Average Error: 43.1 → 23.3
Time: 33.6s
Precision: 64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;i \le -1.05913144671026914 \cdot 10^{-51}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 8143434713.6699142:\\ \;\;\;\;\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\\ \mathbf{elif}\;i \le 2.03754808603946062 \cdot 10^{169}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;i \le -1.05913144671026914 \cdot 10^{-51}:\\
\;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\

\mathbf{elif}\;i \le 8143434713.6699142:\\
\;\;\;\;\left(100 \cdot \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}\right) \cdot n\\

\mathbf{elif}\;i \le 2.03754808603946062 \cdot 10^{169}:\\
\;\;\;\;\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 f(double i, double n) {
        double r101781 = 100.0;
        double r101782 = 1.0;
        double r101783 = i;
        double r101784 = n;
        double r101785 = r101783 / r101784;
        double r101786 = r101782 + r101785;
        double r101787 = pow(r101786, r101784);
        double r101788 = r101787 - r101782;
        double r101789 = r101788 / r101785;
        double r101790 = r101781 * r101789;
        return r101790;
}
↓
double f(double i, double n) {
        double r101791 = i;
        double r101792 = -1.0591314467102691e-51;
        bool r101793 = r101791 <= r101792;
        double r101794 = 100.0;
        double r101795 = 1.0;
        double r101796 = n;
        double r101797 = r101791 / r101796;
        double r101798 = r101795 + r101797;
        double r101799 = pow(r101798, r101796);
        double r101800 = r101799 - r101795;
        double r101801 = r101794 * r101800;
        double r101802 = r101801 / r101797;
        double r101803 = 8143434713.669914;
        bool r101804 = r101791 <= r101803;
        double r101805 = r101795 * r101791;
        double r101806 = 0.5;
        double r101807 = 2.0;
        double r101808 = pow(r101791, r101807);
        double r101809 = r101806 * r101808;
        double r101810 = log(r101795);
        double r101811 = r101810 * r101796;
        double r101812 = r101809 + r101811;
        double r101813 = r101805 + r101812;
        double r101814 = r101808 * r101810;
        double r101815 = r101806 * r101814;
        double r101816 = r101813 - r101815;
        double r101817 = r101816 / r101791;
        double r101818 = r101794 * r101817;
        double r101819 = r101818 * r101796;
        double r101820 = 2.0375480860394606e+169;
        bool r101821 = r101791 <= r101820;
        double r101822 = 1.0;
        double r101823 = r101811 + r101822;
        double r101824 = r101805 + r101823;
        double r101825 = r101824 - r101795;
        double r101826 = r101825 / r101797;
        double r101827 = r101794 * r101826;
        double r101828 = r101821 ? r101802 : r101827;
        double r101829 = r101804 ? r101819 : r101828;
        double r101830 = r101793 ? r101802 : r101829;
        return r101830;
}

Error

Bits error versus i

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original43.1
Target43.0
Herbie23.3
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 3 regimes

  2. if i < -1.0591314467102691e-51 or 8143434713.669914 < i < 2.0375480860394606e+169

    1. Initial program 31.9

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

    3. Applied associate-*r/31.9

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

    if -1.0591314467102691e-51 < i < 8143434713.669914

    1. Initial program 50.6

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

    3. Applied div-inv50.6

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

    4. Applied associate-/r*50.3

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

    5. Taylor expanded around 0 17.0

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

    7. Applied associate-/r/17.0

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

    8. Applied associate-*r*17.0

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

    9. Simplified17.0

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

    if 2.0375480860394606e+169 < i

    1. Initial program 31.6

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

    2. Taylor expanded around 0 35.8

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

  4. Final simplification23.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \le -1.05913144671026914 \cdot 10^{-51}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 8143434713.6699142:\\ \;\;\;\;\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\\ \mathbf{elif}\;i \le 2.03754808603946062 \cdot 10^{169}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019198 
(FPCore (i n)
  :name "Compound Interest"

  :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))))