?

Average Error: 47.5 → 0.6
Time: 18.0s
Precision: binary64
Cost: 22408

?

\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq 10000:\\ \;\;\;\;100 \cdot \left(\frac{\frac{i}{n} \cdot t_0 - \frac{i}{n}}{\frac{i}{n}} \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\ \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
(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 10000.0)
       (* 100.0 (* (/ (- (* (/ i n) t_0) (/ i n)) (/ i n)) (/ n i)))
       (* 100.0 (+ (+ 1.0 n) -1.0))))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}
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 <= 10000.0) {
		tmp = 100.0 * (((((i / n) * t_0) - (i / n)) / (i / n)) * (n / i));
	} else {
		tmp = 100.0 * ((1.0 + n) + -1.0);
	}
	return tmp;
}
public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}
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 <= 10000.0) {
		tmp = 100.0 * (((((i / n) * t_0) - (i / n)) / (i / n)) * (n / i));
	} else {
		tmp = 100.0 * ((1.0 + n) + -1.0);
	}
	return tmp;
}
def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))
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 <= 10000.0:
		tmp = 100.0 * (((((i / n) * t_0) - (i / n)) / (i / n)) * (n / i))
	else:
		tmp = 100.0 * ((1.0 + n) + -1.0)
	return tmp
function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end
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 <= 10000.0)
		tmp = Float64(100.0 * Float64(Float64(Float64(Float64(Float64(i / n) * t_0) - Float64(i / n)) / Float64(i / n)) * Float64(n / i)));
	else
		tmp = Float64(100.0 * Float64(Float64(1.0 + n) + -1.0));
	end
	return tmp
end
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]
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, 10000.0], N[(100.0 * N[(N[(N[(N[(N[(i / n), $MachinePrecision] * t$95$0), $MachinePrecision] - N[(i / n), $MachinePrecision]), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision] * N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(N[(1.0 + n), $MachinePrecision] + -1.0), $MachinePrecision]), $MachinePrecision]]]]]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\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 10000:\\
\;\;\;\;100 \cdot \left(\frac{\frac{i}{n} \cdot t_0 - \frac{i}{n}}{\frac{i}{n}} \cdot \frac{n}{i}\right)\\

\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original47.5
Target47.4
Herbie0.6
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \]

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 46.2

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr0.2

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

    if 0.0 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 1e4

    1. Initial program 2.2

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr2.2

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

    if 1e4 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

    1. Initial program 62.7

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Taylor expanded in i around 0 49.5

      \[\leadsto 100 \cdot \frac{\color{blue}{i}}{\frac{i}{n}} \]
    3. Applied egg-rr1.4

      \[\leadsto 100 \cdot \color{blue}{\left(\left(1 + n\right) - 1\right)} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 0:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 10000:\\ \;\;\;\;100 \cdot \left(\frac{\frac{i}{n} \cdot {\left(1 + \frac{i}{n}\right)}^{n} - \frac{i}{n}}{\frac{i}{n}} \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\ \end{array} \]

Alternatives

Alternative 1
Error13.6
Cost13768
\[\begin{array}{l} \mathbf{if}\;i \leq 1.8 \cdot 10^{-39}:\\ \;\;\;\;n \cdot \left(100 \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;i \leq 1.35 \cdot 10^{+86}:\\ \;\;\;\;100 \cdot \left(\frac{n}{i} \cdot \mathsf{expm1}\left(n \cdot \log \left(\frac{i}{n}\right)\right)\right)\\ \mathbf{elif}\;i \leq 3.8 \cdot 10^{+261}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{100}{\frac{1}{n}}}{1 + i \cdot -0.5}\\ \end{array} \]
Alternative 2
Error13.7
Cost7692
\[\begin{array}{l} \mathbf{if}\;i \leq 1.8 \cdot 10^{-39}:\\ \;\;\;\;n \cdot \left(100 \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;i \leq 1.45 \cdot 10^{+81}:\\ \;\;\;\;100 \cdot \left(\log \left(\frac{i}{n}\right) \cdot \frac{n \cdot n}{i}\right)\\ \mathbf{elif}\;i \leq 3.65 \cdot 10^{+261}:\\ \;\;\;\;100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{100}{\frac{1}{n}}}{1 + i \cdot -0.5}\\ \end{array} \]
Alternative 3
Error14.0
Cost7368
\[\begin{array}{l} \mathbf{if}\;i \leq 1.8 \cdot 10^{-39}:\\ \;\;\;\;n \cdot \left(100 \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{elif}\;i \leq 4.4 \cdot 10^{+155}:\\ \;\;\;\;100 \cdot \left(\log \left(\frac{i}{n}\right) \cdot \frac{n \cdot n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{100}{\frac{1}{n}}}{1 + i \cdot -0.5}\\ \end{array} \]
Alternative 4
Error13.5
Cost7244
\[\begin{array}{l} t_0 := n \cdot \left(100 \cdot \frac{\mathsf{expm1}\left(i\right)}{i}\right)\\ \mathbf{if}\;n \leq -3.1 \cdot 10^{-133}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;n \leq 3.15 \cdot 10^{-282}:\\ \;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\ \mathbf{elif}\;n \leq 2.15 \cdot 10^{-20}:\\ \;\;\;\;\frac{100 \cdot \left(\frac{1}{i + 2} \cdot \left(i \cdot 2\right)\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 5
Error20.7
Cost969
\[\begin{array}{l} \mathbf{if}\;n \leq -3.1 \cdot 10^{-133} \lor \neg \left(n \leq 2.3 \cdot 10^{-280}\right):\\ \;\;\;\;\frac{\frac{100}{\frac{1}{n}}}{1 + i \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\ \end{array} \]
Alternative 6
Error24.2
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -2.9 \cdot 10^{+141} \lor \neg \left(i \leq 10^{+54}\right):\\ \;\;\;\;100 \cdot \left(i \cdot \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 7
Error19.8
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -1.5 \cdot 10^{-40} \lor \neg \left(i \leq 7.2 \cdot 10^{-52}\right):\\ \;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 8
Error28.8
Cost192
\[n \cdot 100 \]

Error

Reproduce?

herbie shell --seed 2023031 
(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))))