Compound Interest

Percentage Accurate: 27.3% → 94.5%

Time: 24.3s

Alternatives: 18

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: 27.3% 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: 94.5% accurate, 0.3× speedup

Math

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

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 INFINITY) (* n (/ (+ (* t_0 100.0) -100.0) 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) / (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 = n * (((t_0 * 100.0) + -100.0) / 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) / (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 = n * (((t_0 * 100.0) + -100.0) / 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) / (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 = n * (((t_0 * 100.0) + -100.0) / 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(Float64(t_0 + -1.0) / 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(n * Float64(Float64(Float64(t_0 * 100.0) + -100.0) / 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[(N[(t$95$0 + -1.0), $MachinePrecision] / 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[(n * N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / i), $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 26.3%

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

      1. sub-neg 26.3%

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

      2. metadata-eval 26.3%

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

   4. Applied egg-rr 26.3%

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

      1. metadata-eval 26.3%

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

      2. sub-neg 26.3%

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

      3. exp-to-pow 25.8%

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

      4. log1p-undefine 47.7%

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

      5. *-commutative 47.7%

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

      6. expm1-undefine 99.2%

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

   6. Simplified 99.2%

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

   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 99.7%

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

      1. associate-/r/ 99.7%

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

      2. associate-*r* 99.7%

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

      3. *-commutative 99.7%

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

      4. associate-*r/ 99.8%

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

      5. sub-neg 99.8%

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

      6. distribute-lft-in 99.9%

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

      7. metadata-eval 99.9%

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

      8. metadata-eval 99.9%

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

      9. metadata-eval 99.9%

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

      10. fma-define 99.8%

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

      11. metadata-eval 99.8%

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

   3. Simplified 99.8%

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

      1. fma-undefine 99.9%

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

      2. *-commutative 99.9%

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

   6. Applied egg-rr 99.9%

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

   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. Add Preprocessing

   3. Taylor expanded in i around 0 86.8%

      \[\leadsto \color{blue}{100 \cdot n} \]
   4. Step-by-step derivation

      1. *-commutative 86.8%

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

   5. Simplified 86.8%

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

4. Final simplification 97.2%

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

Alternative 2: 81.6% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 100 \cdot \frac{n \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{if}\;n \leq -1.6 \cdot 10^{-86}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(i\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;n \leq 1.32 \cdot 10^{-247}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-28}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (* 100.0 (/ (* n (expm1 i)) i))))
   (if (<= n -1.6e-86)
     t_0
     (if (<= n -1.02e-178)
       (* 100.0 (* (expm1 i) (/ n i)))
       (if (<= n 1.32e-247)
         (/ 0.0 (/ i n))
         (if (<= n 1.5e-28) (* 100.0 (/ i (/ i n))) t_0))))))

C

double code(double i, double n) {
	double t_0 = 100.0 * ((n * expm1(i)) / i);
	double tmp;
	if (n <= -1.6e-86) {
		tmp = t_0;
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (expm1(i) * (n / i));
	} else if (n <= 1.32e-247) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.5e-28) {
		tmp = 100.0 * (i / (i / n));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = 100.0 * ((n * Math.expm1(i)) / i);
	double tmp;
	if (n <= -1.6e-86) {
		tmp = t_0;
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (Math.expm1(i) * (n / i));
	} else if (n <= 1.32e-247) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.5e-28) {
		tmp = 100.0 * (i / (i / n));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = 100.0 * ((n * math.expm1(i)) / i)
	tmp = 0
	if n <= -1.6e-86:
		tmp = t_0
	elif n <= -1.02e-178:
		tmp = 100.0 * (math.expm1(i) * (n / i))
	elif n <= 1.32e-247:
		tmp = 0.0 / (i / n)
	elif n <= 1.5e-28:
		tmp = 100.0 * (i / (i / n))
	else:
		tmp = t_0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(100.0 * Float64(Float64(n * expm1(i)) / i))
	tmp = 0.0
	if (n <= -1.6e-86)
		tmp = t_0;
	elseif (n <= -1.02e-178)
		tmp = Float64(100.0 * Float64(expm1(i) * Float64(n / i)));
	elseif (n <= 1.32e-247)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 1.5e-28)
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	else
		tmp = t_0;
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(100.0 * N[(N[(n * N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -1.6e-86], t$95$0, If[LessEqual[n, -1.02e-178], N[(100.0 * N[(N[(Exp[i] - 1), $MachinePrecision] * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.32e-247], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.5e-28], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]]]]

Derivation

1. Split input into 4 regimes

2. if n < -1.60000000000000003e-86 or 1.50000000000000001e-28 < n

   1. Initial program 21.9%

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

      1. div-sub 21.8%

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

      2. clear-num 21.9%

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

      3. sub-neg 21.9%

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

      4. div-inv 21.9%

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

      5. clear-num 21.9%

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

   4. Applied egg-rr 21.9%

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

      1. sub-neg 21.9%

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

   6. Simplified 21.9%

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

      1. associate-*r/ 21.8%

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

      2. sub-div 22.4%

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

      3. +-commutative 22.4%

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

   8. Applied egg-rr 22.4%

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

   9. Taylor expanded in n around inf 42.0%

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

       1. expm1-define 92.0%

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

   11. Simplified 92.0%

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

   if -1.60000000000000003e-86 < n < -1.02000000000000006e-178

   1. Initial program 28.6%

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

   3. Taylor expanded in n around inf 23.8%

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

      1. expm1-define 76.5%

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

   5. Simplified 76.5%

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

      1. div-inv 76.3%

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

      2. clear-num 76.5%

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

   7. Applied egg-rr 76.5%

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

   if -1.02000000000000006e-178 < n < 1.3200000000000001e-247

   1. Initial program 79.9%

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

      1. associate-*r/ 79.9%

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

      2. sub-neg 79.9%

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

      3. distribute-rgt-in 79.9%

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

      4. metadata-eval 79.9%

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

      5. metadata-eval 79.9%

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

   3. Simplified 79.9%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 84.1%

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

   if 1.3200000000000001e-247 < n < 1.50000000000000001e-28

   1. Initial program 16.0%

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

   3. Taylor expanded in i around 0 61.9%

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

4. Final simplification 85.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.6 \cdot 10^{-86}:\\ \;\;\;\;100 \cdot \frac{n \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(i\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;n \leq 1.32 \cdot 10^{-247}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-28}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n \cdot \mathsf{expm1}\left(i\right)}{i}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 74.7% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 100 \cdot \left(\mathsf{expm1}\left(i\right) \cdot \frac{n}{i}\right)\\ \mathbf{if}\;i \leq -3.4 \cdot 10^{+32}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 0.37:\\ \;\;\;\;n \cdot \left(100 + 100 \cdot \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (* 100.0 (* (expm1 i) (/ n i)))))
   (if (<= i -3.4e+32)
     t_0
     (if (<= i -2.35e-5)
       (/ 0.0 (/ i n))
       (if (<= i 0.37)
         (* n (+ 100.0 (* 100.0 (* i (- 0.5 (/ 0.5 n))))))
         t_0)))))

C

double code(double i, double n) {
	double t_0 = 100.0 * (expm1(i) * (n / i));
	double tmp;
	if (i <= -3.4e+32) {
		tmp = t_0;
	} else if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else if (i <= 0.37) {
		tmp = n * (100.0 + (100.0 * (i * (0.5 - (0.5 / n)))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = 100.0 * (Math.expm1(i) * (n / i));
	double tmp;
	if (i <= -3.4e+32) {
		tmp = t_0;
	} else if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else if (i <= 0.37) {
		tmp = n * (100.0 + (100.0 * (i * (0.5 - (0.5 / n)))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = 100.0 * (math.expm1(i) * (n / i))
	tmp = 0
	if i <= -3.4e+32:
		tmp = t_0
	elif i <= -2.35e-5:
		tmp = 0.0 / (i / n)
	elif i <= 0.37:
		tmp = n * (100.0 + (100.0 * (i * (0.5 - (0.5 / n)))))
	else:
		tmp = t_0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(100.0 * Float64(expm1(i) * Float64(n / i)))
	tmp = 0.0
	if (i <= -3.4e+32)
		tmp = t_0;
	elseif (i <= -2.35e-5)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (i <= 0.37)
		tmp = Float64(n * Float64(100.0 + Float64(100.0 * Float64(i * Float64(0.5 - Float64(0.5 / n))))));
	else
		tmp = t_0;
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(100.0 * N[(N[(Exp[i] - 1), $MachinePrecision] * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[i, -3.4e+32], t$95$0, If[LessEqual[i, -2.35e-5], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 0.37], N[(n * N[(100.0 + N[(100.0 * N[(i * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]]]

Derivation

1. Split input into 3 regimes

2. if i < -3.39999999999999979e32 or 0.37 < i

   1. Initial program 50.6%

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

   3. Taylor expanded in n around inf 74.5%

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

      1. expm1-define 74.5%

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

   5. Simplified 74.5%

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

      1. div-inv 74.5%

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

      2. clear-num 74.5%

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

   7. Applied egg-rr 74.5%

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

   if -3.39999999999999979e32 < i < -2.34999999999999986e-5

   1. Initial program 97.9%

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

      1. associate-*r/ 97.9%

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

      2. sub-neg 97.9%

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

      3. distribute-rgt-in 97.8%

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

      4. metadata-eval 97.8%

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

      5. metadata-eval 97.8%

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

   3. Simplified 97.8%

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

   5. Taylor expanded in i around 0 6.5%

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

      1. *-commutative 6.5%

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

   7. Simplified 6.5%

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

   8. Taylor expanded in i around 0 83.9%

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

   if -2.34999999999999986e-5 < i < 0.37

   1. Initial program 7.3%

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

      1. associate-/r/ 7.8%

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

      2. associate-*r* 7.8%

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

      3. *-commutative 7.8%

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

      4. associate-*r/ 7.8%

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

      5. sub-neg 7.8%

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

      6. distribute-lft-in 7.8%

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

      7. metadata-eval 7.8%

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

      8. metadata-eval 7.8%

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

      9. metadata-eval 7.8%

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

      10. fma-define 7.8%

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

      11. metadata-eval 7.8%

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

   3. Simplified 7.8%

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

   5. Taylor expanded in i around 0 85.8%

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

      1. *-commutative 85.8%

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

      2. associate-*r/ 85.8%

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

      3. metadata-eval 85.8%

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

   7. Simplified 85.8%

      \[\leadsto n \cdot \color{blue}{\left(100 + \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right) \cdot 100\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 81.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -3.4 \cdot 10^{+32}:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(i\right) \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 0.37:\\ \;\;\;\;n \cdot \left(100 + 100 \cdot \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\mathsf{expm1}\left(i\right) \cdot \frac{n}{i}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 80.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.02 \cdot 10^{-178} \lor \neg \left(n \leq 1.06 \cdot 10^{-212}\right):\\ \;\;\;\;n \cdot \frac{100 \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.02e-178) (not (<= n 1.06e-212)))
   (* n (/ (* 100.0 (expm1 i)) i))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.02e-178) || !(n <= 1.06e-212)) {
		tmp = n * ((100.0 * expm1(i)) / i);
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1.02e-178) || !(n <= 1.06e-212)) {
		tmp = n * ((100.0 * Math.expm1(i)) / i);
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1.02e-178) or not (n <= 1.06e-212):
		tmp = n * ((100.0 * math.expm1(i)) / i)
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1.02e-178) || !(n <= 1.06e-212))
		tmp = Float64(n * Float64(Float64(100.0 * expm1(i)) / i));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1.02e-178], N[Not[LessEqual[n, 1.06e-212]], $MachinePrecision]], N[(n * N[(N[(100.0 * N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.02000000000000006e-178 or 1.06000000000000004e-212 < n

   1. Initial program 21.6%

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

      1. associate-/r/ 22.0%

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

      2. associate-*r* 22.0%

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

      3. *-commutative 22.0%

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

      4. associate-*r/ 22.0%

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

      5. sub-neg 22.0%

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

      6. distribute-lft-in 22.0%

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

      7. metadata-eval 22.0%

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

      8. metadata-eval 22.0%

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

      9. metadata-eval 22.0%

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

      10. fma-define 22.0%

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

      11. metadata-eval 22.0%

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

   3. Simplified 22.0%

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

   5. Taylor expanded in n around inf 35.2%

      \[\leadsto n \cdot \frac{\color{blue}{100 \cdot e^{i} - 100}}{i} \]
   6. Step-by-step derivation

      1. sub-neg 35.2%

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

      2. metadata-eval 35.2%

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

      3. metadata-eval 35.2%

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

      4. distribute-lft-in 35.2%

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

      5. metadata-eval 35.2%

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

      6. sub-neg 35.2%

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

      7. expm1-define 84.9%

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

   7. Simplified 84.9%

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

   if -1.02000000000000006e-178 < n < 1.06000000000000004e-212

   1. Initial program 67.7%

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

      1. associate-*r/ 67.7%

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

      2. sub-neg 67.7%

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

      3. distribute-rgt-in 67.7%

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

      4. metadata-eval 67.7%

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

      5. metadata-eval 67.7%

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

   3. Simplified 67.7%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 81.0%

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

4. Final simplification 84.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.02 \cdot 10^{-178} \lor \neg \left(n \leq 1.06 \cdot 10^{-212}\right):\\ \;\;\;\;n \cdot \frac{100 \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 80.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.02 \cdot 10^{-178} \lor \neg \left(n \leq 1.06 \cdot 10^{-212}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.02e-178) (not (<= n 1.06e-212)))
   (* 100.0 (* n (/ (expm1 i) i)))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.02e-178) || !(n <= 1.06e-212)) {
		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 <= -1.02e-178) || !(n <= 1.06e-212)) {
		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 <= -1.02e-178) or not (n <= 1.06e-212):
		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 <= -1.02e-178) || !(n <= 1.06e-212))
		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, -1.02e-178], N[Not[LessEqual[n, 1.06e-212]], $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 < -1.02000000000000006e-178 or 1.06000000000000004e-212 < n

   1. Initial program 21.6%

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

   3. Taylor expanded in n around inf 35.2%

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

      1. *-commutative 35.2%

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

      2. associate-/l* 35.2%

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

      3. expm1-define 85.0%

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

   5. Simplified 85.0%

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

   if -1.02000000000000006e-178 < n < 1.06000000000000004e-212

   1. Initial program 67.7%

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

      1. associate-*r/ 67.7%

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

      2. sub-neg 67.7%

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

      3. distribute-rgt-in 67.7%

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

      4. metadata-eval 67.7%

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

      5. metadata-eval 67.7%

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

   3. Simplified 67.7%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 81.0%

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

4. Final simplification 84.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.02 \cdot 10^{-178} \lor \neg \left(n \leq 1.06 \cdot 10^{-212}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 68.4% accurate, 2.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + \left(i \cdot n\right) \cdot 0.16666666666666666\right)\right)}{i}\\ t_1 := 100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{if}\;n \leq -4 \cdot 10^{+64}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;n \leq 1.9 \cdot 10^{-243}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-28}:\\ \;\;\;\;t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0
         (*
          100.0
          (/
           (* i (+ n (* i (+ (* n 0.5) (* (* i n) 0.16666666666666666)))))
           i)))
        (t_1 (* 100.0 (/ i (/ i n)))))
   (if (<= n -4e+64)
     t_0
     (if (<= n -1.02e-178)
       t_1
       (if (<= n 1.9e-243) (/ 0.0 (/ i n)) (if (<= n 1.5e-28) t_1 t_0))))))

C

double code(double i, double n) {
	double t_0 = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	double t_1 = 100.0 * (i / (i / n));
	double tmp;
	if (n <= -4e+64) {
		tmp = t_0;
	} else if (n <= -1.02e-178) {
		tmp = t_1;
	} else if (n <= 1.9e-243) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.5e-28) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = 100.0d0 * ((i * (n + (i * ((n * 0.5d0) + ((i * n) * 0.16666666666666666d0))))) / i)
    t_1 = 100.0d0 * (i / (i / n))
    if (n <= (-4d+64)) then
        tmp = t_0
    else if (n <= (-1.02d-178)) then
        tmp = t_1
    else if (n <= 1.9d-243) then
        tmp = 0.0d0 / (i / n)
    else if (n <= 1.5d-28) then
        tmp = t_1
    else
        tmp = t_0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double t_0 = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	double t_1 = 100.0 * (i / (i / n));
	double tmp;
	if (n <= -4e+64) {
		tmp = t_0;
	} else if (n <= -1.02e-178) {
		tmp = t_1;
	} else if (n <= 1.9e-243) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.5e-28) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i)
	t_1 = 100.0 * (i / (i / n))
	tmp = 0
	if n <= -4e+64:
		tmp = t_0
	elif n <= -1.02e-178:
		tmp = t_1
	elif n <= 1.9e-243:
		tmp = 0.0 / (i / n)
	elif n <= 1.5e-28:
		tmp = t_1
	else:
		tmp = t_0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(100.0 * Float64(Float64(i * Float64(n + Float64(i * Float64(Float64(n * 0.5) + Float64(Float64(i * n) * 0.16666666666666666))))) / i))
	t_1 = Float64(100.0 * Float64(i / Float64(i / n)))
	tmp = 0.0
	if (n <= -4e+64)
		tmp = t_0;
	elseif (n <= -1.02e-178)
		tmp = t_1;
	elseif (n <= 1.9e-243)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 1.5e-28)
		tmp = t_1;
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	t_0 = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	t_1 = 100.0 * (i / (i / n));
	tmp = 0.0;
	if (n <= -4e+64)
		tmp = t_0;
	elseif (n <= -1.02e-178)
		tmp = t_1;
	elseif (n <= 1.9e-243)
		tmp = 0.0 / (i / n);
	elseif (n <= 1.5e-28)
		tmp = t_1;
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(100.0 * N[(N[(i * N[(n + N[(i * N[(N[(n * 0.5), $MachinePrecision] + N[(N[(i * n), $MachinePrecision] * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -4e+64], t$95$0, If[LessEqual[n, -1.02e-178], t$95$1, If[LessEqual[n, 1.9e-243], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.5e-28], t$95$1, t$95$0]]]]]]

Derivation

1. Split input into 3 regimes

2. if n < -4.00000000000000009e64 or 1.50000000000000001e-28 < n

   1. Initial program 19.9%

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

      1. div-sub 19.9%

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

      2. clear-num 20.0%

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

      3. sub-neg 20.0%

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

      4. div-inv 20.0%

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

      5. clear-num 19.9%

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

   4. Applied egg-rr 19.9%

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

      1. sub-neg 19.9%

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

   6. Simplified 19.9%

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

      1. associate-*r/ 19.9%

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

      2. sub-div 20.4%

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

      3. +-commutative 20.4%

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

   8. Applied egg-rr 20.4%

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

   9. Taylor expanded in n around inf 47.5%

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

       1. expm1-define 97.3%

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

   11. Simplified 97.3%

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

   12. Taylor expanded in i around 0 74.2%

       \[\leadsto 100 \cdot \frac{\color{blue}{i \cdot \left(n + i \cdot \left(0.16666666666666666 \cdot \left(i \cdot n\right) + 0.5 \cdot n\right)\right)}}{i} \]

   if -4.00000000000000009e64 < n < -1.02000000000000006e-178 or 1.8999999999999999e-243 < n < 1.50000000000000001e-28

   1. Initial program 24.2%

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

   3. Taylor expanded in i around 0 62.7%

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

   if -1.02000000000000006e-178 < n < 1.8999999999999999e-243

   1. Initial program 79.9%

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

      1. associate-*r/ 79.9%

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

      2. sub-neg 79.9%

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

      3. distribute-rgt-in 79.9%

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

      4. metadata-eval 79.9%

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

      5. metadata-eval 79.9%

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

   3. Simplified 79.9%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 84.1%

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

4. Final simplification 71.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -4 \cdot 10^{+64}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + \left(i \cdot n\right) \cdot 0.16666666666666666\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.9 \cdot 10^{-243}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-28}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + \left(i \cdot n\right) \cdot 0.16666666666666666\right)\right)}{i}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 67.5% accurate, 3.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -3.4 \cdot 10^{+64}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + \left(i \cdot n\right) \cdot 0.16666666666666666\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.2 \cdot 10^{-213}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -3.4e+64)
   (*
    100.0
    (/ (* i (+ n (* i (+ (* n 0.5) (* (* i n) 0.16666666666666666))))) i))
   (if (<= n -1.02e-178)
     (* 100.0 (/ i (/ i n)))
     (if (<= n 3.2e-213)
       (/ 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 <= -3.4e+64) {
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (i / (i / n));
	} else if (n <= 3.2e-213) {
		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 <= (-3.4d+64)) then
        tmp = 100.0d0 * ((i * (n + (i * ((n * 0.5d0) + ((i * n) * 0.16666666666666666d0))))) / i)
    else if (n <= (-1.02d-178)) then
        tmp = 100.0d0 * (i / (i / n))
    else if (n <= 3.2d-213) 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 <= -3.4e+64) {
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (i / (i / n));
	} else if (n <= 3.2e-213) {
		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 <= -3.4e+64:
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i)
	elif n <= -1.02e-178:
		tmp = 100.0 * (i / (i / n))
	elif n <= 3.2e-213:
		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 <= -3.4e+64)
		tmp = Float64(100.0 * Float64(Float64(i * Float64(n + Float64(i * Float64(Float64(n * 0.5) + Float64(Float64(i * n) * 0.16666666666666666))))) / i));
	elseif (n <= -1.02e-178)
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	elseif (n <= 3.2e-213)
		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 <= -3.4e+64)
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + ((i * n) * 0.16666666666666666))))) / i);
	elseif (n <= -1.02e-178)
		tmp = 100.0 * (i / (i / n));
	elseif (n <= 3.2e-213)
		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, -3.4e+64], N[(100.0 * N[(N[(i * N[(n + N[(i * N[(N[(n * 0.5), $MachinePrecision] + N[(N[(i * n), $MachinePrecision] * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, -1.02e-178], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 3.2e-213], 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 4 regimes

2. if n < -3.4000000000000002e64

   1. Initial program 21.3%

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

      1. div-sub 21.3%

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

      2. clear-num 21.4%

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

      3. sub-neg 21.4%

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

      4. div-inv 21.4%

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

      5. clear-num 21.3%

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

   4. Applied egg-rr 21.3%

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

      1. sub-neg 21.3%

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

   6. Simplified 21.3%

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

      1. associate-*r/ 21.3%

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

      2. sub-div 21.9%

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

      3. +-commutative 21.9%

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

   8. Applied egg-rr 21.9%

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

   9. Taylor expanded in n around inf 50.3%

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

       1. expm1-define 96.7%

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

   11. Simplified 96.7%

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

   12. Taylor expanded in i around 0 65.3%

       \[\leadsto 100 \cdot \frac{\color{blue}{i \cdot \left(n + i \cdot \left(0.16666666666666666 \cdot \left(i \cdot n\right) + 0.5 \cdot n\right)\right)}}{i} \]

   if -3.4000000000000002e64 < n < -1.02000000000000006e-178

   1. Initial program 30.2%

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

   3. Taylor expanded in i around 0 63.3%

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

   if -1.02000000000000006e-178 < n < 3.19999999999999972e-213

   1. Initial program 67.7%

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

      1. associate-*r/ 67.7%

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

      2. sub-neg 67.7%

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

      3. distribute-rgt-in 67.7%

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

      4. metadata-eval 67.7%

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

      5. metadata-eval 67.7%

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

   3. Simplified 67.7%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 81.0%

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

   if 3.19999999999999972e-213 < n

   1. Initial program 17.9%

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

   3. Taylor expanded in n around inf 33.5%

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

      1. expm1-define 67.2%

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

   5. Simplified 67.2%

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

   6. Taylor expanded in i around 0 73.5%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(50 \cdot n + i \cdot \left(4.166666666666667 \cdot \left(i \cdot n\right) + 16.666666666666668 \cdot n\right)\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 70.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -3.4 \cdot 10^{+64}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + \left(i \cdot n\right) \cdot 0.16666666666666666\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.2 \cdot 10^{-213}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 67.7% accurate, 3.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -3.3 \cdot 10^{+65}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + i \cdot \left(0.041666666666666664 \cdot \left(i \cdot n\right) + n \cdot 0.16666666666666666\right)\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.06 \cdot 10^{-212}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -3.3e+65)
   (*
    100.0
    (/
     (*
      i
      (+
       n
       (*
        i
        (+
         (* n 0.5)
         (*
          i
          (+ (* 0.041666666666666664 (* i n)) (* n 0.16666666666666666)))))))
     i))
   (if (<= n -1.02e-178)
     (* 100.0 (/ i (/ i n)))
     (if (<= n 1.06e-212)
       (/ 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 <= -3.3e+65) {
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + (i * ((0.041666666666666664 * (i * n)) + (n * 0.16666666666666666))))))) / i);
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (i / (i / n));
	} else if (n <= 1.06e-212) {
		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 <= (-3.3d+65)) then
        tmp = 100.0d0 * ((i * (n + (i * ((n * 0.5d0) + (i * ((0.041666666666666664d0 * (i * n)) + (n * 0.16666666666666666d0))))))) / i)
    else if (n <= (-1.02d-178)) then
        tmp = 100.0d0 * (i / (i / n))
    else if (n <= 1.06d-212) 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 <= -3.3e+65) {
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + (i * ((0.041666666666666664 * (i * n)) + (n * 0.16666666666666666))))))) / i);
	} else if (n <= -1.02e-178) {
		tmp = 100.0 * (i / (i / n));
	} else if (n <= 1.06e-212) {
		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 <= -3.3e+65:
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + (i * ((0.041666666666666664 * (i * n)) + (n * 0.16666666666666666))))))) / i)
	elif n <= -1.02e-178:
		tmp = 100.0 * (i / (i / n))
	elif n <= 1.06e-212:
		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 <= -3.3e+65)
		tmp = Float64(100.0 * Float64(Float64(i * Float64(n + Float64(i * Float64(Float64(n * 0.5) + Float64(i * Float64(Float64(0.041666666666666664 * Float64(i * n)) + Float64(n * 0.16666666666666666))))))) / i));
	elseif (n <= -1.02e-178)
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	elseif (n <= 1.06e-212)
		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 <= -3.3e+65)
		tmp = 100.0 * ((i * (n + (i * ((n * 0.5) + (i * ((0.041666666666666664 * (i * n)) + (n * 0.16666666666666666))))))) / i);
	elseif (n <= -1.02e-178)
		tmp = 100.0 * (i / (i / n));
	elseif (n <= 1.06e-212)
		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, -3.3e+65], N[(100.0 * N[(N[(i * N[(n + N[(i * N[(N[(n * 0.5), $MachinePrecision] + N[(i * N[(N[(0.041666666666666664 * N[(i * n), $MachinePrecision]), $MachinePrecision] + N[(n * 0.16666666666666666), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, -1.02e-178], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.06e-212], 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 4 regimes

2. if n < -3.30000000000000023e65

   1. Initial program 21.3%

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

      1. div-sub 21.3%

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

      2. clear-num 21.4%

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

      3. sub-neg 21.4%

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

      4. div-inv 21.4%

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

      5. clear-num 21.3%

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

   4. Applied egg-rr 21.3%

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

      1. sub-neg 21.3%

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

   6. Simplified 21.3%

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

      1. associate-*r/ 21.3%

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

      2. sub-div 21.9%

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

      3. +-commutative 21.9%

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

   8. Applied egg-rr 21.9%

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

   9. Taylor expanded in n around inf 50.3%

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

       1. expm1-define 96.7%

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

   11. Simplified 96.7%

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

   12. Taylor expanded in i around 0 66.0%

       \[\leadsto 100 \cdot \frac{\color{blue}{i \cdot \left(n + i \cdot \left(0.5 \cdot n + i \cdot \left(0.041666666666666664 \cdot \left(i \cdot n\right) + 0.16666666666666666 \cdot n\right)\right)\right)}}{i} \]

   if -3.30000000000000023e65 < n < -1.02000000000000006e-178

   1. Initial program 30.2%

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

   3. Taylor expanded in i around 0 63.3%

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

   if -1.02000000000000006e-178 < n < 1.06000000000000004e-212

   1. Initial program 67.7%

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

      1. associate-*r/ 67.7%

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

      2. sub-neg 67.7%

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

      3. distribute-rgt-in 67.7%

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

      4. metadata-eval 67.7%

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

      5. metadata-eval 67.7%

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

   3. Simplified 67.7%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 81.0%

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

   if 1.06000000000000004e-212 < n

   1. Initial program 17.9%

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

   3. Taylor expanded in n around inf 33.5%

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

      1. expm1-define 67.2%

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

   5. Simplified 67.2%

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

   6. Taylor expanded in i around 0 73.5%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(50 \cdot n + i \cdot \left(4.166666666666667 \cdot \left(i \cdot n\right) + 16.666666666666668 \cdot n\right)\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 70.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -3.3 \cdot 10^{+65}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + i \cdot \left(n \cdot 0.5 + i \cdot \left(0.041666666666666664 \cdot \left(i \cdot n\right) + n \cdot 0.16666666666666666\right)\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.06 \cdot 10^{-212}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + i \cdot \left(\left(i \cdot n\right) \cdot 4.166666666666667 + n \cdot 16.666666666666668\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 67.8% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 100 \cdot \frac{i}{\frac{i}{n}}\\ t_1 := 100 \cdot \frac{i \cdot \left(n + 0.5 \cdot \left(i \cdot n\right)\right)}{i}\\ \mathbf{if}\;n \leq -3.4 \cdot 10^{+64}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-249}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 7.5 \cdot 10^{-52}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;t\_1\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (* 100.0 (/ i (/ i n))))
        (t_1 (* 100.0 (/ (* i (+ n (* 0.5 (* i n)))) i))))
   (if (<= n -3.4e+64)
     t_1
     (if (<= n -1.02e-178)
       t_0
       (if (<= n 1.5e-249) (/ 0.0 (/ i n)) (if (<= n 7.5e-52) t_0 t_1))))))

C

double code(double i, double n) {
	double t_0 = 100.0 * (i / (i / n));
	double t_1 = 100.0 * ((i * (n + (0.5 * (i * n)))) / i);
	double tmp;
	if (n <= -3.4e+64) {
		tmp = t_1;
	} else if (n <= -1.02e-178) {
		tmp = t_0;
	} else if (n <= 1.5e-249) {
		tmp = 0.0 / (i / n);
	} else if (n <= 7.5e-52) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = 100.0d0 * (i / (i / n))
    t_1 = 100.0d0 * ((i * (n + (0.5d0 * (i * n)))) / i)
    if (n <= (-3.4d+64)) then
        tmp = t_1
    else if (n <= (-1.02d-178)) then
        tmp = t_0
    else if (n <= 1.5d-249) then
        tmp = 0.0d0 / (i / n)
    else if (n <= 7.5d-52) then
        tmp = t_0
    else
        tmp = t_1
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double t_0 = 100.0 * (i / (i / n));
	double t_1 = 100.0 * ((i * (n + (0.5 * (i * n)))) / i);
	double tmp;
	if (n <= -3.4e+64) {
		tmp = t_1;
	} else if (n <= -1.02e-178) {
		tmp = t_0;
	} else if (n <= 1.5e-249) {
		tmp = 0.0 / (i / n);
	} else if (n <= 7.5e-52) {
		tmp = t_0;
	} else {
		tmp = t_1;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = 100.0 * (i / (i / n))
	t_1 = 100.0 * ((i * (n + (0.5 * (i * n)))) / i)
	tmp = 0
	if n <= -3.4e+64:
		tmp = t_1
	elif n <= -1.02e-178:
		tmp = t_0
	elif n <= 1.5e-249:
		tmp = 0.0 / (i / n)
	elif n <= 7.5e-52:
		tmp = t_0
	else:
		tmp = t_1
	return tmp

Julia

function code(i, n)
	t_0 = Float64(100.0 * Float64(i / Float64(i / n)))
	t_1 = Float64(100.0 * Float64(Float64(i * Float64(n + Float64(0.5 * Float64(i * n)))) / i))
	tmp = 0.0
	if (n <= -3.4e+64)
		tmp = t_1;
	elseif (n <= -1.02e-178)
		tmp = t_0;
	elseif (n <= 1.5e-249)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 7.5e-52)
		tmp = t_0;
	else
		tmp = t_1;
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	t_0 = 100.0 * (i / (i / n));
	t_1 = 100.0 * ((i * (n + (0.5 * (i * n)))) / i);
	tmp = 0.0;
	if (n <= -3.4e+64)
		tmp = t_1;
	elseif (n <= -1.02e-178)
		tmp = t_0;
	elseif (n <= 1.5e-249)
		tmp = 0.0 / (i / n);
	elseif (n <= 7.5e-52)
		tmp = t_0;
	else
		tmp = t_1;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(100.0 * N[(N[(i * N[(n + N[(0.5 * N[(i * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -3.4e+64], t$95$1, If[LessEqual[n, -1.02e-178], t$95$0, If[LessEqual[n, 1.5e-249], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 7.5e-52], t$95$0, t$95$1]]]]]]

Derivation

1. Split input into 3 regimes

2. if n < -3.4000000000000002e64 or 7.50000000000000006e-52 < n

   1. Initial program 19.5%

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

      1. div-sub 19.4%

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

      2. clear-num 19.6%

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

      3. sub-neg 19.6%

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

      4. div-inv 19.6%

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

      5. clear-num 19.5%

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

   4. Applied egg-rr 19.5%

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

      1. sub-neg 19.5%

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

   6. Simplified 19.5%

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

      1. associate-*r/ 19.5%

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

      2. sub-div 20.0%

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

      3. +-commutative 20.0%

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

   8. Applied egg-rr 20.0%

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

   9. Taylor expanded in n around inf 46.3%

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

       1. expm1-define 96.0%

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

   11. Simplified 96.0%

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

   12. Taylor expanded in i around 0 69.1%

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

   if -3.4000000000000002e64 < n < -1.02000000000000006e-178 or 1.50000000000000002e-249 < n < 7.50000000000000006e-52

   1. Initial program 25.1%

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

   3. Taylor expanded in i around 0 63.2%

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

   if -1.02000000000000006e-178 < n < 1.50000000000000002e-249

   1. Initial program 79.9%

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

      1. associate-*r/ 79.9%

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

      2. sub-neg 79.9%

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

      3. distribute-rgt-in 79.9%

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

      4. metadata-eval 79.9%

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

      5. metadata-eval 79.9%

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

   3. Simplified 79.9%

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

   5. Taylor expanded in i around 0 52.9%

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

      1. *-commutative 52.9%

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

   7. Simplified 52.9%

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

   8. Taylor expanded in i around 0 84.1%

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

4. Final simplification 68.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -3.4 \cdot 10^{+64}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + 0.5 \cdot \left(i \cdot n\right)\right)}{i}\\ \mathbf{elif}\;n \leq -1.02 \cdot 10^{-178}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.5 \cdot 10^{-249}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 7.5 \cdot 10^{-52}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i \cdot \left(n + 0.5 \cdot \left(i \cdot n\right)\right)}{i}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 63.4% accurate, 5.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + \left(i \cdot n\right) \cdot 16.666666666666668\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -2.35e-5)
   (/ 0.0 (/ i n))
   (+ (* n 100.0) (* i (+ (* n 50.0) (* (* i n) 16.666666666666668))))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (i * ((n * 50.0) + ((i * 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 (i <= (-2.35d-5)) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = (n * 100.0d0) + (i * ((n * 50.0d0) + ((i * n) * 16.666666666666668d0)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (i * ((n * 50.0) + ((i * n) * 16.666666666666668)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -2.35e-5:
		tmp = 0.0 / (i / n)
	else:
		tmp = (n * 100.0) + (i * ((n * 50.0) + ((i * n) * 16.666666666666668)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -2.35e-5)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(Float64(n * 100.0) + Float64(i * Float64(Float64(n * 50.0) + Float64(Float64(i * n) * 16.666666666666668))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -2.35e-5)
		tmp = 0.0 / (i / n);
	else
		tmp = (n * 100.0) + (i * ((n * 50.0) + ((i * n) * 16.666666666666668)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -2.35e-5], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] + N[(i * N[(N[(n * 50.0), $MachinePrecision] + N[(N[(i * n), $MachinePrecision] * 16.666666666666668), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -2.34999999999999986e-5

   1. Initial program 58.7%

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

      1. associate-*r/ 58.8%

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

      2. sub-neg 58.8%

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

      3. distribute-rgt-in 58.7%

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

      4. metadata-eval 58.7%

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

      5. metadata-eval 58.7%

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

   3. Simplified 58.7%

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

   5. Taylor expanded in i around 0 21.5%

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

      1. *-commutative 21.5%

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

   7. Simplified 21.5%

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

   8. Taylor expanded in i around 0 31.3%

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

   if -2.34999999999999986e-5 < i

   1. Initial program 18.9%

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

   3. Taylor expanded in n around inf 24.4%

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

      1. expm1-define 61.7%

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

   5. Simplified 61.7%

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

   6. Taylor expanded in i around 0 73.4%

      \[\leadsto \color{blue}{100 \cdot n + i \cdot \left(16.666666666666668 \cdot \left(i \cdot n\right) + 50 \cdot n\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 64.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot \left(n \cdot 50 + \left(i \cdot n\right) \cdot 16.666666666666668\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 63.1% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -50000 \lor \neg \left(n \leq 7.5 \cdot 10^{-52}\right):\\ \;\;\;\;100 \cdot \frac{i \cdot n}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -50000.0) (not (<= n 7.5e-52)))
   (* 100.0 (/ (* i n) i))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -50000.0) || !(n <= 7.5e-52)) {
		tmp = 100.0 * ((i * n) / i);
	} 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 <= (-50000.0d0)) .or. (.not. (n <= 7.5d-52))) then
        tmp = 100.0d0 * ((i * n) / i)
    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 <= -50000.0) || !(n <= 7.5e-52)) {
		tmp = 100.0 * ((i * n) / i);
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -50000.0) or not (n <= 7.5e-52):
		tmp = 100.0 * ((i * n) / i)
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -50000.0) || !(n <= 7.5e-52))
		tmp = Float64(100.0 * Float64(Float64(i * n) / i));
	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 <= -50000.0) || ~((n <= 7.5e-52)))
		tmp = 100.0 * ((i * n) / i);
	else
		tmp = 100.0 * (i / (i / n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -50000.0], N[Not[LessEqual[n, 7.5e-52]], $MachinePrecision]], N[(100.0 * N[(N[(i * n), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -5e4 or 7.50000000000000006e-52 < n

   1. Initial program 21.0%

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

      1. div-sub 21.0%

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

      2. clear-num 21.1%

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

      3. sub-neg 21.1%

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

      4. div-inv 21.1%

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

      5. clear-num 21.0%

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

   4. Applied egg-rr 21.0%

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

      1. sub-neg 21.0%

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

   6. Simplified 21.0%

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

      1. associate-*r/ 21.0%

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

      2. sub-div 21.5%

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

      3. +-commutative 21.5%

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

   8. Applied egg-rr 21.5%

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

   9. Taylor expanded in i around 0 64.3%

      \[\leadsto 100 \cdot \frac{\color{blue}{i \cdot n}}{i} \]
   10. Step-by-step derivation

       1. *-commutative 64.3%

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

   11. Simplified 64.3%

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

   if -5e4 < n < 7.50000000000000006e-52

   1. Initial program 37.3%

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

   3. Taylor expanded in i around 0 57.2%

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

4. Final simplification 61.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -50000 \lor \neg \left(n \leq 7.5 \cdot 10^{-52}\right):\\ \;\;\;\;100 \cdot \frac{i \cdot n}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 63.5% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -3.4 \cdot 10^{+64} \lor \neg \left(n \leq 2.15 \cdot 10^{-54}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -3.4e+64) (not (<= n 2.15e-54)))
   (* n (+ 100.0 (* i 50.0)))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -3.4e+64) || !(n <= 2.15e-54)) {
		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 <= (-3.4d+64)) .or. (.not. (n <= 2.15d-54))) 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 <= -3.4e+64) || !(n <= 2.15e-54)) {
		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 <= -3.4e+64) or not (n <= 2.15e-54):
		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 <= -3.4e+64) || !(n <= 2.15e-54))
		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 <= -3.4e+64) || ~((n <= 2.15e-54)))
		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, -3.4e+64], N[Not[LessEqual[n, 2.15e-54]], $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 < -3.4000000000000002e64 or 2.15e-54 < n

   1. Initial program 19.5%

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

      1. div-sub 19.4%

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

      2. clear-num 19.6%

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

      3. sub-neg 19.6%

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

      4. div-inv 19.6%

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

      5. clear-num 19.5%

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

   4. Applied egg-rr 19.5%

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

      1. sub-neg 19.5%

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

   6. Simplified 19.5%

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

      1. associate-*r/ 19.5%

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

      2. sub-div 20.0%

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

      3. +-commutative 20.0%

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

   8. Applied egg-rr 20.0%

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

   9. Taylor expanded in n around inf 46.3%

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

       1. expm1-define 96.0%

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

   11. Simplified 96.0%

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

   12. Taylor expanded in i around 0 65.4%

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

       1. associate-*r* 65.4%

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

       2. distribute-rgt-out 65.4%

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

       3. *-commutative 65.4%

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

       4. +-commutative 65.4%

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

   14. Simplified 65.4%

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

   if -3.4000000000000002e64 < n < 2.15e-54

   1. Initial program 37.3%

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

   3. Taylor expanded in i around 0 58.2%

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

4. Final simplification 62.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -3.4 \cdot 10^{+64} \lor \neg \left(n \leq 2.15 \cdot 10^{-54}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 59.5% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -3.5 \cdot 10^{-50}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 1.05:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\left(i \cdot n\right) \cdot 50\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -3.5e-50)
   (* 100.0 (/ i (/ i n)))
   (if (<= i 1.05) (* 100.0 (+ n (* i -0.5))) (* (* i n) 50.0))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -3.5e-50) {
		tmp = 100.0 * (i / (i / n));
	} else if (i <= 1.05) {
		tmp = 100.0 * (n + (i * -0.5));
	} 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 <= (-3.5d-50)) then
        tmp = 100.0d0 * (i / (i / n))
    else if (i <= 1.05d0) then
        tmp = 100.0d0 * (n + (i * (-0.5d0)))
    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 <= -3.5e-50) {
		tmp = 100.0 * (i / (i / n));
	} else if (i <= 1.05) {
		tmp = 100.0 * (n + (i * -0.5));
	} else {
		tmp = (i * n) * 50.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -3.5e-50:
		tmp = 100.0 * (i / (i / n))
	elif i <= 1.05:
		tmp = 100.0 * (n + (i * -0.5))
	else:
		tmp = (i * n) * 50.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -3.5e-50)
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	elseif (i <= 1.05)
		tmp = Float64(100.0 * Float64(n + Float64(i * -0.5)));
	else
		tmp = Float64(Float64(i * n) * 50.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -3.5e-50)
		tmp = 100.0 * (i / (i / n));
	elseif (i <= 1.05)
		tmp = 100.0 * (n + (i * -0.5));
	else
		tmp = (i * n) * 50.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -3.5e-50], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 1.05], N[(100.0 * N[(n + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(i * n), $MachinePrecision] * 50.0), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if i < -3.49999999999999997e-50

   1. Initial program 49.8%

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

   3. Taylor expanded in i around 0 26.9%

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

   if -3.49999999999999997e-50 < i < 1.05000000000000004

   1. Initial program 6.9%

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

   3. Taylor expanded in i around 0 87.8%

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

      1. associate-*r* 87.5%

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

      2. associate-*r/ 87.5%

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

      3. metadata-eval 87.5%

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

   5. Simplified 87.5%

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

   6. Taylor expanded in n around 0 87.5%

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

      1. *-commutative 87.5%

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

   8. Simplified 87.5%

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

   if 1.05000000000000004 < i

   1. Initial program 49.0%

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

   3. Taylor expanded in i around 0 30.8%

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

      1. associate-*r* 30.8%

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

      2. associate-*r/ 30.8%

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

      3. metadata-eval 30.8%

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

   5. Simplified 30.8%

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

   6. Taylor expanded in n around inf 31.3%

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

      1. associate-*r* 31.3%

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

      2. *-commutative 31.3%

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

      3. *-commutative 31.3%

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

   8. Simplified 31.3%

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

   9. Taylor expanded in i around inf 31.3%

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

       1. *-commutative 31.3%

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

   11. Simplified 31.3%

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

4. Final simplification 59.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -3.5 \cdot 10^{-50}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 1.05:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\left(i \cdot n\right) \cdot 50\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 61.1% accurate, 7.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + i \cdot \left(n \cdot 0.5 - 0.5\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -2.35e-5) (/ 0.0 (/ i n)) (* 100.0 (+ n (* i (- (* n 0.5) 0.5))))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = 100.0 * (n + (i * ((n * 0.5) - 0.5)));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= (-2.35d-5)) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = 100.0d0 * (n + (i * ((n * 0.5d0) - 0.5d0)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = 100.0 * (n + (i * ((n * 0.5) - 0.5)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -2.35e-5:
		tmp = 0.0 / (i / n)
	else:
		tmp = 100.0 * (n + (i * ((n * 0.5) - 0.5)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -2.35e-5)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(100.0 * Float64(n + Float64(i * Float64(Float64(n * 0.5) - 0.5))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -2.35e-5)
		tmp = 0.0 / (i / n);
	else
		tmp = 100.0 * (n + (i * ((n * 0.5) - 0.5)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -2.35e-5], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n + N[(i * N[(N[(n * 0.5), $MachinePrecision] - 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -2.34999999999999986e-5

   1. Initial program 58.7%

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

      1. associate-*r/ 58.8%

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

      2. sub-neg 58.8%

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

      3. distribute-rgt-in 58.7%

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

      4. metadata-eval 58.7%

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

      5. metadata-eval 58.7%

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

   3. Simplified 58.7%

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

   5. Taylor expanded in i around 0 21.5%

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

      1. *-commutative 21.5%

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

   7. Simplified 21.5%

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

   8. Taylor expanded in i around 0 31.3%

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

   if -2.34999999999999986e-5 < i

   1. Initial program 18.9%

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

      1. div-sub 18.9%

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

      2. clear-num 16.8%

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

      3. sub-neg 16.8%

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

      4. div-inv 16.8%

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

      5. clear-num 18.9%

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

   4. Applied egg-rr 18.9%

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

      1. sub-neg 18.9%

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

   6. Simplified 18.9%

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

   7. Taylor expanded in i around 0 16.4%

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

      1. associate-*r* 16.8%

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

      2. *-commutative 16.8%

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

      3. associate-*r/ 16.8%

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

      4. metadata-eval 16.8%

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

   9. Simplified 16.8%

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

   10. Taylor expanded in n around 0 12.8%

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

   11. Taylor expanded in i around 0 70.1%

       \[\leadsto 100 \cdot \color{blue}{\left(n + i \cdot \left(0.5 \cdot n - 0.5\right)\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + i \cdot \left(n \cdot 0.5 - 0.5\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 61.2% accurate, 9.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -2.35e-5) (/ 0.0 (/ i n)) (* n (+ 100.0 (* i 50.0)))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -2.35e-5) {
		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 (i <= (-2.35d-5)) 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 (i <= -2.35e-5) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -2.35e-5:
		tmp = 0.0 / (i / n)
	else:
		tmp = n * (100.0 + (i * 50.0))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -2.35e-5)
		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 (i <= -2.35e-5)
		tmp = 0.0 / (i / n);
	else
		tmp = n * (100.0 + (i * 50.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -2.35e-5], 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 2 regimes

2. if i < -2.34999999999999986e-5

   1. Initial program 58.7%

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

      1. associate-*r/ 58.8%

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

      2. sub-neg 58.8%

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

      3. distribute-rgt-in 58.7%

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

      4. metadata-eval 58.7%

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

      5. metadata-eval 58.7%

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

   3. Simplified 58.7%

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

   5. Taylor expanded in i around 0 21.5%

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

      1. *-commutative 21.5%

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

   7. Simplified 21.5%

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

   8. Taylor expanded in i around 0 31.3%

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

   if -2.34999999999999986e-5 < i

   1. Initial program 18.9%

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

      1. div-sub 18.9%

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

      2. clear-num 16.8%

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

      3. sub-neg 16.8%

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

      4. div-inv 16.8%

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

      5. clear-num 18.9%

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

   4. Applied egg-rr 18.9%

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

      1. sub-neg 18.9%

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

   6. Simplified 18.9%

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

      1. associate-*r/ 18.9%

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

      2. sub-div 19.3%

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

      3. +-commutative 19.3%

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

   8. Applied egg-rr 19.3%

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

   9. Taylor expanded in n around inf 24.8%

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

       1. expm1-define 71.4%

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

   11. Simplified 71.4%

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

   12. Taylor expanded in i around 0 70.0%

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

       1. associate-*r* 70.0%

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

       2. distribute-rgt-out 70.0%

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

       3. *-commutative 70.0%

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

       4. +-commutative 70.0%

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

   14. Simplified 70.0%

       \[\leadsto \color{blue}{n \cdot \left(100 + i \cdot 50\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 62.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2.35 \cdot 10^{-5}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 55.6% accurate, 11.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 28500000000000:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;\left(i \cdot n\right) \cdot 50\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 28500000000000.0) (* n 100.0) (* (* i n) 50.0)))

C

double code(double i, double n) {
	double tmp;
	if (i <= 28500000000000.0) {
		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 <= 28500000000000.0d0) 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 <= 28500000000000.0) {
		tmp = n * 100.0;
	} else {
		tmp = (i * n) * 50.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 28500000000000.0:
		tmp = n * 100.0
	else:
		tmp = (i * n) * 50.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 28500000000000.0)
		tmp = Float64(n * 100.0);
	else
		tmp = Float64(Float64(i * n) * 50.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= 28500000000000.0)
		tmp = n * 100.0;
	else
		tmp = (i * n) * 50.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, 28500000000000.0], N[(n * 100.0), $MachinePrecision], N[(N[(i * n), $MachinePrecision] * 50.0), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 2.85e13

   1. Initial program 21.4%

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

   3. Taylor expanded in i around 0 62.6%

      \[\leadsto \color{blue}{100 \cdot n} \]
   4. Step-by-step derivation

      1. *-commutative 62.6%

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

   5. Simplified 62.6%

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

   if 2.85e13 < i

   1. Initial program 47.9%

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

   3. Taylor expanded in i around 0 32.4%

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

      1. associate-*r* 32.4%

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

      2. associate-*r/ 32.4%

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

      3. metadata-eval 32.4%

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

   5. Simplified 32.4%

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

   6. Taylor expanded in n around inf 32.7%

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

      1. associate-*r* 32.7%

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

      2. *-commutative 32.7%

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

      3. *-commutative 32.7%

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

   8. Simplified 32.7%

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

   9. Taylor expanded in i around inf 32.7%

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

       1. *-commutative 32.7%

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

   11. Simplified 32.7%

       \[\leadsto \color{blue}{50 \cdot \left(n \cdot i\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 56.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 28500000000000:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;\left(i \cdot n\right) \cdot 50\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 2.8% accurate, 38.0× speedup

Math

\[\begin{array}{l} \\ i \cdot -50 \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (* i -50.0))

C

double code(double i, double n) {
	return i * -50.0;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = i * (-50.0d0)
end function

Java

public static double code(double i, double n) {
	return i * -50.0;
}

Python

def code(i, n):
	return i * -50.0

Julia

function code(i, n)
	return Float64(i * -50.0)
end

MATLAB

function tmp = code(i, n)
	tmp = i * -50.0;
end

Wolfram

code[i_, n_] := N[(i * -50.0), $MachinePrecision]

Derivation

1. Initial program 27.0%

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

3. Taylor expanded in i around 0 56.5%

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

   1. associate-*r* 56.4%

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

   2. associate-*r/ 56.4%

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

   3. metadata-eval 56.4%

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

5. Simplified 56.4%

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

6. Taylor expanded in n around 0 2.8%

   \[\leadsto \color{blue}{-50 \cdot i} \]
7. Step-by-step derivation

   1. *-commutative 2.8%

      \[\leadsto \color{blue}{i \cdot -50} \]

8. Simplified 2.8%

   \[\leadsto \color{blue}{i \cdot -50} \]

9. Final simplification 2.8%

   \[\leadsto i \cdot -50 \]
10. Add Preprocessing

Alternative 18: 50.3% 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 27.0%

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

3. Taylor expanded in i around 0 50.8%

   \[\leadsto \color{blue}{100 \cdot n} \]
4. Step-by-step derivation

   1. *-commutative 50.8%

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

5. Simplified 50.8%

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

6. Final simplification 50.8%

   \[\leadsto n \cdot 100 \]
7. Add Preprocessing

Developer target: 33.0% 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 2024077 
(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))))