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992233
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location Minneapolis, MN
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visits member for 3 years, 8 months
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I have a Ph.D. with a minor in mathematics and a major in statistics.


Jan
28
revised Finding an Expression for the Difference of Roots of the Quadratic Equation
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Jan
28
comment Find and sample minimum of two exponential distribtions
You're mistaken. $1-e^{-(\lambda_1+\lambda_2)x}$ is the probability that the minimum is LESS than $x$, not that it is greater than $x$. ${}\qquad{}$
Jan
28
comment Which significance test can be used in this instance? - paired, non-normal means
We should stop using the term "off topic" for this sort of thing. It's rude.
Jan
28
comment Evaluation of integral $\int_{-\infty}^{+\infty} xe^{-|x|}\,dx$ is not $0$
@MhenniBenghorbal : But that proves at most that a principal value is $0$. Similarly, you can say that $\displaystyle\int_{-N}^N \frac{x\,dx}{1+x^2}=0$, but $\displaystyle\lim_{N\to\infty}\int_{-N}^N \frac{x\,dx}{1+x^2}$ and $\displaystyle\lim_{N\to\infty}\int_{-N}^{2N} \frac{x\,dx}{1+x^2}$ are two different numbers, both finite. Your method doesn't prove that that doesn't happen with the integral in this question. ${}\qquad{}$
Jan
28
answered Evaluation of integral $\int_{-\infty}^{+\infty} xe^{-|x|}\,dx$ is not $0$
Jan
28
revised Evaluation of integral $\int_{-\infty}^{+\infty} xe^{-|x|}\,dx$ is not $0$
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Jan
28
comment Evaluation of integral $\int_{-\infty}^{+\infty} xe^{-|x|}\,dx$ is not $0$
The value of a definite integral with respect to $x$ will never be a function of $x$. ${}\qquad{}$
Jan
28
revised Limit of $L^p$ norm
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Jan
28
comment Prove that any power of a prime is not a perfect number
@JoshuaMyers : Correct. The sum cannot be divisible by $p$ unless the numerator is divisible by $p$, and the numerator differs by $1$ from something divisible by $p$. But the comment by André Nicolas also suffices. ${}\qquad{}$
Jan
28
answered Find and sample minimum of two exponential distribtions
Jan
28
comment Prove that any power of a prime is not a perfect number
Suppose one has $\text{sum}=1+p+p^2+p^3+p^4+p^5$. Then $p\times\text{sum}=p+p^2+p^3+p^4+p^5+p^6$. Now subtract: $(p\times\text{sum}) - \text{sum}$. All of the terms cancel except $p^6$ and $1$, so you get $p^6-1$. But $(p\times\text{sum}) - \text{sum} = (p-1)\times\text{sum}$. Consequently $(p-1)\times\text{sum}=p^6-1$, so $\text{sum}=\dfrac{p^6-1}{p-1}$. The question is now whether that is equal to $p^6$. ${}\qquad{}$
Jan
28
answered Prove that any power of a prime is not a perfect number
Jan
28
revised how to calculate this line integral $\int_{0}^{2\pi} (16\sin^2 3t +16\cos^2 4t)\sqrt{(144\cos^2 3t +256\sin^2 4t)}dt$
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Jan
28
revised Prove the sequence determined by $a_{n+1}={a_n\over \sin a_n}$ is convergent, and found its limit.
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Jan
28
revised Evaluating $\lim_{n \to \infty} \sum\limits_{k=1}^n \frac{1}{k 2^k} $
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Jan
28
comment Find and sample minimum of two exponential distribtions
It's not clear what you mean by $\arg\min$ in this case.
Jan
28
comment Find and sample minimum of two exponential distribtions
Your assertion about the distribution of $\min\{X_1,X_2\}$ is right if $X_1,X_2$ are independent. ${}\qquad{}$
Jan
28
revised A simple question about Delta Method's demonstration.
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Jan
28
revised Easy Analysis question
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Jan
28
comment Can't figure out $O(n \log n)$ divide-and-conquer algorithm
I also changed"$\text{...}$" to "$\ldots$". Also standard. ${}\qquad{}$