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Pzi
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Homework Statement
Integrate using Monte-Carlo method and evaluate absolute error.
[tex]\int\limits_1^{ + \infty } {\frac{{\sqrt {\ln (x)} }}{{{x^5}}}dx} [/tex]
Homework Equations
Everything about Monte-Carlo integration I guess.
The Attempt at a Solution
[tex] I = \int\limits_1^{ + \infty } {\frac{{\sqrt {\ln (x)} }}{{{x^5}}}dx} = \int\limits_{\pi /4}^{\pi /2} {\frac{{\sqrt {\ln (\tan (t))} }}{{{{(\tan (t))}^5} \cdot {{(\cos (t))}^2}}}dt} [/tex]
I define random variable Y so that its mean would be equal to my integral I.
[tex] {y_k} = \left( {\frac{\pi }{2} - \frac{\pi }{4}} \right) \cdot \frac{{\sqrt {\ln (\tan ({t_k}))} }}{{{{(\tan ({t_k}))}^5} \cdot {{(\cos ({t_k}))}^2}}} = \frac{{\pi \sqrt {\ln (\tan ({t_k}))} }}{{4{{(\tan ({t_k}))}^5} \cdot {{(\cos ({t_k}))}^2}}} [/tex]
[tex] {t_k} = \left( {\frac{\pi }{2} - \frac{\pi }{4}} \right) \cdot {r_k} + \frac{\pi }{4} = \frac{\pi }{4}\left( {{r_k} + 1} \right) [/tex]
r_k - equally likely random numbers between 0 and 1.
I then proceed using 55 random numbers to calculate my approximation.
[tex] {\hat I_n} = \frac{1}{n}\sum\limits_{k = 1}^n {{y_k}} [/tex]
[tex] I \approx {\hat I_{55}} = \frac{1}{{55}}\sum\limits_{k = 1}^{55} {{y_k}} = 0.{\rm{113481}} [/tex]
What about absolute error? It is meant to be like "it is very likely (probability over 0.9 or whatever) that our approximation missed the actual value by no more than *ERROR NUMBER*".