Average Error: 47.3 → 17.3
Time: 14.7s
Precision: 64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;i \le -1.17879906071750874 \cdot 10^{-7}:\\ \;\;\;\;100 \cdot \frac{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 8.74609462986419547 \cdot 10^{-16}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \mathbf{else}:\\ \;\;\;\;\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;i \le -1.17879906071750874 \cdot 10^{-7}:\\
\;\;\;\;100 \cdot \frac{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}}\\

\mathbf{elif}\;i \le 8.74609462986419547 \cdot 10^{-16}:\\
\;\;\;\;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{else}:\\
\;\;\;\;\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n\\

\end{array}
double f(double i, double n) {
        double r116963 = 100.0;
        double r116964 = 1.0;
        double r116965 = i;
        double r116966 = n;
        double r116967 = r116965 / r116966;
        double r116968 = r116964 + r116967;
        double r116969 = pow(r116968, r116966);
        double r116970 = r116969 - r116964;
        double r116971 = r116970 / r116967;
        double r116972 = r116963 * r116971;
        return r116972;
}
↓
double f(double i, double n) {
        double r116973 = i;
        double r116974 = -1.1787990607175087e-07;
        bool r116975 = r116973 <= r116974;
        double r116976 = 100.0;
        double r116977 = 1.0;
        double r116978 = n;
        double r116979 = r116973 / r116978;
        double r116980 = r116977 + r116979;
        double r116981 = pow(r116980, r116978);
        double r116982 = r116981 - r116977;
        double r116983 = exp(r116982);
        double r116984 = log(r116983);
        double r116985 = r116984 / r116979;
        double r116986 = r116976 * r116985;
        double r116987 = 8.746094629864195e-16;
        bool r116988 = r116973 <= r116987;
        double r116989 = r116977 * r116973;
        double r116990 = 0.5;
        double r116991 = 2.0;
        double r116992 = pow(r116973, r116991);
        double r116993 = r116990 * r116992;
        double r116994 = log(r116977);
        double r116995 = r116994 * r116978;
        double r116996 = r116993 + r116995;
        double r116997 = r116989 + r116996;
        double r116998 = r116992 * r116994;
        double r116999 = r116990 * r116998;
        double r117000 = r116997 - r116999;
        double r117001 = r117000 / r116973;
        double r117002 = r117001 * r116978;
        double r117003 = r116976 * r117002;
        double r117004 = r116982 / r116973;
        double r117005 = r116976 * r117004;
        double r117006 = r117005 * r116978;
        double r117007 = r116988 ? r117003 : r117006;
        double r117008 = r116975 ? r116986 : r117007;
        return r117008;
}

Error

Bits error versus i

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original47.3
Target47.0
Herbie17.3
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 3 regimes

  2. if i < -1.1787990607175087e-07

    1. Initial program 29.1

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

    3. Applied add-log-exp29.1

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

    4. Applied add-log-exp29.1

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

    5. Applied diff-log29.1

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

    6. Simplified29.1

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

    if -1.1787990607175087e-07 < i < 8.746094629864195e-16

    1. Initial program 57.8

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

    2. Taylor expanded around 0 27.1

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

    4. Applied associate-/r/9.2

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

    if 8.746094629864195e-16 < i

    1. Initial program 32.5

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

    3. Applied associate-/r/32.5

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

    4. Applied associate-*r*32.5

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

  4. Final simplification17.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \le -1.17879906071750874 \cdot 10^{-7}:\\ \;\;\;\;100 \cdot \frac{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 8.74609462986419547 \cdot 10^{-16}:\\ \;\;\;\;100 \cdot \left(\frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i} \cdot n\right)\\ \mathbf{else}:\\ \;\;\;\;\left(100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{i}\right) \cdot n\\ \end{array}\]

Reproduce

herbie shell --seed 2020100 
(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))))