Compound Interest

Percentage Accurate: 29.4% → 94.6%

Time: 19.5s

Alternatives: 21

Speedup: 38.0×

• could not determine a ground truth

  1. i = -2.0000000000000001e231
  2. n = 1.99999999999999997e77

Specification

Math

\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))

C

double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) - 1.0d0) / (i / n))
end function

Java

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

Python

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

Julia

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

MATLAB

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) - 1.0) / (i / n));
end

Wolfram

code[i_, n_] := N[(100.0 * N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Initial Program: 29.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))

C

double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 100.0d0 * ((((1.0d0 + (i / n)) ** n) - 1.0d0) / (i / n))
end function

Java

public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}

Python

def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))

Julia

function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end

MATLAB

function tmp = code(i, n)
	tmp = 100.0 * ((((1.0 + (i / n)) ^ n) - 1.0) / (i / n));
end

Wolfram

code[i_, n_] := N[(100.0 * N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Alternative 1: 94.6% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t\_0}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 0:\\ \;\;\;\;\frac{100 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (/ (* 100.0 (expm1 (* n (log1p (/ i n))))) (/ i n))
     (if (<= t_1 INFINITY) (* 100.0 (* t_0 (/ n i))) (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = (100.0 * expm1((n * log1p((i / n))))) / (i / n);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = (100.0 * Math.expm1((n * Math.log1p((i / n))))) / (i / n);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = (100.0 * math.expm1((n * math.log1p((i / n))))) / (i / n)
	elif t_1 <= math.inf:
		tmp = 100.0 * (t_0 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(Float64(100.0 * expm1(Float64(n * log1p(Float64(i / n))))) / Float64(i / n));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(t_0 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(N[(100.0 * N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision]), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 25.5%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. sub-neg 25.5%

         \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)}}{\frac{i}{n}} \]

      2. metadata-eval 25.5%

         \[\leadsto 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-1}}{\frac{i}{n}} \]

   4. Applied egg-rr 25.5%

      \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} + -1}}{\frac{i}{n}} \]
   5. Step-by-step derivation

      1. metadata-eval 25.5%

         \[\leadsto 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-1\right)}}{\frac{i}{n}} \]

      2. sub-neg 25.5%

         \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}} \]

      3. exp-to-pow 23.4%

         \[\leadsto 100 \cdot \frac{\color{blue}{e^{\log \left(1 + \frac{i}{n}\right) \cdot n}} - 1}{\frac{i}{n}} \]

      4. log1p-undefine 38.9%

         \[\leadsto 100 \cdot \frac{e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)} \cdot n} - 1}{\frac{i}{n}} \]

      5. *-commutative 38.9%

         \[\leadsto 100 \cdot \frac{e^{\color{blue}{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)}} - 1}{\frac{i}{n}} \]

      6. expm1-undefine 97.4%

         \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

   6. Simplified 97.4%

      \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. associate-*r/ 97.5%

         \[\leadsto \color{blue}{\frac{100 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}} \]

   8. Applied egg-rr 97.5%

      \[\leadsto \color{blue}{\frac{100 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 94.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{100 \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 81.5% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := t\_0 + -1\\ t_2 := \frac{t\_1}{\frac{i}{n}}\\ \mathbf{if}\;t\_2 \leq -1 \cdot 10^{-248}:\\ \;\;\;\;\frac{t\_0 \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{elif}\;t\_2 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;t\_2 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_1 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (+ t_0 -1.0)) (t_2 (/ t_1 (/ i n))))
   (if (<= t_2 -1e-248)
     (/ (+ (* t_0 100.0) -100.0) (/ i n))
     (if (<= t_2 0.0)
       (* 100.0 (* n (/ (expm1 i) i)))
       (if (<= t_2 INFINITY) (* 100.0 (* t_1 (/ n i))) (* n 100.0))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = t_0 + -1.0;
	double t_2 = t_1 / (i / n);
	double tmp;
	if (t_2 <= -1e-248) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else if (t_2 <= 0.0) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else if (t_2 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_1 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = t_0 + -1.0;
	double t_2 = t_1 / (i / n);
	double tmp;
	if (t_2 <= -1e-248) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else if (t_2 <= 0.0) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else if (t_2 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_1 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = t_0 + -1.0
	t_2 = t_1 / (i / n)
	tmp = 0
	if t_2 <= -1e-248:
		tmp = ((t_0 * 100.0) + -100.0) / (i / n)
	elif t_2 <= 0.0:
		tmp = 100.0 * (n * (math.expm1(i) / i))
	elif t_2 <= math.inf:
		tmp = 100.0 * (t_1 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(t_0 + -1.0)
	t_2 = Float64(t_1 / Float64(i / n))
	tmp = 0.0
	if (t_2 <= -1e-248)
		tmp = Float64(Float64(Float64(t_0 * 100.0) + -100.0) / Float64(i / n));
	elseif (t_2 <= 0.0)
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	elseif (t_2 <= Inf)
		tmp = Float64(100.0 * Float64(t_1 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 + -1.0), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$2, -1e-248], N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$2, 0.0], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$2, Infinity], N[(100.0 * N[(t$95$1 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -9.9999999999999998e-249

   1. Initial program 97.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 97.4%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 97.4%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 97.5%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 97.5%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 97.5%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 97.5%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   if -9.9999999999999998e-249 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 19.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 29.4%

      \[\leadsto 100 \cdot \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
   4. Step-by-step derivation

      1. associate-/l* 29.4%

         \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{e^{i} - 1}{i}\right)} \]

      2. expm1-define 76.1%

         \[\leadsto 100 \cdot \left(n \cdot \frac{\color{blue}{\mathsf{expm1}\left(i\right)}}{i}\right) \]

   5. Simplified 76.1%

      \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 80.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-248}:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 81.5% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := t\_0 + -1\\ t_2 := \frac{t\_1}{\frac{i}{n}}\\ \mathbf{if}\;t\_2 \leq -1 \cdot 10^{-248}:\\ \;\;\;\;n \cdot \frac{t\_0 \cdot 100 + -100}{i}\\ \mathbf{elif}\;t\_2 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;t\_2 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_1 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (+ t_0 -1.0)) (t_2 (/ t_1 (/ i n))))
   (if (<= t_2 -1e-248)
     (* n (/ (+ (* t_0 100.0) -100.0) i))
     (if (<= t_2 0.0)
       (* 100.0 (* n (/ (expm1 i) i)))
       (if (<= t_2 INFINITY) (* 100.0 (* t_1 (/ n i))) (* n 100.0))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = t_0 + -1.0;
	double t_2 = t_1 / (i / n);
	double tmp;
	if (t_2 <= -1e-248) {
		tmp = n * (((t_0 * 100.0) + -100.0) / i);
	} else if (t_2 <= 0.0) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else if (t_2 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_1 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = t_0 + -1.0;
	double t_2 = t_1 / (i / n);
	double tmp;
	if (t_2 <= -1e-248) {
		tmp = n * (((t_0 * 100.0) + -100.0) / i);
	} else if (t_2 <= 0.0) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else if (t_2 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_1 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = t_0 + -1.0
	t_2 = t_1 / (i / n)
	tmp = 0
	if t_2 <= -1e-248:
		tmp = n * (((t_0 * 100.0) + -100.0) / i)
	elif t_2 <= 0.0:
		tmp = 100.0 * (n * (math.expm1(i) / i))
	elif t_2 <= math.inf:
		tmp = 100.0 * (t_1 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(t_0 + -1.0)
	t_2 = Float64(t_1 / Float64(i / n))
	tmp = 0.0
	if (t_2 <= -1e-248)
		tmp = Float64(n * Float64(Float64(Float64(t_0 * 100.0) + -100.0) / i));
	elseif (t_2 <= 0.0)
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	elseif (t_2 <= Inf)
		tmp = Float64(100.0 * Float64(t_1 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 + -1.0), $MachinePrecision]}, Block[{t$95$2 = N[(t$95$1 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$2, -1e-248], N[(n * N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$2, 0.0], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$2, Infinity], N[(100.0 * N[(t$95$1 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -9.9999999999999998e-249

   1. Initial program 97.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 97.0%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 97.2%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 97.2%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 97.4%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 97.4%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 97.5%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 97.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 97.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 97.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 97.4%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 97.4%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 97.4%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. fma-undefine 97.5%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + -100}}{i} \]

      2. *-commutative 97.5%

         \[\leadsto n \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100} + -100}{i} \]

   6. Applied egg-rr 97.5%

      \[\leadsto n \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}}{i} \]

   if -9.9999999999999998e-249 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 19.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 29.4%

      \[\leadsto 100 \cdot \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
   4. Step-by-step derivation

      1. associate-/l* 29.4%

         \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{e^{i} - 1}{i}\right)} \]

      2. expm1-define 76.1%

         \[\leadsto 100 \cdot \left(n \cdot \frac{\color{blue}{\mathsf{expm1}\left(i\right)}}{i}\right) \]

   5. Simplified 76.1%

      \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 4 regimes into one program.

4. Final simplification 80.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-248}:\\ \;\;\;\;n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{i}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 81.4% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t\_0}{\frac{i}{n}}\\ t_2 := 100 \cdot \left(t\_0 \cdot \frac{n}{i}\right)\\ \mathbf{if}\;t\_1 \leq -1 \cdot 10^{-49}:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_1 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;t\_2\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0))
        (t_1 (/ t_0 (/ i n)))
        (t_2 (* 100.0 (* t_0 (/ n i)))))
   (if (<= t_1 -1e-49)
     t_2
     (if (<= t_1 0.0)
       (* 100.0 (* n (/ (expm1 i) i)))
       (if (<= t_1 INFINITY) t_2 (* n 100.0))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double t_2 = 100.0 * (t_0 * (n / i));
	double tmp;
	if (t_1 <= -1e-49) {
		tmp = t_2;
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = t_2;
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double t_2 = 100.0 * (t_0 * (n / i));
	double tmp;
	if (t_1 <= -1e-49) {
		tmp = t_2;
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = t_2;
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	t_2 = 100.0 * (t_0 * (n / i))
	tmp = 0
	if t_1 <= -1e-49:
		tmp = t_2
	elif t_1 <= 0.0:
		tmp = 100.0 * (n * (math.expm1(i) / i))
	elif t_1 <= math.inf:
		tmp = t_2
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	t_2 = Float64(100.0 * Float64(t_0 * Float64(n / i)))
	tmp = 0.0
	if (t_1 <= -1e-49)
		tmp = t_2;
	elseif (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	elseif (t_1 <= Inf)
		tmp = t_2;
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(100.0 * N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, -1e-49], t$95$2, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], t$95$2, N[(n * 100.0), $MachinePrecision]]]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -9.99999999999999936e-50 or -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.1%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.1%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.2%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.2%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.2%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 39.1%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 39.4%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 39.4%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 37.8%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 37.8%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 37.8%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.2%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.2%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if -9.99999999999999936e-50 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 22.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 32.2%

      \[\leadsto 100 \cdot \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
   4. Step-by-step derivation

      1. associate-/l* 32.1%

         \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{e^{i} - 1}{i}\right)} \]

      2. expm1-define 77.1%

         \[\leadsto 100 \cdot \left(n \cdot \frac{\color{blue}{\mathsf{expm1}\left(i\right)}}{i}\right) \]

   5. Simplified 77.1%

      \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)} \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 80.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-49}:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 94.6% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t\_0}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 INFINITY) (* 100.0 (* t_0 (/ n i))) (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) / (i / n));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) / (i / n))
	elif t_1 <= math.inf:
		tmp = 100.0 * (t_0 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(t_0 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 25.5%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. sub-neg 25.5%

         \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)}}{\frac{i}{n}} \]

      2. metadata-eval 25.5%

         \[\leadsto 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-1}}{\frac{i}{n}} \]

   4. Applied egg-rr 25.5%

      \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} + -1}}{\frac{i}{n}} \]
   5. Step-by-step derivation

      1. metadata-eval 25.5%

         \[\leadsto 100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-1\right)}}{\frac{i}{n}} \]

      2. sub-neg 25.5%

         \[\leadsto 100 \cdot \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}} \]

      3. exp-to-pow 23.4%

         \[\leadsto 100 \cdot \frac{\color{blue}{e^{\log \left(1 + \frac{i}{n}\right) \cdot n}} - 1}{\frac{i}{n}} \]

      4. log1p-undefine 38.9%

         \[\leadsto 100 \cdot \frac{e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)} \cdot n} - 1}{\frac{i}{n}} \]

      5. *-commutative 38.9%

         \[\leadsto 100 \cdot \frac{e^{\color{blue}{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)}} - 1}{\frac{i}{n}} \]

      6. expm1-undefine 97.4%

         \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

   6. Simplified 97.4%

      \[\leadsto 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 94.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 93.7% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t\_0}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (* n (/ (expm1 (* n (log1p (/ i n)))) i)))
     (if (<= t_1 INFINITY) (* 100.0 (* t_0 (/ n i))) (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (n * (expm1((n * log1p((i / n)))) / i));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (n * (Math.expm1((n * Math.log1p((i / n)))) / i));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 * (n * (math.expm1((n * math.log1p((i / n)))) / i))
	elif t_1 <= math.inf:
		tmp = 100.0 * (t_0 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(n * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / i)));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(t_0 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(n * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 25.5%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. associate-/r/ 25.2%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. add-exp-log 25.2%

         \[\leadsto 100 \cdot \left(\frac{\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1}{i} \cdot n\right) \]

      3. expm1-define 25.2%

         \[\leadsto 100 \cdot \left(\frac{\color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}}{i} \cdot n\right) \]

      4. log-pow 34.6%

         \[\leadsto 100 \cdot \left(\frac{\mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)}{i} \cdot n\right) \]

      5. log1p-define 95.7%

         \[\leadsto 100 \cdot \left(\frac{\mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)}{i} \cdot n\right) \]

   4. Applied egg-rr 95.7%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i} \cdot n\right)} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 93.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 92.9% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t\_0}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 0:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (* (expm1 (* n (log1p (/ i n)))) (/ n i)))
     (if (<= t_1 INFINITY) (* 100.0 (* t_0 (/ n i))) (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) * (n / i));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) * (n / i));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * (t_0 * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) * (n / i))
	elif t_1 <= math.inf:
		tmp = 100.0 * (t_0 * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) * Float64(n / i)));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(t_0 * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < -0.0

   1. Initial program 25.5%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 25.5%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 25.5%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 25.2%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 25.2%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 25.2%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 34.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 95.0%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 95.0%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n)) < +inf.0

   1. Initial program 98.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 98.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 98.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 98.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 36.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 36.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. expm1-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left(e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1\right)}\right) \]

      2. *-commutative 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}} - 1\right)\right) \]

      3. log1p-undefine 35.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n} - 1\right)\right) \]

      4. exp-to-pow 98.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}} - 1\right)\right) \]

   6. Applied egg-rr 98.9%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}\right) \]

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 #s(literal 1 binary64) (/.f64 i n)) n) #s(literal 1 binary64)) (/.f64 i n))

   1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 1.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 1.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 1.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 1.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 1.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 1.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 1.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 1.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 1.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 1.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 3 regimes into one program.

4. Final simplification 92.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left(\left({\left(1 + \frac{i}{n}\right)}^{n} + -1\right) \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 79.1% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 1.06 \cdot 10^{-33}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;i \leq 9.5 \cdot 10^{+158}:\\ \;\;\;\;i \cdot \log \left(e^{n \cdot \left(50 + \frac{100}{i}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n}}{i} - \frac{n}{i}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 1.06e-33)
   (* 100.0 (* n (/ (expm1 i) i)))
   (if (<= i 9.5e+158)
     (* i (log (exp (* n (+ 50.0 (/ 100.0 i))))))
     (* 100.0 (- (* n (/ (pow (+ 1.0 (/ i n)) n) i)) (/ n i))))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 1.06e-33) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else if (i <= 9.5e+158) {
		tmp = i * log(exp((n * (50.0 + (100.0 / i)))));
	} else {
		tmp = 100.0 * ((n * (pow((1.0 + (i / n)), n) / i)) - (n / i));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 1.06e-33) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else if (i <= 9.5e+158) {
		tmp = i * Math.log(Math.exp((n * (50.0 + (100.0 / i)))));
	} else {
		tmp = 100.0 * ((n * (Math.pow((1.0 + (i / n)), n) / i)) - (n / i));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 1.06e-33:
		tmp = 100.0 * (n * (math.expm1(i) / i))
	elif i <= 9.5e+158:
		tmp = i * math.log(math.exp((n * (50.0 + (100.0 / i)))))
	else:
		tmp = 100.0 * ((n * (math.pow((1.0 + (i / n)), n) / i)) - (n / i))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 1.06e-33)
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	elseif (i <= 9.5e+158)
		tmp = Float64(i * log(exp(Float64(n * Float64(50.0 + Float64(100.0 / i))))));
	else
		tmp = Float64(100.0 * Float64(Float64(n * Float64((Float64(1.0 + Float64(i / n)) ^ n) / i)) - Float64(n / i)));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[LessEqual[i, 1.06e-33], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 9.5e+158], N[(i * N[Log[N[Exp[N[(n * N[(50.0 + N[(100.0 / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(N[(n * N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision] - N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if i < 1.0599999999999999e-33

   1. Initial program 19.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 24.5%

      \[\leadsto 100 \cdot \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
   4. Step-by-step derivation

      1. associate-/l* 24.5%

         \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{e^{i} - 1}{i}\right)} \]

      2. expm1-define 83.9%

         \[\leadsto 100 \cdot \left(n \cdot \frac{\color{blue}{\mathsf{expm1}\left(i\right)}}{i}\right) \]

   5. Simplified 83.9%

      \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)} \]

   if 1.0599999999999999e-33 < i < 9.49999999999999913e158

   1. Initial program 47.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.4%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.4%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.4%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.4%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.4%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.4%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 44.9%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 21.0%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 21.0%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 21.0%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in i around inf 12.6%

      \[\leadsto \color{blue}{i \cdot \left(50 \cdot n + 100 \cdot \frac{n}{i}\right)} \]
   10. Step-by-step derivation

       1. add-log-exp 70.8%

          \[\leadsto i \cdot \color{blue}{\log \left(e^{50 \cdot n + 100 \cdot \frac{n}{i}}\right)} \]

       2. associate-*r/ 70.8%

          \[\leadsto i \cdot \log \left(e^{50 \cdot n + \color{blue}{\frac{100 \cdot n}{i}}}\right) \]

       3. associate-*l/ 70.8%

          \[\leadsto i \cdot \log \left(e^{50 \cdot n + \color{blue}{\frac{100}{i} \cdot n}}\right) \]

       4. distribute-rgt-out 70.8%

          \[\leadsto i \cdot \log \left(e^{\color{blue}{n \cdot \left(50 + \frac{100}{i}\right)}}\right) \]

   11. Applied egg-rr 70.8%

       \[\leadsto i \cdot \color{blue}{\log \left(e^{n \cdot \left(50 + \frac{100}{i}\right)}\right)} \]

   if 9.49999999999999913e158 < i

   1. Initial program 67.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 67.8%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 67.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 67.9%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 67.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 67.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 27.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 35.9%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 35.9%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]
   5. Step-by-step derivation

      1. *-commutative 35.9%

         \[\leadsto 100 \cdot \color{blue}{\left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \frac{n}{i}\right)} \]

      2. clear-num 35.9%

         \[\leadsto 100 \cdot \left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \color{blue}{\frac{1}{\frac{i}{n}}}\right) \]

      3. div-inv 35.9%

         \[\leadsto 100 \cdot \color{blue}{\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}} \]

      4. expm1-undefine 31.8%

         \[\leadsto 100 \cdot \frac{\color{blue}{e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1}}{\frac{i}{n}} \]

      5. div-sub 31.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)} \]

      6. *-commutative 31.8%

         \[\leadsto 100 \cdot \left(\frac{e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right) \]

      7. log1p-undefine 23.6%

         \[\leadsto 100 \cdot \left(\frac{e^{\color{blue}{\log \left(1 + \frac{i}{n}\right)} \cdot n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right) \]

      8. exp-to-pow 67.8%

         \[\leadsto 100 \cdot \left(\frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n}}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right) \]

      9. clear-num 67.9%

         \[\leadsto 100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \color{blue}{\frac{n}{i}}\right) \]

   6. Applied egg-rr 67.9%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{n}{i}\right)} \]
   7. Step-by-step derivation

      1. associate-/r/ 68.2%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{i} \cdot n} - \frac{n}{i}\right) \]

   8. Simplified 68.2%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{i} \cdot n - \frac{n}{i}\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 80.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 1.06 \cdot 10^{-33}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;i \leq 9.5 \cdot 10^{+158}:\\ \;\;\;\;i \cdot \log \left(e^{n \cdot \left(50 + \frac{100}{i}\right)}\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n}}{i} - \frac{n}{i}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 79.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -8 \cdot 10^{-244} \lor \neg \left(n \leq 3.1 \cdot 10^{-155}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -8e-244) (not (<= n 3.1e-155)))
   (* 100.0 (* n (/ (expm1 i) i)))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -8e-244) || !(n <= 3.1e-155)) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -8e-244) || !(n <= 3.1e-155)) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -8e-244) or not (n <= 3.1e-155):
		tmp = 100.0 * (n * (math.expm1(i) / i))
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -8e-244) || !(n <= 3.1e-155))
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -8e-244], N[Not[LessEqual[n, 3.1e-155]], $MachinePrecision]], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -7.9999999999999994e-244 or 3.1e-155 < n

   1. Initial program 26.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in n around inf 27.1%

      \[\leadsto 100 \cdot \color{blue}{\frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
   4. Step-by-step derivation

      1. associate-/l* 27.1%

         \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{e^{i} - 1}{i}\right)} \]

      2. expm1-define 79.2%

         \[\leadsto 100 \cdot \left(n \cdot \frac{\color{blue}{\mathsf{expm1}\left(i\right)}}{i}\right) \]

   5. Simplified 79.2%

      \[\leadsto 100 \cdot \color{blue}{\left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)} \]

   if -7.9999999999999994e-244 < n < 3.1e-155

   1. Initial program 46.2%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 46.2%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 46.2%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 46.2%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 46.2%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 46.2%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 46.2%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 68.2%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 77.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -8 \cdot 10^{-244} \lor \neg \left(n \leq 3.1 \cdot 10^{-155}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 65.4% accurate, 3.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.15 \cdot 10^{-163}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + n \cdot 50\right)\\ \mathbf{elif}\;n \leq 8.4 \cdot 10^{-156}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -1.15e-163)
   (+ (* n 100.0) (* i (+ (* 16.666666666666668 (* i n)) (* n 50.0))))
   (if (<= n 8.4e-156)
     (/ 0.0 (/ i n))
     (+
      (* n 100.0)
      (*
       i
       (+
        (* n 50.0)
        (* i (+ (* (* i n) 4.166666666666667) (* n 16.666666666666668)))))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= -1.15e-163) {
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	} else if (n <= 8.4e-156) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (i * ((n * 50.0) + (i * (((i * n) * 4.166666666666667) + (n * 16.666666666666668)))));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= (-1.15d-163)) then
        tmp = (n * 100.0d0) + (i * ((16.666666666666668d0 * (i * n)) + (n * 50.0d0)))
    else if (n <= 8.4d-156) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = (n * 100.0d0) + (i * ((n * 50.0d0) + (i * (((i * n) * 4.166666666666667d0) + (n * 16.666666666666668d0)))))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= -1.15e-163) {
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	} else if (n <= 8.4e-156) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (i * ((n * 50.0) + (i * (((i * n) * 4.166666666666667) + (n * 16.666666666666668)))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= -1.15e-163:
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)))
	elif n <= 8.4e-156:
		tmp = 0.0 / (i / n)
	else:
		tmp = (n * 100.0) + (i * ((n * 50.0) + (i * (((i * n) * 4.166666666666667) + (n * 16.666666666666668)))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= -1.15e-163)
		tmp = Float64(Float64(n * 100.0) + Float64(i * Float64(Float64(16.666666666666668 * Float64(i * n)) + Float64(n * 50.0))));
	elseif (n <= 8.4e-156)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(Float64(n * 100.0) + Float64(i * Float64(Float64(n * 50.0) + Float64(i * Float64(Float64(Float64(i * n) * 4.166666666666667) + Float64(n * 16.666666666666668))))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= -1.15e-163)
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	elseif (n <= 8.4e-156)
		tmp = 0.0 / (i / n);
	else
		tmp = (n * 100.0) + (i * ((n * 50.0) + (i * (((i * n) * 4.166666666666667) + (n * 16.666666666666668)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, -1.15e-163], N[(N[(n * 100.0), $MachinePrecision] + N[(i * N[(N[(16.666666666666668 * N[(i * n), $MachinePrecision]), $MachinePrecision] + N[(n * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 8.4e-156], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] + N[(i * N[(N[(n * 50.0), $MachinePrecision] + N[(i * N[(N[(N[(i * n), $MachinePrecision] * 4.166666666666667), $MachinePrecision] + N[(n * 16.666666666666668), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if n < -1.15e-163

   1. Initial program 30.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 30.6%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 30.6%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 30.6%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 30.6%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 30.6%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 30.6%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 26.1%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 56.6%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + 50 \cdot n\right)} \]

   if -1.15e-163 < n < 8.40000000000000049e-156

   1. Initial program 47.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.3%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.3%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.3%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.3%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 65.7%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]

   if 8.40000000000000049e-156 < n

   1. Initial program 20.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 20.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 20.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 20.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 20.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 27.1%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 78.1%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(50 \cdot n + i \cdot \left(4.166666666666667 \cdot \left(i \cdot n\right) + 16.666666666666668 \cdot n\right)\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 66.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.15 \cdot 10^{-163}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + n \cdot 50\right)\\ \mathbf{elif}\;n \leq 8.4 \cdot 10^{-156}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 64.8% accurate, 4.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -6.4 \cdot 10^{-164}:\\ \;\;\;\;100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \mathbf{elif}\;n \leq 2.9 \cdot 10^{-233}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 0.00047:\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{i} \cdot \left(i \cdot \left(n \cdot \left(100 + i \cdot 50\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -6.4e-164)
   (* 100.0 (+ n (* (* i n) (- 0.5 (/ 0.5 n)))))
   (if (<= n 2.9e-233)
     (/ 0.0 (/ i n))
     (if (<= n 0.00047)
       (* 100.0 (* i (/ n i)))
       (* (/ 1.0 i) (* i (* n (+ 100.0 (* i 50.0)))))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= -6.4e-164) {
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	} else if (n <= 2.9e-233) {
		tmp = 0.0 / (i / n);
	} else if (n <= 0.00047) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = (1.0 / i) * (i * (n * (100.0 + (i * 50.0))));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= (-6.4d-164)) then
        tmp = 100.0d0 * (n + ((i * n) * (0.5d0 - (0.5d0 / n))))
    else if (n <= 2.9d-233) then
        tmp = 0.0d0 / (i / n)
    else if (n <= 0.00047d0) then
        tmp = 100.0d0 * (i * (n / i))
    else
        tmp = (1.0d0 / i) * (i * (n * (100.0d0 + (i * 50.0d0))))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= -6.4e-164) {
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	} else if (n <= 2.9e-233) {
		tmp = 0.0 / (i / n);
	} else if (n <= 0.00047) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = (1.0 / i) * (i * (n * (100.0 + (i * 50.0))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= -6.4e-164:
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))))
	elif n <= 2.9e-233:
		tmp = 0.0 / (i / n)
	elif n <= 0.00047:
		tmp = 100.0 * (i * (n / i))
	else:
		tmp = (1.0 / i) * (i * (n * (100.0 + (i * 50.0))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= -6.4e-164)
		tmp = Float64(100.0 * Float64(n + Float64(Float64(i * n) * Float64(0.5 - Float64(0.5 / n)))));
	elseif (n <= 2.9e-233)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 0.00047)
		tmp = Float64(100.0 * Float64(i * Float64(n / i)));
	else
		tmp = Float64(Float64(1.0 / i) * Float64(i * Float64(n * Float64(100.0 + Float64(i * 50.0)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= -6.4e-164)
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	elseif (n <= 2.9e-233)
		tmp = 0.0 / (i / n);
	elseif (n <= 0.00047)
		tmp = 100.0 * (i * (n / i));
	else
		tmp = (1.0 / i) * (i * (n * (100.0 + (i * 50.0))));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, -6.4e-164], N[(100.0 * N[(n + N[(N[(i * n), $MachinePrecision] * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 2.9e-233], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 0.00047], N[(100.0 * N[(i * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 / i), $MachinePrecision] * N[(i * N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 4 regimes

2. if n < -6.4000000000000001e-164

   1. Initial program 30.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 30.3%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 30.4%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 30.4%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 30.4%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 30.4%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 30.4%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 30.4%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 30.4%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 30.4%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 56.1%

      \[\leadsto \color{blue}{100 \cdot n + 100 \cdot \left(i \cdot \left(n \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)} \]
   6. Step-by-step derivation

      1. distribute-lft-out 56.1%

         \[\leadsto \color{blue}{100 \cdot \left(n + i \cdot \left(n \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)} \]

      2. associate-*r* 56.1%

         \[\leadsto 100 \cdot \left(n + \color{blue}{\left(i \cdot n\right) \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)}\right) \]

      3. associate-*r/ 56.1%

         \[\leadsto 100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \color{blue}{\frac{0.5 \cdot 1}{n}}\right)\right) \]

      4. metadata-eval 56.1%

         \[\leadsto 100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{\color{blue}{0.5}}{n}\right)\right) \]

   7. Simplified 56.1%

      \[\leadsto \color{blue}{100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)} \]

   if -6.4000000000000001e-164 < n < 2.89999999999999982e-233

   1. Initial program 59.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 59.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 59.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 59.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 59.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 59.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 59.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 75.3%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]

   if 2.89999999999999982e-233 < n < 4.69999999999999986e-4

   1. Initial program 20.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 20.4%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 20.4%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 20.4%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 20.4%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 20.4%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 43.3%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 87.6%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 87.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]

   5. Taylor expanded in i around 0 61.2%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{i}\right) \]

   if 4.69999999999999986e-4 < n

   1. Initial program 21.7%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 21.7%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 21.7%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 21.7%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 21.7%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 21.7%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 21.7%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 35.8%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 54.2%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 54.2%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 54.2%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]
   9. Step-by-step derivation

      1. *-un-lft-identity 54.2%

         \[\leadsto \frac{\color{blue}{1 \cdot \left(i \cdot \left(100 + i \cdot 50\right)\right)}}{\frac{i}{n}} \]

      2. div-inv 54.0%

         \[\leadsto \frac{1 \cdot \left(i \cdot \left(100 + i \cdot 50\right)\right)}{\color{blue}{i \cdot \frac{1}{n}}} \]

      3. times-frac 78.9%

         \[\leadsto \color{blue}{\frac{1}{i} \cdot \frac{i \cdot \left(100 + i \cdot 50\right)}{\frac{1}{n}}} \]

      4. +-commutative 78.9%

         \[\leadsto \frac{1}{i} \cdot \frac{i \cdot \color{blue}{\left(i \cdot 50 + 100\right)}}{\frac{1}{n}} \]

      5. fma-define 78.9%

         \[\leadsto \frac{1}{i} \cdot \frac{i \cdot \color{blue}{\mathsf{fma}\left(i, 50, 100\right)}}{\frac{1}{n}} \]

   10. Applied egg-rr 78.9%

       \[\leadsto \color{blue}{\frac{1}{i} \cdot \frac{i \cdot \mathsf{fma}\left(i, 50, 100\right)}{\frac{1}{n}}} \]

   11. Taylor expanded in n around 0 78.9%

       \[\leadsto \frac{1}{i} \cdot \color{blue}{\left(i \cdot \left(n \cdot \left(100 + 50 \cdot i\right)\right)\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 65.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -6.4 \cdot 10^{-164}:\\ \;\;\;\;100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \mathbf{elif}\;n \leq 2.9 \cdot 10^{-233}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 0.00047:\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{i} \cdot \left(i \cdot \left(n \cdot \left(100 + i \cdot 50\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 65.5% accurate, 4.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.5 \cdot 10^{-163} \lor \neg \left(n \leq 2.7 \cdot 10^{-155}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.5e-163) (not (<= n 2.7e-155)))
   (*
    n
    (+
     100.0
     (* i (+ 50.0 (* i (+ 16.666666666666668 (* i 4.166666666666667)))))))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.5e-163) || !(n <= 2.7e-155)) {
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((n <= (-1.5d-163)) .or. (.not. (n <= 2.7d-155))) then
        tmp = n * (100.0d0 + (i * (50.0d0 + (i * (16.666666666666668d0 + (i * 4.166666666666667d0))))))
    else
        tmp = 0.0d0 / (i / n)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1.5e-163) || !(n <= 2.7e-155)) {
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1.5e-163) or not (n <= 2.7e-155):
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))))
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1.5e-163) || !(n <= 2.7e-155))
		tmp = Float64(n * Float64(100.0 + Float64(i * Float64(50.0 + Float64(i * Float64(16.666666666666668 + Float64(i * 4.166666666666667)))))));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -1.5e-163) || ~((n <= 2.7e-155)))
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	else
		tmp = 0.0 / (i / n);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1.5e-163], N[Not[LessEqual[n, 2.7e-155]], $MachinePrecision]], N[(n * N[(100.0 + N[(i * N[(50.0 + N[(i * N[(16.666666666666668 + N[(i * 4.166666666666667), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.5000000000000001e-163 or 2.69999999999999981e-155 < n

   1. Initial program 25.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 25.8%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 25.8%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 25.8%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 25.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 25.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 25.8%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 26.6%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 50.3%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + 4.166666666666667 \cdot i\right)\right)\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 50.3%

         \[\leadsto \frac{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + \color{blue}{i \cdot 4.166666666666667}\right)\right)\right)}{\frac{i}{n}} \]

   8. Simplified 50.3%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in n around 0 66.1%

      \[\leadsto \color{blue}{n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + 4.166666666666667 \cdot i\right)\right)\right)} \]

   if -1.5000000000000001e-163 < n < 2.69999999999999981e-155

   1. Initial program 47.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.3%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.3%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.3%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.3%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 65.7%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 66.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.5 \cdot 10^{-163} \lor \neg \left(n \leq 2.7 \cdot 10^{-155}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 65.4% accurate, 4.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -9.5 \cdot 10^{-162}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + n \cdot 50\right)\\ \mathbf{elif}\;n \leq 9.5 \cdot 10^{-155}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -9.5e-162)
   (+ (* n 100.0) (* i (+ (* 16.666666666666668 (* i n)) (* n 50.0))))
   (if (<= n 9.5e-155)
     (/ 0.0 (/ i n))
     (*
      n
      (+
       100.0
       (* i (+ 50.0 (* i (+ 16.666666666666668 (* i 4.166666666666667))))))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= -9.5e-162) {
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	} else if (n <= 9.5e-155) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= (-9.5d-162)) then
        tmp = (n * 100.0d0) + (i * ((16.666666666666668d0 * (i * n)) + (n * 50.0d0)))
    else if (n <= 9.5d-155) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = n * (100.0d0 + (i * (50.0d0 + (i * (16.666666666666668d0 + (i * 4.166666666666667d0))))))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= -9.5e-162) {
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	} else if (n <= 9.5e-155) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= -9.5e-162:
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)))
	elif n <= 9.5e-155:
		tmp = 0.0 / (i / n)
	else:
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= -9.5e-162)
		tmp = Float64(Float64(n * 100.0) + Float64(i * Float64(Float64(16.666666666666668 * Float64(i * n)) + Float64(n * 50.0))));
	elseif (n <= 9.5e-155)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(n * Float64(100.0 + Float64(i * Float64(50.0 + Float64(i * Float64(16.666666666666668 + Float64(i * 4.166666666666667)))))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= -9.5e-162)
		tmp = (n * 100.0) + (i * ((16.666666666666668 * (i * n)) + (n * 50.0)));
	elseif (n <= 9.5e-155)
		tmp = 0.0 / (i / n);
	else
		tmp = n * (100.0 + (i * (50.0 + (i * (16.666666666666668 + (i * 4.166666666666667))))));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, -9.5e-162], N[(N[(n * 100.0), $MachinePrecision] + N[(i * N[(N[(16.666666666666668 * N[(i * n), $MachinePrecision]), $MachinePrecision] + N[(n * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 9.5e-155], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(n * N[(100.0 + N[(i * N[(50.0 + N[(i * N[(16.666666666666668 + N[(i * 4.166666666666667), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if n < -9.5000000000000004e-162

   1. Initial program 30.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 30.6%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 30.6%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 30.6%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 30.6%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 30.6%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 30.6%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 26.1%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 56.6%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + 50 \cdot n\right)} \]

   if -9.5000000000000004e-162 < n < 9.50000000000000024e-155

   1. Initial program 47.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.3%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.3%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.3%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.3%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 65.7%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]

   if 9.50000000000000024e-155 < n

   1. Initial program 20.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 20.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 20.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 20.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 20.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 27.1%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 61.4%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + 4.166666666666667 \cdot i\right)\right)\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 61.4%

         \[\leadsto \frac{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + \color{blue}{i \cdot 4.166666666666667}\right)\right)\right)}{\frac{i}{n}} \]

   8. Simplified 61.4%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in n around 0 78.1%

      \[\leadsto \color{blue}{n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + 4.166666666666667 \cdot i\right)\right)\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 66.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -9.5 \cdot 10^{-162}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + n \cdot 50\right)\\ \mathbf{elif}\;n \leq 9.5 \cdot 10^{-155}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot \left(50 + i \cdot \left(16.666666666666668 + i \cdot 4.166666666666667\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 61.6% accurate, 6.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.22 \cdot 10^{-163}:\\ \;\;\;\;100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \mathbf{elif}\;n \leq 8.8 \cdot 10^{-156}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -1.22e-163)
   (* 100.0 (+ n (* (* i n) (- 0.5 (/ 0.5 n)))))
   (if (<= n 8.8e-156) (/ 0.0 (/ i n)) (* n (+ 100.0 (* i 50.0))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= -1.22e-163) {
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	} else if (n <= 8.8e-156) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= (-1.22d-163)) then
        tmp = 100.0d0 * (n + ((i * n) * (0.5d0 - (0.5d0 / n))))
    else if (n <= 8.8d-156) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = n * (100.0d0 + (i * 50.0d0))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= -1.22e-163) {
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	} else if (n <= 8.8e-156) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= -1.22e-163:
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))))
	elif n <= 8.8e-156:
		tmp = 0.0 / (i / n)
	else:
		tmp = n * (100.0 + (i * 50.0))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= -1.22e-163)
		tmp = Float64(100.0 * Float64(n + Float64(Float64(i * n) * Float64(0.5 - Float64(0.5 / n)))));
	elseif (n <= 8.8e-156)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= -1.22e-163)
		tmp = 100.0 * (n + ((i * n) * (0.5 - (0.5 / n))));
	elseif (n <= 8.8e-156)
		tmp = 0.0 / (i / n);
	else
		tmp = n * (100.0 + (i * 50.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, -1.22e-163], N[(100.0 * N[(n + N[(N[(i * n), $MachinePrecision] * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 8.8e-156], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if n < -1.22000000000000003e-163

   1. Initial program 30.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 30.3%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 30.4%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 30.4%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 30.4%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 30.4%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 30.4%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 30.4%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 30.4%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 30.4%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 30.4%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 56.1%

      \[\leadsto \color{blue}{100 \cdot n + 100 \cdot \left(i \cdot \left(n \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)} \]
   6. Step-by-step derivation

      1. distribute-lft-out 56.1%

         \[\leadsto \color{blue}{100 \cdot \left(n + i \cdot \left(n \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)} \]

      2. associate-*r* 56.1%

         \[\leadsto 100 \cdot \left(n + \color{blue}{\left(i \cdot n\right) \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)}\right) \]

      3. associate-*r/ 56.1%

         \[\leadsto 100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \color{blue}{\frac{0.5 \cdot 1}{n}}\right)\right) \]

      4. metadata-eval 56.1%

         \[\leadsto 100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{\color{blue}{0.5}}{n}\right)\right) \]

   7. Simplified 56.1%

      \[\leadsto \color{blue}{100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)} \]

   if -1.22000000000000003e-163 < n < 8.7999999999999996e-156

   1. Initial program 47.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.3%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.3%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.3%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.3%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 65.7%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]

   if 8.7999999999999996e-156 < n

   1. Initial program 20.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 20.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 20.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 20.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 20.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 20.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 27.1%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 55.4%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 55.4%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 55.4%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in n around 0 69.2%

      \[\leadsto \color{blue}{n \cdot \left(100 + 50 \cdot i\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 62.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.22 \cdot 10^{-163}:\\ \;\;\;\;100 \cdot \left(n + \left(i \cdot n\right) \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \mathbf{elif}\;n \leq 8.8 \cdot 10^{-156}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 61.6% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -4.5 \cdot 10^{-163} \lor \neg \left(n \leq 5.4 \cdot 10^{-152}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -4.5e-163) (not (<= n 5.4e-152)))
   (* n (+ 100.0 (* i 50.0)))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -4.5e-163) || !(n <= 5.4e-152)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((n <= (-4.5d-163)) .or. (.not. (n <= 5.4d-152))) then
        tmp = n * (100.0d0 + (i * 50.0d0))
    else
        tmp = 0.0d0 / (i / n)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -4.5e-163) || !(n <= 5.4e-152)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -4.5e-163) or not (n <= 5.4e-152):
		tmp = n * (100.0 + (i * 50.0))
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -4.5e-163) || !(n <= 5.4e-152))
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -4.5e-163) || ~((n <= 5.4e-152)))
		tmp = n * (100.0 + (i * 50.0));
	else
		tmp = 0.0 / (i / n);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -4.5e-163], N[Not[LessEqual[n, 5.4e-152]], $MachinePrecision]], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -4.4999999999999997e-163 or 5.39999999999999997e-152 < n

   1. Initial program 25.8%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 25.8%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 25.8%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 25.8%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 25.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 25.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 25.8%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 26.6%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 47.6%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 47.6%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 47.6%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in n around 0 62.0%

      \[\leadsto \color{blue}{n \cdot \left(100 + 50 \cdot i\right)} \]

   if -4.4999999999999997e-163 < n < 5.39999999999999997e-152

   1. Initial program 47.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 47.3%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 47.3%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 47.3%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 47.3%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 47.3%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 65.7%

      \[\leadsto \frac{\color{blue}{1} \cdot 100 + -100}{\frac{i}{n}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -4.5 \cdot 10^{-163} \lor \neg \left(n \leq 5.4 \cdot 10^{-152}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 62.0% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.95 \cdot 10^{+22} \lor \neg \left(n \leq 0.00047\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.95e+22) (not (<= n 0.00047)))
   (* n (+ 100.0 (* i 50.0)))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.95e+22) || !(n <= 0.00047)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((n <= (-1.95d+22)) .or. (.not. (n <= 0.00047d0))) then
        tmp = n * (100.0d0 + (i * 50.0d0))
    else
        tmp = 100.0d0 * (i / (i / n))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1.95e+22) || !(n <= 0.00047)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1.95e+22) or not (n <= 0.00047):
		tmp = n * (100.0 + (i * 50.0))
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1.95e+22) || !(n <= 0.00047))
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	else
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -1.95e+22) || ~((n <= 0.00047)))
		tmp = n * (100.0 + (i * 50.0));
	else
		tmp = 100.0 * (i / (i / n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1.95e+22], N[Not[LessEqual[n, 0.00047]], $MachinePrecision]], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.9500000000000001e22 or 4.69999999999999986e-4 < n

   1. Initial program 28.7%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 28.8%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 28.8%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 28.8%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 28.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 28.8%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 28.8%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 34.3%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 40.6%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 40.6%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 40.6%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in n around 0 61.6%

      \[\leadsto \color{blue}{n \cdot \left(100 + 50 \cdot i\right)} \]

   if -1.9500000000000001e22 < n < 4.69999999999999986e-4

   1. Initial program 31.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in i around 0 60.8%

      \[\leadsto 100 \cdot \frac{\color{blue}{i}}{\frac{i}{n}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 61.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.95 \cdot 10^{+22} \lor \neg \left(n \leq 0.00047\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 56.3% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -5 \cdot 10^{-15} \lor \neg \left(i \leq 3.1 \cdot 10^{-56}\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= i -5e-15) (not (<= i 3.1e-56)))
   (* 100.0 (/ i (/ i n)))
   (* n 100.0)))

C

double code(double i, double n) {
	double tmp;
	if ((i <= -5e-15) || !(i <= 3.1e-56)) {
		tmp = 100.0 * (i / (i / n));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((i <= (-5d-15)) .or. (.not. (i <= 3.1d-56))) then
        tmp = 100.0d0 * (i / (i / n))
    else
        tmp = n * 100.0d0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((i <= -5e-15) || !(i <= 3.1e-56)) {
		tmp = 100.0 * (i / (i / n));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (i <= -5e-15) or not (i <= 3.1e-56):
		tmp = 100.0 * (i / (i / n))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((i <= -5e-15) || !(i <= 3.1e-56))
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((i <= -5e-15) || ~((i <= 3.1e-56)))
		tmp = 100.0 * (i / (i / n));
	else
		tmp = n * 100.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[i, -5e-15], N[Not[LessEqual[i, 3.1e-56]], $MachinePrecision]], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -4.99999999999999999e-15 or 3.09999999999999987e-56 < i

   1. Initial program 53.6%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing

   3. Taylor expanded in i around 0 24.8%

      \[\leadsto 100 \cdot \frac{\color{blue}{i}}{\frac{i}{n}} \]

   if -4.99999999999999999e-15 < i < 3.09999999999999987e-56

   1. Initial program 8.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 8.5%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 8.5%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 8.5%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 8.5%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 8.5%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 8.5%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 8.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 8.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 8.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 8.5%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 8.5%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 8.5%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 87.2%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 2 regimes into one program.

4. Final simplification 57.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -5 \cdot 10^{-15} \lor \neg \left(i \leq 3.1 \cdot 10^{-56}\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 55.3% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2 \cdot 10^{+112} \lor \neg \left(i \leq 1.42 \cdot 10^{-21}\right):\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= i -2e+112) (not (<= i 1.42e-21)))
   (* 100.0 (* i (/ n i)))
   (* n 100.0)))

C

double code(double i, double n) {
	double tmp;
	if ((i <= -2e+112) || !(i <= 1.42e-21)) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((i <= (-2d+112)) .or. (.not. (i <= 1.42d-21))) then
        tmp = 100.0d0 * (i * (n / i))
    else
        tmp = n * 100.0d0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((i <= -2e+112) || !(i <= 1.42e-21)) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (i <= -2e+112) or not (i <= 1.42e-21):
		tmp = 100.0 * (i * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((i <= -2e+112) || !(i <= 1.42e-21))
		tmp = Float64(100.0 * Float64(i * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((i <= -2e+112) || ~((i <= 1.42e-21)))
		tmp = 100.0 * (i * (n / i));
	else
		tmp = n * 100.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[i, -2e+112], N[Not[LessEqual[i, 1.42e-21]], $MachinePrecision]], N[(100.0 * N[(i * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -1.9999999999999999e112 or 1.42e-21 < i

   1. Initial program 61.3%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 61.3%

         \[\leadsto 100 \cdot \color{blue}{\frac{1}{\frac{\frac{i}{n}}{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}} \]

      2. associate-/r/ 61.3%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{1}{\frac{i}{n}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right)} \]

      3. clear-num 60.8%

         \[\leadsto 100 \cdot \left(\color{blue}{\frac{n}{i}} \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)\right) \]

      4. add-exp-log 60.8%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \left(\color{blue}{e^{\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)}} - 1\right)\right) \]

      5. expm1-define 60.8%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{\mathsf{expm1}\left(\log \left({\left(1 + \frac{i}{n}\right)}^{n}\right)\right)}\right) \]

      6. log-pow 43.7%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)\right) \]

      7. log1p-define 65.4%

         \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)\right) \]

   4. Applied egg-rr 65.4%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)\right)} \]

   5. Taylor expanded in i around 0 20.6%

      \[\leadsto 100 \cdot \left(\frac{n}{i} \cdot \color{blue}{i}\right) \]

   if -1.9999999999999999e112 < i < 1.42e-21

   1. Initial program 11.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 11.8%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 11.8%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 11.8%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 11.8%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 11.8%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 11.8%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 11.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 11.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 11.8%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 11.8%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 11.8%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 11.8%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 77.7%

      \[\leadsto \color{blue}{100 \cdot n} \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2 \cdot 10^{+112} \lor \neg \left(i \leq 1.42 \cdot 10^{-21}\right):\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 19: 53.9% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 1.15 \cdot 10^{+23}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;i \cdot \left(n \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 1.15e+23) (* n 100.0) (* i (* n 50.0))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 1.15e+23) {
		tmp = n * 100.0;
	} else {
		tmp = i * (n * 50.0);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= 1.15d+23) then
        tmp = n * 100.0d0
    else
        tmp = i * (n * 50.0d0)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 1.15e+23) {
		tmp = n * 100.0;
	} else {
		tmp = i * (n * 50.0);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 1.15e+23:
		tmp = n * 100.0
	else:
		tmp = i * (n * 50.0)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 1.15e+23)
		tmp = Float64(n * 100.0);
	else
		tmp = Float64(i * Float64(n * 50.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= 1.15e+23)
		tmp = n * 100.0;
	else
		tmp = i * (n * 50.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, 1.15e+23], N[(n * 100.0), $MachinePrecision], N[(i * N[(n * 50.0), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 1.15e23

   1. Initial program 21.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 21.4%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 21.5%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 21.5%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 21.5%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 21.5%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 21.5%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 21.5%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 21.5%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 21.5%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 63.8%

      \[\leadsto \color{blue}{100 \cdot n} \]

   if 1.15e23 < i

   1. Initial program 61.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 61.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 61.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 61.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 61.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 61.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 61.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 36.9%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 25.5%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 25.5%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 25.5%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in i around inf 19.0%

      \[\leadsto \color{blue}{50 \cdot \left(i \cdot n\right)} \]
   10. Step-by-step derivation

       1. *-commutative 19.0%

          \[\leadsto \color{blue}{\left(i \cdot n\right) \cdot 50} \]

       2. associate-*r* 19.0%

          \[\leadsto \color{blue}{i \cdot \left(n \cdot 50\right)} \]

   11. Simplified 19.0%

       \[\leadsto \color{blue}{i \cdot \left(n \cdot 50\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 54.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 1.15 \cdot 10^{+23}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;i \cdot \left(n \cdot 50\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 20: 53.9% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 1.15 \cdot 10^{+23}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;50 \cdot \left(i \cdot n\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 1.15e+23) (* n 100.0) (* 50.0 (* i n))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 1.15e+23) {
		tmp = n * 100.0;
	} else {
		tmp = 50.0 * (i * n);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= 1.15d+23) then
        tmp = n * 100.0d0
    else
        tmp = 50.0d0 * (i * n)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 1.15e+23) {
		tmp = n * 100.0;
	} else {
		tmp = 50.0 * (i * n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 1.15e+23:
		tmp = n * 100.0
	else:
		tmp = 50.0 * (i * n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 1.15e+23)
		tmp = Float64(n * 100.0);
	else
		tmp = Float64(50.0 * Float64(i * n));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= 1.15e+23)
		tmp = n * 100.0;
	else
		tmp = 50.0 * (i * n);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, 1.15e+23], N[(n * 100.0), $MachinePrecision], N[(50.0 * N[(i * n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 1.15e23

   1. Initial program 21.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-/r/ 21.4%

         \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

      2. associate-*r* 21.5%

         \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

      3. *-commutative 21.5%

         \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

      4. associate-*r/ 21.5%

         \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

      5. sub-neg 21.5%

         \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

      6. distribute-lft-in 21.5%

         \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

      7. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

      8. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

      9. metadata-eval 21.5%

         \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

      10. fma-define 21.5%

          \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

      11. metadata-eval 21.5%

          \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

   3. Simplified 21.5%

      \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
   4. Add Preprocessing

   5. Taylor expanded in i around 0 63.8%

      \[\leadsto \color{blue}{100 \cdot n} \]

   if 1.15e23 < i

   1. Initial program 61.1%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
   2. Step-by-step derivation

      1. associate-*r/ 61.1%

         \[\leadsto \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}} \]

      2. sub-neg 61.1%

         \[\leadsto \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{\frac{i}{n}} \]

      3. distribute-rgt-in 61.1%

         \[\leadsto \frac{\color{blue}{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \left(-1\right) \cdot 100}}{\frac{i}{n}} \]

      4. metadata-eval 61.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-1} \cdot 100}{\frac{i}{n}} \]

      5. metadata-eval 61.1%

         \[\leadsto \frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + \color{blue}{-100}}{\frac{i}{n}} \]

   3. Simplified 61.1%

      \[\leadsto \color{blue}{\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}} \]
   4. Add Preprocessing

   5. Taylor expanded in n around inf 36.9%

      \[\leadsto \frac{\color{blue}{100 \cdot e^{i} - 100}}{\frac{i}{n}} \]

   6. Taylor expanded in i around 0 25.5%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + 50 \cdot i\right)}}{\frac{i}{n}} \]
   7. Step-by-step derivation

      1. *-commutative 25.5%

         \[\leadsto \frac{i \cdot \left(100 + \color{blue}{i \cdot 50}\right)}{\frac{i}{n}} \]

   8. Simplified 25.5%

      \[\leadsto \frac{\color{blue}{i \cdot \left(100 + i \cdot 50\right)}}{\frac{i}{n}} \]

   9. Taylor expanded in i around inf 19.0%

      \[\leadsto \color{blue}{50 \cdot \left(i \cdot n\right)} \]
   10. Step-by-step derivation

       1. *-commutative 19.0%

          \[\leadsto 50 \cdot \color{blue}{\left(n \cdot i\right)} \]

   11. Simplified 19.0%

       \[\leadsto \color{blue}{50 \cdot \left(n \cdot i\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 54.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 1.15 \cdot 10^{+23}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;50 \cdot \left(i \cdot n\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 21: 48.9% accurate, 38.0× speedup

Math

\[\begin{array}{l} \\ n \cdot 100 \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (* n 100.0))

C

double code(double i, double n) {
	return n * 100.0;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = n * 100.0d0
end function

Java

public static double code(double i, double n) {
	return n * 100.0;
}

Python

def code(i, n):
	return n * 100.0

Julia

function code(i, n)
	return Float64(n * 100.0)
end

MATLAB

function tmp = code(i, n)
	tmp = n * 100.0;
end

Wolfram

code[i_, n_] := N[(n * 100.0), $MachinePrecision]

Derivation

1. Initial program 29.9%

   \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
2. Step-by-step derivation

   1. associate-/r/ 30.0%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i} \cdot n\right)} \]

   2. associate-*r* 30.0%

      \[\leadsto \color{blue}{\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n} \]

   3. *-commutative 30.0%

      \[\leadsto \color{blue}{n \cdot \left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right)} \]

   4. associate-*r/ 30.0%

      \[\leadsto n \cdot \color{blue}{\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{i}} \]

   5. sub-neg 30.0%

      \[\leadsto n \cdot \frac{100 \cdot \color{blue}{\left({\left(1 + \frac{i}{n}\right)}^{n} + \left(-1\right)\right)}}{i} \]

   6. distribute-lft-in 30.0%

      \[\leadsto n \cdot \frac{\color{blue}{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \left(-1\right)}}{i} \]

   7. metadata-eval 30.0%

      \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + 100 \cdot \color{blue}{-1}}{i} \]

   8. metadata-eval 30.0%

      \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-100}}{i} \]

   9. metadata-eval 30.0%

      \[\leadsto n \cdot \frac{100 \cdot {\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{\left(-100\right)}}{i} \]

   10. fma-define 30.0%

       \[\leadsto n \cdot \frac{\color{blue}{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}}{i} \]

   11. metadata-eval 30.0%

       \[\leadsto n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, \color{blue}{-100}\right)}{i} \]

3. Simplified 30.0%

   \[\leadsto \color{blue}{n \cdot \frac{\mathsf{fma}\left(100, {\left(1 + \frac{i}{n}\right)}^{n}, -100\right)}{i}} \]
4. Add Preprocessing

5. Taylor expanded in i around 0 51.0%

   \[\leadsto \color{blue}{100 \cdot n} \]

6. Final simplification 51.0%

   \[\leadsto n \cdot 100 \]
7. Add Preprocessing

Developer target: 34.3% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 + \frac{i}{n}\\ 100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;t\_0 = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log t\_0}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ 1.0 (/ i n))))
   (*
    100.0
    (/
     (-
      (exp
       (*
        n
        (if (== t_0 1.0)
          (/ i n)
          (/ (* (/ i n) (log t_0)) (- (+ (/ i n) 1.0) 1.0)))))
      1.0)
     (/ i n)))))

C

double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + (i / n)
    if (t_0 == 1.0d0) then
        tmp = i / n
    else
        tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0d0) - 1.0d0)
    end if
    code = 100.0d0 * ((exp((n * tmp)) - 1.0d0) / (i / n))
end function

Java

public static double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * Math.log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((Math.exp((n * tmp)) - 1.0) / (i / n));
}

Python

def code(i, n):
	t_0 = 1.0 + (i / n)
	tmp = 0
	if t_0 == 1.0:
		tmp = i / n
	else:
		tmp = ((i / n) * math.log(t_0)) / (((i / n) + 1.0) - 1.0)
	return 100.0 * ((math.exp((n * tmp)) - 1.0) / (i / n))

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n))
	tmp = 0.0
	if (t_0 == 1.0)
		tmp = Float64(i / n);
	else
		tmp = Float64(Float64(Float64(i / n) * log(t_0)) / Float64(Float64(Float64(i / n) + 1.0) - 1.0));
	end
	return Float64(100.0 * Float64(Float64(exp(Float64(n * tmp)) - 1.0) / Float64(i / n)))
end

MATLAB

function tmp_2 = code(i, n)
	t_0 = 1.0 + (i / n);
	tmp = 0.0;
	if (t_0 == 1.0)
		tmp = i / n;
	else
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	end
	tmp_2 = 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision]}, N[(100.0 * N[(N[(N[Exp[N[(n * If[Equal[t$95$0, 1.0], N[(i / n), $MachinePrecision], N[(N[(N[(i / n), $MachinePrecision] * N[Log[t$95$0], $MachinePrecision]), $MachinePrecision] / N[(N[(N[(i / n), $MachinePrecision] + 1.0), $MachinePrecision] - 1.0), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Reproduce

herbie shell --seed 2024111 
(FPCore (i n)
  :name "Compound Interest"
  :precision binary64

  :alt
  (* 100.0 (/ (- (exp (* n (if (== (+ 1.0 (/ i n)) 1.0) (/ i n) (/ (* (/ i n) (log (+ 1.0 (/ i n)))) (- (+ (/ i n) 1.0) 1.0))))) 1.0) (/ i n)))

  (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))