?

Average Accuracy: 53.5% → 98.3%
Time: 15.7s
Precision: binary64
Cost: 32448

?

\[ \begin{array}{c}[a, b] = \mathsf{sort}([a, b])\\ \end{array} \]
\[\log \left(e^{a} + e^{b}\right) \]
\[\mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
(FPCore (a b)
 :precision binary64
 (+ (log1p (exp a)) (/ b (expm1 (log1p (+ (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 / expm1(log1p((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.expm1(Math.log1p((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.expm1(math.log1p((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 / expm1(log1p(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[(Exp[N[Log[1 + N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]], $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\log \left(e^{a} + e^{b}\right)
\mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 53.5%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 98.1%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Simplified98.3%

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

    [Start]98.1

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

    log1p-def [=>]98.3

    \[ \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Applied egg-rr98.3%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(1 + e^{a}\right)\right)}} \]
    Proof

    [Start]98.3

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

    expm1-log1p-u [=>]98.3

    \[ \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(1 + e^{a}\right)\right)}} \]
  5. Final simplification98.3%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) + \frac{b}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{a} + 1\right)\right)} \]

Alternatives

Alternative 1
Accuracy98.3%
Cost19648
\[\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \]
Alternative 2
Accuracy97.8%
Cost19392
\[\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right) \]
Alternative 3
Accuracy95.7%
Cost12992
\[\mathsf{log1p}\left(e^{a} + b\right) \]
Alternative 4
Accuracy50.3%
Cost12864
\[\mathsf{log1p}\left(e^{a}\right) \]
Alternative 5
Accuracy49.0%
Cost6720
\[b \cdot 0.5 + \log 2 \]
Alternative 6
Accuracy48.7%
Cost6592
\[\log \left(b + 2\right) \]
Alternative 7
Accuracy48.8%
Cost6592
\[\mathsf{log1p}\left(b + 1\right) \]
Alternative 8
Accuracy48.3%
Cost6464
\[\log 2 \]
Alternative 9
Accuracy2.6%
Cost192
\[a \cdot 0.5 \]

Error

Reproduce?

herbie shell --seed 2023151 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))