?

Average Accuracy: 38.5% → 100.0%
Time: 3.2s
Precision: binary64
Cost: 6464

?

\[\log \left(1 + x\right) \]
\[\mathsf{log1p}\left(x\right) \]
(FPCore (x) :precision binary64 (log (+ 1.0 x)))
(FPCore (x) :precision binary64 (log1p x))
double code(double x) {
	return log((1.0 + x));
}
double code(double x) {
	return log1p(x);
}
public static double code(double x) {
	return Math.log((1.0 + x));
}
public static double code(double x) {
	return Math.log1p(x);
}
def code(x):
	return math.log((1.0 + x))
def code(x):
	return math.log1p(x)
function code(x)
	return log(Float64(1.0 + x))
end
function code(x)
	return log1p(x)
end
code[x_] := N[Log[N[(1.0 + x), $MachinePrecision]], $MachinePrecision]
code[x_] := N[Log[1 + x], $MachinePrecision]
\log \left(1 + x\right)
\mathsf{log1p}\left(x\right)

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original38.5%
Target99.6%
Herbie100.0%
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array} \]

Derivation?

  1. Initial program 38.5%

    \[\log \left(1 + x\right) \]
  2. Simplified100.0%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(x\right)} \]
    Proof

    [Start]38.5

    \[ \log \left(1 + x\right) \]

    log1p-def [=>]100.0

    \[ \color{blue}{\mathsf{log1p}\left(x\right)} \]
  3. Final simplification100.0%

    \[\leadsto \mathsf{log1p}\left(x\right) \]

Reproduce?

herbie shell --seed 2023122 
(FPCore (x)
  :name "ln(1 + x)"
  :precision binary64

  :herbie-target
  (if (== (+ 1.0 x) 1.0) x (/ (* x (log (+ 1.0 x))) (- (+ 1.0 x) 1.0)))

  (log (+ 1.0 x)))