Average Error: 43.0 → 21.6
Time: 25.1s
Precision: 64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;i \le -0.003542899967172804479714764980258223658893:\\ \;\;\;\;100 \cdot \frac{\frac{{\left(1 + \frac{i}{n}\right)}^{\left(2 \cdot n\right)} - 1 \cdot 1}{{\left(1 + \frac{i}{n}\right)}^{n} + 1}}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 1.164724267812725200244017287332098931074:\\ \;\;\;\;\frac{100}{i} \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{\frac{1}{n}}\\ \mathbf{elif}\;i \le 5.534188978290047690365799985685838194202 \cdot 10^{103}:\\ \;\;\;\;100 \cdot \frac{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;i \le -0.003542899967172804479714764980258223658893:\\
\;\;\;\;100 \cdot \frac{\frac{{\left(1 + \frac{i}{n}\right)}^{\left(2 \cdot n\right)} - 1 \cdot 1}{{\left(1 + \frac{i}{n}\right)}^{n} + 1}}{\frac{i}{n}}\\

\mathbf{elif}\;i \le 1.164724267812725200244017287332098931074:\\
\;\;\;\;\frac{100}{i} \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)}{\frac{1}{n}}\\

\mathbf{elif}\;i \le 5.534188978290047690365799985685838194202 \cdot 10^{103}:\\
\;\;\;\;100 \cdot \frac{\log \left(e^{{\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 r130206 = 100.0;
        double r130207 = 1.0;
        double r130208 = i;
        double r130209 = n;
        double r130210 = r130208 / r130209;
        double r130211 = r130207 + r130210;
        double r130212 = pow(r130211, r130209);
        double r130213 = r130212 - r130207;
        double r130214 = r130213 / r130210;
        double r130215 = r130206 * r130214;
        return r130215;
}
↓
double f(double i, double n) {
        double r130216 = i;
        double r130217 = -0.0035428999671728045;
        bool r130218 = r130216 <= r130217;
        double r130219 = 100.0;
        double r130220 = 1.0;
        double r130221 = n;
        double r130222 = r130216 / r130221;
        double r130223 = r130220 + r130222;
        double r130224 = 2.0;
        double r130225 = r130224 * r130221;
        double r130226 = pow(r130223, r130225);
        double r130227 = r130220 * r130220;
        double r130228 = r130226 - r130227;
        double r130229 = pow(r130223, r130221);
        double r130230 = r130229 + r130220;
        double r130231 = r130228 / r130230;
        double r130232 = r130231 / r130222;
        double r130233 = r130219 * r130232;
        double r130234 = 1.1647242678127252;
        bool r130235 = r130216 <= r130234;
        double r130236 = r130219 / r130216;
        double r130237 = r130220 * r130216;
        double r130238 = 0.5;
        double r130239 = pow(r130216, r130224);
        double r130240 = r130238 * r130239;
        double r130241 = log(r130220);
        double r130242 = r130241 * r130221;
        double r130243 = r130240 + r130242;
        double r130244 = r130237 + r130243;
        double r130245 = r130239 * r130241;
        double r130246 = r130238 * r130245;
        double r130247 = r130244 - r130246;
        double r130248 = 1.0;
        double r130249 = r130248 / r130221;
        double r130250 = r130247 / r130249;
        double r130251 = r130236 * r130250;
        double r130252 = 5.534188978290048e+103;
        bool r130253 = r130216 <= r130252;
        double r130254 = r130229 - r130220;
        double r130255 = exp(r130254);
        double r130256 = log(r130255);
        double r130257 = r130256 / r130222;
        double r130258 = r130219 * r130257;
        double r130259 = r130242 + r130248;
        double r130260 = r130237 + r130259;
        double r130261 = r130260 - r130220;
        double r130262 = r130261 / r130222;
        double r130263 = r130219 * r130262;
        double r130264 = r130253 ? r130258 : r130263;
        double r130265 = r130235 ? r130251 : r130264;
        double r130266 = r130218 ? r130233 : r130265;
        return r130266;
}

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
Target42.4
Herbie21.6
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 4 regimes

  2. if i < -0.0035428999671728045

    1. Initial program 29.5

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

    3. Applied flip--29.5

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

    4. Simplified29.4

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

    if -0.0035428999671728045 < i < 1.1647242678127252

    1. Initial program 50.3

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

    2. Taylor expanded around 0 33.5

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

    4. Applied div-inv33.5

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

    5. Applied *-un-lft-identity33.5

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

    6. Applied times-frac15.4

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

    7. Applied associate-*r*15.7

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

    8. Simplified15.7

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

    if 1.1647242678127252 < i < 5.534188978290048e+103

    1. Initial program 33.7

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

    3. Applied add-log-exp33.7

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

    4. Applied add-log-exp34.0

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

    5. Applied diff-log34.0

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

    6. Simplified34.0

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

    if 5.534188978290048e+103 < i

    1. Initial program 32.2

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

    2. Taylor expanded around 0 36.1

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

  4. Final simplification21.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \le -0.003542899967172804479714764980258223658893:\\ \;\;\;\;100 \cdot \frac{\frac{{\left(1 + \frac{i}{n}\right)}^{\left(2 \cdot n\right)} - 1 \cdot 1}{{\left(1 + \frac{i}{n}\right)}^{n} + 1}}{\frac{i}{n}}\\ \mathbf{elif}\;i \le 1.164724267812725200244017287332098931074:\\ \;\;\;\;\frac{100}{i} \cdot \frac{\left(1 \cdot i + \left(0.5 \cdot {i}^{2} + \log 1 \cdot n\right)\right) - 0.5 \cdot \left({i}^{2} \cdot \log 1\right)}{\frac{1}{n}}\\ \mathbf{elif}\;i \le 5.534188978290047690365799985685838194202 \cdot 10^{103}:\\ \;\;\;\;100 \cdot \frac{\log \left(e^{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{\left(1 \cdot i + \left(\log 1 \cdot n + 1\right)\right) - 1}{\frac{i}{n}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019209 
(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))))