ln(1 + x)

Average Error: 39.3 → 0.2

Time: 12.7s

Precision: binary64

Cost: 20484

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.002:\\ \;\;\;\;-0.5 \cdot {x}^{2} + \left(0.3333333333333333 \cdot {x}^{3} + \left(-0.25 \cdot {x}^{4} + x\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

FPCore

(FPCore (x) :precision binary64 (log (+ 1.0 x)))

↓

(FPCore (x)
 :precision binary64
 (if (<= (+ 1.0 x) 1.002)
   (+
    (* -0.5 (pow x 2.0))
    (+ (* 0.3333333333333333 (pow x 3.0)) (+ (* -0.25 (pow x 4.0)) x)))
   (log (+ 1.0 x))))

C

double code(double x) {
	return log((1.0 + x));
}

↓

double code(double x) {
	double tmp;
	if ((1.0 + x) <= 1.002) {
		tmp = (-0.5 * pow(x, 2.0)) + ((0.3333333333333333 * pow(x, 3.0)) + ((-0.25 * pow(x, 4.0)) + x));
	} else {
		tmp = log((1.0 + x));
	}
	return tmp;
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = log((1.0d0 + x))
end function

↓

real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if ((1.0d0 + x) <= 1.002d0) then
        tmp = ((-0.5d0) * (x ** 2.0d0)) + ((0.3333333333333333d0 * (x ** 3.0d0)) + (((-0.25d0) * (x ** 4.0d0)) + x))
    else
        tmp = log((1.0d0 + x))
    end if
    code = tmp
end function

Java

public static double code(double x) {
	return Math.log((1.0 + x));
}

↓

public static double code(double x) {
	double tmp;
	if ((1.0 + x) <= 1.002) {
		tmp = (-0.5 * Math.pow(x, 2.0)) + ((0.3333333333333333 * Math.pow(x, 3.0)) + ((-0.25 * Math.pow(x, 4.0)) + x));
	} else {
		tmp = Math.log((1.0 + x));
	}
	return tmp;
}

Python

def code(x):
	return math.log((1.0 + x))

↓

def code(x):
	tmp = 0
	if (1.0 + x) <= 1.002:
		tmp = (-0.5 * math.pow(x, 2.0)) + ((0.3333333333333333 * math.pow(x, 3.0)) + ((-0.25 * math.pow(x, 4.0)) + x))
	else:
		tmp = math.log((1.0 + x))
	return tmp

Julia

function code(x)
	return log(Float64(1.0 + x))
end

↓

function code(x)
	tmp = 0.0
	if (Float64(1.0 + x) <= 1.002)
		tmp = Float64(Float64(-0.5 * (x ^ 2.0)) + Float64(Float64(0.3333333333333333 * (x ^ 3.0)) + Float64(Float64(-0.25 * (x ^ 4.0)) + x)));
	else
		tmp = log(Float64(1.0 + x));
	end
	return tmp
end

MATLAB

function tmp = code(x)
	tmp = log((1.0 + x));
end

↓

function tmp_2 = code(x)
	tmp = 0.0;
	if ((1.0 + x) <= 1.002)
		tmp = (-0.5 * (x ^ 2.0)) + ((0.3333333333333333 * (x ^ 3.0)) + ((-0.25 * (x ^ 4.0)) + x));
	else
		tmp = log((1.0 + x));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := N[Log[N[(1.0 + x), $MachinePrecision]], $MachinePrecision]

↓

code[x_] := If[LessEqual[N[(1.0 + x), $MachinePrecision], 1.002], N[(N[(-0.5 * N[Power[x, 2.0], $MachinePrecision]), $MachinePrecision] + N[(N[(0.3333333333333333 * N[Power[x, 3.0], $MachinePrecision]), $MachinePrecision] + N[(N[(-0.25 * N[Power[x, 4.0], $MachinePrecision]), $MachinePrecision] + x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Log[N[(1.0 + x), $MachinePrecision]], $MachinePrecision]]

Target

Original	39.3
Target	0.2
Herbie	0.2

\[\begin{array}{l} \mathbf{if}\;1 + x = 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot \log \left(1 + x\right)}{\left(1 + x\right) - 1}\\ \end{array} \]

Derivation

1. Split input into 2 regimes

2. if (+.f64 1 x) < 1.002

   1. Initial program 58.8

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

   2. Taylor expanded in x around 0 0.2

      \[\leadsto \color{blue}{-0.5 \cdot {x}^{2} + \left(0.3333333333333333 \cdot {x}^{3} + \left(-0.25 \cdot {x}^{4} + x\right)\right)} \]

   if 1.002 < (+.f64 1 x)

   1. Initial program 0.1

      \[\log \left(1 + x\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 0.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;1 + x \leq 1.002:\\ \;\;\;\;-0.5 \cdot {x}^{2} + \left(0.3333333333333333 \cdot {x}^{3} + \left(-0.25 \cdot {x}^{4} + x\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

Alternatives

Alternative 1
Error	0.2
Cost	13956

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.0002:\\ \;\;\;\;\left(\left(x + \left(x + {x}^{3} \cdot 0.6666666666666666\right)\right) + \left(-{x}^{2}\right)\right) \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

Alternative 2
Error	0.2
Cost	13764

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.0002:\\ \;\;\;\;-0.5 \cdot {x}^{2} + \left(0.3333333333333333 \cdot {x}^{3} + x\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

Alternative 3
Error	0.3
Cost	7044

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.00002:\\ \;\;\;\;-0.5 \cdot {x}^{2} + x\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

Alternative 4
Error	0.7
Cost	6852

\[\begin{array}{l} \mathbf{if}\;1 + x \leq 1.0000002:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + x\right)\\ \end{array} \]

Alternative 5
Error	20.8
Cost	64

\[x \]

Reproduce

herbie shell --seed 2023067 
(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)))