Compound Interest

Percentage Accurate: 28.1% → 94.0%

Time: 25.5s

Alternatives: 13

Speedup: 38.0×

• could not determine a ground truth

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

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 28.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 94.0% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t\_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 2 \cdot 10^{-170}:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i} + n \cdot \frac{-1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 2e-170)
     (/ (* (expm1 (* n (log1p (/ i n)))) 100.0) (/ i n))
     (if (<= t_1 INFINITY)
       (* 100.0 (+ (* t_0 (/ n i)) (* n (/ -1.0 i))))
       (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-170) {
		tmp = (expm1((n * log1p((i / n)))) * 100.0) / (i / n);
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-170) {
		tmp = (Math.expm1((n * Math.log1p((i / n)))) * 100.0) / (i / n);
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = (t_0 + -1.0) / (i / n)
	tmp = 0
	if t_1 <= 2e-170:
		tmp = (math.expm1((n * math.log1p((i / n)))) * 100.0) / (i / n)
	elif t_1 <= math.inf:
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 2e-170)
		tmp = Float64(Float64(expm1(Float64(n * log1p(Float64(i / n)))) * 100.0) / Float64(i / n));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(Float64(t_0 * Float64(n / i)) + Float64(n * Float64(-1.0 / i))));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 2e-170], N[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * 100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision] + N[(n * N[(-1.0 / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 1.99999999999999997e-170

   1. Initial program 28.4%

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

      1. associate-*r/ 28.4%

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

      2. *-commutative 28.4%

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

      3. add-exp-log 28.4%

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

      4. expm1-define 28.4%

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

      5. log-pow 39.4%

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

      6. log1p-define 97.5%

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

   4. Applied egg-rr 97.5%

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

   if 1.99999999999999997e-170 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < +inf.0

   1. Initial program 99.7%

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

      1. div-sub 99.7%

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

      2. clear-num 99.7%

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

      3. div-inv 99.8%

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

      4. cancel-sign-sub-inv 99.8%

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

      5. div-inv 99.8%

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

      6. clear-num 99.8%

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

   4. Applied egg-rr 99.8%

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

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

   1. Initial program 0.0%

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

   3. Taylor expanded in i around 0 81.0%

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

      1. *-commutative 81.0%

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

   5. Simplified 81.0%

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

4. Final simplification 94.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 2 \cdot 10^{-170}:\\ \;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} \cdot \frac{n}{i} + n \cdot \frac{-1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 92.1% accurate, 0.3× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t\_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t\_1 \leq 2 \cdot 10^{-170}:\\ \;\;\;\;\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \left(100 \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;t\_1 \leq \infty:\\ \;\;\;\;100 \cdot \left(t\_0 \cdot \frac{n}{i} + n \cdot \frac{-1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 2e-170)
     (* (expm1 (* n (log1p (/ i n)))) (* 100.0 (/ n i)))
     (if (<= t_1 INFINITY)
       (* 100.0 (+ (* t_0 (/ n i)) (* n (/ -1.0 i))))
       (* n 100.0)))))

C

double code(double i, double n) {
	double t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-170) {
		tmp = expm1((n * log1p((i / n)))) * (100.0 * (n / i));
	} else if (t_1 <= ((double) INFINITY)) {
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-170) {
		tmp = Math.expm1((n * Math.log1p((i / n)))) * (100.0 * (n / i));
	} else if (t_1 <= Double.POSITIVE_INFINITY) {
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = (t_0 + -1.0) / (i / n)
	tmp = 0
	if t_1 <= 2e-170:
		tmp = math.expm1((n * math.log1p((i / n)))) * (100.0 * (n / i))
	elif t_1 <= math.inf:
		tmp = 100.0 * ((t_0 * (n / i)) + (n * (-1.0 / i)))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 2e-170)
		tmp = Float64(expm1(Float64(n * log1p(Float64(i / n)))) * Float64(100.0 * Float64(n / i)));
	elseif (t_1 <= Inf)
		tmp = Float64(100.0 * Float64(Float64(t_0 * Float64(n / i)) + Float64(n * Float64(-1.0 / i))));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 2e-170], N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * N[(100.0 * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, Infinity], N[(100.0 * N[(N[(t$95$0 * N[(n / i), $MachinePrecision]), $MachinePrecision] + N[(n * N[(-1.0 / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 1.99999999999999997e-170

   1. Initial program 28.4%

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

      1. associate-*r/ 28.4%

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

      2. sub-neg 28.4%

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

      3. distribute-lft-in 28.4%

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

      4. metadata-eval 28.4%

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

      5. metadata-eval 28.4%

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

   3. Simplified 28.4%

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

      1. metadata-eval 28.4%

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

      2. metadata-eval 28.4%

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

      3. distribute-lft-in 28.4%

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

      4. sub-neg 28.4%

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

      5. associate-*r/ 28.4%

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

      6. *-commutative 28.4%

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

      7. div-inv 28.4%

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

      8. clear-num 28.4%

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

      9. associate-*l* 28.3%

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

      10. add-exp-log 28.3%

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

      11. expm1-define 28.3%

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

      12. log-pow 39.5%

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

      13. log1p-define 96.5%

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

   6. Applied egg-rr 96.5%

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

   if 1.99999999999999997e-170 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < +inf.0

   1. Initial program 99.7%

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

      1. div-sub 99.7%

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

      2. clear-num 99.7%

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

      3. div-inv 99.8%

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

      4. cancel-sign-sub-inv 99.8%

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

      5. div-inv 99.8%

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

      6. clear-num 99.8%

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

   4. Applied egg-rr 99.8%

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

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

   1. Initial program 0.0%

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

   3. Taylor expanded in i around 0 81.0%

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

      1. *-commutative 81.0%

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

   5. Simplified 81.0%

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

4. Final simplification 93.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 2 \cdot 10^{-170}:\\ \;\;\;\;\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot \left(100 \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq \infty:\\ \;\;\;\;100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} \cdot \frac{n}{i} + n \cdot \frac{-1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 78.1% accurate, 0.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \log \left(\frac{i}{n}\right)\\ t_1 := 100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{if}\;n \leq -3.9 \cdot 10^{-83}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;n \leq -9.5 \cdot 10^{-189}:\\ \;\;\;\;100 \cdot \frac{n \cdot t\_0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.08 \cdot 10^{-120}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.38 \cdot 10^{-27}:\\ \;\;\;\;100 \cdot \left({n}^{2} \cdot \frac{t\_0}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;t\_1\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (log (/ i n))) (t_1 (* 100.0 (* n (/ (expm1 i) i)))))
   (if (<= n -3.9e-83)
     t_1
     (if (<= n -9.5e-189)
       (* 100.0 (/ (* n t_0) (/ i n)))
       (if (<= n 1.08e-120)
         (/ 0.0 (/ i n))
         (if (<= n 1.38e-27) (* 100.0 (* (pow n 2.0) (/ t_0 i))) t_1))))))

C

double code(double i, double n) {
	double t_0 = log((i / n));
	double t_1 = 100.0 * (n * (expm1(i) / i));
	double tmp;
	if (n <= -3.9e-83) {
		tmp = t_1;
	} else if (n <= -9.5e-189) {
		tmp = 100.0 * ((n * t_0) / (i / n));
	} else if (n <= 1.08e-120) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.38e-27) {
		tmp = 100.0 * (pow(n, 2.0) * (t_0 / i));
	} else {
		tmp = t_1;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = Math.log((i / n));
	double t_1 = 100.0 * (n * (Math.expm1(i) / i));
	double tmp;
	if (n <= -3.9e-83) {
		tmp = t_1;
	} else if (n <= -9.5e-189) {
		tmp = 100.0 * ((n * t_0) / (i / n));
	} else if (n <= 1.08e-120) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.38e-27) {
		tmp = 100.0 * (Math.pow(n, 2.0) * (t_0 / i));
	} else {
		tmp = t_1;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = math.log((i / n))
	t_1 = 100.0 * (n * (math.expm1(i) / i))
	tmp = 0
	if n <= -3.9e-83:
		tmp = t_1
	elif n <= -9.5e-189:
		tmp = 100.0 * ((n * t_0) / (i / n))
	elif n <= 1.08e-120:
		tmp = 0.0 / (i / n)
	elif n <= 1.38e-27:
		tmp = 100.0 * (math.pow(n, 2.0) * (t_0 / i))
	else:
		tmp = t_1
	return tmp

Julia

function code(i, n)
	t_0 = log(Float64(i / n))
	t_1 = Float64(100.0 * Float64(n * Float64(expm1(i) / i)))
	tmp = 0.0
	if (n <= -3.9e-83)
		tmp = t_1;
	elseif (n <= -9.5e-189)
		tmp = Float64(100.0 * Float64(Float64(n * t_0) / Float64(i / n)));
	elseif (n <= 1.08e-120)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 1.38e-27)
		tmp = Float64(100.0 * Float64((n ^ 2.0) * Float64(t_0 / i)));
	else
		tmp = t_1;
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[Log[N[(i / n), $MachinePrecision]], $MachinePrecision]}, Block[{t$95$1 = N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -3.9e-83], t$95$1, If[LessEqual[n, -9.5e-189], N[(100.0 * N[(N[(n * t$95$0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.08e-120], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.38e-27], N[(100.0 * N[(N[Power[n, 2.0], $MachinePrecision] * N[(t$95$0 / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$1]]]]]]

Derivation

1. Split input into 4 regimes

2. if n < -3.9e-83 or 1.38e-27 < n

   1. Initial program 21.7%

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

   3. Taylor expanded in n around inf 38.6%

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

      1. *-commutative 38.6%

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

      2. associate-/l* 38.5%

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

      3. expm1-define 88.7%

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

   5. Simplified 88.7%

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

   if -3.9e-83 < n < -9.499999999999999e-189

   1. Initial program 36.8%

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

   3. Taylor expanded in i around inf 36.5%

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

   4. Taylor expanded in n around 0 0.0%

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

      1. neg-mul-1 0.0%

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

      2. sub-neg 0.0%

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

      3. log-div 84.7%

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

   6. Simplified 84.7%

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

   if -9.499999999999999e-189 < n < 1.0800000000000001e-120

   1. Initial program 59.7%

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

      1. associate-*r/ 59.7%

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

      2. sub-neg 59.7%

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

      3. distribute-lft-in 59.7%

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

      4. metadata-eval 59.7%

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

      5. metadata-eval 59.7%

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

   3. Simplified 59.7%

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

   5. Taylor expanded in i around 0 77.2%

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

   if 1.0800000000000001e-120 < n < 1.38e-27

   1. Initial program 14.5%

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

      1. associate-*r/ 14.5%

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

      2. sub-neg 14.5%

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

      3. distribute-lft-in 14.5%

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

      4. metadata-eval 14.5%

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

      5. metadata-eval 14.5%

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

   3. Simplified 14.5%

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

      1. metadata-eval 14.5%

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

      2. metadata-eval 14.5%

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

      3. distribute-lft-in 14.5%

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

      4. sub-neg 14.5%

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

      5. associate-*r/ 14.5%

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

      6. *-commutative 14.5%

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

      7. div-inv 14.5%

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

      8. clear-num 14.6%

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

      9. associate-*l* 14.6%

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

      10. add-exp-log 14.6%

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

      11. expm1-define 14.6%

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

      12. log-pow 63.6%

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

      13. log1p-define 89.1%

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

   6. Applied egg-rr 89.1%

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

      1. *-commutative 89.1%

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

      2. associate-*r* 89.1%

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

      3. associate-*r/ 84.2%

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

   8. Applied egg-rr 84.2%

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

   9. Taylor expanded in n around 0 68.1%

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

       1. associate-/l* 68.0%

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

       2. neg-mul-1 68.0%

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

       3. sub-neg 68.0%

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

       4. log-div 57.8%

          \[\leadsto 100 \cdot \left({n}^{2} \cdot \frac{\color{blue}{\log \left(\frac{i}{n}\right)}}{i}\right) \]

   11. Simplified 57.8%

       \[\leadsto \color{blue}{100 \cdot \left({n}^{2} \cdot \frac{\log \left(\frac{i}{n}\right)}{i}\right)} \]
3. Recombined 4 regimes into one program.

4. Final simplification 84.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -3.9 \cdot 10^{-83}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;n \leq -9.5 \cdot 10^{-189}:\\ \;\;\;\;100 \cdot \frac{n \cdot \log \left(\frac{i}{n}\right)}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.08 \cdot 10^{-120}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.38 \cdot 10^{-27}:\\ \;\;\;\;100 \cdot \left({n}^{2} \cdot \frac{\log \left(\frac{i}{n}\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 78.1% accurate, 0.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ t_1 := 100 \cdot \frac{n \cdot \log \left(\frac{i}{n}\right)}{\frac{i}{n}}\\ \mathbf{if}\;n \leq -4.3 \cdot 10^{-83}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;n \leq -1.7 \cdot 10^{-190}:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;n \leq 3.8 \cdot 10^{-120}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.3 \cdot 10^{-27}:\\ \;\;\;\;t\_1\\ \mathbf{else}:\\ \;\;\;\;t\_0\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (let* ((t_0 (* 100.0 (* n (/ (expm1 i) i))))
        (t_1 (* 100.0 (/ (* n (log (/ i n))) (/ i n)))))
   (if (<= n -4.3e-83)
     t_0
     (if (<= n -1.7e-190)
       t_1
       (if (<= n 3.8e-120) (/ 0.0 (/ i n)) (if (<= n 1.3e-27) t_1 t_0))))))

C

double code(double i, double n) {
	double t_0 = 100.0 * (n * (expm1(i) / i));
	double t_1 = 100.0 * ((n * log((i / n))) / (i / n));
	double tmp;
	if (n <= -4.3e-83) {
		tmp = t_0;
	} else if (n <= -1.7e-190) {
		tmp = t_1;
	} else if (n <= 3.8e-120) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.3e-27) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double t_0 = 100.0 * (n * (Math.expm1(i) / i));
	double t_1 = 100.0 * ((n * Math.log((i / n))) / (i / n));
	double tmp;
	if (n <= -4.3e-83) {
		tmp = t_0;
	} else if (n <= -1.7e-190) {
		tmp = t_1;
	} else if (n <= 3.8e-120) {
		tmp = 0.0 / (i / n);
	} else if (n <= 1.3e-27) {
		tmp = t_1;
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

def code(i, n):
	t_0 = 100.0 * (n * (math.expm1(i) / i))
	t_1 = 100.0 * ((n * math.log((i / n))) / (i / n))
	tmp = 0
	if n <= -4.3e-83:
		tmp = t_0
	elif n <= -1.7e-190:
		tmp = t_1
	elif n <= 3.8e-120:
		tmp = 0.0 / (i / n)
	elif n <= 1.3e-27:
		tmp = t_1
	else:
		tmp = t_0
	return tmp

Julia

function code(i, n)
	t_0 = Float64(100.0 * Float64(n * Float64(expm1(i) / i)))
	t_1 = Float64(100.0 * Float64(Float64(n * log(Float64(i / n))) / Float64(i / n)))
	tmp = 0.0
	if (n <= -4.3e-83)
		tmp = t_0;
	elseif (n <= -1.7e-190)
		tmp = t_1;
	elseif (n <= 3.8e-120)
		tmp = Float64(0.0 / Float64(i / n));
	elseif (n <= 1.3e-27)
		tmp = t_1;
	else
		tmp = t_0;
	end
	return tmp
end

Wolfram

code[i_, n_] := Block[{t$95$0 = N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(100.0 * N[(N[(n * N[Log[N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[n, -4.3e-83], t$95$0, If[LessEqual[n, -1.7e-190], t$95$1, If[LessEqual[n, 3.8e-120], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 1.3e-27], t$95$1, t$95$0]]]]]]

Derivation

1. Split input into 3 regimes

2. if n < -4.30000000000000033e-83 or 1.30000000000000009e-27 < n

   1. Initial program 21.7%

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

   3. Taylor expanded in n around inf 38.6%

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

      1. *-commutative 38.6%

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

      2. associate-/l* 38.5%

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

      3. expm1-define 88.7%

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

   5. Simplified 88.7%

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

   if -4.30000000000000033e-83 < n < -1.69999999999999991e-190 or 3.7999999999999997e-120 < n < 1.30000000000000009e-27

   1. Initial program 24.0%

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

   3. Taylor expanded in i around inf 23.4%

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

   4. Taylor expanded in n around 0 39.2%

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

      1. neg-mul-1 39.2%

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

      2. sub-neg 39.2%

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

      3. log-div 69.1%

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

   6. Simplified 69.1%

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

   if -1.69999999999999991e-190 < n < 3.7999999999999997e-120

   1. Initial program 59.7%

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

      1. associate-*r/ 59.7%

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

      2. sub-neg 59.7%

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

      3. distribute-lft-in 59.7%

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

      4. metadata-eval 59.7%

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

      5. metadata-eval 59.7%

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

   3. Simplified 59.7%

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

   5. Taylor expanded in i around 0 77.2%

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

4. Final simplification 84.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -4.3 \cdot 10^{-83}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;n \leq -1.7 \cdot 10^{-190}:\\ \;\;\;\;100 \cdot \frac{n \cdot \log \left(\frac{i}{n}\right)}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 3.8 \cdot 10^{-120}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{elif}\;n \leq 1.3 \cdot 10^{-27}:\\ \;\;\;\;100 \cdot \frac{n \cdot \log \left(\frac{i}{n}\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 76.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -2.1 \cdot 10^{-8} \lor \neg \left(i \leq 2 \cdot 10^{-20}\right):\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= i -2.1e-8) (not (<= i 2e-20)))
   (* 100.0 (/ (expm1 i) (/ i n)))
   (* n (+ 100.0 (* i 50.0)))))

C

double code(double i, double n) {
	double tmp;
	if ((i <= -2.1e-8) || !(i <= 2e-20)) {
		tmp = 100.0 * (expm1(i) / (i / n));
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((i <= -2.1e-8) || !(i <= 2e-20)) {
		tmp = 100.0 * (Math.expm1(i) / (i / n));
	} else {
		tmp = n * (100.0 + (i * 50.0));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (i <= -2.1e-8) or not (i <= 2e-20):
		tmp = 100.0 * (math.expm1(i) / (i / n))
	else:
		tmp = n * (100.0 + (i * 50.0))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((i <= -2.1e-8) || !(i <= 2e-20))
		tmp = Float64(100.0 * Float64(expm1(i) / Float64(i / n)));
	else
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[i, -2.1e-8], N[Not[LessEqual[i, 2e-20]], $MachinePrecision]], N[(100.0 * N[(N[(Exp[i] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -2.09999999999999994e-8 or 1.99999999999999989e-20 < i

   1. Initial program 45.8%

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

   3. Taylor expanded in n around inf 59.2%

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

      1. expm1-define 59.5%

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

   5. Simplified 59.5%

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

   if -2.09999999999999994e-8 < i < 1.99999999999999989e-20

   1. Initial program 11.4%

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

   3. Taylor expanded in i around 0 80.7%

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

   4. Taylor expanded in n around inf 81.4%

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

      1. *-commutative 81.4%

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

   6. Simplified 81.4%

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

4. Final simplification 70.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -2.1 \cdot 10^{-8} \lor \neg \left(i \leq 2 \cdot 10^{-20}\right):\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 80.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -6.3 \cdot 10^{-156} \lor \neg \left(n \leq 2.5 \cdot 10^{-87}\right):\\ \;\;\;\;n \cdot \frac{100 \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -6.3e-156) (not (<= n 2.5e-87)))
   (* n (/ (* 100.0 (expm1 i)) i))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -6.3e-156) || !(n <= 2.5e-87)) {
		tmp = n * ((100.0 * expm1(i)) / i);
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -6.3e-156) || !(n <= 2.5e-87)) {
		tmp = n * ((100.0 * Math.expm1(i)) / i);
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -6.3e-156) or not (n <= 2.5e-87):
		tmp = n * ((100.0 * math.expm1(i)) / i)
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -6.3e-156) || !(n <= 2.5e-87))
		tmp = Float64(n * Float64(Float64(100.0 * expm1(i)) / i));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -6.3e-156], N[Not[LessEqual[n, 2.5e-87]], $MachinePrecision]], N[(n * N[(N[(100.0 * N[(Exp[i] - 1), $MachinePrecision]), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -6.3000000000000001e-156 or 2.50000000000000021e-87 < n

   1. Initial program 21.2%

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

      1. associate-*r/ 21.2%

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

      2. sub-neg 21.2%

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

      3. distribute-lft-in 21.2%

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

      4. metadata-eval 21.2%

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

      5. metadata-eval 21.2%

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

   3. Simplified 21.2%

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

   5. Taylor expanded in n around inf 35.1%

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

      1. *-commutative 35.1%

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

   7. Simplified 35.1%

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

   8. Taylor expanded in i around inf 35.5%

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

      1. fma-neg 35.4%

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

      2. metadata-eval 35.4%

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

      3. associate-*r/ 35.5%

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

      4. fma-undefine 35.5%

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

      5. metadata-eval 35.5%

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

      6. distribute-lft-in 35.5%

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

      7. metadata-eval 35.5%

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

      8. sub-neg 35.5%

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

      9. expm1-define 83.0%

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

   10. Simplified 83.0%

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

   if -6.3000000000000001e-156 < n < 2.50000000000000021e-87

   1. Initial program 54.4%

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

      1. associate-*r/ 54.4%

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

      2. sub-neg 54.4%

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

      3. distribute-lft-in 54.4%

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

      4. metadata-eval 54.4%

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

      5. metadata-eval 54.4%

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

   3. Simplified 54.4%

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

   5. Taylor expanded in i around 0 69.8%

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

4. Final simplification 80.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -6.3 \cdot 10^{-156} \lor \neg \left(n \leq 2.5 \cdot 10^{-87}\right):\\ \;\;\;\;n \cdot \frac{100 \cdot \mathsf{expm1}\left(i\right)}{i}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 80.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1 \cdot 10^{-158} \lor \neg \left(n \leq 2.8 \cdot 10^{-87}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1e-158) (not (<= n 2.8e-87)))
   (* 100.0 (* n (/ (expm1 i) i)))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1e-158) || !(n <= 2.8e-87)) {
		tmp = 100.0 * (n * (expm1(i) / i));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1e-158) || !(n <= 2.8e-87)) {
		tmp = 100.0 * (n * (Math.expm1(i) / i));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1e-158) or not (n <= 2.8e-87):
		tmp = 100.0 * (n * (math.expm1(i) / i))
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1e-158) || !(n <= 2.8e-87))
		tmp = Float64(100.0 * Float64(n * Float64(expm1(i) / i)));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1e-158], N[Not[LessEqual[n, 2.8e-87]], $MachinePrecision]], N[(100.0 * N[(n * N[(N[(Exp[i] - 1), $MachinePrecision] / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.00000000000000006e-158 or 2.8000000000000001e-87 < n

   1. Initial program 21.2%

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

   3. Taylor expanded in n around inf 35.4%

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

      1. *-commutative 35.4%

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

      2. associate-/l* 35.4%

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

      3. expm1-define 83.0%

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

   5. Simplified 83.0%

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

   if -1.00000000000000006e-158 < n < 2.8000000000000001e-87

   1. Initial program 54.4%

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

      1. associate-*r/ 54.4%

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

      2. sub-neg 54.4%

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

      3. distribute-lft-in 54.4%

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

      4. metadata-eval 54.4%

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

      5. metadata-eval 54.4%

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

   3. Simplified 54.4%

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

   5. Taylor expanded in i around 0 69.8%

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

4. Final simplification 80.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1 \cdot 10^{-158} \lor \neg \left(n \leq 2.8 \cdot 10^{-87}\right):\\ \;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 61.1% accurate, 6.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.7 \cdot 10^{-73}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{elif}\;n \leq 2.8 \cdot 10^{-87}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + 50 \cdot \left(i \cdot n\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= n -1.7e-73)
   (* n (+ 100.0 (* i 50.0)))
   (if (<= n 2.8e-87) (/ 0.0 (/ i n)) (+ (* n 100.0) (* 50.0 (* i n))))))

C

double code(double i, double n) {
	double tmp;
	if (n <= -1.7e-73) {
		tmp = n * (100.0 + (i * 50.0));
	} else if (n <= 2.8e-87) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (50.0 * (i * n));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (n <= (-1.7d-73)) then
        tmp = n * (100.0d0 + (i * 50.0d0))
    else if (n <= 2.8d-87) then
        tmp = 0.0d0 / (i / n)
    else
        tmp = (n * 100.0d0) + (50.0d0 * (i * n))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (n <= -1.7e-73) {
		tmp = n * (100.0 + (i * 50.0));
	} else if (n <= 2.8e-87) {
		tmp = 0.0 / (i / n);
	} else {
		tmp = (n * 100.0) + (50.0 * (i * n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if n <= -1.7e-73:
		tmp = n * (100.0 + (i * 50.0))
	elif n <= 2.8e-87:
		tmp = 0.0 / (i / n)
	else:
		tmp = (n * 100.0) + (50.0 * (i * n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (n <= -1.7e-73)
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	elseif (n <= 2.8e-87)
		tmp = Float64(0.0 / Float64(i / n));
	else
		tmp = Float64(Float64(n * 100.0) + Float64(50.0 * Float64(i * n)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (n <= -1.7e-73)
		tmp = n * (100.0 + (i * 50.0));
	elseif (n <= 2.8e-87)
		tmp = 0.0 / (i / n);
	else
		tmp = (n * 100.0) + (50.0 * (i * n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[n, -1.7e-73], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[n, 2.8e-87], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision], N[(N[(n * 100.0), $MachinePrecision] + N[(50.0 * N[(i * n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if n < -1.7000000000000001e-73

   1. Initial program 24.2%

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

   3. Taylor expanded in i around 0 58.6%

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

   4. Taylor expanded in n around inf 58.6%

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

      1. *-commutative 58.6%

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

   6. Simplified 58.6%

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

   if -1.7000000000000001e-73 < n < 2.8000000000000001e-87

   1. Initial program 51.2%

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

      1. associate-*r/ 51.2%

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

      2. sub-neg 51.2%

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

      3. distribute-lft-in 51.2%

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

      4. metadata-eval 51.2%

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

      5. metadata-eval 51.2%

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

   3. Simplified 51.2%

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

   5. Taylor expanded in i around 0 64.3%

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

   if 2.8000000000000001e-87 < n

   1. Initial program 17.7%

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

   3. Taylor expanded in i around 0 64.0%

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

   4. Taylor expanded in n around inf 64.0%

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

      1. *-commutative 64.0%

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

   6. Simplified 64.0%

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

4. Final simplification 62.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.7 \cdot 10^{-73}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{elif}\;n \leq 2.8 \cdot 10^{-87}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + 50 \cdot \left(i \cdot n\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 54.5% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -6.2 \cdot 10^{+234} \lor \neg \left(i \leq 5 \cdot 10^{-14}\right):\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= i -6.2e+234) (not (<= i 5e-14)))
   (* 100.0 (* i (/ n i)))
   (* n 100.0)))

C

double code(double i, double n) {
	double tmp;
	if ((i <= -6.2e+234) || !(i <= 5e-14)) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((i <= (-6.2d+234)) .or. (.not. (i <= 5d-14))) then
        tmp = 100.0d0 * (i * (n / i))
    else
        tmp = n * 100.0d0
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((i <= -6.2e+234) || !(i <= 5e-14)) {
		tmp = 100.0 * (i * (n / i));
	} else {
		tmp = n * 100.0;
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (i <= -6.2e+234) or not (i <= 5e-14):
		tmp = 100.0 * (i * (n / i))
	else:
		tmp = n * 100.0
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((i <= -6.2e+234) || !(i <= 5e-14))
		tmp = Float64(100.0 * Float64(i * Float64(n / i)));
	else
		tmp = Float64(n * 100.0);
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((i <= -6.2e+234) || ~((i <= 5e-14)))
		tmp = 100.0 * (i * (n / i));
	else
		tmp = n * 100.0;
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[i, -6.2e+234], N[Not[LessEqual[i, 5e-14]], $MachinePrecision]], N[(100.0 * N[(i * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(n * 100.0), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if i < -6.19999999999999979e234 or 5.0000000000000002e-14 < i

   1. Initial program 47.2%

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

   3. Taylor expanded in i around 0 22.0%

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

      1. clear-num 22.0%

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

      2. associate-/r/ 22.0%

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

      3. clear-num 18.0%

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

   5. Applied egg-rr 18.0%

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

   if -6.19999999999999979e234 < i < 5.0000000000000002e-14

   1. Initial program 18.2%

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

   3. Taylor expanded in i around 0 64.8%

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

      1. *-commutative 64.8%

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

   5. Simplified 64.8%

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

4. Final simplification 49.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -6.2 \cdot 10^{+234} \lor \neg \left(i \leq 5 \cdot 10^{-14}\right):\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 62.5% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.45 \cdot 10^{+33} \lor \neg \left(n \leq 0.23\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.45e+33) (not (<= n 0.23)))
   (* n (+ 100.0 (* i 50.0)))
   (* 100.0 (/ i (/ i n)))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.45e+33) || !(n <= 0.23)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((n <= (-1.45d+33)) .or. (.not. (n <= 0.23d0))) then
        tmp = n * (100.0d0 + (i * 50.0d0))
    else
        tmp = 100.0d0 * (i / (i / n))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1.45e+33) || !(n <= 0.23)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1.45e+33) or not (n <= 0.23):
		tmp = n * (100.0 + (i * 50.0))
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1.45e+33) || !(n <= 0.23))
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	else
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -1.45e+33) || ~((n <= 0.23)))
		tmp = n * (100.0 + (i * 50.0));
	else
		tmp = 100.0 * (i / (i / n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1.45e+33], N[Not[LessEqual[n, 0.23]], $MachinePrecision]], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.45000000000000012e33 or 0.23000000000000001 < n

   1. Initial program 19.8%

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

   3. Taylor expanded in i around 0 65.5%

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

   4. Taylor expanded in n around inf 65.5%

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

      1. *-commutative 65.5%

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

   6. Simplified 65.5%

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

   if -1.45000000000000012e33 < n < 0.23000000000000001

   1. Initial program 40.7%

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

   3. Taylor expanded in i around 0 43.5%

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

4. Final simplification 56.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.45 \cdot 10^{+33} \lor \neg \left(n \leq 0.23\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 61.1% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq -1.7 \cdot 10^{-73} \lor \neg \left(n \leq 2.5 \cdot 10^{-87}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (or (<= n -1.7e-73) (not (<= n 2.5e-87)))
   (* n (+ 100.0 (* i 50.0)))
   (/ 0.0 (/ i n))))

C

double code(double i, double n) {
	double tmp;
	if ((n <= -1.7e-73) || !(n <= 2.5e-87)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if ((n <= (-1.7d-73)) .or. (.not. (n <= 2.5d-87))) then
        tmp = n * (100.0d0 + (i * 50.0d0))
    else
        tmp = 0.0d0 / (i / n)
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if ((n <= -1.7e-73) || !(n <= 2.5e-87)) {
		tmp = n * (100.0 + (i * 50.0));
	} else {
		tmp = 0.0 / (i / n);
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if (n <= -1.7e-73) or not (n <= 2.5e-87):
		tmp = n * (100.0 + (i * 50.0))
	else:
		tmp = 0.0 / (i / n)
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if ((n <= -1.7e-73) || !(n <= 2.5e-87))
		tmp = Float64(n * Float64(100.0 + Float64(i * 50.0)));
	else
		tmp = Float64(0.0 / Float64(i / n));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if ((n <= -1.7e-73) || ~((n <= 2.5e-87)))
		tmp = n * (100.0 + (i * 50.0));
	else
		tmp = 0.0 / (i / n);
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[Or[LessEqual[n, -1.7e-73], N[Not[LessEqual[n, 2.5e-87]], $MachinePrecision]], N[(n * N[(100.0 + N[(i * 50.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(0.0 / N[(i / n), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < -1.7000000000000001e-73 or 2.50000000000000021e-87 < n

   1. Initial program 20.7%

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

   3. Taylor expanded in i around 0 61.5%

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

   4. Taylor expanded in n around inf 61.6%

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

      1. *-commutative 61.6%

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

   6. Simplified 61.6%

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

   if -1.7000000000000001e-73 < n < 2.50000000000000021e-87

   1. Initial program 51.2%

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

      1. associate-*r/ 51.2%

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

      2. sub-neg 51.2%

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

      3. distribute-lft-in 51.2%

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

      4. metadata-eval 51.2%

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

      5. metadata-eval 51.2%

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

   3. Simplified 51.2%

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

   5. Taylor expanded in i around 0 64.3%

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

4. Final simplification 62.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;n \leq -1.7 \cdot 10^{-73} \lor \neg \left(n \leq 2.5 \cdot 10^{-87}\right):\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 56.6% accurate, 6.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;i \leq -5 \cdot 10^{+69}:\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;i \leq 10^{-6}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \end{array} \]

FPCore

(FPCore (i n)
 :precision binary64
 (if (<= i -5e+69)
   (* 100.0 (* i (/ n i)))
   (if (<= i 1e-6) (* n 100.0) (* 100.0 (/ i (/ i n))))))

C

double code(double i, double n) {
	double tmp;
	if (i <= -5e+69) {
		tmp = 100.0 * (i * (n / i));
	} else if (i <= 1e-6) {
		tmp = n * 100.0;
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Fortran

real(8) function code(i, n)
    real(8), intent (in) :: i
    real(8), intent (in) :: n
    real(8) :: tmp
    if (i <= (-5d+69)) then
        tmp = 100.0d0 * (i * (n / i))
    else if (i <= 1d-6) then
        tmp = n * 100.0d0
    else
        tmp = 100.0d0 * (i / (i / n))
    end if
    code = tmp
end function

Java

public static double code(double i, double n) {
	double tmp;
	if (i <= -5e+69) {
		tmp = 100.0 * (i * (n / i));
	} else if (i <= 1e-6) {
		tmp = n * 100.0;
	} else {
		tmp = 100.0 * (i / (i / n));
	}
	return tmp;
}

Python

def code(i, n):
	tmp = 0
	if i <= -5e+69:
		tmp = 100.0 * (i * (n / i))
	elif i <= 1e-6:
		tmp = n * 100.0
	else:
		tmp = 100.0 * (i / (i / n))
	return tmp

Julia

function code(i, n)
	tmp = 0.0
	if (i <= -5e+69)
		tmp = Float64(100.0 * Float64(i * Float64(n / i)));
	elseif (i <= 1e-6)
		tmp = Float64(n * 100.0);
	else
		tmp = Float64(100.0 * Float64(i / Float64(i / n)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(i, n)
	tmp = 0.0;
	if (i <= -5e+69)
		tmp = 100.0 * (i * (n / i));
	elseif (i <= 1e-6)
		tmp = n * 100.0;
	else
		tmp = 100.0 * (i / (i / n));
	end
	tmp_2 = tmp;
end

Wolfram

code[i_, n_] := If[LessEqual[i, -5e+69], N[(100.0 * N[(i * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[i, 1e-6], N[(n * 100.0), $MachinePrecision], N[(100.0 * N[(i / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if i < -5.00000000000000036e69

   1. Initial program 56.8%

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

   3. Taylor expanded in i around 0 17.7%

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

      1. clear-num 17.7%

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

      2. associate-/r/ 17.7%

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

      3. clear-num 17.7%

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

   5. Applied egg-rr 17.7%

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

   if -5.00000000000000036e69 < i < 9.99999999999999955e-7

   1. Initial program 13.8%

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

   3. Taylor expanded in i around 0 74.3%

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

      1. *-commutative 74.3%

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

   5. Simplified 74.3%

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

   if 9.99999999999999955e-7 < i

   1. Initial program 42.5%

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

   3. Taylor expanded in i around 0 18.5%

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

4. Final simplification 50.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -5 \cdot 10^{+69}:\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{elif}\;i \leq 10^{-6}:\\ \;\;\;\;n \cdot 100\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 49.6% accurate, 38.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 27.9%

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

3. Taylor expanded in i around 0 45.1%

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

   1. *-commutative 45.1%

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

5. Simplified 45.1%

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

6. Final simplification 45.1%

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

Developer target: 34.2% 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 2024039 
(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))))