2log (problem 3.3.6)

Percentage Accurate: 53.6% → 99.9%

Time: 8.4s

Alternatives: 6

Speedup: 2.0×

Specification

Math

\[\begin{array}{l} \\ \log \left(N + 1\right) - \log N \end{array} \]

FPCore

(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))

C

double code(double N) {
	return log((N + 1.0)) - log(N);
}

Fortran

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

Java

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

Python

def code(N):
	return math.log((N + 1.0)) - math.log(N)

Julia

function code(N)
	return Float64(log(Float64(N + 1.0)) - log(N))
end

MATLAB

function tmp = code(N)
	tmp = log((N + 1.0)) - log(N);
end

Wolfram

code[N_] := N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision]

Initial Program: 53.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \log \left(N + 1\right) - \log N \end{array} \]

FPCore

(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))

C

double code(double N) {
	return log((N + 1.0)) - log(N);
}

Fortran

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

Java

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

Python

def code(N):
	return math.log((N + 1.0)) - math.log(N)

Julia

function code(N)
	return Float64(log(Float64(N + 1.0)) - log(N))
end

MATLAB

function tmp = code(N)
	tmp = log((N + 1.0)) - log(N);
end

Wolfram

code[N_] := N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.9% accurate, 0.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\frac{0.3333333333333333}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 5e-6)
   (+ (/ 0.3333333333333333 (pow N 3.0)) (/ (- 1.0 (/ 0.5 N)) N))
   (log (/ (+ N 1.0) N))))

C

double code(double N) {
	double tmp;
	if ((log((N + 1.0)) - log(N)) <= 5e-6) {
		tmp = (0.3333333333333333 / pow(N, 3.0)) + ((1.0 - (0.5 / N)) / N);
	} else {
		tmp = log(((N + 1.0) / N));
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((log((n + 1.0d0)) - log(n)) <= 5d-6) then
        tmp = (0.3333333333333333d0 / (n ** 3.0d0)) + ((1.0d0 - (0.5d0 / n)) / n)
    else
        tmp = log(((n + 1.0d0) / n))
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if ((Math.log((N + 1.0)) - Math.log(N)) <= 5e-6) {
		tmp = (0.3333333333333333 / Math.pow(N, 3.0)) + ((1.0 - (0.5 / N)) / N);
	} else {
		tmp = Math.log(((N + 1.0) / N));
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if (math.log((N + 1.0)) - math.log(N)) <= 5e-6:
		tmp = (0.3333333333333333 / math.pow(N, 3.0)) + ((1.0 - (0.5 / N)) / N)
	else:
		tmp = math.log(((N + 1.0) / N))
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (Float64(log(Float64(N + 1.0)) - log(N)) <= 5e-6)
		tmp = Float64(Float64(0.3333333333333333 / (N ^ 3.0)) + Float64(Float64(1.0 - Float64(0.5 / N)) / N));
	else
		tmp = log(Float64(Float64(N + 1.0) / N));
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if ((log((N + 1.0)) - log(N)) <= 5e-6)
		tmp = (0.3333333333333333 / (N ^ 3.0)) + ((1.0 - (0.5 / N)) / N);
	else
		tmp = log(((N + 1.0) / N));
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N[(N[Log[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[Log[N], $MachinePrecision]), $MachinePrecision], 5e-6], N[(N[(0.3333333333333333 / N[Power[N, 3.0], $MachinePrecision]), $MachinePrecision] + N[(N[(1.0 - N[(0.5 / N), $MachinePrecision]), $MachinePrecision] / N), $MachinePrecision]), $MachinePrecision], N[Log[N[(N[(N + 1.0), $MachinePrecision] / N), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 5.00000000000000041e-6

   1. Initial program 6.6%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 6.9%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 6.9%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   4. Taylor expanded in N around inf 100.0%

      \[\leadsto \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{{N}^{3}} + \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}} \]
   5. Step-by-step derivation

      1. associate--l+ 100.0%

         \[\leadsto \color{blue}{0.3333333333333333 \cdot \frac{1}{{N}^{3}} + \left(\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\right)} \]

      2. associate-*r/ 100.0%

         \[\leadsto \color{blue}{\frac{0.3333333333333333 \cdot 1}{{N}^{3}}} + \left(\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      3. metadata-eval 100.0%

         \[\leadsto \frac{\color{blue}{0.3333333333333333}}{{N}^{3}} + \left(\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\right) \]

      4. unpow2 100.0%

         \[\leadsto \frac{0.3333333333333333}{{N}^{3}} + \left(\frac{1}{N} - 0.5 \cdot \frac{1}{\color{blue}{N \cdot N}}\right) \]

      5. associate-*r/ 100.0%

         \[\leadsto \frac{0.3333333333333333}{{N}^{3}} + \left(\frac{1}{N} - \color{blue}{\frac{0.5 \cdot 1}{N \cdot N}}\right) \]

      6. metadata-eval 100.0%

         \[\leadsto \frac{0.3333333333333333}{{N}^{3}} + \left(\frac{1}{N} - \frac{\color{blue}{0.5}}{N \cdot N}\right) \]

      7. associate-/l/ 100.0%

         \[\leadsto \frac{0.3333333333333333}{{N}^{3}} + \left(\frac{1}{N} - \color{blue}{\frac{\frac{0.5}{N}}{N}}\right) \]

      8. div-sub 100.0%

         \[\leadsto \frac{0.3333333333333333}{{N}^{3}} + \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]

   6. Simplified 100.0%

      \[\leadsto \color{blue}{\frac{0.3333333333333333}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}} \]

   if 5.00000000000000041e-6 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

   1. Initial program 100.0%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 100.0%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 100.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 5 \cdot 10^{-6}:\\ \;\;\;\;\frac{0.3333333333333333}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array} \]

Alternative 2: 99.8% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;N \leq 165000:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= N 165000.0) (log (/ (+ N 1.0) N)) (/ (- 1.0 (/ 0.5 N)) N)))

C

double code(double N) {
	double tmp;
	if (N <= 165000.0) {
		tmp = log(((N + 1.0) / N));
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 165000.0d0) then
        tmp = log(((n + 1.0d0) / n))
    else
        tmp = (1.0d0 - (0.5d0 / n)) / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 165000.0) {
		tmp = Math.log(((N + 1.0) / N));
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 165000.0:
		tmp = math.log(((N + 1.0) / N))
	else:
		tmp = (1.0 - (0.5 / N)) / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 165000.0)
		tmp = log(Float64(Float64(N + 1.0) / N));
	else
		tmp = Float64(Float64(1.0 - Float64(0.5 / N)) / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 165000.0)
		tmp = log(((N + 1.0) / N));
	else
		tmp = (1.0 - (0.5 / N)) / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 165000.0], N[Log[N[(N[(N + 1.0), $MachinePrecision] / N), $MachinePrecision]], $MachinePrecision], N[(N[(1.0 - N[(0.5 / N), $MachinePrecision]), $MachinePrecision] / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 165000

   1. Initial program 100.0%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 100.0%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 100.0%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   if 165000 < N

   1. Initial program 6.6%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 6.9%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 6.9%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   4. Taylor expanded in N around inf 99.8%

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 99.8%

         \[\leadsto \frac{1}{N} - 0.5 \cdot \frac{1}{\color{blue}{N \cdot N}} \]

      2. associate-*r/ 99.8%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{0.5 \cdot 1}{N \cdot N}} \]

      3. metadata-eval 99.8%

         \[\leadsto \frac{1}{N} - \frac{\color{blue}{0.5}}{N \cdot N} \]

      4. associate-/l/ 99.8%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{\frac{0.5}{N}}{N}} \]

      5. div-sub 99.8%

         \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]

   6. Simplified 99.8%

      \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 165000:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]

Alternative 3: 99.1% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;N \leq 0.9:\\ \;\;\;\;N - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= N 0.9) (- N (log N)) (/ (- 1.0 (/ 0.5 N)) N)))

C

double code(double N) {
	double tmp;
	if (N <= 0.9) {
		tmp = N - log(N);
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 0.9d0) then
        tmp = n - log(n)
    else
        tmp = (1.0d0 - (0.5d0 / n)) / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 0.9) {
		tmp = N - Math.log(N);
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 0.9:
		tmp = N - math.log(N)
	else:
		tmp = (1.0 - (0.5 / N)) / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 0.9)
		tmp = Float64(N - log(N));
	else
		tmp = Float64(Float64(1.0 - Float64(0.5 / N)) / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 0.9)
		tmp = N - log(N);
	else
		tmp = (1.0 - (0.5 / N)) / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 0.9], N[(N - N[Log[N], $MachinePrecision]), $MachinePrecision], N[(N[(1.0 - N[(0.5 / N), $MachinePrecision]), $MachinePrecision] / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 0.900000000000000022

   1. Initial program 100.0%

      \[\log \left(N + 1\right) - \log N \]

   2. Taylor expanded in N around 0 97.9%

      \[\leadsto \color{blue}{N + -1 \cdot \log N} \]
   3. Step-by-step derivation

      1. neg-mul-1 97.9%

         \[\leadsto N + \color{blue}{\left(-\log N\right)} \]

      2. unsub-neg 97.9%

         \[\leadsto \color{blue}{N - \log N} \]

   4. Simplified 97.9%

      \[\leadsto \color{blue}{N - \log N} \]

   if 0.900000000000000022 < N

   1. Initial program 7.3%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 7.7%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 7.7%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   4. Taylor expanded in N around inf 99.2%

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 99.2%

         \[\leadsto \frac{1}{N} - 0.5 \cdot \frac{1}{\color{blue}{N \cdot N}} \]

      2. associate-*r/ 99.2%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{0.5 \cdot 1}{N \cdot N}} \]

      3. metadata-eval 99.2%

         \[\leadsto \frac{1}{N} - \frac{\color{blue}{0.5}}{N \cdot N} \]

      4. associate-/l/ 99.2%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{\frac{0.5}{N}}{N}} \]

      5. div-sub 99.2%

         \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]

   6. Simplified 99.2%

      \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 0.9:\\ \;\;\;\;N - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]

Alternative 4: 98.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;N \leq 0.68:\\ \;\;\;\;-\log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \end{array} \]

FPCore

(FPCore (N)
 :precision binary64
 (if (<= N 0.68) (- (log N)) (/ (- 1.0 (/ 0.5 N)) N)))

C

double code(double N) {
	double tmp;
	if (N <= 0.68) {
		tmp = -log(N);
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 0.68d0) then
        tmp = -log(n)
    else
        tmp = (1.0d0 - (0.5d0 / n)) / n
    end if
    code = tmp
end function

Java

public static double code(double N) {
	double tmp;
	if (N <= 0.68) {
		tmp = -Math.log(N);
	} else {
		tmp = (1.0 - (0.5 / N)) / N;
	}
	return tmp;
}

Python

def code(N):
	tmp = 0
	if N <= 0.68:
		tmp = -math.log(N)
	else:
		tmp = (1.0 - (0.5 / N)) / N
	return tmp

Julia

function code(N)
	tmp = 0.0
	if (N <= 0.68)
		tmp = Float64(-log(N));
	else
		tmp = Float64(Float64(1.0 - Float64(0.5 / N)) / N);
	end
	return tmp
end

MATLAB

function tmp_2 = code(N)
	tmp = 0.0;
	if (N <= 0.68)
		tmp = -log(N);
	else
		tmp = (1.0 - (0.5 / N)) / N;
	end
	tmp_2 = tmp;
end

Wolfram

code[N_] := If[LessEqual[N, 0.68], (-N[Log[N], $MachinePrecision]), N[(N[(1.0 - N[(0.5 / N), $MachinePrecision]), $MachinePrecision] / N), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if N < 0.680000000000000049

   1. Initial program 100.0%

      \[\log \left(N + 1\right) - \log N \]

   2. Taylor expanded in N around 0 96.6%

      \[\leadsto \color{blue}{-1 \cdot \log N} \]
   3. Step-by-step derivation

      1. neg-mul-1 96.6%

         \[\leadsto \color{blue}{-\log N} \]

   4. Simplified 96.6%

      \[\leadsto \color{blue}{-\log N} \]

   if 0.680000000000000049 < N

   1. Initial program 7.3%

      \[\log \left(N + 1\right) - \log N \]
   2. Step-by-step derivation

      1. diff-log 7.7%

         \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   3. Applied egg-rr 7.7%

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)} \]

   4. Taylor expanded in N around inf 99.2%

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}} \]
   5. Step-by-step derivation

      1. unpow2 99.2%

         \[\leadsto \frac{1}{N} - 0.5 \cdot \frac{1}{\color{blue}{N \cdot N}} \]

      2. associate-*r/ 99.2%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{0.5 \cdot 1}{N \cdot N}} \]

      3. metadata-eval 99.2%

         \[\leadsto \frac{1}{N} - \frac{\color{blue}{0.5}}{N \cdot N} \]

      4. associate-/l/ 99.2%

         \[\leadsto \frac{1}{N} - \color{blue}{\frac{\frac{0.5}{N}}{N}} \]

      5. div-sub 99.2%

         \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]

   6. Simplified 99.2%

      \[\leadsto \color{blue}{\frac{1 - \frac{0.5}{N}}{N}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 97.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 0.68:\\ \;\;\;\;-\log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1 - \frac{0.5}{N}}{N}\\ \end{array} \]

Alternative 5: 52.2% accurate, 68.3× speedup

Math

\[\begin{array}{l} \\ \frac{1}{N} \end{array} \]

FPCore

(FPCore (N) :precision binary64 (/ 1.0 N))

C

double code(double N) {
	return 1.0 / N;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = 1.0d0 / n
end function

Java

public static double code(double N) {
	return 1.0 / N;
}

Python

def code(N):
	return 1.0 / N

Julia

function code(N)
	return Float64(1.0 / N)
end

MATLAB

function tmp = code(N)
	tmp = 1.0 / N;
end

Wolfram

code[N_] := N[(1.0 / N), $MachinePrecision]

Derivation

1. Initial program 54.0%

   \[\log \left(N + 1\right) - \log N \]

2. Taylor expanded in N around inf 51.6%

   \[\leadsto \color{blue}{\frac{1}{N}} \]

3. Final simplification 51.6%

   \[\leadsto \frac{1}{N} \]

Alternative 6: 4.6% accurate, 205.0× speedup

Math

\[\begin{array}{l} \\ N \end{array} \]

FPCore

(FPCore (N) :precision binary64 N)

C

double code(double N) {
	return N;
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = n
end function

Java

public static double code(double N) {
	return N;
}

Python

def code(N):
	return N

Julia

function code(N)
	return N
end

MATLAB

function tmp = code(N)
	tmp = N;
end

Wolfram

code[N_] := N

Derivation

1. Initial program 54.0%

   \[\log \left(N + 1\right) - \log N \]

2. Taylor expanded in N around 0 51.3%

   \[\leadsto \color{blue}{N + -1 \cdot \log N} \]
3. Step-by-step derivation

   1. neg-mul-1 51.3%

      \[\leadsto N + \color{blue}{\left(-\log N\right)} \]

   2. unsub-neg 51.3%

      \[\leadsto \color{blue}{N - \log N} \]

4. Simplified 51.3%

   \[\leadsto \color{blue}{N - \log N} \]

5. Taylor expanded in N around inf 4.7%

   \[\leadsto \color{blue}{N} \]

6. Final simplification 4.7%

   \[\leadsto N \]

Reproduce

herbie shell --seed 2023279 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))