?

Average Accuracy: 22.1% → 82.2%
Time: 20.1s
Precision: binary64
Cost: 20740

?

\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
\[\begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 5 \cdot 10^{+14}:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
(FPCore (i n)
 :precision binary64
 (if (<= (/ (+ (pow (+ 1.0 (/ i n)) n) -1.0) (/ i n)) 5e+14)
   (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
   (+ (+ 1.0 (* n 100.0)) -1.0)))
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 (((pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= 5e+14) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else {
		tmp = (1.0 + (n * 100.0)) + -1.0;
	}
	return tmp;
}
public static double code(double i, double n) {
	return 100.0 * ((Math.pow((1.0 + (i / n)), n) - 1.0) / (i / n));
}
public static double code(double i, double n) {
	double tmp;
	if (((Math.pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= 5e+14) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) / (i / n));
	} else {
		tmp = (1.0 + (n * 100.0)) + -1.0;
	}
	return tmp;
}
def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))
def code(i, n):
	tmp = 0
	if ((math.pow((1.0 + (i / n)), n) + -1.0) / (i / n)) <= 5e+14:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) / (i / n))
	else:
		tmp = (1.0 + (n * 100.0)) + -1.0
	return tmp
function code(i, n)
	return Float64(100.0 * Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) - 1.0) / Float64(i / n)))
end
function code(i, n)
	tmp = 0.0
	if (Float64(Float64((Float64(1.0 + Float64(i / n)) ^ n) + -1.0) / Float64(i / n)) <= 5e+14)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	else
		tmp = Float64(Float64(1.0 + Float64(n * 100.0)) + -1.0);
	end
	return tmp
end
code[i_, n_] := N[(100.0 * N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] - 1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
code[i_, n_] := If[LessEqual[N[(N[(N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision] + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], 5e+14], N[(100.0 * N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 + N[(n * 100.0), $MachinePrecision]), $MachinePrecision] + -1.0), $MachinePrecision]]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\begin{array}{l}
\mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 5 \cdot 10^{+14}:\\
\;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\

\mathbf{else}:\\
\;\;\;\;\left(1 + n \cdot 100\right) + -1\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original22.1%
Target21.4%
Herbie82.2%
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \]

Derivation?

  1. Split input into 2 regimes
  2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 5e14

    1. Initial program 27.4%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr82.6%

      \[\leadsto 100 \cdot \color{blue}{\left(\frac{{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}}\right)} \]
      Proof

      [Start]27.4

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

      add-cube-cbrt [=>]27.4

      \[ 100 \cdot \frac{\color{blue}{\left(\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1} \cdot \sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right) \cdot \sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}}{\frac{i}{n}} \]

      *-un-lft-identity [=>]27.4

      \[ 100 \cdot \frac{\left(\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1} \cdot \sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right) \cdot \sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\color{blue}{1 \cdot \frac{i}{n}}} \]

      times-frac [=>]27.4

      \[ 100 \cdot \color{blue}{\left(\frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1} \cdot \sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right)} \]

      pow2 [=>]27.4

      \[ 100 \cdot \left(\frac{\color{blue}{{\left(\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}\right)}^{2}}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right) \]

      pow-to-exp [=>]25.8

      \[ 100 \cdot \left(\frac{{\left(\sqrt[3]{\color{blue}{e^{\log \left(1 + \frac{i}{n}\right) \cdot n}} - 1}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right) \]

      expm1-def [=>]25.8

      \[ 100 \cdot \left(\frac{{\left(\sqrt[3]{\color{blue}{\mathsf{expm1}\left(\log \left(1 + \frac{i}{n}\right) \cdot n\right)}}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right) \]

      *-commutative [=>]25.8

      \[ 100 \cdot \left(\frac{{\left(\sqrt[3]{\mathsf{expm1}\left(\color{blue}{n \cdot \log \left(1 + \frac{i}{n}\right)}\right)}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right) \]

      log1p-udef [<=]25.6

      \[ 100 \cdot \left(\frac{{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{{\left(1 + \frac{i}{n}\right)}^{n} - 1}}{\frac{i}{n}}\right) \]
    3. Simplified83.6%

      \[\leadsto 100 \cdot \color{blue}{\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}} \]
      Proof

      [Start]82.6

      \[ 100 \cdot \left(\frac{{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}\right)}^{2}}{1} \cdot \frac{\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}}\right) \]

      /-rgt-identity [=>]82.6

      \[ 100 \cdot \left(\color{blue}{{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}\right)}^{2}} \cdot \frac{\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}}\right) \]

      associate-*r/ [=>]82.5

      \[ 100 \cdot \color{blue}{\frac{{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}\right)}^{2} \cdot \sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}}} \]

      unpow2 [=>]82.5

      \[ 100 \cdot \frac{\color{blue}{\left(\sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)} \cdot \sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}\right)} \cdot \sqrt[3]{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

      rem-3cbrt-lft [=>]83.6

      \[ 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]

    if 5e14 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n))

    1. Initial program 0.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Taylor expanded in n around inf 1.3%

      \[\leadsto \color{blue}{100 \cdot \frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
    3. Simplified61.4%

      \[\leadsto \color{blue}{\frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}} \cdot 100} \]
      Proof

      [Start]1.3

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

      *-commutative [=>]1.3

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

      associate-/l* [=>]1.3

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

      expm1-def [=>]61.4

      \[ \frac{n}{\frac{i}{\color{blue}{\mathsf{expm1}\left(i\right)}}} \cdot 100 \]
    4. Applied egg-rr61.3%

      \[\leadsto \color{blue}{\frac{1}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n \cdot 100}}} \]
      Proof

      [Start]61.4

      \[ \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}} \cdot 100 \]

      associate-*l/ [=>]61.4

      \[ \color{blue}{\frac{n \cdot 100}{\frac{i}{\mathsf{expm1}\left(i\right)}}} \]

      clear-num [=>]61.3

      \[ \color{blue}{\frac{1}{\frac{\frac{i}{\mathsf{expm1}\left(i\right)}}{n \cdot 100}}} \]
    5. Taylor expanded in i around 0 63.3%

      \[\leadsto \frac{1}{\color{blue}{\frac{0.01}{n}}} \]
    6. Applied egg-rr82.1%

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

      [Start]63.3

      \[ \frac{1}{\frac{0.01}{n}} \]

      expm1-log1p-u [=>]39.5

      \[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{1}{\frac{0.01}{n}}\right)\right)} \]

      expm1-udef [=>]58.1

      \[ \color{blue}{e^{\mathsf{log1p}\left(\frac{1}{\frac{0.01}{n}}\right)} - 1} \]

      log1p-udef [=>]58.1

      \[ e^{\color{blue}{\log \left(1 + \frac{1}{\frac{0.01}{n}}\right)}} - 1 \]

      add-exp-log [<=]81.9

      \[ \color{blue}{\left(1 + \frac{1}{\frac{0.01}{n}}\right)} - 1 \]

      clear-num [<=]82.0

      \[ \left(1 + \color{blue}{\frac{n}{0.01}}\right) - 1 \]

      div-inv [=>]82.1

      \[ \left(1 + \color{blue}{n \cdot \frac{1}{0.01}}\right) - 1 \]

      metadata-eval [=>]82.1

      \[ \left(1 + n \cdot \color{blue}{100}\right) - 1 \]
  3. Recombined 2 regimes into one program.
  4. Final simplification83.2%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 5 \cdot 10^{+14}:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \end{array} \]

Alternatives

Alternative 1
Accuracy69.1%
Cost6980
\[\begin{array}{l} \mathbf{if}\;i \leq -25000000:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{n}}\\ \mathbf{elif}\;i \leq -4.4 \cdot 10^{-143} \lor \neg \left(i \leq 4.4 \cdot 10^{-79}\right):\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + 100 \cdot \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \end{array} \]
Alternative 2
Accuracy69.5%
Cost6980
\[\begin{array}{l} \mathbf{if}\;i \leq 2.45 \cdot 10^{-67}:\\ \;\;\;\;100 \cdot \frac{n}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \end{array} \]
Alternative 3
Accuracy60.2%
Cost1097
\[\begin{array}{l} \mathbf{if}\;i \leq -2.1 \cdot 10^{-142} \lor \neg \left(i \leq 1.25 \cdot 10^{-78}\right):\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + n \cdot \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \end{array} \]
Alternative 4
Accuracy60.2%
Cost1097
\[\begin{array}{l} \mathbf{if}\;i \leq -1.6 \cdot 10^{-142} \lor \neg \left(i \leq 2.2 \cdot 10^{-79}\right):\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + 100 \cdot \left(i \cdot \left(0.5 - \frac{0.5}{n}\right)\right)\right)\\ \end{array} \]
Alternative 5
Accuracy56.1%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -2 \cdot 10^{+103} \lor \neg \left(i \leq 1.24 \cdot 10^{-42}\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 6
Accuracy57.0%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -1.45 \lor \neg \left(i \leq 3300000000000\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
Alternative 7
Accuracy60.2%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -4.4 \cdot 10^{-143} \lor \neg \left(i \leq 1.1 \cdot 10^{-79}\right):\\ \;\;\;\;\left(1 + n \cdot 100\right) + -1\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot -50\\ \end{array} \]
Alternative 8
Accuracy2.8%
Cost192
\[i \cdot -50 \]
Alternative 9
Accuracy48.8%
Cost192
\[n \cdot 100 \]

Error

Reproduce?

herbie shell --seed 2023157 
(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))))