?

Average Accuracy: 24.7% → 99.1%
Time: 22.1s
Precision: binary64
Cost: 21896

?

\[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
\[\begin{array}{l} t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\ t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\ \mathbf{if}\;t_1 \leq 2 \cdot 10^{-286}:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;t_1 \leq 1:\\ \;\;\;\;100 \cdot \left(\frac{t_0}{\frac{i}{n}} - \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{1}{n} + \frac{i}{n} \cdot -0.5}\\ \end{array} \]
(FPCore (i n)
 :precision binary64
 (* 100.0 (/ (- (pow (+ 1.0 (/ i n)) n) 1.0) (/ i n))))
(FPCore (i n)
 :precision binary64
 (let* ((t_0 (pow (+ 1.0 (/ i n)) n)) (t_1 (/ (+ t_0 -1.0) (/ i n))))
   (if (<= t_1 2e-286)
     (* 100.0 (/ (expm1 (* n (log1p (/ i n)))) (/ i n)))
     (if (<= t_1 1.0)
       (* 100.0 (- (/ t_0 (/ i n)) (/ n i)))
       (/ 100.0 (+ (/ 1.0 n) (* (/ i n) -0.5)))))))
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 t_0 = pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-286) {
		tmp = 100.0 * (expm1((n * log1p((i / n)))) / (i / n));
	} else if (t_1 <= 1.0) {
		tmp = 100.0 * ((t_0 / (i / n)) - (n / i));
	} else {
		tmp = 100.0 / ((1.0 / n) + ((i / n) * -0.5));
	}
	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 t_0 = Math.pow((1.0 + (i / n)), n);
	double t_1 = (t_0 + -1.0) / (i / n);
	double tmp;
	if (t_1 <= 2e-286) {
		tmp = 100.0 * (Math.expm1((n * Math.log1p((i / n)))) / (i / n));
	} else if (t_1 <= 1.0) {
		tmp = 100.0 * ((t_0 / (i / n)) - (n / i));
	} else {
		tmp = 100.0 / ((1.0 / n) + ((i / n) * -0.5));
	}
	return tmp;
}
def code(i, n):
	return 100.0 * ((math.pow((1.0 + (i / n)), n) - 1.0) / (i / n))
def code(i, n):
	t_0 = math.pow((1.0 + (i / n)), n)
	t_1 = (t_0 + -1.0) / (i / n)
	tmp = 0
	if t_1 <= 2e-286:
		tmp = 100.0 * (math.expm1((n * math.log1p((i / n)))) / (i / n))
	elif t_1 <= 1.0:
		tmp = 100.0 * ((t_0 / (i / n)) - (n / i))
	else:
		tmp = 100.0 / ((1.0 / n) + ((i / n) * -0.5))
	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)
	t_0 = Float64(1.0 + Float64(i / n)) ^ n
	t_1 = Float64(Float64(t_0 + -1.0) / Float64(i / n))
	tmp = 0.0
	if (t_1 <= 2e-286)
		tmp = Float64(100.0 * Float64(expm1(Float64(n * log1p(Float64(i / n)))) / Float64(i / n)));
	elseif (t_1 <= 1.0)
		tmp = Float64(100.0 * Float64(Float64(t_0 / Float64(i / n)) - Float64(n / i)));
	else
		tmp = Float64(100.0 / Float64(Float64(1.0 / n) + Float64(Float64(i / n) * -0.5)));
	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_] := Block[{t$95$0 = N[Power[N[(1.0 + N[(i / n), $MachinePrecision]), $MachinePrecision], n], $MachinePrecision]}, Block[{t$95$1 = N[(N[(t$95$0 + -1.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, 2e-286], 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], If[LessEqual[t$95$1, 1.0], N[(100.0 * N[(N[(t$95$0 / N[(i / n), $MachinePrecision]), $MachinePrecision] - N[(n / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 / N[(N[(1.0 / n), $MachinePrecision] + N[(N[(i / n), $MachinePrecision] * -0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]
100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq 2 \cdot 10^{-286}:\\
\;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\

\mathbf{elif}\;t_1 \leq 1:\\
\;\;\;\;100 \cdot \left(\frac{t_0}{\frac{i}{n}} - \frac{n}{i}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{100}{\frac{1}{n} + \frac{i}{n} \cdot -0.5}\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original24.7%
Target25.9%
Herbie99.1%
\[100 \cdot \frac{e^{n \cdot \begin{array}{l} \mathbf{if}\;1 + \frac{i}{n} = 1:\\ \;\;\;\;\frac{i}{n}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{i}{n} \cdot \log \left(1 + \frac{i}{n}\right)}{\left(\frac{i}{n} + 1\right) - 1}\\ \end{array}} - 1}{\frac{i}{n}} \]

Derivation?

  1. Split input into 3 regimes
  2. if (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 2.0000000000000001e-286

    1. Initial program 26.7%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr99.6%

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

      [Start]26.7

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

      *-un-lft-identity [=>]26.7

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

      pow-to-exp [=>]26.7

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

      expm1-def [=>]39.0

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

      *-commutative [=>]39.0

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

      log1p-def [=>]99.6

      \[ 100 \cdot \frac{1 \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)}{\frac{i}{n}} \]
    3. Simplified99.6%

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

      [Start]99.6

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

      *-lft-identity [=>]99.6

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

    if 2.0000000000000001e-286 < (/.f64 (-.f64 (pow.f64 (+.f64 1 (/.f64 i n)) n) 1) (/.f64 i n)) < 1

    1. Initial program 95.9%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Applied egg-rr11.8%

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

      [Start]95.9

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

      *-un-lft-identity [=>]95.9

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

      pow-to-exp [=>]8.3

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

      expm1-def [=>]11.6

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

      *-commutative [=>]11.6

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

      log1p-def [=>]11.8

      \[ 100 \cdot \frac{1 \cdot \mathsf{expm1}\left(n \cdot \color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right)}\right)}{\frac{i}{n}} \]
    3. Simplified11.8%

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

      [Start]11.8

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

      *-lft-identity [=>]11.8

      \[ 100 \cdot \frac{\color{blue}{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}}{\frac{i}{n}} \]
    4. Applied egg-rr95.8%

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

      [Start]11.8

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

      expm1-udef [=>]8.3

      \[ 100 \cdot \frac{\color{blue}{e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)} - 1}}{\frac{i}{n}} \]

      div-sub [=>]8.1

      \[ 100 \cdot \color{blue}{\left(\frac{e^{n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right)} \]

      *-commutative [=>]8.1

      \[ 100 \cdot \left(\frac{e^{\color{blue}{\mathsf{log1p}\left(\frac{i}{n}\right) \cdot n}}}{\frac{i}{n}} - \frac{1}{\frac{i}{n}}\right) \]

      log1p-udef [=>]8.1

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

      exp-to-pow [=>]95.8

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

      +-commutative [=>]95.8

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

      clear-num [<=]95.8

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

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

    1. Initial program 2.0%

      \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}} \]
    2. Simplified3.5%

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

      [Start]2.0

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

      associate-/r/ [=>]3.5

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

      *-commutative [=>]3.5

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

      *-rgt-identity [<=]3.5

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

      associate-*l* [=>]3.5

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

      *-lft-identity [=>]3.5

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

      sub-neg [=>]3.5

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

      metadata-eval [=>]3.5

      \[ 100 \cdot \left(n \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} + \color{blue}{-1}}{i}\right) \]
    3. Taylor expanded in n around inf 3.5%

      \[\leadsto \color{blue}{100 \cdot \frac{n \cdot \left(e^{i} - 1\right)}{i}} \]
    4. Simplified78.3%

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

      [Start]3.5

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

      associate-*r/ [=>]3.5

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

      associate-/l* [=>]3.5

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

      expm1-def [=>]78.3

      \[ \frac{100}{\frac{i}{n \cdot \color{blue}{\mathsf{expm1}\left(i\right)}}} \]
    5. Taylor expanded in i around 0 97.9%

      \[\leadsto \frac{100}{\color{blue}{\frac{1}{n} + -0.5 \cdot \frac{i}{n}}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification99.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 2 \cdot 10^{-286}:\\ \;\;\;\;100 \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{n}}\\ \mathbf{elif}\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \leq 1:\\ \;\;\;\;100 \cdot \left(\frac{{\left(1 + \frac{i}{n}\right)}^{n}}{\frac{i}{n}} - \frac{n}{i}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{1}{n} + \frac{i}{n} \cdot -0.5}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy81.9%
Cost6980
\[\begin{array}{l} \mathbf{if}\;i \leq 9.8 \cdot 10^{-111}:\\ \;\;\;\;n \cdot \frac{100}{\frac{i}{\mathsf{expm1}\left(i\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} + \frac{-0.5}{n}\right)}\\ \end{array} \]
Alternative 2
Accuracy74.7%
Cost1353
\[\begin{array}{l} \mathbf{if}\;i \leq -2.4 \cdot 10^{-128} \lor \neg \left(i \leq 2.15 \cdot 10^{-134}\right):\\ \;\;\;\;\frac{100}{\frac{1}{n} + i \cdot \left(\frac{0.5}{n \cdot n} + \frac{-0.5}{n}\right)}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100 + i \cdot -50\\ \end{array} \]
Alternative 3
Accuracy68.5%
Cost841
\[\begin{array}{l} \mathbf{if}\;i \leq -0.41 \lor \neg \left(i \leq 1700\right):\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;100 \cdot \left(n + i \cdot \left(n \cdot 0.5\right)\right)\\ \end{array} \]
Alternative 4
Accuracy68.6%
Cost836
\[\begin{array}{l} \mathbf{if}\;i \leq 0.28:\\ \;\;\;\;\frac{100}{\frac{1}{n} + \frac{i}{n} \cdot -0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \end{array} \]
Alternative 5
Accuracy64.2%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -1 \cdot 10^{+78} \lor \neg \left(i \leq 1.2 \cdot 10^{-11}\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot 100\\ \end{array} \]
Alternative 6
Accuracy64.8%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -1.45 \lor \neg \left(i \leq 240\right):\\ \;\;\;\;100 \cdot \frac{i}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
Alternative 7
Accuracy68.5%
Cost713
\[\begin{array}{l} \mathbf{if}\;i \leq -2 \lor \neg \left(i \leq 1700\right):\\ \;\;\;\;\frac{0}{\frac{i}{n}}\\ \mathbf{else}:\\ \;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\ \end{array} \]
Alternative 8
Accuracy3.0%
Cost192
\[i \cdot -50 \]
Alternative 9
Accuracy56.3%
Cost192
\[n \cdot 100 \]

Error

Reproduce?

herbie shell --seed 2023140 
(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))))