?

Average Error: 60.26% → 0%
Time: 3.1s
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

Original60.26%
Target0.38%
Herbie0%
\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array} \]

Derivation?

  1. Initial program 60.26

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

  2. Simplified0

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

    [Start]60.26

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

    log1p-def [=>]0

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

  3. Final simplification0

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

Reproduce?

herbie shell --seed 2023102 
(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)))