Average Error: 47.5 → 11.6
Time: 25.5s
Precision: binary64
\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}\]
↓
\[\begin{array}{l} \mathbf{if}\;i \leq -0.00018743494896283588:\\ \;\;\;\;100 \cdot \frac{e^{i} - 1}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 0.0006241617893643045:\\ \;\;\;\;100 \cdot \left(\left(0.5 \cdot \left(i \cdot n\right) + \left(0.16666666666666666 \cdot \left(n \cdot {i}^{2}\right) + \left(n + 0.3333333333333333 \cdot \frac{{i}^{2}}{n}\right)\right)\right) - \left(0.5 \cdot {i}^{2} + i \cdot 0.5\right)\right)\\ \mathbf{elif}\;i \leq 5.375287223785503 \cdot 10^{+191}:\\ \;\;\;\;100 \cdot \frac{\left(0.5 \cdot \left({\log n}^{2} \cdot {n}^{2}\right) + \left(n \cdot \log i + \left(\frac{{n}^{2}}{i} + \left(\frac{\log i \cdot {n}^{3}}{i} + \left(0.5 \cdot \left({\log n}^{2} \cdot \left(\log i \cdot {n}^{3}\right)\right) + \left(0.5 \cdot \left({n}^{2} \cdot {\log i}^{2}\right) + 0.16666666666666666 \cdot \left({n}^{3} \cdot {\log i}^{3}\right)\right)\right)\right)\right)\right)\right) - \left(0.5 \cdot \frac{{n}^{3}}{{i}^{2}} + \left(n \cdot \log n + \left(\log n \cdot \left({n}^{2} \cdot \log i\right) + \left(\frac{\log n \cdot {n}^{3}}{i} + \left(0.16666666666666666 \cdot \left({n}^{3} \cdot {\log n}^{3}\right) + 0.5 \cdot \left(\log n \cdot \left({n}^{3} \cdot {\log i}^{2}\right)\right)\right)\right)\right)\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 2.018574178061804 \cdot 10^{+227}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{0}{\frac{i}{n}}\\ \end{array}\]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
↓
\begin{array}{l}
\mathbf{if}\;i \leq -0.00018743494896283588:\\
\;\;\;\;100 \cdot \frac{e^{i} - 1}{\frac{i}{n}}\\

\mathbf{elif}\;i \leq 0.0006241617893643045:\\
\;\;\;\;100 \cdot \left(\left(0.5 \cdot \left(i \cdot n\right) + \left(0.16666666666666666 \cdot \left(n \cdot {i}^{2}\right) + \left(n + 0.3333333333333333 \cdot \frac{{i}^{2}}{n}\right)\right)\right) - \left(0.5 \cdot {i}^{2} + i \cdot 0.5\right)\right)\\

\mathbf{elif}\;i \leq 5.375287223785503 \cdot 10^{+191}:\\
\;\;\;\;100 \cdot \frac{\left(0.5 \cdot \left({\log n}^{2} \cdot {n}^{2}\right) + \left(n \cdot \log i + \left(\frac{{n}^{2}}{i} + \left(\frac{\log i \cdot {n}^{3}}{i} + \left(0.5 \cdot \left({\log n}^{2} \cdot \left(\log i \cdot {n}^{3}\right)\right) + \left(0.5 \cdot \left({n}^{2} \cdot {\log i}^{2}\right) + 0.16666666666666666 \cdot \left({n}^{3} \cdot {\log i}^{3}\right)\right)\right)\right)\right)\right)\right) - \left(0.5 \cdot \frac{{n}^{3}}{{i}^{2}} + \left(n \cdot \log n + \left(\log n \cdot \left({n}^{2} \cdot \log i\right) + \left(\frac{\log n \cdot {n}^{3}}{i} + \left(0.16666666666666666 \cdot \left({n}^{3} \cdot {\log n}^{3}\right) + 0.5 \cdot \left(\log n \cdot \left({n}^{3} \cdot {\log i}^{2}\right)\right)\right)\right)\right)\right)\right)}{\frac{i}{n}}\\

\mathbf{elif}\;i \leq 2.018574178061804 \cdot 10^{+227}:\\
\;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\

\mathbf{else}:\\
\;\;\;\;100 \cdot \frac{0}{\frac{i}{n}}\\

\end{array}
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
↓
(FPCore (i n)
 :precision binary64
 (if (<= i -0.00018743494896283588)
   (* 100.0 (/ (- (exp i) 1.0) (/ i n)))
   (if (<= i 0.0006241617893643045)
     (*
      100.0
      (-
       (+
        (* 0.5 (* i n))
        (+
         (* 0.16666666666666666 (* n (pow i 2.0)))
         (+ n (* 0.3333333333333333 (/ (pow i 2.0) n)))))
       (+ (* 0.5 (pow i 2.0)) (* i 0.5))))
     (if (<= i 5.375287223785503e+191)
       (*
        100.0
        (/
         (-
          (+
           (* 0.5 (* (pow (log n) 2.0) (pow n 2.0)))
           (+
            (* n (log i))
            (+
             (/ (pow n 2.0) i)
             (+
              (/ (* (log i) (pow n 3.0)) i)
              (+
               (* 0.5 (* (pow (log n) 2.0) (* (log i) (pow n 3.0))))
               (+
                (* 0.5 (* (pow n 2.0) (pow (log i) 2.0)))
                (* 0.16666666666666666 (* (pow n 3.0) (pow (log i) 3.0)))))))))
          (+
           (* 0.5 (/ (pow n 3.0) (pow i 2.0)))
           (+
            (* n (log n))
            (+
             (* (log n) (* (pow n 2.0) (log i)))
             (+
              (/ (* (log n) (pow n 3.0)) i)
              (+
               (* 0.16666666666666666 (* (pow n 3.0) (pow (log n) 3.0)))
               (* 0.5 (* (log n) (* (pow n 3.0) (pow (log i) 2.0))))))))))
         (/ i n)))
       (if (<= i 2.018574178061804e+227)
         (/ (* 100.0 (- (pow (+ 1.0 (/ i n)) n) 1.0)) (/ i n))
         (* 100.0 (/ 0.0 (/ i n))))))))
double code(double i, double n) {
	return 100.0 * ((pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}
↓
double code(double i, double n) {
	double tmp;
	if (i <= -0.00018743494896283588) {
		tmp = 100.0 * ((exp(i) - 1.0) / (i / n));
	} else if (i <= 0.0006241617893643045) {
		tmp = 100.0 * (((0.5 * (i * n)) + ((0.16666666666666666 * (n * pow(i, 2.0))) + (n + (0.3333333333333333 * (pow(i, 2.0) / n))))) - ((0.5 * pow(i, 2.0)) + (i * 0.5)));
	} else if (i <= 5.375287223785503e+191) {
		tmp = 100.0 * ((((0.5 * (pow(log(n), 2.0) * pow(n, 2.0))) + ((n * log(i)) + ((pow(n, 2.0) / i) + (((log(i) * pow(n, 3.0)) / i) + ((0.5 * (pow(log(n), 2.0) * (log(i) * pow(n, 3.0)))) + ((0.5 * (pow(n, 2.0) * pow(log(i), 2.0))) + (0.16666666666666666 * (pow(n, 3.0) * pow(log(i), 3.0))))))))) - ((0.5 * (pow(n, 3.0) / pow(i, 2.0))) + ((n * log(n)) + ((log(n) * (pow(n, 2.0) * log(i))) + (((log(n) * pow(n, 3.0)) / i) + ((0.16666666666666666 * (pow(n, 3.0) * pow(log(n), 3.0))) + (0.5 * (log(n) * (pow(n, 3.0) * pow(log(i), 2.0)))))))))) / (i / n));
	} else if (i <= 2.018574178061804e+227) {
		tmp = (100.0 * (pow((1.0 + (i / n)), n) - 1.0)) / (i / n);
	} else {
		tmp = 100.0 * (0.0 / (i / n));
	}
	return tmp;
}

Error

Bits error versus i

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original47.5
Target47.1
Herbie11.6
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}}\]

Derivation

  1. Split input into 5 regimes

  2. if i < -1.8743494896283588e-4

    1. Initial program 28.4

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

    2. Taylor expanded around inf 11.4

      \[\leadsto 100 \cdot \frac{\color{blue}{e^{i}} - 1}{\frac{i}{n}}\]

    if -1.8743494896283588e-4 < i < 6.24161789364304528e-4

    1. Initial program 58.0

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

    2. Taylor expanded around 0 9.3

      \[\leadsto 100 \cdot \color{blue}{\left(\left(0.5 \cdot \left(i \cdot n\right) + \left(0.16666666666666666 \cdot \left({i}^{2} \cdot n\right) + \left(0.3333333333333333 \cdot \frac{{i}^{2}}{n} + n\right)\right)\right) - \left(0.5 \cdot {i}^{2} + 0.5 \cdot i\right)\right)}\]

    if 6.24161789364304528e-4 < i < 5.37528722378550314e191

    1. Initial program 31.1

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

    2. Taylor expanded around 0 15.7

      \[\leadsto 100 \cdot \frac{\color{blue}{\left(0.5 \cdot \left({\log n}^{2} \cdot {n}^{2}\right) + \left(\log i \cdot n + \left(\frac{{n}^{2}}{i} + \left(\frac{\log i \cdot {n}^{3}}{i} + \left(0.5 \cdot \left({\log n}^{2} \cdot \left({n}^{3} \cdot \log i\right)\right) + \left(0.5 \cdot \left({\log i}^{2} \cdot {n}^{2}\right) + 0.16666666666666666 \cdot \left({n}^{3} \cdot {\log i}^{3}\right)\right)\right)\right)\right)\right)\right) - \left(0.5 \cdot \frac{{n}^{3}}{{i}^{2}} + \left(\log n \cdot n + \left(\log n \cdot \left(\log i \cdot {n}^{2}\right) + \left(\frac{\log n \cdot {n}^{3}}{i} + \left(0.16666666666666666 \cdot \left({\log n}^{3} \cdot {n}^{3}\right) + 0.5 \cdot \left(\log n \cdot \left({n}^{3} \cdot {\log i}^{2}\right)\right)\right)\right)\right)\right)\right)}}{\frac{i}{n}}\]

    if 5.37528722378550314e191 < i < 2.0185741780618041e227

    1. Initial program 33.2

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

    3. Applied associate-*r/_binary64_138433.1

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

    if 2.0185741780618041e227 < i

    1. Initial program 33.8

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

    2. Taylor expanded around 0 29.0

      \[\leadsto 100 \cdot \frac{\color{blue}{1} - 1}{\frac{i}{n}}\]
  3. Recombined 5 regimes into one program.

  4. Final simplification11.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;i \leq -0.00018743494896283588:\\ \;\;\;\;100 \cdot \frac{e^{i} - 1}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 0.0006241617893643045:\\ \;\;\;\;100 \cdot \left(\left(0.5 \cdot \left(i \cdot n\right) + \left(0.16666666666666666 \cdot \left(n \cdot {i}^{2}\right) + \left(n + 0.3333333333333333 \cdot \frac{{i}^{2}}{n}\right)\right)\right) - \left(0.5 \cdot {i}^{2} + i \cdot 0.5\right)\right)\\ \mathbf{elif}\;i \leq 5.375287223785503 \cdot 10^{+191}:\\ \;\;\;\;100 \cdot \frac{\left(0.5 \cdot \left({\log n}^{2} \cdot {n}^{2}\right) + \left(n \cdot \log i + \left(\frac{{n}^{2}}{i} + \left(\frac{\log i \cdot {n}^{3}}{i} + \left(0.5 \cdot \left({\log n}^{2} \cdot \left(\log i \cdot {n}^{3}\right)\right) + \left(0.5 \cdot \left({n}^{2} \cdot {\log i}^{2}\right) + 0.16666666666666666 \cdot \left({n}^{3} \cdot {\log i}^{3}\right)\right)\right)\right)\right)\right)\right) - \left(0.5 \cdot \frac{{n}^{3}}{{i}^{2}} + \left(n \cdot \log n + \left(\log n \cdot \left({n}^{2} \cdot \log i\right) + \left(\frac{\log n \cdot {n}^{3}}{i} + \left(0.16666666666666666 \cdot \left({n}^{3} \cdot {\log n}^{3}\right) + 0.5 \cdot \left(\log n \cdot \left({n}^{3} \cdot {\log i}^{2}\right)\right)\right)\right)\right)\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq 2.018574178061804 \cdot 10^{+227}:\\ \;\;\;\;\frac{100 \cdot \left({\left(1 + \frac{i}{n}\right)}^{n} - 1\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \frac{0}{\frac{i}{n}}\\ \end{array}\]

Reproduce

herbie shell --seed 2021045 
(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))))