?

Average Accuracy: 54.1% → 98.4%
Time: 15.4s
Precision: binary64
Cost: 19648

?

\[ \begin{array}{c}[a, b] = \mathsf{sort}([a, b])\\ \end{array} \]
\[\log \left(e^{a} + e^{b}\right) \]
\[\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
(FPCore (a b) :precision binary64 (+ (log1p (exp a)) (/ b (+ (exp a) 1.0))))
double code(double a, double b) {
	return log((exp(a) + exp(b)));
}
double code(double a, double b) {
	return log1p(exp(a)) + (b / (exp(a) + 1.0));
}
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
public static double code(double a, double b) {
	return Math.log1p(Math.exp(a)) + (b / (Math.exp(a) + 1.0));
}
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
def code(a, b):
	return math.log1p(math.exp(a)) + (b / (math.exp(a) + 1.0))
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
function code(a, b)
	return Float64(log1p(exp(a)) + Float64(b / Float64(exp(a) + 1.0)))
end
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
code[a_, b_] := N[(N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision] + N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\log \left(e^{a} + e^{b}\right)
\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 54.1%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 73.8%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Simplified74.2%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
    Step-by-step derivation

    [Start]73.8

    \[ \log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}} \]

    log1p-def [=>]74.2

    \[ \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Final simplification74.2%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \]

Alternatives

Alternative 1
Accuracy96.7%
Cost25928
\[\begin{array}{l} \mathbf{if}\;e^{b} \leq 1.005:\\ \;\;\;\;\mathsf{log1p}\left(e^{a} + b\right)\\ \mathbf{elif}\;e^{b} \leq 5 \cdot 10^{+219}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \mathbf{else}:\\ \;\;\;\;b \cdot 0.5 + \log 2\\ \end{array} \]
Alternative 2
Accuracy97.1%
Cost19524
\[\begin{array}{l} \mathbf{if}\;b \leq 1.15 \cdot 10^{-8}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a} + b\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(e^{a} + e^{b}\right)\\ \end{array} \]
Alternative 3
Accuracy57.8%
Cost19396
\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{a}\right)\\ \end{array} \]
Alternative 4
Accuracy57.6%
Cost19396
\[\begin{array}{l} \mathbf{if}\;e^{a} \leq 0:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\mathsf{log1p}\left(e^{b}\right)\\ \end{array} \]
Alternative 5
Accuracy98.1%
Cost19392
\[\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right) \]
Alternative 6
Accuracy11.9%
Cost7108
\[\begin{array}{l} \mathbf{if}\;a \leq 1.4:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{b + 2} + \log \left(b + 2\right)\\ \end{array} \]
Alternative 7
Accuracy11.9%
Cost6852
\[\begin{array}{l} \mathbf{if}\;a \leq 1:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + \left(a + 2\right)\right)\\ \end{array} \]
Alternative 8
Accuracy11.9%
Cost6724
\[\begin{array}{l} \mathbf{if}\;a \leq 86:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\log \left(b + 2\right)\\ \end{array} \]
Alternative 9
Accuracy11.9%
Cost6596
\[\begin{array}{l} \mathbf{if}\;a \leq 105:\\ \;\;\;\;b \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\log 2\\ \end{array} \]
Alternative 10
Accuracy11.9%
Cost192
\[b \cdot 0.5 \]

Error

Reproduce?

herbie shell --seed 2023159 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))