Average Error: 30.2 → 1.1
Time: 5.5s
Precision: binary64
\[[a, b]=\mathsf{sort}([a, b])\]
\[\log \left(e^{a} + e^{b}\right) \]
↓
\[\begin{array}{l} t_0 := 1 + e^{a}\\ t_1 := 0.5 - \frac{0.5}{t_0}\\ \log t_0 + \left(\frac{b}{t_0} + \log \left(1 + \left(\frac{\left(b \cdot b\right) \cdot t_1}{t_0} + 0.5 \cdot \frac{{t_1}^{2} \cdot {b}^{4}}{{t_0}^{2}}\right)\right)\right) \end{array} \]
\log \left(e^{a} + e^{b}\right)
↓
\begin{array}{l}
t_0 := 1 + e^{a}\\
t_1 := 0.5 - \frac{0.5}{t_0}\\
\log t_0 + \left(\frac{b}{t_0} + \log \left(1 + \left(\frac{\left(b \cdot b\right) \cdot t_1}{t_0} + 0.5 \cdot \frac{{t_1}^{2} \cdot {b}^{4}}{{t_0}^{2}}\right)\right)\right)
\end{array}

(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
↓
(FPCore (a b)
 :precision binary64
 (let* ((t_0 (+ 1.0 (exp a))) (t_1 (- 0.5 (/ 0.5 t_0))))
   (+
    (log t_0)
    (+
     (/ b t_0)
     (log
      (+
       1.0
       (+
        (/ (* (* b b) t_1) t_0)
        (* 0.5 (/ (* (pow t_1 2.0) (pow b 4.0)) (pow t_0 2.0))))))))))
double code(double a, double b) {
	return log(exp(a) + exp(b));
}
↓
double code(double a, double b) {
	double t_0 = 1.0 + exp(a);
	double t_1 = 0.5 - (0.5 / t_0);
	return log(t_0) + ((b / t_0) + log(1.0 + ((((b * b) * t_1) / t_0) + (0.5 * ((pow(t_1, 2.0) * pow(b, 4.0)) / pow(t_0, 2.0))))));
}

Error

Bits error versus a

Bits error versus b

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 30.2

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

  2. Taylor expanded around 0 1.3

    \[\leadsto \color{blue}{\left(0.5 \cdot \frac{{b}^{2}}{1 + e^{a}} + \left(\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}\right)\right) - 0.5 \cdot \frac{{b}^{2}}{{\left(1 + e^{a}\right)}^{2}}} \]

  3. Simplified1.1

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \left(\frac{b}{1 + e^{a}} \cdot b\right) \cdot \left(0.5 + \frac{-0.5}{1 + e^{a}}\right)\right)} \]
  4. Using strategy rm

  5. Applied add-log-exp_binary641.2

    \[\leadsto \log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \color{blue}{\log \left(e^{\left(\frac{b}{1 + e^{a}} \cdot b\right) \cdot \left(0.5 + \frac{-0.5}{1 + e^{a}}\right)}\right)}\right) \]

  6. Simplified1.2

    \[\leadsto \log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \log \color{blue}{\left({\left(e^{0.5 - \frac{0.5}{1 + e^{a}}}\right)}^{\left(\frac{b \cdot b}{1 + e^{a}}\right)}\right)}\right) \]

  7. Taylor expanded around 0 1.1

    \[\leadsto \log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \log \color{blue}{\left(1 + \left(0.5 \cdot \frac{{\log \left(e^{0.5 - 0.5 \cdot \frac{1}{1 + e^{a}}}\right)}^{2} \cdot {b}^{4}}{{\left(1 + e^{a}\right)}^{2}} + \frac{\log \left(e^{0.5 - 0.5 \cdot \frac{1}{1 + e^{a}}}\right) \cdot {b}^{2}}{1 + e^{a}}\right)\right)}\right) \]

  8. Simplified1.1

    \[\leadsto \log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \log \color{blue}{\left(1 + \left(\frac{\left(b \cdot b\right) \cdot \left(0.5 - \frac{0.5}{1 + e^{a}}\right)}{1 + e^{a}} + 0.5 \cdot \frac{{\left(0.5 - \frac{0.5}{1 + e^{a}}\right)}^{2} \cdot {b}^{4}}{{\left(1 + e^{a}\right)}^{2}}\right)\right)}\right) \]

  9. Final simplification1.1

    \[\leadsto \log \left(1 + e^{a}\right) + \left(\frac{b}{1 + e^{a}} + \log \left(1 + \left(\frac{\left(b \cdot b\right) \cdot \left(0.5 - \frac{0.5}{1 + e^{a}}\right)}{1 + e^{a}} + 0.5 \cdot \frac{{\left(0.5 - \frac{0.5}{1 + e^{a}}\right)}^{2} \cdot {b}^{4}}{{\left(1 + e^{a}\right)}^{2}}\right)\right)\right) \]

Reproduce

herbie shell --seed 2021204 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))