symmetry log of sum of exp

Percentage Accurate: 53.9% → 98.2%
Time: 13.8s
Alternatives: 10
Speedup: 1.5×

Specification

?
\[\begin{array}{l} \\ \log \left(e^{a} + e^{b}\right) \end{array} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
double code(double a, double b) {
	return log((exp(a) + exp(b)));
}
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((exp(a) + exp(b)))
end function
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
function tmp = code(a, b)
	tmp = log((exp(a) + exp(b)));
end
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\log \left(e^{a} + e^{b}\right)
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 10 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 53.9% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \log \left(e^{a} + e^{b}\right) \end{array} \]
(FPCore (a b) :precision binary64 (log (+ (exp a) (exp b))))
double code(double a, double b) {
	return log((exp(a) + exp(b)));
}
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((exp(a) + exp(b)))
end function
public static double code(double a, double b) {
	return Math.log((Math.exp(a) + Math.exp(b)));
}
def code(a, b):
	return math.log((math.exp(a) + math.exp(b)))
function code(a, b)
	return log(Float64(exp(a) + exp(b)))
end
function tmp = code(a, b)
	tmp = log((exp(a) + exp(b)));
end
code[a_, b_] := N[Log[N[(N[Exp[a], $MachinePrecision] + N[Exp[b], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}

\\
\log \left(e^{a} + e^{b}\right)
\end{array}

Alternative 1: 98.2% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1} \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (+ (log1p (exp a)) (/ b (+ (exp a) 1.0))))
assert(a < b);
double code(double a, double b) {
	return log1p(exp(a)) + (b / (exp(a) + 1.0));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p(Math.exp(a)) + (b / (Math.exp(a) + 1.0));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p(math.exp(a)) + (b / (math.exp(a) + 1.0))
a, b = sort([a, b])
function code(a, b)
	return Float64(log1p(exp(a)) + Float64(b / Float64(exp(a) + 1.0)))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision] + N[(b / N[(N[Exp[a], $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a}\right) + \frac{b}{e^{a} + 1}
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 81.1%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  3. Step-by-step derivation
    1. log1p-def81.1%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} + \frac{b}{1 + e^{a}} \]
  4. Simplified81.1%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right) + \frac{b}{1 + e^{a}}} \]
  5. Final simplification81.1%

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

Alternative 2: 97.7% accurate, 1.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (+ (exp a) (expm1 b))))
assert(a < b);
double code(double a, double b) {
	return log1p((exp(a) + expm1(b)));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p((Math.exp(a) + Math.expm1(b)));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p((math.exp(a) + math.expm1(b)))
a, b = sort([a, b])
function code(a, b)
	return log1p(Float64(exp(a) + expm1(b)))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[1 + N[(N[Exp[a], $MachinePrecision] + N[(Exp[b] - 1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Step-by-step derivation
    1. add-sqr-sqrt56.2%

      \[\leadsto \log \color{blue}{\left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. log-prod56.7%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  3. Applied egg-rr56.7%

    \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}}\right) + \log \left(\sqrt{e^{a} + e^{b}}\right)} \]
  4. Step-by-step derivation
    1. log-prod56.2%

      \[\leadsto \color{blue}{\log \left(\sqrt{e^{a} + e^{b}} \cdot \sqrt{e^{a} + e^{b}}\right)} \]
    2. rem-square-sqrt57.6%

      \[\leadsto \log \color{blue}{\left(e^{a} + e^{b}\right)} \]
    3. log1p-expm157.6%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(\mathsf{expm1}\left(\log \left(e^{a} + e^{b}\right)\right)\right)} \]
    4. expm1-def57.6%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{\log \left(e^{a} + e^{b}\right)} - 1}\right) \]
    5. rem-exp-log57.6%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{\left(e^{a} + e^{b}\right)} - 1\right) \]
    6. associate--l+57.6%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{a} + \left(e^{b} - 1\right)}\right) \]
    7. expm1-def83.7%

      \[\leadsto \mathsf{log1p}\left(e^{a} + \color{blue}{\mathsf{expm1}\left(b\right)}\right) \]
  5. Simplified83.7%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a} + \mathsf{expm1}\left(b\right)\right)} \]
  6. Final simplification83.7%

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

Alternative 3: 95.8% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a} + b\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (+ (exp a) b)))
assert(a < b);
double code(double a, double b) {
	return log1p((exp(a) + b));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p((Math.exp(a) + b));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p((math.exp(a) + b))
a, b = sort([a, b])
function code(a, b)
	return log1p(Float64(exp(a) + b))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[1 + N[(N[Exp[a], $MachinePrecision] + b), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a} + b\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.4%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Step-by-step derivation
    1. log1p-def80.4%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(b + e^{a}\right)} \]
    2. +-commutative80.4%

      \[\leadsto \mathsf{log1p}\left(\color{blue}{e^{a} + b}\right) \]
  4. Applied egg-rr80.4%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a} + b\right)} \]
  5. Final simplification80.4%

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

Alternative 4: 51.0% accurate, 1.5× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \mathsf{log1p}\left(e^{a}\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log1p (exp a)))
assert(a < b);
double code(double a, double b) {
	return log1p(exp(a));
}
assert a < b;
public static double code(double a, double b) {
	return Math.log1p(Math.exp(a));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log1p(math.exp(a))
a, b = sort([a, b])
function code(a, b)
	return log1p(exp(a))
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[1 + N[Exp[a], $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\mathsf{log1p}\left(e^{a}\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.7%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
  3. Step-by-step derivation
    1. log1p-def54.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  4. Simplified54.7%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  5. Final simplification54.7%

    \[\leadsto \mathsf{log1p}\left(e^{a}\right) \]

Alternative 5: 49.6% accurate, 2.7× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log (+ 2.0 (+ a (+ b (* 0.5 (* a a)))))))
assert(a < b);
double code(double a, double b) {
	return log((2.0 + (a + (b + (0.5 * (a * a))))));
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((2.0d0 + (a + (b + (0.5d0 * (a * a))))))
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log((2.0 + (a + (b + (0.5 * (a * a))))));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log((2.0 + (a + (b + (0.5 * (a * a))))))
a, b = sort([a, b])
function code(a, b)
	return log(Float64(2.0 + Float64(a + Float64(b + Float64(0.5 * Float64(a * a))))))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log((2.0 + (a + (b + (0.5 * (a * a))))));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[N[(2.0 + N[(a + N[(b + N[(0.5 * N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.4%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Taylor expanded in a around 0 53.2%

    \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(b + 0.5 \cdot {a}^{2}\right)\right)\right)} \]
  4. Step-by-step derivation
    1. +-commutative53.2%

      \[\leadsto \log \left(2 + \left(a + \color{blue}{\left(0.5 \cdot {a}^{2} + b\right)}\right)\right) \]
    2. unpow253.2%

      \[\leadsto \log \left(2 + \left(a + \left(0.5 \cdot \color{blue}{\left(a \cdot a\right)} + b\right)\right)\right) \]
  5. Simplified53.2%

    \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(0.5 \cdot \left(a \cdot a\right) + b\right)\right)\right)} \]
  6. Final simplification53.2%

    \[\leadsto \log \left(2 + \left(a + \left(b + 0.5 \cdot \left(a \cdot a\right)\right)\right)\right) \]

Alternative 6: 49.2% accurate, 2.8× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log \left(2 + \left(a + 0.5 \cdot \left(a \cdot a\right)\right)\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log (+ 2.0 (+ a (* 0.5 (* a a))))))
assert(a < b);
double code(double a, double b) {
	return log((2.0 + (a + (0.5 * (a * a)))));
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((2.0d0 + (a + (0.5d0 * (a * a)))))
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log((2.0 + (a + (0.5 * (a * a)))));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log((2.0 + (a + (0.5 * (a * a)))))
a, b = sort([a, b])
function code(a, b)
	return log(Float64(2.0 + Float64(a + Float64(0.5 * Float64(a * a)))))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log((2.0 + (a + (0.5 * (a * a)))));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[N[(2.0 + N[(a + N[(0.5 * N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log \left(2 + \left(a + 0.5 \cdot \left(a \cdot a\right)\right)\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.4%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Taylor expanded in a around 0 53.2%

    \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(b + 0.5 \cdot {a}^{2}\right)\right)\right)} \]
  4. Step-by-step derivation
    1. +-commutative53.2%

      \[\leadsto \log \left(2 + \left(a + \color{blue}{\left(0.5 \cdot {a}^{2} + b\right)}\right)\right) \]
    2. unpow253.2%

      \[\leadsto \log \left(2 + \left(a + \left(0.5 \cdot \color{blue}{\left(a \cdot a\right)} + b\right)\right)\right) \]
  5. Simplified53.2%

    \[\leadsto \log \color{blue}{\left(2 + \left(a + \left(0.5 \cdot \left(a \cdot a\right) + b\right)\right)\right)} \]
  6. Taylor expanded in b around 0 53.6%

    \[\leadsto \color{blue}{\log \left(2 + \left(a + 0.5 \cdot {a}^{2}\right)\right)} \]
  7. Step-by-step derivation
    1. unpow253.6%

      \[\leadsto \log \left(2 + \left(a + 0.5 \cdot \color{blue}{\left(a \cdot a\right)}\right)\right) \]
  8. Simplified53.6%

    \[\leadsto \color{blue}{\log \left(2 + \left(a + 0.5 \cdot \left(a \cdot a\right)\right)\right)} \]
  9. Final simplification53.6%

    \[\leadsto \log \left(2 + \left(a + 0.5 \cdot \left(a \cdot a\right)\right)\right) \]

Alternative 7: 49.9% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ b \cdot 0.5 + \log 2 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (+ (* b 0.5) (log 2.0)))
assert(a < b);
double code(double a, double b) {
	return (b * 0.5) + log(2.0);
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = (b * 0.5d0) + log(2.0d0)
end function
assert a < b;
public static double code(double a, double b) {
	return (b * 0.5) + Math.log(2.0);
}
[a, b] = sort([a, b])
def code(a, b):
	return (b * 0.5) + math.log(2.0)
a, b = sort([a, b])
function code(a, b)
	return Float64(Float64(b * 0.5) + log(2.0))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = (b * 0.5) + log(2.0);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(N[(b * 0.5), $MachinePrecision] + N[Log[2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
b \cdot 0.5 + \log 2
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in a around 0 54.4%

    \[\leadsto \color{blue}{\log \left(1 + e^{b}\right)} \]
  3. Step-by-step derivation
    1. log1p-def54.4%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  4. Simplified54.4%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{b}\right)} \]
  5. Taylor expanded in b around 0 53.5%

    \[\leadsto \color{blue}{\log 2 + 0.5 \cdot b} \]
  6. Step-by-step derivation
    1. +-commutative53.5%

      \[\leadsto \color{blue}{0.5 \cdot b + \log 2} \]
  7. Simplified53.5%

    \[\leadsto \color{blue}{0.5 \cdot b + \log 2} \]
  8. Final simplification53.5%

    \[\leadsto b \cdot 0.5 + \log 2 \]

Alternative 8: 49.6% accurate, 2.9× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log \left(b + 2\right) \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log (+ b 2.0)))
assert(a < b);
double code(double a, double b) {
	return log((b + 2.0));
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log((b + 2.0d0))
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log((b + 2.0));
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log((b + 2.0))
a, b = sort([a, b])
function code(a, b)
	return log(Float64(b + 2.0))
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log((b + 2.0));
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[N[(b + 2.0), $MachinePrecision]], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log \left(b + 2\right)
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.4%

    \[\leadsto \log \color{blue}{\left(1 + \left(b + e^{a}\right)\right)} \]
  3. Taylor expanded in a around 0 52.9%

    \[\leadsto \color{blue}{\log \left(2 + b\right)} \]
  4. Final simplification52.9%

    \[\leadsto \log \left(b + 2\right) \]

Alternative 9: 49.1% accurate, 3.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ \log 2 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (log 2.0))
assert(a < b);
double code(double a, double b) {
	return log(2.0);
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = log(2.0d0)
end function
assert a < b;
public static double code(double a, double b) {
	return Math.log(2.0);
}
[a, b] = sort([a, b])
def code(a, b):
	return math.log(2.0)
a, b = sort([a, b])
function code(a, b)
	return log(2.0)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = log(2.0);
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[Log[2.0], $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
\log 2
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.7%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
  3. Step-by-step derivation
    1. log1p-def54.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  4. Simplified54.7%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  5. Taylor expanded in a around 0 53.4%

    \[\leadsto \color{blue}{\log 2} \]
  6. Final simplification53.4%

    \[\leadsto \log 2 \]

Alternative 10: 2.6% accurate, 101.0× speedup?

\[\begin{array}{l} [a, b] = \mathsf{sort}([a, b])\\ \\ a \cdot 0.5 \end{array} \]
NOTE: a and b should be sorted in increasing order before calling this function.
(FPCore (a b) :precision binary64 (* a 0.5))
assert(a < b);
double code(double a, double b) {
	return a * 0.5;
}
NOTE: a and b should be sorted in increasing order before calling this function.
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = a * 0.5d0
end function
assert a < b;
public static double code(double a, double b) {
	return a * 0.5;
}
[a, b] = sort([a, b])
def code(a, b):
	return a * 0.5
a, b = sort([a, b])
function code(a, b)
	return Float64(a * 0.5)
end
a, b = num2cell(sort([a, b])){:}
function tmp = code(a, b)
	tmp = a * 0.5;
end
NOTE: a and b should be sorted in increasing order before calling this function.
code[a_, b_] := N[(a * 0.5), $MachinePrecision]
\begin{array}{l}
[a, b] = \mathsf{sort}([a, b])\\
\\
a \cdot 0.5
\end{array}
Derivation
  1. Initial program 57.6%

    \[\log \left(e^{a} + e^{b}\right) \]
  2. Taylor expanded in b around 0 54.7%

    \[\leadsto \color{blue}{\log \left(1 + e^{a}\right)} \]
  3. Step-by-step derivation
    1. log1p-def54.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  4. Simplified54.7%

    \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{a}\right)} \]
  5. Taylor expanded in a around 0 53.3%

    \[\leadsto \color{blue}{\log 2 + 0.5 \cdot a} \]
  6. Step-by-step derivation
    1. *-commutative53.3%

      \[\leadsto \log 2 + \color{blue}{a \cdot 0.5} \]
  7. Simplified53.3%

    \[\leadsto \color{blue}{\log 2 + a \cdot 0.5} \]
  8. Taylor expanded in a around inf 6.2%

    \[\leadsto \color{blue}{0.5 \cdot a} \]
  9. Step-by-step derivation
    1. *-commutative6.2%

      \[\leadsto \color{blue}{a \cdot 0.5} \]
  10. Simplified6.2%

    \[\leadsto \color{blue}{a \cdot 0.5} \]
  11. Final simplification6.2%

    \[\leadsto a \cdot 0.5 \]

Reproduce

?
herbie shell --seed 2023271 
(FPCore (a b)
  :name "symmetry log of sum of exp"
  :precision binary64
  (log (+ (exp a) (exp b))))