Average Error: 0.0 → 0.0
Time: 979.0ms
Precision: binary64
Cost: 320
\[x + x \cdot x \]
\[x + x \cdot x \]
(FPCore (x) :precision binary64 (+ x (* x x)))
(FPCore (x) :precision binary64 (+ x (* x x)))
double code(double x) {
	return x + (x * x);
}
double code(double x) {
	return x + (x * x);
}
real(8) function code(x)
    real(8), intent (in) :: x
    code = x + (x * x)
end function
real(8) function code(x)
    real(8), intent (in) :: x
    code = x + (x * x)
end function
public static double code(double x) {
	return x + (x * x);
}
public static double code(double x) {
	return x + (x * x);
}
def code(x):
	return x + (x * x)
def code(x):
	return x + (x * x)
function code(x)
	return Float64(x + Float64(x * x))
end
function code(x)
	return Float64(x + Float64(x * x))
end
function tmp = code(x)
	tmp = x + (x * x);
end
function tmp = code(x)
	tmp = x + (x * x);
end
code[x_] := N[(x + N[(x * x), $MachinePrecision]), $MachinePrecision]
code[x_] := N[(x + N[(x * x), $MachinePrecision]), $MachinePrecision]
x + x \cdot x
x + x \cdot x

Error

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.0

    \[x + x \cdot x \]
  2. Final simplification0.0

    \[\leadsto x + x \cdot x \]

Alternatives

Alternative 1
Error1.8
Cost456
\[\begin{array}{l} \mathbf{if}\;x \leq -3640.325933075684:\\ \;\;\;\;x \cdot x\\ \mathbf{elif}\;x \leq 0.0029421815227809534:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;x \cdot x\\ \end{array} \]
Alternative 2
Error0.0
Cost320
\[x \cdot \left(x + 1\right) \]
Alternative 3
Error21.2
Cost64
\[x \]

Error

Reproduce

herbie shell --seed 2022325 
(FPCore (x)
  :name "Main:bigenough1 from B"
  :precision binary64
  (+ x (* x x)))