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network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 4 months |
| seen | Apr 14 at 14:08 | |
| stats | profile views | 34 |
undergrad physics student
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Jun 16 |
accepted | Chain relation in three variables |
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Jun 16 |
asked | Evaluation of a product of sines |
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Apr 15 |
asked | Chain relation in three variables |
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Apr 11 |
accepted | Improper integral of $\dfrac{x}{e^x-1}$ |
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Apr 10 |
asked | Improper integral of $\dfrac{x}{e^x-1}$ |
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Apr 10 |
awarded | Editor |
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Apr 10 |
revised |
Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$ deleted 1 characters in body |
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Apr 5 |
accepted | Probability of throwing the same multiset twice in a row with six dice |
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Mar 28 |
comment |
Probability of throwing the same multiset twice in a row with six dice Yes, it helps in the sense that now it is immediate to conclude, but is there some slick way to do this? I mean, how can one produce your table during an examination, when time is limited? (this is part of an exercise taken from an exam paper in probability) |
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Mar 28 |
asked | Probability of throwing the same multiset twice in a row with six dice |
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Mar 9 |
accepted | Random walk problem |
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Mar 8 |
comment |
Random walk problem About your second hint: the condition we need is that the number of left moves is equal for the two drunks, or that the number of left moves of the first + the number of right moves of the second is equal to N. The number of all possible walks is $4^N$. Call $n$ the number of left moves of the first drunk. We can choose these $n$ moves in $(N\,\,\,\, n)$ ways. The number of left moves of the second drunk must be $n$ too and so the probability is $\sum_{n=0}^N=(N \,\,\,\, n)^2/4^N$, that is $4^{-N}(2N\,\,\,\, N)$. Is this right? Thanks |
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Mar 8 |
comment |
Random walk problem Yes, it is just what I thought and following that idea I found the summation above. Can you rewrite it in a nicer form? |
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Mar 8 |
accepted | How to calculate the principal part of improper integral? |
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Mar 8 |
accepted | Find locus of points in the plane |
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Mar 8 |
accepted | Evaluate improper integral $(\cos(2x)-1)/x^2$ |
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Mar 8 |
asked | Random walk problem |
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Feb 19 |
comment |
Evaluate improper integral $(\cos(2x)-1)/x^2$ @GEdgar: oh yes, thanks for pointing that out. |
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Feb 19 |
comment |
Evaluate improper integral $(\cos(2x)-1)/x^2$ @GEdgar: the absolute value along the large semicircle is less then $(2\pi R)$ times the maximum of the absolute value of the integrand, that is $2/R^2$. As $R\to\infty$, the all thing goes to zero... |
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Feb 19 |
comment |
Evaluate improper integral $(\cos(2x)-1)/x^2$ Now we are rescued by the half residue theorem that states that the last integral is $\pi i$ times the residue in zero, which is $2i$. Change the sign because the semicircle is oriented negatively. So, the integral is $-2\pi$. Done. Yet, can you state more precisely where my previous argument goes wrong? |
