Average Error: 47.7 → 17.2
Time: 13.9s
Precision: 64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
\[\begin{array}{l} \mathbf{if}\;i \le -0.00377737376123091695:\\ \;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\ \mathbf{elif}\;i \le 4.4408729626575344 \cdot 10^{-11}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \mathbf{elif}\;i \le 4.12968528167230101 \cdot 10^{131}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\begin{array}{l}
\mathbf{if}\;i \le -0.00377737376123091695:\\
\;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\

\mathbf{elif}\;i \le 4.4408729626575344 \cdot 10^{-11}:\\
\;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\

\mathbf{elif}\;i \le 4.12968528167230101 \cdot 10^{131}:\\
\;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\

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

\end{array}
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 VAR;
	if ((i <= -0.003777373761230917)) {
		VAR = (100.0 * ((pow((1.0 + (i / n)), n) / (i / n)) - (1.0 / (i / n))));
	} else {
		double VAR_1;
		if ((i <= 4.4408729626575344e-11)) {
			VAR_1 = (100.0 * (((((1.0 * i) + ((0.5 * pow(i, 2.0)) + (log(1.0) * n))) - (0.5 * (pow(i, 2.0) * log(1.0)))) / i) * n));
		} else {
			double VAR_2;
			if ((i <= 4.129685281672301e+131)) {
				VAR_2 = ((100.0 * (pow((1.0 + (i / n)), n) - 1.0)) / (i / n));
			} else {
				VAR_2 = (100.0 * ((((1.0 * i) + ((log(1.0) * n) + 1.0)) - 1.0) / (i / n)));
			}
			VAR_1 = VAR_2;
		}
		VAR = VAR_1;
	}
	return VAR;
}

Error

Bits error versus i

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original47.7
Target47.5
Herbie17.2
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 4 regimes
  2. if i < -0.003777373761230917

    1. Initial program 28.4

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
    2. Using strategy rm
    3. Applied div-sub28.4

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

    if -0.003777373761230917 < i < 4.4408729626575344e-11

    1. Initial program 58.3

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
    2. Taylor expanded around 0 26.3

      \[\leadsto 100 \cdot \frac{\color{blue}{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}}{\frac{i}{n}}\]
    3. Using strategy rm
    4. Applied associate-/r/8.9

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

    if 4.4408729626575344e-11 < i < 4.129685281672301e+131

    1. Initial program 33.6

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
    2. Using strategy rm
    3. Applied associate-*r/33.6

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

    if 4.129685281672301e+131 < i

    1. Initial program 30.9

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
    2. Taylor expanded around 0 36.4

      \[\leadsto 100 \cdot \frac{\color{blue}{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right)} - 1}{\frac{i}{n}}\]
  3. Recombined 4 regimes into one program.
  4. Final simplification17.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \le -0.00377737376123091695:\\ \;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\ \mathbf{elif}\;i \le 4.4408729626575344 \cdot 10^{-11}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \mathbf{elif}\;i \le 4.12968528167230101 \cdot 10^{131}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020091 
(FPCore (i n)
  :name "Compound Interest"
  :precision binary64

  :herbie-target
  (* 100 (/ (- (exp (* n (if (== (+ 1 (/ i n)) 1) (/ i n) (/ (* (/ i n) (log (+ 1 (/ i n)))) (- (+ (/ i n) 1) 1))))) 1) (/ i n)))

  (* 100 (/ (- (pow (+ 1 (/ i n)) n) 1) (/ i n))))