdot a\right)}^{4}}{a} \cdot \frac{20}{{b}^{7}}, \frac{-2 \cdot \left(a \cdot \left({c}^{3} \cdot a\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{b \cdot \left(b \cdot b\right)}{c}} \]

Alternative 3: 96.9% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \left(\frac{-2 \cdot \left(a \cdot \left({c}^{3} \cdot a\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (-
  (- (/ (* -2.0 (* a (* (pow c 3.0) a))) (pow b 5.0)) (/ c b))
  (* a (/ c (/ (pow b 3.0) c)))))

C

double code(double a, double b, double c) {
	return (((-2.0 * (a * (pow(c, 3.0) * a))) / pow(b, 5.0)) - (c / b)) - (a * (c / (pow(b, 3.0) / c)));
}

Fortran

real(8) function code(a, b, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    code = ((((-2.0d0) * (a * ((c ** 3.0d0) * a))) / (b ** 5.0d0)) - (c / b)) - (a * (c / ((b ** 3.0d0) / c)))
end function

Java

public static double code(double a, double b, double c) {
	return (((-2.0 * (a * (Math.pow(c, 3.0) * a))) / Math.pow(b, 5.0)) - (c / b)) - (a * (c / (Math.pow(b, 3.0) / c)));
}

Python

def code(a, b, c):
	return (((-2.0 * (a * (math.pow(c, 3.0) * a))) / math.pow(b, 5.0)) - (c / b)) - (a * (c / (math.pow(b, 3.0) / c)))

Julia

function code(a, b, c)
	return Float64(Float64(Float64(Float64(-2.0 * Float64(a * Float64((c ^ 3.0) * a))) / (b ^ 5.0)) - Float64(c / b)) - Float64(a * Float64(c / Float64((b ^ 3.0) / c))))
end

MATLAB

function tmp = code(a, b, c)
	tmp = (((-2.0 * (a * ((c ^ 3.0) * a))) / (b ^ 5.0)) - (c / b)) - (a * (c / ((b ^ 3.0) / c)));
end

Wolfram

code[a_, b_, c_] := N[(N[(N[(N[(-2.0 * N[(a * N[(N[Power[c, 3.0], $MachinePrecision] * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Power[b, 5.0], $MachinePrecision]), $MachinePrecision] - N[(c / b), $MachinePrecision]), $MachinePrecision] - N[(a * N[(c / N[(N[Power[b, 3.0], $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.3%

   \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]

2. Taylor expanded in b around inf 97.0%

   \[\leadsto \color{blue}{-1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)} \]
3. Step-by-step derivation

   1. +-commutative 97.0%

      \[\leadsto \color{blue}{\left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right) + -1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}}} \]

   2. mul-1-neg 97.0%

      \[\leadsto \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right) + \color{blue}{\left(-\frac{{c}^{2} \cdot a}{{b}^{3}}\right)} \]

   3. unsub-neg 97.0%

      \[\leadsto \color{blue}{\left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}}} \]

   4. +-commutative 97.0%

      \[\leadsto \color{blue}{\left(-2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}} + -1 \cdot \frac{c}{b}\right)} - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   5. mul-1-neg 97.0%

      \[\leadsto \left(-2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}} + \color{blue}{\left(-\frac{c}{b}\right)}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   6. unsub-neg 97.0%

      \[\leadsto \color{blue}{\left(-2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}} - \frac{c}{b}\right)} - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   7. associate-*r/ 97.0%

      \[\leadsto \left(\color{blue}{\frac{-2 \cdot \left({c}^{3} \cdot {a}^{2}\right)}{{b}^{5}}} - \frac{c}{b}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   8. *-commutative 97.0%

      \[\leadsto \left(\frac{-2 \cdot \color{blue}{\left({a}^{2} \cdot {c}^{3}\right)}}{{b}^{5}} - \frac{c}{b}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   9. unpow2 97.0%

      \[\leadsto \left(\frac{-2 \cdot \left(\color{blue}{\left(a \cdot a\right)} \cdot {c}^{3}\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   10. associate-*l* 97.0%

       \[\leadsto \left(\frac{-2 \cdot \color{blue}{\left(a \cdot \left(a \cdot {c}^{3}\right)\right)}}{{b}^{5}} - \frac{c}{b}\right) - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   11. associate-/l* 97.0%

       \[\leadsto \left(\frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \color{blue}{\frac{{c}^{2}}{\frac{{b}^{3}}{a}}} \]

   12. associate-/r/ 97.0%

       \[\leadsto \left(\frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \color{blue}{\frac{{c}^{2}}{{b}^{3}} \cdot a} \]

4. Simplified 97.0%

   \[\leadsto \color{blue}{\left(\frac{-2 \cdot \left(a \cdot \left(a \cdot {c}^{3}\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a} \]

5. Final simplification 97.0%

   \[\leadsto \left(\frac{-2 \cdot \left(a \cdot \left({c}^{3} \cdot a\right)\right)}{{b}^{5}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \]

Alternative 4: 95.3% accurate, 7.3× speedup

Math

\[\begin{array}{l} \\ \frac{-c}{b} - \frac{c \cdot a}{\frac{b \cdot \left(b \cdot b\right)}{c}} \end{array} \]

FPCore

(FPCore (a b c)
 :precision binary64
 (- (/ (- c) b) (/ (* c a) (/ (* b (* b b)) c))))

C

double code(double a, double b, double c) {
	return (-c / b) - ((c * a) / ((b * (b * b)) / c));
}

Fortran

real(8) function code(a, b, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    code = (-c / b) - ((c * a) / ((b * (b * b)) / c))
end function

Java

public static double code(double a, double b, double c) {
	return (-c / b) - ((c * a) / ((b * (b * b)) / c));
}

Python

def code(a, b, c):
	return (-c / b) - ((c * a) / ((b * (b * b)) / c))

Julia

function code(a, b, c)
	return Float64(Float64(Float64(-c) / b) - Float64(Float64(c * a) / Float64(Float64(b * Float64(b * b)) / c)))
end

MATLAB

function tmp = code(a, b, c)
	tmp = (-c / b) - ((c * a) / ((b * (b * b)) / c));
end

Wolfram

code[a_, b_, c_] := N[(N[((-c) / b), $MachinePrecision] - N[(N[(c * a), $MachinePrecision] / N[(N[(b * N[(b * b), $MachinePrecision]), $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.3%

   \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]

2. Taylor expanded in b around inf 95.3%

   \[\leadsto \color{blue}{-1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + -1 \cdot \frac{c}{b}} \]
3. Step-by-step derivation

   1. +-commutative 95.3%

      \[\leadsto \color{blue}{-1 \cdot \frac{c}{b} + -1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}}} \]

   2. mul-1-neg 95.3%

      \[\leadsto -1 \cdot \frac{c}{b} + \color{blue}{\left(-\frac{{c}^{2} \cdot a}{{b}^{3}}\right)} \]

   3. unsub-neg 95.3%

      \[\leadsto \color{blue}{-1 \cdot \frac{c}{b} - \frac{{c}^{2} \cdot a}{{b}^{3}}} \]

   4. mul-1-neg 95.3%

      \[\leadsto \color{blue}{\left(-\frac{c}{b}\right)} - \frac{{c}^{2} \cdot a}{{b}^{3}} \]

   5. associate-/l* 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \color{blue}{\frac{{c}^{2}}{\frac{{b}^{3}}{a}}} \]

   6. associate-/r/ 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \color{blue}{\frac{{c}^{2}}{{b}^{3}} \cdot a} \]

   7. unpow2 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \frac{\color{blue}{c \cdot c}}{{b}^{3}} \cdot a \]

   8. associate-/l* 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \color{blue}{\frac{c}{\frac{{b}^{3}}{c}}} \cdot a \]

4. Simplified 95.3%

   \[\leadsto \color{blue}{\left(-\frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a} \]
5. Step-by-step derivation

   1. associate-*l/ 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \color{blue}{\frac{c \cdot a}{\frac{{b}^{3}}{c}}} \]

6. Applied egg-rr 95.3%

   \[\leadsto \left(-\frac{c}{b}\right) - \color{blue}{\frac{c \cdot a}{\frac{{b}^{3}}{c}}} \]
7. Step-by-step derivation

   1. unpow3 95.3%

      \[\leadsto \left(-\frac{c}{b}\right) - \frac{c \cdot a}{\frac{\color{blue}{\left(b \cdot b\right) \cdot b}}{c}} \]

8. Applied egg-rr 95.3%

   \[\leadsto \left(-\frac{c}{b}\right) - \frac{c \cdot a}{\frac{\color{blue}{\left(b \cdot b\right) \cdot b}}{c}} \]

9. Final simplification 95.3%

   \[\leadsto \frac{-c}{b} - \frac{c \cdot a}{\frac{b \cdot \left(b \cdot b\right)}{c}} \]

Alternative 5: 90.5% accurate, 29.0× speedup

Math

\[\begin{array}{l} \\ \frac{-c}{b} \end{array} \]

FPCore

(FPCore (a b c) :precision binary64 (/ (- c) b))

C

double code(double a, double b, double c) {
	return -c / b;
}

Fortran

real(8) function code(a, b, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    code = -c / b
end function

Java

public static double code(double a, double b, double c) {
	return -c / b;
}

Python

def code(a, b, c):
	return -c / b

Julia

function code(a, b, c)
	return Float64(Float64(-c) / b)
end

MATLAB

function tmp = code(a, b, c)
	tmp = -c / b;
end

Wolfram

code[a_, b_, c_] := N[((-c) / b), $MachinePrecision]

Derivation

1. Initial program 18.3%

   \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]

2. Taylor expanded in b around inf 90.3%

   \[\leadsto \color{blue}{-1 \cdot \frac{c}{b}} \]
3. Step-by-step derivation

   1. mul-1-neg 90.3%

      \[\leadsto \color{blue}{-\frac{c}{b}} \]

4. Simplified 90.3%

   \[\leadsto \color{blue}{-\frac{c}{b}} \]

5. Final simplification 90.3%

   \[\leadsto \frac{-c}{b} \]

Reproduce

herbie shell --seed 2023274 
(FPCore (a b c)
  :name "Quadratic roots, wide range"
  :precision binary64
  :pre (and (and (and (< 4.930380657631324e-32 a) (< a 2.028240960365167e+31)) (and (< 4.930380657631324e-32 b) (< b 2.028240960365167e+31))) (and (< 4.930380657631324e-32 c) (< c 2.028240960365167e+31)))
  (/ (+ (- b) (sqrt (- (* b b) (* (* 4.0 a) c)))) (* 2.0 a)))