Average Error: 43.0 → 22.9
Time: 14.0s
Precision: 64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;n \le -7.76693767745361788861631953913126532955 \cdot 10^{94}:\\ \;\;\;\;{\left(\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\right)}^{1}\\ \mathbf{elif}\;n \le -9.519350012649904306163831713839410413139 \cdot 10^{-251}:\\ \;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\ \mathbf{elif}\;n \le 1.613414883038850832801467887537631845006 \cdot 10^{-130}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\right)}^{1}\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;n \le -7.76693767745361788861631953913126532955 \cdot 10^{94}:\\
\;\;\;\;{\left(\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\right)}^{1}\\

\mathbf{elif}\;n \le -9.519350012649904306163831713839410413139 \cdot 10^{-251}:\\
\;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\

\mathbf{elif}\;n \le 1.613414883038850832801467887537631845006 \cdot 10^{-130}:\\
\;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\

\mathbf{else}:\\
\;\;\;\;{\left(\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\right)}^{1}\\

\end{array}
double f(double i, double n) {
        double r151916 = 100.0;
        double r151917 = 1.0;
        double r151918 = i;
        double r151919 = n;
        double r151920 = r151918 / r151919;
        double r151921 = r151917 + r151920;
        double r151922 = pow(r151921, r151919);
        double r151923 = r151922 - r151917;
        double r151924 = r151923 / r151920;
        double r151925 = r151916 * r151924;
        return r151925;
}
↓
double f(double i, double n) {
        double r151926 = n;
        double r151927 = -7.766937677453618e+94;
        bool r151928 = r151926 <= r151927;
        double r151929 = 100.0;
        double r151930 = 1.0;
        double r151931 = i;
        double r151932 = r151930 * r151931;
        double r151933 = 0.5;
        double r151934 = 2.0;
        double r151935 = pow(r151931, r151934);
        double r151936 = r151933 * r151935;
        double r151937 = log(r151930);
        double r151938 = r151937 * r151926;
        double r151939 = r151936 + r151938;
        double r151940 = r151932 + r151939;
        double r151941 = r151935 * r151937;
        double r151942 = r151933 * r151941;
        double r151943 = r151940 - r151942;
        double r151944 = r151943 / r151931;
        double r151945 = r151929 * r151944;
        double r151946 = r151945 * r151926;
        double r151947 = 1.0;
        double r151948 = pow(r151946, r151947);
        double r151949 = -9.519350012649904e-251;
        bool r151950 = r151926 <= r151949;
        double r151951 = r151931 / r151926;
        double r151952 = r151930 + r151951;
        double r151953 = pow(r151952, r151926);
        double r151954 = r151953 / r151951;
        double r151955 = r151930 / r151951;
        double r151956 = r151954 - r151955;
        double r151957 = r151929 * r151956;
        double r151958 = 1.6134148830388508e-130;
        bool r151959 = r151926 <= r151958;
        double r151960 = r151938 + r151947;
        double r151961 = r151932 + r151960;
        double r151962 = r151961 - r151930;
        double r151963 = r151962 / r151951;
        double r151964 = r151929 * r151963;
        double r151965 = r151959 ? r151964 : r151948;
        double r151966 = r151950 ? r151957 : r151965;
        double r151967 = r151928 ? r151948 : r151966;
        return r151967;
}

Error

Bits error versus i

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original43.0
Target42.8
Herbie22.9
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 3 regimes

  2. if n < -7.766937677453618e+94 or 1.6134148830388508e-130 < n

    1. Initial program 55.3

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

    3. Applied associate-/r/55.0

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

    4. Applied associate-*r*55.0

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

    5. Taylor expanded around 0 21.5

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

    7. Applied pow121.5

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

    8. Applied pow121.5

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

    9. Applied pow121.5

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

    10. Applied pow-prod-down21.5

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

    11. Applied pow-prod-down21.5

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

    if -7.766937677453618e+94 < n < -9.519350012649904e-251

    1. Initial program 23.8

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

    3. Applied div-sub23.9

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

    if -9.519350012649904e-251 < n < 1.6134148830388508e-130

    1. Initial program 35.4

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

    2. Taylor expanded around 0 26.0

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

  4. Final simplification22.9

    \[\leadsto \begin{array}{l} \mathbf{if}\;n \le -7.76693767745361788861631953913126532955 \cdot 10^{94}:\\ \;\;\;\;{\left(\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\right)}^{1}\\ \mathbf{elif}\;n \le -9.519350012649904306163831713839410413139 \cdot 10^{-251}:\\ \;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)\\ \mathbf{elif}\;n \le 1.613414883038850832801467887537631845006 \cdot 10^{-130}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\left(100 \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{i}\right) \cdot n\right)}^{1}\\ \end{array}\]

Reproduce

herbie shell --seed 2019353 
(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))))