Compound Interest

Percentage Accurate: 29.0% → 98.0%

Time: 26.5s

Alternatives: 21

Speedup: 38.0×

• could not determine a ground truth

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

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 29.0% 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: 98.0% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\frac{t_0 \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

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)
     (/ (* (expm1 (* n (log1p (/ i n)))) 100.0) (/ i n))
     (if (<= t_1 INFINITY)
       (/ (+ (* t_0 100.0) -100.0) (/ i n))
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

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 = (expm1((n * log1p((i / n)))) * 100.0) / (i / n);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (Math.expm1((n * Math.log1p((i / n)))) * 100.0) / (i / n);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (math.expm1((n * math.log1p((i / n)))) * 100.0) / (i / n)
	elif t_1 <= math.inf:
		tmp = ((t_0 * 100.0) + -100.0) / (i / n)
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	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(Float64(expm1(Float64(n * log1p(Float64(i / n)))) * 100.0) / Float64(i / n));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(Float64(t_0 * 100.0) + -100.0) / Float64(i / n));
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	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[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * 100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

      1. associate-*r/ 20.2%

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

      2. *-commutative 20.2%

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

      3. pow-to-exp 19.1%

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

      4. expm1-def 29.2%

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

      5. *-commutative 29.2%

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

      6. log1p-udef 98.6%

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

   3. Applied egg-rr 98.6%

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

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.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/ 99.9%

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

      2. sub-neg 99.9%

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

      3. distribute-lft-in 99.9%

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

      4. fma-def 99.9%

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

      5. metadata-eval 99.9%

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

      6. metadata-eval 99.9%

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

   3. Simplified 99.9%

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

      1. fma-udef 99.9%

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

      2. *-commutative 99.9%

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

   5. Applied egg-rr 99.9%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 98.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 2: 83.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_0 \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;t_0 \leq \infty:\\ \;\;\;\;t_0 \cdot 100\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) (/ i n))))
   (if (<= t_0 0.0)
     (/ 100.0 (/ (/ i (expm1 i)) n))
     (if (<= t_0 INFINITY)
       (* t_0 100.0)
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

C

double code(double i, double n) {
	double t_0 = (pow((1.0 + (i / n)), n) + -1.0) / (i / n);
	double tmp;
	if (t_0 <= 0.0) {
		tmp = 100.0 / ((i / expm1(i)) / n);
	} else if (t_0 <= ((double) INFINITY)) {
		tmp = t_0 * 100.0;
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = (Math.pow((1.0 + (i / n)), n) + -1.0) / (i / n);
	double tmp;
	if (t_0 <= 0.0) {
		tmp = 100.0 / ((i / Math.expm1(i)) / n);
	} else if (t_0 <= Double.POSITIVE_INFINITY) {
		tmp = t_0 * 100.0;
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = (math.pow((1.0 + (i / n)), n) + -1.0) / (i / n)
	tmp = 0
	if t_0 <= 0.0:
		tmp = 100.0 / ((i / math.expm1(i)) / n)
	elif t_0 <= math.inf:
		tmp = t_0 * 100.0
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	return tmp

Julia

function code(i, n)
	t_0 = Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_0 <= 0.0)
		tmp = Float64(100.0 / Float64(Float64(i / expm1(i)) / n));
	elseif (t_0 <= Inf)
		tmp = Float64(t_0 * 100.0);
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$0, 0.0], N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, Infinity], N[(t$95$0 * 100.0), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

   2. Taylor expanded in n around inf 38.0%

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

      1. *-commutative 38.0%

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

      2. associate-/l* 38.0%

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

      3. expm1-def 83.3%

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

   4. Simplified 83.3%

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

      1. *-commutative 83.3%

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

      2. clear-num 83.5%

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

      3. un-div-inv 83.5%

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

   6. Applied egg-rr 83.5%

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

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

   1. Initial program 99.7%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 87.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \cdot 100\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 3: 83.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n} + -1\\ t_1 := \frac{t_0}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{t_0}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ (pow (+ 1.0 (/ i n)) n) -1.0)) (t_1 (/ t_0 (/ i n))))
   (if (<= t_1 0.0)
     (/ 100.0 (/ (/ i (expm1 i)) n))
     (if (<= t_1 INFINITY)
       (* (* n 100.0) (/ t_0 i))
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 / ((i / expm1(i)) / n);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = (n * 100.0) * (t_0 / i);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n) + -1.0;
	double t_1 = t_0 / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 / ((i / Math.expm1(i)) / n);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = (n * 100.0) * (t_0 / i);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n) + -1.0
	t_1 = t_0 / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 / ((i / math.expm1(i)) / n)
	elif t_1 <= math.inf:
		tmp = (n * 100.0) * (t_0 / i)
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	return tmp

Julia

function code(i, n)
	t_0 = Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0)
	t_1 = Float64(t_0 / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 / Float64(Float64(i / expm1(i)) / n));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(n * 100.0) * Float64(t_0 / i));
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(n * 100.0), $MachinePrecision] * N[(t$95$0 / i), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

   2. Taylor expanded in n around inf 38.0%

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

      1. *-commutative 38.0%

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

      2. associate-/l* 38.0%

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

      3. expm1-def 83.3%

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

   4. Simplified 83.3%

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

      1. *-commutative 83.3%

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

      2. clear-num 83.5%

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

      3. un-div-inv 83.5%

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

   6. Applied egg-rr 83.5%

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

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.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. *-commutative 99.7%

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

      2. associate-/r/ 99.6%

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

      3. associate-*l* 99.8%

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

      4. sub-neg 99.8%

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

      5. metadata-eval 99.8%

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

   3. Simplified 99.8%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 87.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 4: 83.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\frac{t_0 \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

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 (/ (/ i (expm1 i)) n))
     (if (<= t_1 INFINITY)
       (/ (+ (* t_0 100.0) -100.0) (/ i n))
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

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 / ((i / expm1(i)) / n);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 / ((i / Math.expm1(i)) / n);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 / ((i / math.expm1(i)) / n)
	elif t_1 <= math.inf:
		tmp = ((t_0 * 100.0) + -100.0) / (i / n)
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	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(Float64(i / expm1(i)) / n));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(Float64(t_0 * 100.0) + -100.0) / Float64(i / n));
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	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[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

   2. Taylor expanded in n around inf 38.0%

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

      1. *-commutative 38.0%

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

      2. associate-/l* 38.0%

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

      3. expm1-def 83.3%

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

   4. Simplified 83.3%

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

      1. *-commutative 83.3%

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

      2. clear-num 83.5%

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

      3. un-div-inv 83.5%

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

   6. Applied egg-rr 83.5%

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

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.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/ 99.9%

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

      2. sub-neg 99.9%

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

      3. distribute-lft-in 99.9%

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

      4. fma-def 99.9%

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

      5. metadata-eval 99.9%

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

      6. metadata-eval 99.9%

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

   3. Simplified 99.9%

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

      1. fma-udef 99.9%

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

      2. *-commutative 99.9%

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

   5. Applied egg-rr 99.9%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 87.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 5: 96.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;\left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100\right) \cdot \frac{n}{i}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\frac{t_0 \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

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)
     (* (* (expm1 (* n (log1p (/ i n)))) 100.0) (/ n i))
     (if (<= t_1 INFINITY)
       (/ (+ (* t_0 100.0) -100.0) (/ i n))
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

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 = (expm1((n * log1p((i / n)))) * 100.0) * (n / i);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (Math.expm1((n * Math.log1p((i / n)))) * 100.0) * (n / i);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (math.expm1((n * math.log1p((i / n)))) * 100.0) * (n / i)
	elif t_1 <= math.inf:
		tmp = ((t_0 * 100.0) + -100.0) / (i / n)
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	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(Float64(expm1(Float64(n * log1p(Float64(i / n)))) * 100.0) * Float64(n / i));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(Float64(t_0 * 100.0) + -100.0) / Float64(i / n));
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	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[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * 100.0), $MachinePrecision] * N[(n / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

      1. associate-*r/ 20.2%

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

      2. sub-neg 20.2%

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

      3. distribute-lft-in 20.2%

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

      4. fma-def 20.2%

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

      5. metadata-eval 20.2%

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

      6. metadata-eval 20.2%

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

   3. Simplified 20.2%

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

      1. clear-num 20.1%

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

      2. associate-/r/ 20.1%

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

      3. clear-num 19.8%

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

      4. fma-udef 19.8%

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

      5. metadata-eval 19.8%

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

      6. distribute-lft-in 19.8%

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

      7. metadata-eval 19.8%

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

      8. sub-neg 19.8%

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

      9. *-commutative 19.8%

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

      10. pow-to-exp 18.7%

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

      11. expm1-def 28.8%

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

      12. *-commutative 28.8%

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

      13. log1p-udef 96.2%

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

   5. Applied egg-rr 96.2%

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

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.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/ 99.9%

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

      2. sub-neg 99.9%

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

      3. distribute-lft-in 99.9%

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

      4. fma-def 99.9%

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

      5. metadata-eval 99.9%

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

      6. metadata-eval 99.9%

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

   3. Simplified 99.9%

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

      1. fma-udef 99.9%

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

      2. *-commutative 99.9%

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

   5. Applied egg-rr 99.9%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 97.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\left(\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100\right) \cdot \frac{n}{i}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 6: 97.3% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i} \cdot \left(n \cdot 100\right)\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;\frac{t_0 \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \end{array} \]

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)
     (* (/ (expm1 (* n (log1p (/ i n)))) i) (* n 100.0))
     (if (<= t_1 INFINITY)
       (/ (+ (* t_0 100.0) -100.0) (/ i n))
       (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))))))

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 = (expm1((n * log1p((i / n)))) / i) * (n * 100.0);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (Math.expm1((n * Math.log1p((i / n)))) / i) * (n * 100.0);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = ((t_0 * 100.0) + -100.0) / (i / n);
	} else {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	}
	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 = (math.expm1((n * math.log1p((i / n)))) / i) * (n * 100.0)
	elif t_1 <= math.inf:
		tmp = ((t_0 * 100.0) + -100.0) / (i / n)
	else:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	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(Float64(expm1(Float64(n * log1p(Float64(i / n)))) / i) * Float64(n * 100.0));
	elseif (t_1 <= Inf)
		tmp = Float64(Float64(Float64(t_0 * 100.0) + -100.0) / Float64(i / n));
	else
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	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[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / i), $MachinePrecision] * N[(n * 100.0), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(N[(N[(t$95$0 * 100.0), $MachinePrecision] + -100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -0.0

   1. Initial program 20.1%

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

      1. *-commutative 20.1%

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

      2. associate-/r/ 19.8%

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

      3. associate-*l* 19.8%

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

      4. sub-neg 19.8%

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

      5. metadata-eval 19.8%

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

   3. Simplified 19.8%

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

      1. expm1-log1p-u 18.8%

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

      2. expm1-udef 14.1%

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

      3. metadata-eval 14.1%

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

      4. sub-neg 14.1%

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

      5. pow-to-exp 14.1%

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

      6. expm1-def 17.3%

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

      7. *-commutative 17.3%

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

      8. log1p-udef 75.7%

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

   5. Applied egg-rr 75.7%

      \[\leadsto \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)} - 1\right)} \cdot \left(n \cdot 100\right) \]
   6. Step-by-step derivation

      1. expm1-def 97.3%

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

      2. expm1-log1p 97.3%

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

   7. Simplified 97.3%

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

   if -0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.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/ 99.9%

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

      2. sub-neg 99.9%

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

      3. distribute-lft-in 99.9%

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

      4. fma-def 99.9%

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

      5. metadata-eval 99.9%

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

      6. metadata-eval 99.9%

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

   3. Simplified 99.9%

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

      1. fma-udef 99.9%

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

      2. *-commutative 99.9%

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

   5. Applied egg-rr 99.9%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

   1. Initial program 0.0%

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

   2. Taylor expanded in n around inf 1.9%

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

      1. *-commutative 1.9%

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

      2. associate-/l* 1.9%

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

      3. expm1-def 84.0%

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

   4. Simplified 84.0%

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

      1. *-commutative 84.0%

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

      2. clear-num 83.8%

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

      3. un-div-inv 83.8%

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

   6. Applied egg-rr 83.8%

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

   7. Taylor expanded in i around 0 99.8%

      \[\leadsto \frac{100}{\color{blue}{-0.5 \cdot \frac{i}{n} + \frac{1}{n}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i} \cdot \left(n \cdot 100\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} \cdot 100 + -100}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \end{array} \]

Alternative 7: 78.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 9.5 \cdot 10^{+176}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\frac{n}{i} \cdot \left(-1 + {\left(\frac{i}{n}\right)}^{n}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 9.5e+176)
   (/ 100.0 (/ (/ i (expm1 i)) n))
   (* 100.0 (* (/ n i) (+ -1.0 (pow (/ i n) n))))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 9.5e+176) {
		tmp = 100.0 / ((i / expm1(i)) / n);
	} else {
		tmp = 100.0 * ((n / i) * (-1.0 + pow((i / n), n)));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 9.5e+176) {
		tmp = 100.0 / ((i / Math.expm1(i)) / n);
	} else {
		tmp = 100.0 * ((n / i) * (-1.0 + Math.pow((i / n), n)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 9.5e+176:
		tmp = 100.0 / ((i / math.expm1(i)) / n)
	else:
		tmp = 100.0 * ((n / i) * (-1.0 + math.pow((i / n), n)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 9.5e+176)
		tmp = Float64(100.0 / Float64(Float64(i / expm1(i)) / n));
	else
		tmp = Float64(100.0 * Float64(Float64(n / i) * Float64(-1.0 + (Float64(i / n) ^ n))));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[LessEqual[i, 9.5e+176], N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(N[(n / i), $MachinePrecision] * N[(-1.0 + N[Power[N[(i / n), $MachinePrecision], n], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 9.4999999999999995e176

   1. Initial program 20.2%

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

   2. Taylor expanded in n around inf 33.7%

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

      1. *-commutative 33.7%

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

      2. associate-/l* 33.7%

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

      3. expm1-def 85.6%

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

   4. Simplified 85.6%

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

      1. *-commutative 85.6%

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

      2. clear-num 85.7%

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

      3. un-div-inv 85.7%

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

   6. Applied egg-rr 85.7%

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

   if 9.4999999999999995e176 < i

   1. Initial program 63.0%

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

      1. associate-*r/ 63.2%

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

      2. *-commutative 63.2%

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

      3. pow-to-exp 26.9%

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

      4. expm1-def 36.4%

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

      5. *-commutative 36.4%

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

      6. log1p-udef 40.8%

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

   3. Applied egg-rr 40.8%

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

   4. Taylor expanded in i around inf 45.0%

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

      1. associate-/l* 45.0%

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

      2. associate-/r/ 45.0%

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

      3. sub-neg 45.0%

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

      4. metadata-eval 45.0%

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

      5. +-commutative 45.0%

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

      6. *-commutative 45.0%

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

      7. exp-prod 22.3%

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

   6. Simplified 63.0%

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

4. Final simplification 83.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 9.5 \cdot 10^{+176}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\frac{n}{i} \cdot \left(-1 + {\left(\frac{i}{n}\right)}^{n}\right)\right)\\ \end{array} \]

Alternative 8: 78.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 9.5 \cdot 10^{+176}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{-1 + {\left(\frac{i}{n}\right)}^{n}}{i}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 9.5e+176)
   (/ 100.0 (/ (/ i (expm1 i)) n))
   (* (* n 100.0) (/ (+ -1.0 (pow (/ i n) n)) i))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 9.5e+176) {
		tmp = 100.0 / ((i / expm1(i)) / n);
	} else {
		tmp = (n * 100.0) * ((-1.0 + pow((i / n), n)) / i);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 9.5e+176) {
		tmp = 100.0 / ((i / Math.expm1(i)) / n);
	} else {
		tmp = (n * 100.0) * ((-1.0 + Math.pow((i / n), n)) / i);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 9.5e+176:
		tmp = 100.0 / ((i / math.expm1(i)) / n)
	else:
		tmp = (n * 100.0) * ((-1.0 + math.pow((i / n), n)) / i)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 9.5e+176)
		tmp = Float64(100.0 / Float64(Float64(i / expm1(i)) / n));
	else
		tmp = Float64(Float64(n * 100.0) * Float64(Float64(-1.0 + (Float64(i / n) ^ n)) / i));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[LessEqual[i, 9.5e+176], N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] * N[(N[(-1.0 + N[Power[N[(i / n), $MachinePrecision], n], $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 9.4999999999999995e176

   1. Initial program 20.2%

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

   2. Taylor expanded in n around inf 33.7%

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

      1. *-commutative 33.7%

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

      2. associate-/l* 33.7%

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

      3. expm1-def 85.6%

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

   4. Simplified 85.6%

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

      1. *-commutative 85.6%

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

      2. clear-num 85.7%

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

      3. un-div-inv 85.7%

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

   6. Applied egg-rr 85.7%

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

   if 9.4999999999999995e176 < i

   1. Initial program 63.0%

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

      1. *-commutative 63.0%

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

      2. associate-/r/ 63.3%

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

      3. associate-*l* 63.5%

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

      4. sub-neg 63.5%

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

      5. metadata-eval 63.5%

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

   3. Simplified 63.5%

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

   4. Taylor expanded in i around inf 63.4%

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

4. Final simplification 83.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 9.5 \cdot 10^{+176}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{-1 + {\left(\frac{i}{n}\right)}^{n}}{i}\\ \end{array} \]

Alternative 9: 76.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -1.6 \cdot 10^{-70} \lor \neg \left(i \leq 2.7 \cdot 10^{-27}\right):\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= i -1.6e-70) (not (<= i 2.7e-27)))
   (* 100.0 (/ (expm1 i) (/ i n)))
   (* (* n 100.0) (+ 1.0 (* i (- 0.5 (/ 0.5 n)))))))

C

double code(double i, double n) {
	double tmp;
	if ((i <= -1.6e-70) || !(i <= 2.7e-27)) {
		tmp = 100.0 * (expm1(i) / (i / n));
	} else {
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((i <= -1.6e-70) || !(i <= 2.7e-27)) {
		tmp = 100.0 * (Math.expm1(i) / (i / n));
	} else {
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (i <= -1.6e-70) or not (i <= 2.7e-27):
		tmp = 100.0 * (math.expm1(i) / (i / n))
	else:
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((i <= -1.6e-70) || !(i <= 2.7e-27))
		tmp = Float64(100.0 * Float64(expm1(i) / Float64(i / n)));
	else
		tmp = Float64(Float64(n * 100.0) * Float64(1.0 + Float64(i * Float64(0.5 - Float64(0.5 / n)))));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[i, -1.6e-70], N[Not[LessEqual[i, 2.7e-27]], $MachinePrecision]], N[(100.0 * N[(N[(Exp[i] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] * N[(1.0 + N[(i * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -1.5999999999999999e-70 or 2.69999999999999989e-27 < i

   1. Initial program 44.1%

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

   2. Taylor expanded in n around inf 63.2%

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

      1. expm1-def 67.8%

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

   4. Simplified 67.8%

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

   if -1.5999999999999999e-70 < i < 2.69999999999999989e-27

   1. Initial program 4.9%

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

      1. *-commutative 4.9%

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

      2. associate-/r/ 5.4%

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

      3. associate-*l* 5.4%

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

      4. sub-neg 5.4%

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

      5. metadata-eval 5.4%

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

   3. Simplified 5.4%

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

   4. Taylor expanded in i around 0 93.7%

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

      1. associate-*r/ 93.7%

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

      2. metadata-eval 93.7%

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

   6. Simplified 93.7%

      \[\leadsto \color{blue}{\left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)} \cdot \left(n \cdot 100\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 81.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -1.6 \cdot 10^{-70} \lor \neg \left(i \leq 2.7 \cdot 10^{-27}\right):\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \end{array} \]

Alternative 10: 77.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 10^{+98}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{\left(n \cdot 100\right) \cdot \left(\left(n \cdot n\right) \cdot 10000\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i 1e+98)
   (/ 100.0 (/ (/ i (expm1 i)) n))
   (cbrt (* (* n 100.0) (* (* n n) 10000.0)))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 1e+98) {
		tmp = 100.0 / ((i / expm1(i)) / n);
	} else {
		tmp = cbrt(((n * 100.0) * ((n * n) * 10000.0)));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 1e+98) {
		tmp = 100.0 / ((i / Math.expm1(i)) / n);
	} else {
		tmp = Math.cbrt(((n * 100.0) * ((n * n) * 10000.0)));
	}
	return tmp;
}

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 1e+98)
		tmp = Float64(100.0 / Float64(Float64(i / expm1(i)) / n));
	else
		tmp = cbrt(Float64(Float64(n * 100.0) * Float64(Float64(n * n) * 10000.0)));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[LessEqual[i, 1e+98], N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision], N[Power[N[(N[(n * 100.0), $MachinePrecision] * N[(N[(n * n), $MachinePrecision] * 10000.0), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 9.99999999999999998e97

   1. Initial program 18.1%

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

   2. Taylor expanded in n around inf 31.4%

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

      1. *-commutative 31.4%

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

      2. associate-/l* 31.4%

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

      3. expm1-def 87.6%

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

   4. Simplified 87.6%

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

      1. *-commutative 87.6%

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

      2. clear-num 87.7%

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

      3. un-div-inv 87.7%

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

   6. Applied egg-rr 87.7%

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

   if 9.99999999999999998e97 < i

   1. Initial program 55.1%

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

   2. Taylor expanded in n around inf 43.2%

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

      1. *-commutative 43.2%

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

      2. associate-/l* 43.2%

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

      3. expm1-def 43.2%

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

   4. Simplified 43.2%

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

      1. *-commutative 43.2%

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

      2. clear-num 43.2%

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

      3. un-div-inv 43.2%

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

   6. Applied egg-rr 43.2%

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

   7. Taylor expanded in i around 0 4.5%

      \[\leadsto \frac{100}{\color{blue}{\frac{1}{n}}} \]
   8. Step-by-step derivation

      1. associate-/r/ 4.5%

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

      2. metadata-eval 4.5%

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

      3. *-commutative 4.5%

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

      4. add-cbrt-cube 58.4%

         \[\leadsto \color{blue}{\sqrt[3]{\left(\left(n \cdot 100\right) \cdot \left(n \cdot 100\right)\right) \cdot \left(n \cdot 100\right)}} \]

   9. Applied egg-rr 58.4%

      \[\leadsto \color{blue}{\sqrt[3]{\left(\left(n \cdot 100\right) \cdot \left(n \cdot 100\right)\right) \cdot \left(n \cdot 100\right)}} \]
   10. Step-by-step derivation

       1. associate-*l* 58.4%

          \[\leadsto \sqrt[3]{\color{blue}{\left(n \cdot 100\right) \cdot \left(\left(n \cdot 100\right) \cdot \left(n \cdot 100\right)\right)}} \]

       2. *-commutative 58.4%

          \[\leadsto \sqrt[3]{\color{blue}{\left(100 \cdot n\right)} \cdot \left(\left(n \cdot 100\right) \cdot \left(n \cdot 100\right)\right)} \]

       3. swap-sqr 58.4%

          \[\leadsto \sqrt[3]{\left(100 \cdot n\right) \cdot \color{blue}{\left(\left(n \cdot n\right) \cdot \left(100 \cdot 100\right)\right)}} \]

       4. metadata-eval 58.4%

          \[\leadsto \sqrt[3]{\left(100 \cdot n\right) \cdot \left(\left(n \cdot n\right) \cdot \color{blue}{10000}\right)} \]

   11. Simplified 58.4%

       \[\leadsto \color{blue}{\sqrt[3]{\left(100 \cdot n\right) \cdot \left(\left(n \cdot n\right) \cdot 10000\right)}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 83.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 10^{+98}:\\ \;\;\;\;\frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{\left(n \cdot 100\right) \cdot \left(\left(n \cdot n\right) \cdot 10000\right)}\\ \end{array} \]

Alternative 11: 76.3% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ 100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}} \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (* 100.0 (/ n (/ i (expm1 i)))))

C

double code(double i, double n) {
	return 100.0 * (n / (i / expm1(i)));
}

Java

public static double code(double i, double n) {
	return 100.0 * (n / (i / Math.expm1(i)));
}

Python

def code(i, n):
	return 100.0 * (n / (i / math.expm1(i)))

Julia

function code(i, n)
	return Float64(100.0 * Float64(n / Float64(i / expm1(i))))
end

Wolfram

code[i_, n_] := N[(100.0 * N[(n / N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 23.9%

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

2. Taylor expanded in n around inf 33.2%

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

   1. *-commutative 33.2%

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

   2. associate-/l* 33.2%

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

   3. expm1-def 80.6%

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

4. Simplified 80.6%

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

5. Final simplification 80.6%

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

Alternative 12: 76.3% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \frac{100}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n}} \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (/ 100.0 (/ (/ i (expm1 i)) n)))

C

double code(double i, double n) {
	return 100.0 / ((i / expm1(i)) / n);
}

Java

public static double code(double i, double n) {
	return 100.0 / ((i / Math.expm1(i)) / n);
}

Python

def code(i, n):
	return 100.0 / ((i / math.expm1(i)) / n)

Julia

function code(i, n)
	return Float64(100.0 / Float64(Float64(i / expm1(i)) / n))
end

Wolfram

code[i_, n_] := N[(100.0 / N[(N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / n), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 23.9%

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

2. Taylor expanded in n around inf 33.2%

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

   1. *-commutative 33.2%

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

   2. associate-/l* 33.2%

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

   3. expm1-def 80.6%

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

4. Simplified 80.6%

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

   1. *-commutative 80.6%

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

   2. clear-num 80.8%

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

   3. un-div-inv 80.7%

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

6. Applied egg-rr 80.7%

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

7. Final simplification 80.7%

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

Alternative 13: 62.3% accurate, 7.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 3.4 \cdot 10^{-87}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + \left(0.5 - \frac{0.5}{n}\right) \cdot \left(i \cdot n\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 3.4e-87)
   (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))
   (* 100.0 (+ n (* (- 0.5 (/ 0.5 n)) (* i n))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 3.4e-87) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = 100.0 * (n + ((0.5 - (0.5 / n)) * (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-87) then
        tmp = 100.0d0 / (((i / n) * (-0.5d0)) + (1.0d0 / n))
    else
        tmp = 100.0d0 * (n + ((0.5d0 - (0.5d0 / n)) * (i * n)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 3.4e-87) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = 100.0 * (n + ((0.5 - (0.5 / n)) * (i * n)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 3.4e-87:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	else:
		tmp = 100.0 * (n + ((0.5 - (0.5 / n)) * (i * n)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 3.4e-87)
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	else
		tmp = Float64(100.0 * Float64(n + Float64(Float64(0.5 - Float64(0.5 / n)) * Float64(i * n))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 3.4e-87)
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	else
		tmp = 100.0 * (n + ((0.5 - (0.5 / n)) * (i * n)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 3.4e-87], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n + N[(N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision] * N[(i * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 3.3999999999999999e-87

   1. Initial program 27.1%

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

   2. Taylor expanded in n around inf 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

      3. expm1-def 75.1%

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

   4. Simplified 75.1%

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

      1. *-commutative 75.1%

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

      2. clear-num 75.4%

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

      3. un-div-inv 75.3%

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

   6. Applied egg-rr 75.3%

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

   7. Taylor expanded in i around 0 63.2%

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

   if 3.3999999999999999e-87 < n

   1. Initial program 18.0%

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

   2. Taylor expanded in i around 0 68.9%

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

      1. associate-*r* 68.9%

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

      2. associate-*r/ 68.9%

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

      3. metadata-eval 68.9%

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

   4. Simplified 68.9%

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

4. Final simplification 65.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 3.4 \cdot 10^{-87}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + \left(0.5 - \frac{0.5}{n}\right) \cdot \left(i \cdot n\right)\right)\\ \end{array} \]

Alternative 14: 62.3% accurate, 7.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 3.4 \cdot 10^{-87}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 3.4e-87)
   (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))
   (* (* n 100.0) (+ 1.0 (* i (- 0.5 (/ 0.5 n)))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 3.4e-87) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / 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-87) then
        tmp = 100.0d0 / (((i / n) * (-0.5d0)) + (1.0d0 / n))
    else
        tmp = (n * 100.0d0) * (1.0d0 + (i * (0.5d0 - (0.5d0 / n))))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 3.4e-87) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 3.4e-87:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	else:
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 3.4e-87)
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	else
		tmp = Float64(Float64(n * 100.0) * Float64(1.0 + Float64(i * Float64(0.5 - Float64(0.5 / n)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 3.4e-87)
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	else
		tmp = (n * 100.0) * (1.0 + (i * (0.5 - (0.5 / n))));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 3.4e-87], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] * N[(1.0 + N[(i * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 3.3999999999999999e-87

   1. Initial program 27.1%

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

   2. Taylor expanded in n around inf 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

      3. expm1-def 75.1%

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

   4. Simplified 75.1%

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

      1. *-commutative 75.1%

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

      2. clear-num 75.4%

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

      3. un-div-inv 75.3%

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

   6. Applied egg-rr 75.3%

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

   7. Taylor expanded in i around 0 63.2%

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

   if 3.3999999999999999e-87 < n

   1. Initial program 18.0%

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

      1. *-commutative 18.0%

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

      2. associate-/r/ 18.4%

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

      3. associate-*l* 18.4%

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

      4. sub-neg 18.4%

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

      5. metadata-eval 18.4%

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

   3. Simplified 18.4%

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

   4. Taylor expanded in i around 0 68.9%

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

      1. associate-*r/ 68.9%

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

      2. metadata-eval 68.9%

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

   6. Simplified 68.9%

      \[\leadsto \color{blue}{\left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)} \cdot \left(n \cdot 100\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 65.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 3.4 \cdot 10^{-87}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \left(1 + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\\ \end{array} \]

Alternative 15: 63.7% accurate, 7.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 1.55 \cdot 10^{-16}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{i + 0.5 \cdot \left(i \cdot i\right)}{i}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 1.55e-16)
   (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))
   (* (* n 100.0) (/ (+ i (* 0.5 (* i i))) i))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 1.55e-16) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = (n * 100.0) * ((i + (0.5 * (i * i))) / i);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.55d-16) then
        tmp = 100.0d0 / (((i / n) * (-0.5d0)) + (1.0d0 / n))
    else
        tmp = (n * 100.0d0) * ((i + (0.5d0 * (i * i))) / i)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 1.55e-16) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = (n * 100.0) * ((i + (0.5 * (i * i))) / i);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 1.55e-16:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	else:
		tmp = (n * 100.0) * ((i + (0.5 * (i * i))) / i)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 1.55e-16)
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	else
		tmp = Float64(Float64(n * 100.0) * Float64(Float64(i + Float64(0.5 * Float64(i * i))) / i));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 1.55e-16)
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	else
		tmp = (n * 100.0) * ((i + (0.5 * (i * i))) / i);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 1.55e-16], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] * N[(N[(i + N[(0.5 * N[(i * i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.55e-16

   1. Initial program 25.6%

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

   2. Taylor expanded in n around inf 28.1%

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

      1. *-commutative 28.1%

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

      2. associate-/l* 28.1%

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

      3. expm1-def 73.6%

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

   4. Simplified 73.6%

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

      1. *-commutative 73.6%

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

      2. clear-num 73.9%

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

      3. un-div-inv 73.8%

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

   6. Applied egg-rr 73.8%

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

   7. Taylor expanded in i around 0 62.8%

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

   if 1.55e-16 < n

   1. Initial program 19.8%

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

      1. *-commutative 19.8%

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

      2. associate-/r/ 20.2%

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

      3. associate-*l* 20.3%

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

      4. sub-neg 20.3%

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

      5. metadata-eval 20.3%

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

   3. Simplified 20.3%

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

   4. Taylor expanded in i around 0 21.3%

      \[\leadsto \frac{\color{blue}{\left(1 + \left(i + {i}^{2} \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)} + -1}{i} \cdot \left(n \cdot 100\right) \]
   5. Step-by-step derivation

      1. unpow2 21.3%

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

      2. associate-*r/ 21.3%

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

      3. metadata-eval 21.3%

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

   6. Simplified 21.3%

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

   7. Taylor expanded in n around inf 72.8%

      \[\leadsto \frac{\color{blue}{i + 0.5 \cdot {i}^{2}}}{i} \cdot \left(n \cdot 100\right) \]
   8. Step-by-step derivation

      1. unpow2 72.8%

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

   9. Simplified 72.8%

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

4. Final simplification 65.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 1.55 \cdot 10^{-16}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(n \cdot 100\right) \cdot \frac{i + 0.5 \cdot \left(i \cdot i\right)}{i}\\ \end{array} \]

Alternative 16: 61.9% accurate, 8.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 1.8 \cdot 10^{+104}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 1.8e+104)
   (/ 100.0 (+ (* (/ i n) -0.5) (/ 1.0 n)))
   (* n (+ 100.0 (* i 50.0)))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 1.8e+104) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.8d+104) then
        tmp = 100.0d0 / (((i / n) * (-0.5d0)) + (1.0d0 / n))
    else
        tmp = n * (100.0d0 + (i * 50.0d0))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 1.8e+104) {
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 1.8e+104:
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n))
	else:
		tmp = n * (100.0 + (i * 50.0))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 1.8e+104)
		tmp = Float64(100.0 / Float64(Float64(Float64(i / n) * -0.5) + Float64(1.0 / n)));
	else
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 1.8e+104)
		tmp = 100.0 / (((i / n) * -0.5) + (1.0 / n));
	else
		tmp = n * (100.0 + (i * 50.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 1.8e+104], N[(100.0 / N[(N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision] + N[(1.0 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.8e104

   1. Initial program 25.1%

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

   2. Taylor expanded in n around inf 28.3%

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

      1. *-commutative 28.3%

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

      2. associate-/l* 28.3%

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

      3. expm1-def 76.0%

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

   4. Simplified 76.0%

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

      1. *-commutative 76.0%

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

      2. clear-num 76.2%

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

      3. un-div-inv 76.1%

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

   6. Applied egg-rr 76.1%

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

   7. Taylor expanded in i around 0 63.9%

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

   if 1.8e104 < n

   1. Initial program 19.6%

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

   2. Taylor expanded in n around inf 51.7%

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

      1. *-commutative 51.7%

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

      2. associate-/l* 51.7%

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

      3. expm1-def 98.0%

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

   4. Simplified 98.0%

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

   5. Taylor expanded in i around 0 69.6%

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

      1. associate-*r* 69.6%

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

      2. distribute-rgt-out 69.6%

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

   7. Simplified 69.6%

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

4. Final simplification 65.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 1.8 \cdot 10^{+104}:\\ \;\;\;\;\frac{100}{\frac{i}{n} \cdot -0.5 + \frac{1}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]

Alternative 17: 62.2% accurate, 10.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1 \cdot 10^{+82} \lor \neg \left(n \leq 1.5 \cdot 10^{-16}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1e+82) (not (<= n 1.5e-16)))
   (* n (+ 100.0 (* i 50.0)))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1e+82) || !(n <= 1.5e-16)) {
		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 <= (-1d+82)) .or. (.not. (n <= 1.5d-16))) 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 <= -1e+82) || !(n <= 1.5e-16)) {
		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 <= -1e+82) or not (n <= 1.5e-16):
		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 <= -1e+82) || !(n <= 1.5e-16))
		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 <= -1e+82) || ~((n <= 1.5e-16)))
		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, -1e+82], N[Not[LessEqual[n, 1.5e-16]], $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 < -9.9999999999999996e81 or 1.49999999999999997e-16 < n

   1. Initial program 20.0%

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

   2. Taylor expanded in n around inf 44.2%

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

      1. *-commutative 44.2%

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

      2. associate-/l* 44.2%

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

      3. expm1-def 94.1%

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

   4. Simplified 94.1%

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

   5. Taylor expanded in i around 0 64.0%

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

      1. associate-*r* 64.0%

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

      2. distribute-rgt-out 64.0%

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

   7. Simplified 64.0%

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

   if -9.9999999999999996e81 < n < 1.49999999999999997e-16

   1. Initial program 28.4%

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

   2. Taylor expanded in i around 0 65.1%

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

4. Final simplification 64.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1 \cdot 10^{+82} \lor \neg \left(n \leq 1.5 \cdot 10^{-16}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 18: 58.5% accurate, 10.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2.05 \cdot 10^{-70}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 1.2:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -2.05e-70)
   (* 100.0 (/ i (/ i n)))
   (if (<= i 1.2) (* 100.0 (+ n (* i -0.5))) (* n (* i 50.0)))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -2.05e-70) {
		tmp = 100.0 * (i / (i / n));
	} else if (i <= 1.2) {
		tmp = 100.0 * (n + (i * -0.5));
	} else {
		tmp = n * (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 <= (-2.05d-70)) then
        tmp = 100.0d0 * (i / (i / n))
    else if (i <= 1.2d0) then
        tmp = 100.0d0 * (n + (i * (-0.5d0)))
    else
        tmp = n * (i * 50.0d0)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -2.05e-70) {
		tmp = 100.0 * (i / (i / n));
	} else if (i <= 1.2) {
		tmp = 100.0 * (n + (i * -0.5));
	} else {
		tmp = n * (i * 50.0);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -2.05e-70:
		tmp = 100.0 * (i / (i / n))
	elif i <= 1.2:
		tmp = 100.0 * (n + (i * -0.5))
	else:
		tmp = n * (i * 50.0)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -2.05e-70)
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	elseif (i <= 1.2)
		tmp = Float64(100.0 * Float64(n + Float64(i * -0.5)));
	else
		tmp = Float64(n * Float64(i * 50.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -2.05e-70)
		tmp = 100.0 * (i / (i / n));
	elseif (i <= 1.2)
		tmp = 100.0 * (n + (i * -0.5));
	else
		tmp = n * (i * 50.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -2.05e-70], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 1.2], N[(100.0 * N[(n + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * N[(i * 50.0), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if i < -2.04999999999999989e-70

   1. Initial program 43.9%

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

   2. Taylor expanded in i around 0 27.5%

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

   if -2.04999999999999989e-70 < i < 1.19999999999999996

   1. Initial program 4.8%

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

   2. Taylor expanded in i around 0 93.0%

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

      1. associate-*r* 92.7%

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

      2. associate-*r/ 92.7%

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

      3. metadata-eval 92.7%

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

   4. Simplified 92.7%

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

   5. Taylor expanded in n around 0 92.3%

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

      1. *-commutative 92.3%

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

   7. Simplified 92.3%

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

   if 1.19999999999999996 < i

   1. Initial program 47.9%

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

      1. *-commutative 47.9%

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

      2. associate-/r/ 48.1%

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

      3. associate-*l* 48.2%

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

      4. sub-neg 48.2%

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

      5. metadata-eval 48.2%

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

   3. Simplified 48.2%

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

   4. Taylor expanded in n around inf 55.0%

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

      1. expm1-def 55.0%

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

   6. Simplified 55.0%

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

   7. Taylor expanded in i around 0 24.7%

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

      1. *-commutative 24.7%

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

   9. Simplified 24.7%

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

   10. Taylor expanded in i around inf 24.7%

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

       1. *-commutative 24.7%

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

       2. *-commutative 24.7%

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

       3. associate-*l* 24.7%

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

   12. Simplified 24.7%

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

4. Final simplification 61.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2.05 \cdot 10^{-70}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 1.2:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \end{array} \]

Alternative 19: 54.6% accurate, 16.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq 2:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (if (<= i 2.0) (* n 100.0) (* n (* i 50.0))))

C

double code(double i, double n) {
	double tmp;
	if (i <= 2.0) {
		tmp = n * 100.0;
	} else {
		tmp = n * (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 <= 2.0d0) then
        tmp = n * 100.0d0
    else
        tmp = n * (i * 50.0d0)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= 2.0) {
		tmp = n * 100.0;
	} else {
		tmp = n * (i * 50.0);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= 2.0:
		tmp = n * 100.0
	else:
		tmp = n * (i * 50.0)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= 2.0)
		tmp = Float64(n * 100.0);
	else
		tmp = Float64(n * Float64(i * 50.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= 2.0)
		tmp = n * 100.0;
	else
		tmp = n * (i * 50.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, 2.0], N[(n * 100.0), $MachinePrecision], N[(n * N[(i * 50.0), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < 2

   1. Initial program 17.0%

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

   2. Taylor expanded in i around 0 68.5%

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

      1. *-commutative 68.5%

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

   4. Simplified 68.5%

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

   if 2 < i

   1. Initial program 47.9%

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

      1. *-commutative 47.9%

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

      2. associate-/r/ 48.1%

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

      3. associate-*l* 48.2%

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

      4. sub-neg 48.2%

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

      5. metadata-eval 48.2%

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

   3. Simplified 48.2%

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

   4. Taylor expanded in n around inf 55.0%

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

      1. expm1-def 55.0%

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

   6. Simplified 55.0%

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

   7. Taylor expanded in i around 0 24.7%

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

      1. *-commutative 24.7%

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

   9. Simplified 24.7%

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

   10. Taylor expanded in i around inf 24.7%

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

       1. *-commutative 24.7%

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

       2. *-commutative 24.7%

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

       3. associate-*l* 24.7%

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

   12. Simplified 24.7%

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

4. Final simplification 58.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq 2:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \end{array} \]

Alternative 20: 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 23.9%

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

2. Taylor expanded in i around 0 59.0%

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

   1. associate-*r* 58.9%

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

   2. associate-*r/ 58.9%

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

   3. metadata-eval 58.9%

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

4. Simplified 58.9%

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

5. Taylor expanded in n around 0 3.0%

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

   1. *-commutative 3.0%

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

7. Simplified 3.0%

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

8. Final simplification 3.0%

   \[\leadsto i \cdot -50 \]

Alternative 21: 48.9% accurate, 38.0× speedup

Math

\[\begin{array}{l} \\ n \cdot 100 \end{array} \]

FPCore

(FPCore (i n) :precision binary64 (* n 100.0))

C

double code(double i, double n) {
	return n * 100.0;
}

Fortran

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

Java

public static double code(double i, double n) {
	return n * 100.0;
}

Python

def code(i, n):
	return n * 100.0

Julia

function code(i, n)
	return Float64(n * 100.0)
end

MATLAB

function tmp = code(i, n)
	tmp = n * 100.0;
end

Wolfram

code[i_, n_] := N[(n * 100.0), $MachinePrecision]

Derivation

1. Initial program 23.9%

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

2. Taylor expanded in i around 0 54.3%

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

   1. *-commutative 54.3%

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

4. Simplified 54.3%

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

5. Final simplification 54.3%

   \[\leadsto n \cdot 100 \]

Developer target: 35.0% 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 2023278 
(FPCore (i n)
  :name "Compound Interest"
  :precision binary64

  :herbie-target
  (* 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))))