Compound Interest

Percentage Accurate: 27.9% → 98.1%

Time: 18.5s

Alternatives: 15

Speedup: 16.0×

• could not determine a ground truth

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

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 27.9% 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: 98.1% accurate, 0.3× speedup

Math

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

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 INFINITY)
       (* n (/ (fma 100.0 t_0 -100.0) i))
       (* 100.0 (/ n (+ 1.0 (* i -0.5))))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = n * (fma(100.0, t_0, -100.0) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	return tmp;
}

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= Inf)
		tmp = Float64(n * Float64(fma(100.0, t_0, -100.0) / i));
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(n * N[(N[(100.0 * t$95$0 + -100.0), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $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 24.9%

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

      1. *-un-lft-identity 24.9%

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

      2. pow-to-exp 22.8%

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

      3. expm1-def 35.8%

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

      4. *-commutative 35.8%

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

      5. log1p-udef 97.1%

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

   3. Applied egg-rr 97.1%

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

      1. *-lft-identity 97.1%

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

   5. Simplified 97.1%

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

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

   1. Initial program 98.1%

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

      1. associate-/r/ 98.2%

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

      2. associate-*r* 98.0%

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

      3. *-commutative 98.0%

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

      4. associate-*r/ 98.2%

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

      5. sub-neg 98.2%

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

      6. distribute-lft-in 98.2%

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

      7. fma-def 98.2%

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

      8. metadata-eval 98.2%

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

      9. metadata-eval 98.2%

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

   3. Simplified 98.2%

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

   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 71.9%

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

   4. Simplified 71.9%

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

   5. Taylor expanded in i around 0 100.0%

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

      1. *-commutative 100.0%

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

   7. Simplified 100.0%

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

4. Final simplification 97.7%

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

Alternative 2: 84.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_0 \leq -1 \cdot 10^{-305}:\\ \;\;\;\;n \cdot \frac{-100 + 100 \cdot {\left(\frac{i}{n}\right)}^{n}}{i}\\ \mathbf{elif}\;t_0 \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;t_0 \leq \infty:\\ \;\;\;\;t_0 \cdot 100\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) (/ i n))))
   (if (<= t_0 -1e-305)
     (* n (/ (+ -100.0 (* 100.0 (pow (/ i n) n))) i))
     (if (<= t_0 0.0)
       (* 100.0 (/ n (/ i (expm1 i))))
       (if (<= t_0 INFINITY)
         (* t_0 100.0)
         (* 100.0 (/ n (+ 1.0 (* i -0.5)))))))))

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 <= -1e-305) {
		tmp = n * ((-100.0 + (100.0 * pow((i / n), n))) / i);
	} else if (t_0 <= 0.0) {
		tmp = 100.0 * (n / (i / expm1(i)));
	} else if (t_0 <= ((double) INFINITY)) {
		tmp = t_0 * 100.0;
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305) {
		tmp = n * ((-100.0 + (100.0 * Math.pow((i / n), n))) / i);
	} else if (t_0 <= 0.0) {
		tmp = 100.0 * (n / (i / Math.expm1(i)));
	} else if (t_0 <= Double.POSITIVE_INFINITY) {
		tmp = t_0 * 100.0;
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305:
		tmp = n * ((-100.0 + (100.0 * math.pow((i / n), n))) / i)
	elif t_0 <= 0.0:
		tmp = 100.0 * (n / (i / math.expm1(i)))
	elif t_0 <= math.inf:
		tmp = t_0 * 100.0
	else:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	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 <= -1e-305)
		tmp = Float64(n * Float64(Float64(-100.0 + Float64(100.0 * (Float64(i / n) ^ n))) / i));
	elseif (t_0 <= 0.0)
		tmp = Float64(100.0 * Float64(n / Float64(i / expm1(i))));
	elseif (t_0 <= Inf)
		tmp = Float64(t_0 * 100.0);
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	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, -1e-305], N[(n * N[(N[(-100.0 + N[(100.0 * N[Power[N[(i / n), $MachinePrecision], n], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 0.0], N[(100.0 * N[(n / N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, Infinity], N[(t$95$0 * 100.0), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -9.99999999999999996e-306

   1. Initial program 97.9%

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

   2. Taylor expanded in i around inf 97.9%

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

      1. clear-num 97.9%

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

      2. un-div-inv 98.2%

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

      3. sub-neg 98.2%

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

      4. metadata-eval 98.2%

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

   4. Applied egg-rr 98.2%

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

      1. associate-/l* 98.2%

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

      2. associate-/r/ 98.1%

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

      3. distribute-lft-in 98.3%

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

      4. metadata-eval 98.3%

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

   6. Simplified 98.3%

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

   if -9.99999999999999996e-306 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 0.0

   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 30.1%

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

      1. *-commutative 30.1%

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

      2. associate-/l* 30.1%

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

      3. expm1-def 76.9%

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

   4. Simplified 76.9%

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

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

   1. Initial program 98.1%

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

   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 71.9%

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

   4. Simplified 71.9%

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

   5. Taylor expanded in i around 0 100.0%

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

      1. *-commutative 100.0%

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

   7. Simplified 100.0%

      \[\leadsto \frac{n}{\color{blue}{1 + i \cdot -0.5}} \cdot 100 \]
3. Recombined 4 regimes into one program.

4. Final simplification 84.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-305}:\\ \;\;\;\;n \cdot \frac{-100 + 100 \cdot {\left(\frac{i}{n}\right)}^{n}}{i}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \cdot 100\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

Alternative 3: 84.5% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -1 \cdot 10^{-305}:\\ \;\;\;\;n \cdot \frac{-100 + 100 \cdot {\left(\frac{i}{n}\right)}^{n}}{i}\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;n \cdot \frac{-100 + t_0 \cdot 100}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \end{array} \]

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 -1e-305)
     (* n (/ (+ -100.0 (* 100.0 (pow (/ i n) n))) i))
     (if (<= t_1 0.0)
       (* 100.0 (/ n (/ i (expm1 i))))
       (if (<= t_1 INFINITY)
         (* n (/ (+ -100.0 (* t_0 100.0)) i))
         (* 100.0 (/ n (+ 1.0 (* i -0.5)))))))))

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 <= -1e-305) {
		tmp = n * ((-100.0 + (100.0 * pow((i / n), n))) / i);
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n / (i / expm1(i)));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305) {
		tmp = n * ((-100.0 + (100.0 * Math.pow((i / n), n))) / i);
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n / (i / Math.expm1(i)));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305:
		tmp = n * ((-100.0 + (100.0 * math.pow((i / n), n))) / i)
	elif t_1 <= 0.0:
		tmp = 100.0 * (n / (i / math.expm1(i)))
	elif t_1 <= math.inf:
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i)
	else:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	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 <= -1e-305)
		tmp = Float64(n * Float64(Float64(-100.0 + Float64(100.0 * (Float64(i / n) ^ n))) / i));
	elseif (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(n / Float64(i / expm1(i))));
	elseif (t_1 <= Inf)
		tmp = Float64(n * Float64(Float64(-100.0 + Float64(t_0 * 100.0)) / i));
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	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, -1e-305], N[(n * N[(N[(-100.0 + N[(100.0 * N[Power[N[(i / n), $MachinePrecision], n], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(n / N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(n * N[(N[(-100.0 + N[(t$95$0 * 100.0), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -9.99999999999999996e-306

   1. Initial program 97.9%

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

   2. Taylor expanded in i around inf 97.9%

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

      1. clear-num 97.9%

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

      2. un-div-inv 98.2%

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

      3. sub-neg 98.2%

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

      4. metadata-eval 98.2%

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

   4. Applied egg-rr 98.2%

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

      1. associate-/l* 98.2%

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

      2. associate-/r/ 98.1%

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

      3. distribute-lft-in 98.3%

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

      4. metadata-eval 98.3%

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

   6. Simplified 98.3%

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

   if -9.99999999999999996e-306 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 0.0

   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 30.1%

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

      1. *-commutative 30.1%

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

      2. associate-/l* 30.1%

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

      3. expm1-def 76.9%

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

   4. Simplified 76.9%

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

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

   1. Initial program 98.1%

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

      1. associate-/r/ 98.2%

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

      2. associate-*r* 98.0%

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

      3. *-commutative 98.0%

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

      4. associate-*r/ 98.2%

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

      5. sub-neg 98.2%

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

      6. distribute-lft-in 98.2%

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

      7. fma-def 98.2%

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

      8. metadata-eval 98.2%

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

      9. metadata-eval 98.2%

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

   3. Simplified 98.2%

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

      1. fma-udef 98.2%

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

      2. *-commutative 98.2%

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

   5. Applied egg-rr 98.2%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 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 71.9%

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

   4. Simplified 71.9%

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

   5. Taylor expanded in i around 0 100.0%

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

      1. *-commutative 100.0%

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

   7. Simplified 100.0%

      \[\leadsto \frac{n}{\color{blue}{1 + i \cdot -0.5}} \cdot 100 \]
3. Recombined 4 regimes into one program.

4. Final simplification 84.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-305}:\\ \;\;\;\;n \cdot \frac{-100 + 100 \cdot {\left(\frac{i}{n}\right)}^{n}}{i}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;n \cdot \frac{-100 + {\left(1 + \frac{i}{n}\right)}^{n} \cdot 100}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

Alternative 4: 84.8% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq -1 \cdot 10^{-305}:\\ \;\;\;\;100 \cdot \left(t_0 \cdot \frac{n}{i} - \frac{n}{i}\right)\\ \mathbf{elif}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;t_1 \leq \infty:\\ \;\;\;\;n \cdot \frac{-100 + t_0 \cdot 100}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \end{array} \]

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 -1e-305)
     (* 100.0 (- (* t_0 (/ n i)) (/ n i)))
     (if (<= t_1 0.0)
       (* 100.0 (/ n (/ i (expm1 i))))
       (if (<= t_1 INFINITY)
         (* n (/ (+ -100.0 (* t_0 100.0)) i))
         (* 100.0 (/ n (+ 1.0 (* i -0.5)))))))))

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 <= -1e-305) {
		tmp = 100.0 * ((t_0 * (n / i)) - (n / i));
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n / (i / expm1(i)));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305) {
		tmp = 100.0 * ((t_0 * (n / i)) - (n / i));
	} else if (t_1 <= 0.0) {
		tmp = 100.0 * (n / (i / Math.expm1(i)));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	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 <= -1e-305:
		tmp = 100.0 * ((t_0 * (n / i)) - (n / i))
	elif t_1 <= 0.0:
		tmp = 100.0 * (n / (i / math.expm1(i)))
	elif t_1 <= math.inf:
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i)
	else:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	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 <= -1e-305)
		tmp = Float64(100.0 * Float64(Float64(t_0 * Float64(n / i)) - Float64(n / i)));
	elseif (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(n / Float64(i / expm1(i))));
	elseif (t_1 <= Inf)
		tmp = Float64(n * Float64(Float64(-100.0 + Float64(t_0 * 100.0)) / i));
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	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, -1e-305], N[(100.0 * N[(N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision] - N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(n / N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(n * N[(N[(-100.0 + N[(t$95$0 * 100.0), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < -9.99999999999999996e-306

   1. Initial program 97.9%

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

      1. div-sub 98.4%

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

      2. clear-num 98.1%

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

      3. sub-neg 98.1%

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

      4. div-inv 98.1%

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

      5. clear-num 98.6%

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

   3. Applied egg-rr 98.6%

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

      1. sub-neg 98.6%

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

   5. Simplified 98.6%

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

   if -9.99999999999999996e-306 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 0.0

   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 30.1%

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

      1. *-commutative 30.1%

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

      2. associate-/l* 30.1%

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

      3. expm1-def 76.9%

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

   4. Simplified 76.9%

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

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

   1. Initial program 98.1%

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

      1. associate-/r/ 98.2%

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

      2. associate-*r* 98.0%

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

      3. *-commutative 98.0%

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

      4. associate-*r/ 98.2%

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

      5. sub-neg 98.2%

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

      6. distribute-lft-in 98.2%

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

      7. fma-def 98.2%

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

      8. metadata-eval 98.2%

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

      9. metadata-eval 98.2%

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

   3. Simplified 98.2%

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

      1. fma-udef 98.2%

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

      2. *-commutative 98.2%

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

   5. Applied egg-rr 98.2%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 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 71.9%

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

   4. Simplified 71.9%

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

   5. Taylor expanded in i around 0 100.0%

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

      1. *-commutative 100.0%

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

   7. Simplified 100.0%

      \[\leadsto \frac{n}{\color{blue}{1 + i \cdot -0.5}} \cdot 100 \]
3. Recombined 4 regimes into one program.

4. Final simplification 84.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq -1 \cdot 10^{-305}:\\ \;\;\;\;100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} \cdot \frac{n}{i} - \frac{n}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;n \cdot \frac{-100 + {\left(1 + \frac{i}{n}\right)}^{n} \cdot 100}{i}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \end{array} \]

Alternative 5: 98.1% accurate, 0.3× speedup

Math

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

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 0.0)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 INFINITY)
       (* n (/ (+ -100.0 (* t_0 100.0)) i))
       (* 100.0 (/ n (+ 1.0 (* i -0.5))))))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 0.0) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) / (i / n));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i);
	} else {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = (t_0 + -1.0) / (i / n)
	tmp = 0
	if t_1 <= 0.0:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) / (i / n))
	elif t_1 <= math.inf:
		tmp = n * ((-100.0 + (t_0 * 100.0)) / i)
	else:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	return tmp

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 0.0)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= Inf)
		tmp = Float64(n * Float64(Float64(-100.0 + Float64(t_0 * 100.0)) / i));
	else
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 0.0], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(n * N[(N[(-100.0 + N[(t$95$0 * 100.0), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $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 24.9%

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

      1. *-un-lft-identity 24.9%

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

      2. pow-to-exp 22.8%

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

      3. expm1-def 35.8%

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

      4. *-commutative 35.8%

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

      5. log1p-udef 97.1%

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

   3. Applied egg-rr 97.1%

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

      1. *-lft-identity 97.1%

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

   5. Simplified 97.1%

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

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

   1. Initial program 98.1%

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

      1. associate-/r/ 98.2%

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

      2. associate-*r* 98.0%

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

      3. *-commutative 98.0%

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

      4. associate-*r/ 98.2%

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

      5. sub-neg 98.2%

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

      6. distribute-lft-in 98.2%

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

      7. fma-def 98.2%

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

      8. metadata-eval 98.2%

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

      9. metadata-eval 98.2%

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

   3. Simplified 98.2%

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

      1. fma-udef 98.2%

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

      2. *-commutative 98.2%

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

   5. Applied egg-rr 98.2%

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

   if +inf.0 < (/.f64 (-.f64 (pow.f64 (+.f64 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 71.9%

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

   4. Simplified 71.9%

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

   5. Taylor expanded in i around 0 100.0%

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

      1. *-commutative 100.0%

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

   7. Simplified 100.0%

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

4. Final simplification 97.7%

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

Alternative 6: 84.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -0.00012 \lor \neg \left(n \leq 0.006\right):\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{n \cdot \left(n \cdot 10000\right)}{n \cdot 100 - 100 \cdot \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -0.00012) (not (<= n 0.006)))
   (* 100.0 (/ n (/ i (expm1 i))))
   (/
    (* n (* n 10000.0))
    (- (* n 100.0) (* 100.0 (* n (* i (+ 0.5 (/ -0.5 n)))))))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -0.00012) || !(n <= 0.006)) {
		tmp = 100.0 * (n / (i / expm1(i)));
	} else {
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -0.00012) || !(n <= 0.006)) {
		tmp = 100.0 * (n / (i / Math.expm1(i)));
	} else {
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -0.00012) or not (n <= 0.006):
		tmp = 100.0 * (n / (i / math.expm1(i)))
	else:
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -0.00012) || !(n <= 0.006))
		tmp = Float64(100.0 * Float64(n / Float64(i / expm1(i))));
	else
		tmp = Float64(Float64(n * Float64(n * 10000.0)) / Float64(Float64(n * 100.0) - Float64(100.0 * Float64(n * Float64(i * Float64(0.5 + Float64(-0.5 / n)))))));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -0.00012], N[Not[LessEqual[n, 0.006]], $MachinePrecision]], N[(100.0 * N[(n / N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(n * N[(n * 10000.0), $MachinePrecision]), $MachinePrecision] / N[(N[(n * 100.0), $MachinePrecision] - N[(100.0 * N[(n * N[(i * N[(0.5 + N[(-0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.20000000000000003e-4 or 0.0060000000000000001 < n

   1. Initial program 25.0%

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

   2. Taylor expanded in n around inf 36.7%

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

      1. *-commutative 36.7%

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

      2. associate-/l* 36.7%

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

      3. expm1-def 92.5%

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

   4. Simplified 92.5%

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

   if -1.20000000000000003e-4 < n < 0.0060000000000000001

   1. Initial program 32.8%

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

   2. Taylor expanded in i around 0 32.6%

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

      1. distribute-rgt-in 32.6%

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

      2. flip-+ 23.5%

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

   4. Applied egg-rr 23.5%

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

   5. Taylor expanded in i around 0 63.0%

      \[\leadsto \frac{\color{blue}{10000 \cdot {n}^{2}}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]
   6. Step-by-step derivation

      1. *-commutative 63.0%

         \[\leadsto \frac{\color{blue}{{n}^{2} \cdot 10000}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

      2. unpow2 63.0%

         \[\leadsto \frac{\color{blue}{\left(n \cdot n\right)} \cdot 10000}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

      3. associate-*l* 63.4%

         \[\leadsto \frac{\color{blue}{n \cdot \left(n \cdot 10000\right)}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

   7. Simplified 63.4%

      \[\leadsto \frac{\color{blue}{n \cdot \left(n \cdot 10000\right)}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]
3. Recombined 2 regimes into one program.

4. Final simplification 81.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -0.00012 \lor \neg \left(n \leq 0.006\right):\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{n \cdot \left(n \cdot 10000\right)}{n \cdot 100 - 100 \cdot \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right)}\\ \end{array} \]

Alternative 7: 84.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{i}{\mathsf{expm1}\left(i\right)}\\ \mathbf{if}\;n \leq -0.00058:\\ \;\;\;\;\frac{n}{\frac{t_0}{100}}\\ \mathbf{elif}\;n \leq 0.006:\\ \;\;\;\;\frac{n \cdot \left(n \cdot 10000\right)}{n \cdot 100 - 100 \cdot \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{t_0}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (/ i (expm1 i))))
   (if (<= n -0.00058)
     (/ n (/ t_0 100.0))
     (if (<= n 0.006)
       (/
        (* n (* n 10000.0))
        (- (* n 100.0) (* 100.0 (* n (* i (+ 0.5 (/ -0.5 n)))))))
       (* 100.0 (/ n t_0))))))

C

double code(double i, double n) {
	double t_0 = i / expm1(i);
	double tmp;
	if (n <= -0.00058) {
		tmp = n / (t_0 / 100.0);
	} else if (n <= 0.006) {
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))));
	} else {
		tmp = 100.0 * (n / t_0);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = i / Math.expm1(i);
	double tmp;
	if (n <= -0.00058) {
		tmp = n / (t_0 / 100.0);
	} else if (n <= 0.006) {
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))));
	} else {
		tmp = 100.0 * (n / t_0);
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = i / math.expm1(i)
	tmp = 0
	if n <= -0.00058:
		tmp = n / (t_0 / 100.0)
	elif n <= 0.006:
		tmp = (n * (n * 10000.0)) / ((n * 100.0) - (100.0 * (n * (i * (0.5 + (-0.5 / n))))))
	else:
		tmp = 100.0 * (n / t_0)
	return tmp

Julia

function code(i, n)
	t_0 = Float64(i / expm1(i))
	tmp = 0.0
	if (n <= -0.00058)
		tmp = Float64(n / Float64(t_0 / 100.0));
	elseif (n <= 0.006)
		tmp = Float64(Float64(n * Float64(n * 10000.0)) / Float64(Float64(n * 100.0) - Float64(100.0 * Float64(n * Float64(i * Float64(0.5 + Float64(-0.5 / n)))))));
	else
		tmp = Float64(100.0 * Float64(n / t_0));
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(i / N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -0.00058], N[(n / N[(t$95$0 / 100.0), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 0.006], N[(N[(n * N[(n * 10000.0), $MachinePrecision]), $MachinePrecision] / N[(N[(n * 100.0), $MachinePrecision] - N[(100.0 * N[(n * N[(i * N[(0.5 + N[(-0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n / t$95$0), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if n < -5.8e-4

   1. Initial program 21.8%

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

   2. Taylor expanded in n around inf 32.7%

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

      1. *-commutative 32.7%

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

      2. associate-/l* 32.7%

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

      3. expm1-def 90.8%

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

   4. Simplified 90.8%

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

      1. associate-*l/ 90.8%

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

      2. associate-/l* 90.9%

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

   6. Applied egg-rr 90.9%

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

   if -5.8e-4 < n < 0.0060000000000000001

   1. Initial program 32.8%

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

   2. Taylor expanded in i around 0 32.6%

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

      1. distribute-rgt-in 32.6%

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

      2. flip-+ 23.5%

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

   4. Applied egg-rr 23.5%

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

   5. Taylor expanded in i around 0 63.0%

      \[\leadsto \frac{\color{blue}{10000 \cdot {n}^{2}}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]
   6. Step-by-step derivation

      1. *-commutative 63.0%

         \[\leadsto \frac{\color{blue}{{n}^{2} \cdot 10000}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

      2. unpow2 63.0%

         \[\leadsto \frac{\color{blue}{\left(n \cdot n\right)} \cdot 10000}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

      3. associate-*l* 63.4%

         \[\leadsto \frac{\color{blue}{n \cdot \left(n \cdot 10000\right)}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

   7. Simplified 63.4%

      \[\leadsto \frac{\color{blue}{n \cdot \left(n \cdot 10000\right)}}{n \cdot 100 - \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right) \cdot 100} \]

   if 0.0060000000000000001 < n

   1. Initial program 28.9%

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

   2. Taylor expanded in n around inf 41.6%

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

      1. *-commutative 41.6%

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

      2. associate-/l* 41.6%

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

      3. expm1-def 94.4%

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

   4. Simplified 94.4%

      \[\leadsto \color{blue}{\frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}} \cdot 100} \]
3. Recombined 3 regimes into one program.

4. Final simplification 81.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -0.00058:\\ \;\;\;\;\frac{n}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{100}}\\ \mathbf{elif}\;n \leq 0.006:\\ \;\;\;\;\frac{n \cdot \left(n \cdot 10000\right)}{n \cdot 100 - 100 \cdot \left(n \cdot \left(i \cdot \left(0.5 + \frac{-0.5}{n}\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \end{array} \]

Alternative 8: 69.9% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -6.6 \cdot 10^{+121} \lor \neg \left(n \leq 0.006\right):\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{0.5}{n}\right)\right) + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -6.6e+121) (not (<= n 0.006)))
   (*
    100.0
    (+
     n
     (*
      n
      (+
       (*
        (* i i)
        (+ (/ 0.3333333333333333 (* n n)) (- 0.16666666666666666 (/ 0.5 n))))
       (* i (- 0.5 (/ 0.5 n)))))))
   (/ 100.0 (+ (/ 1.0 n) (* i (- (/ 0.5 (* n n)) (/ 0.5 n)))))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -6.6e+121) || !(n <= 0.006)) {
		tmp = 100.0 * (n + (n * (((i * i) * ((0.3333333333333333 / (n * n)) + (0.16666666666666666 - (0.5 / n)))) + (i * (0.5 - (0.5 / n))))));
	} else {
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (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 <= (-6.6d+121)) .or. (.not. (n <= 0.006d0))) then
        tmp = 100.0d0 * (n + (n * (((i * i) * ((0.3333333333333333d0 / (n * n)) + (0.16666666666666666d0 - (0.5d0 / n)))) + (i * (0.5d0 - (0.5d0 / n))))))
    else
        tmp = 100.0d0 / ((1.0d0 / n) + (i * ((0.5d0 / (n * n)) - (0.5d0 / n))))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -6.6e+121) || !(n <= 0.006)) {
		tmp = 100.0 * (n + (n * (((i * i) * ((0.3333333333333333 / (n * n)) + (0.16666666666666666 - (0.5 / n)))) + (i * (0.5 - (0.5 / n))))));
	} else {
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -6.6e+121) or not (n <= 0.006):
		tmp = 100.0 * (n + (n * (((i * i) * ((0.3333333333333333 / (n * n)) + (0.16666666666666666 - (0.5 / n)))) + (i * (0.5 - (0.5 / n))))))
	else:
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -6.6e+121) || !(n <= 0.006))
		tmp = Float64(100.0 * Float64(n + Float64(n * Float64(Float64(Float64(i * i) * Float64(Float64(0.3333333333333333 / Float64(n * n)) + Float64(0.16666666666666666 - Float64(0.5 / n)))) + Float64(i * Float64(0.5 - Float64(0.5 / n)))))));
	else
		tmp = Float64(100.0 / Float64(Float64(1.0 / n) + Float64(i * Float64(Float64(0.5 / Float64(n * n)) - Float64(0.5 / n)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -6.6e+121) || ~((n <= 0.006)))
		tmp = 100.0 * (n + (n * (((i * i) * ((0.3333333333333333 / (n * n)) + (0.16666666666666666 - (0.5 / n)))) + (i * (0.5 - (0.5 / n))))));
	else
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -6.6e+121], N[Not[LessEqual[n, 0.006]], $MachinePrecision]], N[(100.0 * N[(n + N[(n * N[(N[(N[(i * i), $MachinePrecision] * N[(N[(0.3333333333333333 / N[(n * n), $MachinePrecision]), $MachinePrecision] + N[(0.16666666666666666 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(i * N[(0.5 - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(1.0 / n), $MachinePrecision] + N[(i * N[(N[(0.5 / N[(n * n), $MachinePrecision]), $MachinePrecision] - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -6.59999999999999958e121 or 0.0060000000000000001 < n

   1. Initial program 22.5%

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

   2. Taylor expanded in i around 0 79.5%

      \[\leadsto 100 \cdot \color{blue}{\left(n + \left(n \cdot \left({i}^{2} \cdot \left(\left(0.3333333333333333 \cdot \frac{1}{{n}^{2}} + 0.16666666666666666\right) - 0.5 \cdot \frac{1}{n}\right)\right) + n \cdot \left(i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right)\right)} \]
   3. Step-by-step derivation

      1. distribute-lft-out 79.8%

         \[\leadsto 100 \cdot \left(n + \color{blue}{n \cdot \left({i}^{2} \cdot \left(\left(0.3333333333333333 \cdot \frac{1}{{n}^{2}} + 0.16666666666666666\right) - 0.5 \cdot \frac{1}{n}\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)}\right) \]

      2. unpow2 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\color{blue}{\left(i \cdot i\right)} \cdot \left(\left(0.3333333333333333 \cdot \frac{1}{{n}^{2}} + 0.16666666666666666\right) - 0.5 \cdot \frac{1}{n}\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      3. associate--l+ 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{{n}^{2}} + \left(0.16666666666666666 - 0.5 \cdot \frac{1}{n}\right)\right)} + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      4. associate-*r/ 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\color{blue}{\frac{0.3333333333333333 \cdot 1}{{n}^{2}}} + \left(0.16666666666666666 - 0.5 \cdot \frac{1}{n}\right)\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      5. metadata-eval 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{\color{blue}{0.3333333333333333}}{{n}^{2}} + \left(0.16666666666666666 - 0.5 \cdot \frac{1}{n}\right)\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      6. unpow2 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{\color{blue}{n \cdot n}} + \left(0.16666666666666666 - 0.5 \cdot \frac{1}{n}\right)\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      7. associate-*r/ 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \color{blue}{\frac{0.5 \cdot 1}{n}}\right)\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      8. metadata-eval 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{\color{blue}{0.5}}{n}\right)\right) + i \cdot \left(0.5 - 0.5 \cdot \frac{1}{n}\right)\right)\right) \]

      9. associate-*r/ 79.8%

         \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{0.5}{n}\right)\right) + i \cdot \left(0.5 - \color{blue}{\frac{0.5 \cdot 1}{n}}\right)\right)\right) \]

      10. metadata-eval 79.8%

          \[\leadsto 100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{0.5}{n}\right)\right) + i \cdot \left(0.5 - \frac{\color{blue}{0.5}}{n}\right)\right)\right) \]

   4. Simplified 79.8%

      \[\leadsto 100 \cdot \color{blue}{\left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{0.5}{n}\right)\right) + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)} \]

   if -6.59999999999999958e121 < n < 0.0060000000000000001

   1. Initial program 32.6%

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

      1. clear-num 32.6%

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

      2. un-div-inv 32.6%

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

      3. associate-/l/ 32.7%

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

      4. pow-to-exp 28.4%

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

      5. expm1-def 38.2%

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

      6. *-commutative 38.2%

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

      7. log1p-udef 67.1%

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

   3. Applied egg-rr 67.1%

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

   4. Taylor expanded in i around 0 65.1%

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

      1. associate-*r/ 65.1%

         \[\leadsto \frac{100}{\frac{1}{n} + i \cdot \left(\color{blue}{\frac{0.5 \cdot 1}{{n}^{2}}} - 0.5 \cdot \frac{1}{n}\right)} \]

      2. metadata-eval 65.1%

         \[\leadsto \frac{100}{\frac{1}{n} + i \cdot \left(\frac{\color{blue}{0.5}}{{n}^{2}} - 0.5 \cdot \frac{1}{n}\right)} \]

      3. unpow2 65.1%

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

      4. associate-*r/ 65.1%

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

      5. metadata-eval 65.1%

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

   6. Simplified 65.1%

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

4. Final simplification 71.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -6.6 \cdot 10^{+121} \lor \neg \left(n \leq 0.006\right):\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(\left(i \cdot i\right) \cdot \left(\frac{0.3333333333333333}{n \cdot n} + \left(0.16666666666666666 - \frac{0.5}{n}\right)\right) + i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)}\\ \end{array} \]

Alternative 9: 67.7% accurate, 6.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 500:\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 500.0)
   (/ 100.0 (+ (/ 1.0 n) (* i (- (/ 0.5 (* n n)) (/ 0.5 n)))))
   (/ n (/ (- 100.0 (* i 50.0)) (- 10000.0 (* (* i i) 2500.0))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 500.0) {
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))));
	} else {
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 500.0d0) then
        tmp = 100.0d0 / ((1.0d0 / n) + (i * ((0.5d0 / (n * n)) - (0.5d0 / n))))
    else
        tmp = n / ((100.0d0 - (i * 50.0d0)) / (10000.0d0 - ((i * i) * 2500.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 500.0) {
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))));
	} else {
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 500.0:
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))))
	else:
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 500.0)
		tmp = Float64(100.0 / Float64(Float64(1.0 / n) + Float64(i * Float64(Float64(0.5 / Float64(n * n)) - Float64(0.5 / n)))));
	else
		tmp = Float64(n / Float64(Float64(100.0 - Float64(i * 50.0)) / Float64(10000.0 - Float64(Float64(i * i) * 2500.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 500.0)
		tmp = 100.0 / ((1.0 / n) + (i * ((0.5 / (n * n)) - (0.5 / n))));
	else
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 500.0], N[(100.0 / N[(N[(1.0 / n), $MachinePrecision] + N[(i * N[(N[(0.5 / N[(n * n), $MachinePrecision]), $MachinePrecision] - N[(0.5 / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n / N[(N[(100.0 - N[(i * 50.0), $MachinePrecision]), $MachinePrecision] / N[(10000.0 - N[(N[(i * i), $MachinePrecision] * 2500.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 500

   1. Initial program 27.3%

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

      1. clear-num 27.3%

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

      2. un-div-inv 27.3%

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

      3. associate-/l/ 27.6%

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

      4. pow-to-exp 23.3%

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

      5. expm1-def 30.6%

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

      6. *-commutative 30.6%

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

      7. log1p-udef 66.6%

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

   3. Applied egg-rr 66.6%

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

   4. Taylor expanded in i around 0 64.3%

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

      1. associate-*r/ 64.3%

         \[\leadsto \frac{100}{\frac{1}{n} + i \cdot \left(\color{blue}{\frac{0.5 \cdot 1}{{n}^{2}}} - 0.5 \cdot \frac{1}{n}\right)} \]

      2. metadata-eval 64.3%

         \[\leadsto \frac{100}{\frac{1}{n} + i \cdot \left(\frac{\color{blue}{0.5}}{{n}^{2}} - 0.5 \cdot \frac{1}{n}\right)} \]

      3. unpow2 64.3%

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

      4. associate-*r/ 64.3%

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

      5. metadata-eval 64.3%

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

   6. Simplified 64.3%

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

   if 500 < n

   1. Initial program 29.7%

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

   2. Taylor expanded in i around 0 67.8%

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

      1. distribute-rgt-in 67.8%

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

      2. flip-+ 48.6%

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

   4. Applied egg-rr 48.6%

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

   5. Taylor expanded in n around inf 82.1%

      \[\leadsto \color{blue}{\frac{n \cdot \left(10000 - 2500 \cdot {i}^{2}\right)}{100 - 50 \cdot i}} \]
   6. Step-by-step derivation

      1. associate-/l* 76.9%

         \[\leadsto \color{blue}{\frac{n}{\frac{100 - 50 \cdot i}{10000 - 2500 \cdot {i}^{2}}}} \]

      2. *-commutative 76.9%

         \[\leadsto \frac{n}{\frac{100 - \color{blue}{i \cdot 50}}{10000 - 2500 \cdot {i}^{2}}} \]

      3. *-commutative 76.9%

         \[\leadsto \frac{n}{\frac{100 - i \cdot 50}{10000 - \color{blue}{{i}^{2} \cdot 2500}}} \]

      4. unpow2 76.9%

         \[\leadsto \frac{n}{\frac{100 - i \cdot 50}{10000 - \color{blue}{\left(i \cdot i\right)} \cdot 2500}} \]

   7. Simplified 76.9%

      \[\leadsto \color{blue}{\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 67.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 500:\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} - \frac{0.5}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}\\ \end{array} \]

Alternative 10: 63.3% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 1.06 \cdot 10^{-76}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 1.06e-76)
   (* 100.0 (/ n (+ 1.0 (* i -0.5))))
   (/ n (/ (- 100.0 (* i 50.0)) (- 10000.0 (* (* i i) 2500.0))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 1.06e-76) {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	} else {
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.06d-76) then
        tmp = 100.0d0 * (n / (1.0d0 + (i * (-0.5d0))))
    else
        tmp = n / ((100.0d0 - (i * 50.0d0)) / (10000.0d0 - ((i * i) * 2500.0d0)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 1.06e-76) {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	} else {
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 1.06e-76:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	else:
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 1.06e-76)
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	else
		tmp = Float64(n / Float64(Float64(100.0 - Float64(i * 50.0)) / Float64(10000.0 - Float64(Float64(i * i) * 2500.0))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 1.06e-76)
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	else
		tmp = n / ((100.0 - (i * 50.0)) / (10000.0 - ((i * i) * 2500.0)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 1.06e-76], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n / N[(N[(100.0 - N[(i * 50.0), $MachinePrecision]), $MachinePrecision] / N[(10000.0 - N[(N[(i * i), $MachinePrecision] * 2500.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.06000000000000003e-76

   1. Initial program 28.1%

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

   2. Taylor expanded in n around inf 27.6%

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

      1. *-commutative 27.6%

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

      2. associate-/l* 27.6%

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

      3. expm1-def 67.2%

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

   4. Simplified 67.2%

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

   5. Taylor expanded in i around 0 61.6%

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

      1. *-commutative 61.6%

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

   7. Simplified 61.6%

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

   if 1.06000000000000003e-76 < n

   1. Initial program 27.7%

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

   2. Taylor expanded in i around 0 66.0%

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

      1. distribute-rgt-in 66.0%

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

      2. flip-+ 48.8%

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

   4. Applied egg-rr 48.8%

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

   5. Taylor expanded in n around inf 79.0%

      \[\leadsto \color{blue}{\frac{n \cdot \left(10000 - 2500 \cdot {i}^{2}\right)}{100 - 50 \cdot i}} \]
   6. Step-by-step derivation

      1. associate-/l* 74.4%

         \[\leadsto \color{blue}{\frac{n}{\frac{100 - 50 \cdot i}{10000 - 2500 \cdot {i}^{2}}}} \]

      2. *-commutative 74.4%

         \[\leadsto \frac{n}{\frac{100 - \color{blue}{i \cdot 50}}{10000 - 2500 \cdot {i}^{2}}} \]

      3. *-commutative 74.4%

         \[\leadsto \frac{n}{\frac{100 - i \cdot 50}{10000 - \color{blue}{{i}^{2} \cdot 2500}}} \]

      4. unpow2 74.4%

         \[\leadsto \frac{n}{\frac{100 - i \cdot 50}{10000 - \color{blue}{\left(i \cdot i\right)} \cdot 2500}} \]

   7. Simplified 74.4%

      \[\leadsto \color{blue}{\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 65.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 1.06 \cdot 10^{-76}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{n}{\frac{100 - i \cdot 50}{10000 - \left(i \cdot i\right) \cdot 2500}}\\ \end{array} \]

Alternative 11: 60.1% accurate, 8.6× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -0.76:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 0.088:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{elif}\;i \leq 7.5 \cdot 10^{+108}:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 1.9 \cdot 10^{+244}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -0.76)
   0.0
   (if (<= i 0.088)
     (* 100.0 (+ n (* i -0.5)))
     (if (<= i 7.5e+108) 0.0 (if (<= i 1.9e+244) (* n (* i 50.0)) 0.0)))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -0.76) {
		tmp = 0.0;
	} else if (i <= 0.088) {
		tmp = 100.0 * (n + (i * -0.5));
	} else if (i <= 7.5e+108) {
		tmp = 0.0;
	} else if (i <= 1.9e+244) {
		tmp = n * (i * 50.0);
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= (-0.76d0)) then
        tmp = 0.0d0
    else if (i <= 0.088d0) then
        tmp = 100.0d0 * (n + (i * (-0.5d0)))
    else if (i <= 7.5d+108) then
        tmp = 0.0d0
    else if (i <= 1.9d+244) then
        tmp = n * (i * 50.0d0)
    else
        tmp = 0.0d0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -0.76) {
		tmp = 0.0;
	} else if (i <= 0.088) {
		tmp = 100.0 * (n + (i * -0.5));
	} else if (i <= 7.5e+108) {
		tmp = 0.0;
	} else if (i <= 1.9e+244) {
		tmp = n * (i * 50.0);
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -0.76:
		tmp = 0.0
	elif i <= 0.088:
		tmp = 100.0 * (n + (i * -0.5))
	elif i <= 7.5e+108:
		tmp = 0.0
	elif i <= 1.9e+244:
		tmp = n * (i * 50.0)
	else:
		tmp = 0.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -0.76)
		tmp = 0.0;
	elseif (i <= 0.088)
		tmp = Float64(100.0 * Float64(n + Float64(i * -0.5)));
	elseif (i <= 7.5e+108)
		tmp = 0.0;
	elseif (i <= 1.9e+244)
		tmp = Float64(n * Float64(i * 50.0));
	else
		tmp = 0.0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -0.76)
		tmp = 0.0;
	elseif (i <= 0.088)
		tmp = 100.0 * (n + (i * -0.5));
	elseif (i <= 7.5e+108)
		tmp = 0.0;
	elseif (i <= 1.9e+244)
		tmp = n * (i * 50.0);
	else
		tmp = 0.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -0.76], 0.0, If[LessEqual[i, 0.088], N[(100.0 * N[(n + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 7.5e+108], 0.0, If[LessEqual[i, 1.9e+244], N[(n * N[(i * 50.0), $MachinePrecision]), $MachinePrecision], 0.0]]]]

Derivation

1. Split input into 3 regimes

2. if i < -0.76000000000000001 or 0.087999999999999995 < i < 7.50000000000000039e108 or 1.89999999999999991e244 < i

   1. Initial program 53.4%

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

   2. Taylor expanded in i around 0 36.7%

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

   3. Taylor expanded in i around 0 36.7%

      \[\leadsto 100 \cdot \color{blue}{0} \]

   if -0.76000000000000001 < i < 0.087999999999999995

   1. Initial program 7.3%

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

   2. Taylor expanded in i around 0 84.7%

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

   3. Taylor expanded in n around 0 84.2%

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

      1. *-commutative 84.2%

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

   5. Simplified 84.2%

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

   if 7.50000000000000039e108 < i < 1.89999999999999991e244

   1. Initial program 60.4%

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

   2. Taylor expanded in n around inf 67.1%

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

      1. *-commutative 67.1%

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

      2. associate-/l* 67.1%

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

      3. expm1-def 67.1%

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

   4. Simplified 67.1%

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

   5. Taylor expanded in i around 0 37.1%

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

   6. Taylor expanded in i around inf 37.1%

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

      1. *-commutative 37.1%

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

      2. associate-*l* 37.1%

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

   8. Simplified 37.1%

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

4. Final simplification 63.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -0.76:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 0.088:\\ \;\;\;\;100 \cdot \left(n + i \cdot -0.5\right)\\ \mathbf{elif}\;i \leq 7.5 \cdot 10^{+108}:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 1.9 \cdot 10^{+244}:\\ \;\;\;\;n \cdot \left(i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 12: 62.2% accurate, 8.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -2.15 \cdot 10^{+68} \lor \neg \left(n \leq 4.9 \cdot 10^{-8}\right):\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -2.15e+68) (not (<= n 4.9e-8)))
   (* 100.0 (+ n (* n (* i 0.5))))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -2.15e+68) || !(n <= 4.9e-8)) {
		tmp = 100.0 * (n + (n * (i * 0.5)));
	} 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 <= (-2.15d+68)) .or. (.not. (n <= 4.9d-8))) then
        tmp = 100.0d0 * (n + (n * (i * 0.5d0)))
    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 <= -2.15e+68) || !(n <= 4.9e-8)) {
		tmp = 100.0 * (n + (n * (i * 0.5)));
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -2.15e+68) or not (n <= 4.9e-8):
		tmp = 100.0 * (n + (n * (i * 0.5)))
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -2.15e+68) || !(n <= 4.9e-8))
		tmp = Float64(100.0 * Float64(n + Float64(n * Float64(i * 0.5))));
	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 <= -2.15e+68) || ~((n <= 4.9e-8)))
		tmp = 100.0 * (n + (n * (i * 0.5)));
	else
		tmp = 100.0 * (i / (i / n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -2.15e+68], N[Not[LessEqual[n, 4.9e-8]], $MachinePrecision]], N[(100.0 * N[(n + N[(n * N[(i * 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -2.1500000000000001e68 or 4.9000000000000002e-8 < n

   1. Initial program 24.8%

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

   2. Taylor expanded in i around 0 65.5%

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

   3. Taylor expanded in n around inf 65.5%

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

      1. *-commutative 65.5%

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

   5. Simplified 65.5%

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

   if -2.1500000000000001e68 < n < 4.9000000000000002e-8

   1. Initial program 31.5%

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

   2. Taylor expanded in i around 0 60.2%

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

4. Final simplification 63.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -2.15 \cdot 10^{+68} \lor \neg \left(n \leq 4.9 \cdot 10^{-8}\right):\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]

Alternative 13: 62.0% accurate, 10.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 1.06 \cdot 10^{-76}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n 1.06e-76)
   (* 100.0 (/ n (+ 1.0 (* i -0.5))))
   (* 100.0 (+ n (* n (* i 0.5))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= 1.06e-76) {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	} else {
		tmp = 100.0 * (n + (n * (i * 0.5)));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= 1.06d-76) then
        tmp = 100.0d0 * (n / (1.0d0 + (i * (-0.5d0))))
    else
        tmp = 100.0d0 * (n + (n * (i * 0.5d0)))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= 1.06e-76) {
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	} else {
		tmp = 100.0 * (n + (n * (i * 0.5)));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= 1.06e-76:
		tmp = 100.0 * (n / (1.0 + (i * -0.5)))
	else:
		tmp = 100.0 * (n + (n * (i * 0.5)))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= 1.06e-76)
		tmp = Float64(100.0 * Float64(n / Float64(1.0 + Float64(i * -0.5))));
	else
		tmp = Float64(100.0 * Float64(n + Float64(n * Float64(i * 0.5))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= 1.06e-76)
		tmp = 100.0 * (n / (1.0 + (i * -0.5)));
	else
		tmp = 100.0 * (n + (n * (i * 0.5)));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, 1.06e-76], N[(100.0 * N[(n / N[(1.0 + N[(i * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(n + N[(n * N[(i * 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.06000000000000003e-76

   1. Initial program 28.1%

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

   2. Taylor expanded in n around inf 27.6%

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

      1. *-commutative 27.6%

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

      2. associate-/l* 27.6%

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

      3. expm1-def 67.2%

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

   4. Simplified 67.2%

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

   5. Taylor expanded in i around 0 61.6%

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

      1. *-commutative 61.6%

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

   7. Simplified 61.6%

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

   if 1.06000000000000003e-76 < n

   1. Initial program 27.7%

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

   2. Taylor expanded in i around 0 66.0%

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

   3. Taylor expanded in n around inf 66.2%

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

      1. *-commutative 66.2%

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

   5. Simplified 66.2%

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

4. Final simplification 63.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq 1.06 \cdot 10^{-76}:\\ \;\;\;\;100 \cdot \frac{n}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot 0.5\right)\right)\\ \end{array} \]

Alternative 14: 59.2% accurate, 16.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -165000:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 0.085:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -165000.0) 0.0 (if (<= i 0.085) (* n 100.0) 0.0)))

C

double code(double i, double n) {
	double tmp;
	if (i <= -165000.0) {
		tmp = 0.0;
	} else if (i <= 0.085) {
		tmp = n * 100.0;
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= (-165000.0d0)) then
        tmp = 0.0d0
    else if (i <= 0.085d0) then
        tmp = n * 100.0d0
    else
        tmp = 0.0d0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -165000.0) {
		tmp = 0.0;
	} else if (i <= 0.085) {
		tmp = n * 100.0;
	} else {
		tmp = 0.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -165000.0:
		tmp = 0.0
	elif i <= 0.085:
		tmp = n * 100.0
	else:
		tmp = 0.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -165000.0)
		tmp = 0.0;
	elseif (i <= 0.085)
		tmp = Float64(n * 100.0);
	else
		tmp = 0.0;
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -165000.0)
		tmp = 0.0;
	elseif (i <= 0.085)
		tmp = n * 100.0;
	else
		tmp = 0.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -165000.0], 0.0, If[LessEqual[i, 0.085], N[(n * 100.0), $MachinePrecision], 0.0]]

Derivation

1. Split input into 2 regimes

2. if i < -165000 or 0.0850000000000000061 < i

   1. Initial program 55.3%

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

   2. Taylor expanded in i around 0 30.8%

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

   3. Taylor expanded in i around 0 30.8%

      \[\leadsto 100 \cdot \color{blue}{0} \]

   if -165000 < i < 0.0850000000000000061

   1. Initial program 7.3%

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

   2. Taylor expanded in i around 0 84.1%

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

      1. *-commutative 84.1%

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

   4. Simplified 84.1%

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

4. Final simplification 61.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -165000:\\ \;\;\;\;0\\ \mathbf{elif}\;i \leq 0.085:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;0\\ \end{array} \]

Alternative 15: 17.5% accurate, 114.0× speedup

Math

\[\begin{array}{l} \\ 0 \end{array} \]

FPCore

(FPCore (i n) :precision binary64 0.0)

C

double code(double i, double n) {
	return 0.0;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    code = 0.0d0
end function

Java

public static double code(double i, double n) {
	return 0.0;
}

Python

def code(i, n):
	return 0.0

Julia

function code(i, n)
	return 0.0
end

MATLAB

function tmp = code(i, n)
	tmp = 0.0;
end

Wolfram

code[i_, n_] := 0.0

Derivation

1. Initial program 28.0%

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

2. Taylor expanded in i around 0 17.4%

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

3. Taylor expanded in i around 0 17.7%

   \[\leadsto 100 \cdot \color{blue}{0} \]

4. Final simplification 17.7%

   \[\leadsto 0 \]

Developer target: 34.3% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 1 + \frac{i}{n}\\ 100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;t_0 = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log t_0}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (+ 1.0 (/ i n))))
   (*
    100.0
    (/
     (-
      (exp
       (*
        n
        (if (== t_0 1.0)
          (/ i n)
          (/ (* (/ i n) (log t_0)) (- (+ (/ i n) 1.0) 1.0)))))
      1.0)
     (/ i n)))))

C

double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: t_0
    real(8) :: tmp
    t_0 = 1.0d0 + (i / n)
    if (t_0 == 1.0d0) then
        tmp = i / n
    else
        tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0d0) - 1.0d0)
    end if
    code = 100.0d0 * ((exp((n * tmp)) - 1.0d0) / (i / n))
end function

Java

public static double code(double i, double n) {
	double t_0 = 1.0 + (i / n);
	double tmp;
	if (t_0 == 1.0) {
		tmp = i / n;
	} else {
		tmp = ((i / n) * Math.log(t_0)) / (((i / n) + 1.0) - 1.0);
	}
	return 100.0 * ((Math.exp((n * tmp)) - 1.0) / (i / n));
}

Python

def code(i, n):
	t_0 = 1.0 + (i / n)
	tmp = 0
	if t_0 == 1.0:
		tmp = i / n
	else:
		tmp = ((i / n) * math.log(t_0)) / (((i / n) + 1.0) - 1.0)
	return 100.0 * ((math.exp((n * tmp)) - 1.0) / (i / n))

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n))
	tmp = 0.0
	if (t_0 == 1.0)
		tmp = Float64(i / n);
	else
		tmp = Float64(Float64(Float64(i / n) * log(t_0)) / Float64(Float64(Float64(i / n) + 1.0) - 1.0));
	end
	return Float64(100.0 * Float64(Float64(exp(Float64(n * tmp)) - 1.0) / Float64(i / n)))
end

MATLAB

function tmp_2 = code(i, n)
	t_0 = 1.0 + (i / n);
	tmp = 0.0;
	if (t_0 == 1.0)
		tmp = i / n;
	else
		tmp = ((i / n) * log(t_0)) / (((i / n) + 1.0) - 1.0);
	end
	tmp_2 = 100.0 * ((exp((n * tmp)) - 1.0) / (i / n));
end

Wolfram

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

Reproduce

herbie shell --seed 2023240 
(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))))