Math FPCore C Java Python Julia Wolfram TeX \[100 \cdot \frac{{\left(1 + \frac{i}{n}\right)}^{n} - 1}{\frac{i}{n}}
\]
↓
\[\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq 0:\\
\;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\
\mathbf{elif}\;t_1 \leq 100000000:\\
\;\;\;\;n \cdot \left(\frac{t_0}{i \cdot 0.01} + \frac{-100}{i}\right)\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\end{array}
\]
(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 0.0)
(/ (* (expm1 (* n (log1p (/ i n)))) 100.0) (/ i n))
(if (<= t_1 100000000.0)
(* n (+ (/ t_0 (* i 0.01)) (/ -100.0 i)))
(* 100.0 (+ (+ 1.0 n) -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 t_0 = pow((1.0 + (i / n)), n);
double t_1 = (t_0 + -1.0) / (i / n);
double tmp;
if (t_1 <= 0.0) {
tmp = (expm1((n * log1p((i / n)))) * 100.0) / (i / n);
} else if (t_1 <= 100000000.0) {
tmp = n * ((t_0 / (i * 0.01)) + (-100.0 / i));
} else {
tmp = 100.0 * ((1.0 + n) + -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 t_0 = Math.pow((1.0 + (i / n)), n);
double t_1 = (t_0 + -1.0) / (i / n);
double tmp;
if (t_1 <= 0.0) {
tmp = (Math.expm1((n * Math.log1p((i / n)))) * 100.0) / (i / n);
} else if (t_1 <= 100000000.0) {
tmp = n * ((t_0 / (i * 0.01)) + (-100.0 / i));
} else {
tmp = 100.0 * ((1.0 + n) + -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):
t_0 = math.pow((1.0 + (i / n)), n)
t_1 = (t_0 + -1.0) / (i / n)
tmp = 0
if t_1 <= 0.0:
tmp = (math.expm1((n * math.log1p((i / n)))) * 100.0) / (i / n)
elif t_1 <= 100000000.0:
tmp = n * ((t_0 / (i * 0.01)) + (-100.0 / i))
else:
tmp = 100.0 * ((1.0 + n) + -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)
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 <= 0.0)
tmp = Float64(Float64(expm1(Float64(n * log1p(Float64(i / n)))) * 100.0) / Float64(i / n));
elseif (t_1 <= 100000000.0)
tmp = Float64(n * Float64(Float64(t_0 / Float64(i * 0.01)) + Float64(-100.0 / i)));
else
tmp = Float64(100.0 * Float64(Float64(1.0 + n) + -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_] := 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, 0.0], N[(N[(N[(Exp[N[(n * N[Log[1 + N[(i / n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] * 100.0), $MachinePrecision] / N[(i / n), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$1, 100000000.0], N[(n * N[(N[(t$95$0 / N[(i * 0.01), $MachinePrecision]), $MachinePrecision] + N[(-100.0 / i), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(100.0 * N[(N[(1.0 + n), $MachinePrecision] + -1.0), $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 0:\\
\;\;\;\;\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right) \cdot 100}{\frac{i}{n}}\\
\mathbf{elif}\;t_1 \leq 100000000:\\
\;\;\;\;n \cdot \left(\frac{t_0}{i \cdot 0.01} + \frac{-100}{i}\right)\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\end{array}
Alternatives Alternative 1 Error 1.0 Cost 21896
\[\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq 0:\\
\;\;\;\;100 \cdot \left(n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{i}\right)\\
\mathbf{elif}\;t_1 \leq 100000000:\\
\;\;\;\;n \cdot \left(\frac{t_0}{i \cdot 0.01} + \frac{-100}{i}\right)\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\end{array}
\]
Alternative 2 Error 1.0 Cost 21896
\[\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq 0:\\
\;\;\;\;n \cdot \frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{\frac{i}{100}}\\
\mathbf{elif}\;t_1 \leq 100000000:\\
\;\;\;\;n \cdot \left(\frac{t_0}{i \cdot 0.01} + \frac{-100}{i}\right)\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\end{array}
\]
Alternative 3 Error 1.0 Cost 21896
\[\begin{array}{l}
t_0 := {\left(1 + \frac{i}{n}\right)}^{n}\\
t_1 := \frac{t_0 + -1}{\frac{i}{n}}\\
\mathbf{if}\;t_1 \leq 0:\\
\;\;\;\;\frac{n}{\frac{i}{\frac{\mathsf{expm1}\left(n \cdot \mathsf{log1p}\left(\frac{i}{n}\right)\right)}{0.01}}}\\
\mathbf{elif}\;t_1 \leq 100000000:\\
\;\;\;\;n \cdot \left(\frac{t_0}{i \cdot 0.01} + \frac{-100}{i}\right)\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\end{array}
\]
Alternative 4 Error 11.6 Cost 7560
\[\begin{array}{l}
\mathbf{if}\;i \leq 20:\\
\;\;\;\;n \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{100}}\\
\mathbf{elif}\;i \leq 4.8 \cdot 10^{+184}:\\
\;\;\;\;\frac{{\left(1 + \frac{i}{n}\right)}^{n} + -1}{\frac{i}{n}} \cdot 100\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(\left(1 + \frac{n}{i}\right) + -1\right) - \frac{n}{i}\right)\\
\end{array}
\]
Alternative 5 Error 12.0 Cost 6980
\[\begin{array}{l}
\mathbf{if}\;i \leq 20:\\
\;\;\;\;n \cdot \frac{\mathsf{expm1}\left(i\right)}{\frac{i}{100}}\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(\left(\left(1 + \frac{n}{i}\right) + -1\right) - \frac{n}{i}\right)\\
\end{array}
\]
Alternative 6 Error 17.6 Cost 1228
\[\begin{array}{l}
t_0 := 100 \cdot \left(\left(\left(1 + \frac{n}{i}\right) + -1\right) - \frac{n}{i}\right)\\
\mathbf{if}\;i \leq -3.25 \cdot 10^{+240}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;i \leq -5 \cdot 10^{-62}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\mathbf{elif}\;i \leq 1.36 \cdot 10^{+35}:\\
\;\;\;\;50 \cdot \left(i \cdot n\right) + n \cdot 100\\
\mathbf{else}:\\
\;\;\;\;t_0\\
\end{array}
\]
Alternative 7 Error 19.5 Cost 1032
\[\begin{array}{l}
\mathbf{if}\;i \leq -3.1 \cdot 10^{-60}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\mathbf{elif}\;i \leq 7.4:\\
\;\;\;\;50 \cdot \left(i \cdot n\right) + n \cdot 100\\
\mathbf{else}:\\
\;\;\;\;100 \cdot \left(n \cdot \frac{-1}{-i} - \frac{n}{i}\right)\\
\end{array}
\]
Alternative 8 Error 19.2 Cost 840
\[\begin{array}{l}
\mathbf{if}\;i \leq -5.1 \cdot 10^{-62}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\mathbf{elif}\;i \leq 10.8:\\
\;\;\;\;50 \cdot \left(i \cdot n\right) + n \cdot 100\\
\mathbf{else}:\\
\;\;\;\;0\\
\end{array}
\]
Alternative 9 Error 19.8 Cost 712
\[\begin{array}{l}
t_0 := 100 \cdot \left(\left(1 + n\right) + -1\right)\\
\mathbf{if}\;i \leq -8.2 \cdot 10^{-61}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;i \leq 9.5 \cdot 10^{-169}:\\
\;\;\;\;n \cdot 100\\
\mathbf{else}:\\
\;\;\;\;t_0\\
\end{array}
\]
Alternative 10 Error 19.2 Cost 712
\[\begin{array}{l}
\mathbf{if}\;i \leq -1 \cdot 10^{-59}:\\
\;\;\;\;100 \cdot \left(\left(1 + n\right) + -1\right)\\
\mathbf{elif}\;i \leq 6:\\
\;\;\;\;n \cdot \left(100 + i \cdot 50\right)\\
\mathbf{else}:\\
\;\;\;\;0\\
\end{array}
\]
Alternative 11 Error 20.1 Cost 456
\[\begin{array}{l}
\mathbf{if}\;i \leq -2.25 \cdot 10^{+24}:\\
\;\;\;\;0\\
\mathbf{elif}\;i \leq 4:\\
\;\;\;\;n \cdot 100\\
\mathbf{else}:\\
\;\;\;\;0\\
\end{array}
\]
Alternative 12 Error 50.9 Cost 64
\[0
\]