(New page: Regarding Seraj's solution, i believe the solution would be <math>\sqrt(A) cos(''2pi*''drand48())</math>) |
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Regarding Seraj's solution, i believe the solution would be | Regarding Seraj's solution, i believe the solution would be | ||
− | <math>\sqrt(A) cos( | + | <math>\sqrt(A) cos(2pi*drand48())</math> |
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+ | where A is still the exponential RV calculated in 1b. | ||
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+ | - Yes you are correct, I forgot the 2pi factor to change the range of the uniform random variable from [0-1] to [1-2pi]. | ||
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+ | thank you for pointing it out. | ||
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+ | Seraj |
Latest revision as of 15:07, 21 October 2008
Regarding Seraj's solution, i believe the solution would be
$ \sqrt(A) cos(2pi*drand48()) $
where A is still the exponential RV calculated in 1b.
- Yes you are correct, I forgot the 2pi factor to change the range of the uniform random variable from [0-1] to [1-2pi].
thank you for pointing it out.
Seraj