Average Error: 47.0 → 16.0
Time: 15.3s
Precision: binary64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;i \le -0.39780581747886845 \lor \neg \left(i \le 1.5120834945497763\right):\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;i \le -0.39780581747886845 \lor \neg \left(i \le 1.5120834945497763\right):\\
\;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\

\mathbf{else}:\\
\;\;\;\;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)\\

\end{array}
double code(double i, double n) {
	return ((double) (100.0 * ((double) (((double) (((double) pow(((double) (1.0 + ((double) (i / n)))), n)) - 1.0)) / ((double) (i / n))))));
}
↓
double code(double i, double n) {
	double VAR;
	if (((i <= -0.39780581747886845) || !(i <= 1.5120834945497763))) {
		VAR = ((double) (((double) (100.0 * ((double) (((double) pow(((double) (1.0 + ((double) (i / n)))), n)) - 1.0)))) / ((double) (i / n))));
	} else {
		VAR = ((double) (100.0 * ((double) (((double) (((double) (((double) (((double) (1.0 * i)) + ((double) (((double) (0.5 * ((double) pow(i, 2.0)))) + ((double) (((double) log(1.0)) * n)))))) - ((double) (0.5 * ((double) (((double) pow(i, 2.0)) * ((double) log(1.0)))))))) / i)) * n))));
	}
	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

Original47.0
Target47.6
Herbie16.0
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 2 regimes

  2. if i < -0.39780581747886845 or 1.5120834945497763 < i

    1. Initial program 27.3

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

    3. Applied associate-*r/27.3

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

    if -0.39780581747886845 < i < 1.5120834945497763

    1. Initial program 58.0

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

    2. Taylor expanded around 0 26.9

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

    4. Applied associate-/r/9.6

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

  4. Final simplification16.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \le -0.39780581747886845 \lor \neg \left(i \le 1.5120834945497763\right):\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(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))))