?

Average Error: 8.25% → 0.14%
Time: 5.9s
Precision: binary64
Cost: 713

?

\[x \cdot \left(1 + y \cdot y\right) \]
\[\begin{array}{l} \mathbf{if}\;y \leq -5.8 \cdot 10^{+132} \lor \neg \left(y \leq 1950000000\right):\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;x + x \cdot \left(y \cdot y\right)\\ \end{array} \]
(FPCore (x y) :precision binary64 (* x (+ 1.0 (* y y))))
(FPCore (x y)
 :precision binary64
 (if (or (<= y -5.8e+132) (not (<= y 1950000000.0)))
   (* y (* y x))
   (+ x (* x (* y y)))))
double code(double x, double y) {
	return x * (1.0 + (y * y));
}
double code(double x, double y) {
	double tmp;
	if ((y <= -5.8e+132) || !(y <= 1950000000.0)) {
		tmp = y * (y * x);
	} else {
		tmp = x + (x * (y * y));
	}
	return tmp;
}
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * (1.0d0 + (y * y))
end function
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((y <= (-5.8d+132)) .or. (.not. (y <= 1950000000.0d0))) then
        tmp = y * (y * x)
    else
        tmp = x + (x * (y * y))
    end if
    code = tmp
end function
public static double code(double x, double y) {
	return x * (1.0 + (y * y));
}
public static double code(double x, double y) {
	double tmp;
	if ((y <= -5.8e+132) || !(y <= 1950000000.0)) {
		tmp = y * (y * x);
	} else {
		tmp = x + (x * (y * y));
	}
	return tmp;
}
def code(x, y):
	return x * (1.0 + (y * y))
def code(x, y):
	tmp = 0
	if (y <= -5.8e+132) or not (y <= 1950000000.0):
		tmp = y * (y * x)
	else:
		tmp = x + (x * (y * y))
	return tmp
function code(x, y)
	return Float64(x * Float64(1.0 + Float64(y * y)))
end
function code(x, y)
	tmp = 0.0
	if ((y <= -5.8e+132) || !(y <= 1950000000.0))
		tmp = Float64(y * Float64(y * x));
	else
		tmp = Float64(x + Float64(x * Float64(y * y)));
	end
	return tmp
end
function tmp = code(x, y)
	tmp = x * (1.0 + (y * y));
end
function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y <= -5.8e+132) || ~((y <= 1950000000.0)))
		tmp = y * (y * x);
	else
		tmp = x + (x * (y * y));
	end
	tmp_2 = tmp;
end
code[x_, y_] := N[(x * N[(1.0 + N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
code[x_, y_] := If[Or[LessEqual[y, -5.8e+132], N[Not[LessEqual[y, 1950000000.0]], $MachinePrecision]], N[(y * N[(y * x), $MachinePrecision]), $MachinePrecision], N[(x + N[(x * N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
x \cdot \left(1 + y \cdot y\right)
\begin{array}{l}
\mathbf{if}\;y \leq -5.8 \cdot 10^{+132} \lor \neg \left(y \leq 1950000000\right):\\
\;\;\;\;y \cdot \left(y \cdot x\right)\\

\mathbf{else}:\\
\;\;\;\;x + x \cdot \left(y \cdot y\right)\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original8.25%
Target0.13%
Herbie0.14%
\[x + \left(x \cdot y\right) \cdot y \]

Derivation?

  1. Split input into 2 regimes
  2. if y < -5.7999999999999997e132 or 1.95e9 < y

    1. Initial program 38.87

      \[x \cdot \left(1 + y \cdot y\right) \]
    2. Taylor expanded in y around inf 38.87

      \[\leadsto \color{blue}{{y}^{2} \cdot x} \]
    3. Simplified0.38

      \[\leadsto \color{blue}{y \cdot \left(y \cdot x\right)} \]
      Proof

      [Start]38.87

      \[ {y}^{2} \cdot x \]

      unpow2 [=>]38.87

      \[ \color{blue}{\left(y \cdot y\right)} \cdot x \]

      associate-*l* [=>]0.38

      \[ \color{blue}{y \cdot \left(y \cdot x\right)} \]

    if -5.7999999999999997e132 < y < 1.95e9

    1. Initial program 0.08

      \[x \cdot \left(1 + y \cdot y\right) \]
    2. Applied egg-rr0.07

      \[\leadsto \color{blue}{x \cdot \left(y \cdot y\right) + x} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.14

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq -5.8 \cdot 10^{+132} \lor \neg \left(y \leq 1950000000\right):\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;x + x \cdot \left(y \cdot y\right)\\ \end{array} \]

Alternatives

Alternative 1
Error0.39%
Cost13376
\[\frac{\mathsf{hypot}\left(1, y\right)}{\frac{1}{\mathsf{hypot}\left(1, y\right) \cdot x}} \]
Alternative 2
Error0.14%
Cost713
\[\begin{array}{l} \mathbf{if}\;y \leq -6 \cdot 10^{+132} \lor \neg \left(y \leq 100000000\right):\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(1 + y \cdot y\right)\\ \end{array} \]
Alternative 3
Error9.65%
Cost580
\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(y \cdot y\right)\\ \end{array} \]
Alternative 4
Error1.55%
Cost580
\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 0.01:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \end{array} \]
Alternative 5
Error32.23%
Cost64
\[x \]

Error

Reproduce?

herbie shell --seed 2023115 
(FPCore (x y)
  :name "Numeric.Integration.TanhSinh:everywhere from integration-0.2.1"
  :precision binary64

  :herbie-target
  (+ x (* (* x y) y))

  (* x (+ 1.0 (* y y))))