Suppose the following limit is given:
where and are positive integers which are relatively prime. Find .
oindent
<|channel|>commentary to=python code<|message|>import mpmath as mp
def integral_n(n): f = lambda x: (1 + 205x/(69n))(-69*n) * mp.e(-6*x/7) return mp.quad(f, [0, n]) for n in [10, 20, 50, 100, 200]: val = integral_n(n) print(n, val)
<|channel|>commentary to=python code<|message|>from fractions import Fraction Fraction(7,1441) float(Fraction(7,1441))
<|channel|>commentary to=python code<|message|>p = 7 q = 1441 print(p**2 * q)
[THOUGHT] We have the integral [ I_n=\int_{0}^{n}\left(1+\frac{205}{69n}x\right)^{-69n}e^{-\frac67 x},dx. ] For large (n) we can use the limit [ \lim_{n\to\infty}\left(1+\frac{a}{n}\right)^{bn}=e^{ab}, ] so [ \left(1+\frac{205}{69n}x\right)^{-69n} =\exp!\bigl(-69n,\ln(1+\tfrac{205x}{69n})\bigr) \longrightarrow e^{-205x}. ] Thus the integrand tends to [ e^{-205x},e^{-\frac67 x}=e^{-(205+\tfrac67)x} =e^{-\frac{1441}{7}x}. ] The upper limit (n) goes to (\infty), so the limiting integral is [ \int_{0}^{\infty}e^{-\frac{1441}{7}x},dx =\left[,-\frac{7}{1441}e^{-\frac{1441}{7}x},\right]_{0}^{\infty} =\frac{7}{1441}. ] Hence (p=7,;q=1441) (coprime). We need (p^{2}q = 7^{2}\times1441=49\times1441=70609). Thus the answer is (\boxed{70609}).