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| bio | website | |
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| visits | member for | 2 years, 4 months |
| seen | 7 hours ago | |
| stats | profile views | 15,003 |
As somebody used to say:
Does research. Smokes. Battles administration. Smokes. Wishes he could stop battling administration so that he could have more time to do research. Smokes some more.
The same. Except I do not smoke.
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Jun 14 |
comment |
Differential equation, perturbation method Are you asked to prove that $y_0(x)=\frac12x^2$ and, for every $\epsilon\gt0$, $y_\epsilon(x)\gt y_0(x)$ for $x\gt0$ and $y_\epsilon(x)\lt y_0(x)$ for $x\lt0$? |
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Jun 14 |
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Proving there are no integer solutions for $3x^2=9+y^3$ In the "alternative proof", the implication $3^2\mid x^2-3\implies 3^2\mid x^2$ is dubious since $3^2$ is not a divisor of $3$. The "initial proof" is correct. |
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Jun 13 |
revised |
Continued fraction fallacy: $1=2?$ added 11 characters in body |
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Jun 13 |
revised |
Finding $\lim\limits_{n \to \infty} \sum\limits_{k=0}^n { n \choose k}^{-1}$ deleted 5 characters in body |
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Jun 13 |
answered | What does $\lim \limits_{n\rightarrow \infty }\sum \limits_{k=0}^{n} {n \choose k}^{-1}$ converge to (if it converges)? |
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Jun 13 |
comment |
$X,Y$ i.i.d., $X$ and $(X+Y)/\sqrt{2}$ have the same dist., then show that $X$ has a normal distribution One does not assume that $X$, $Y$ and $(X+Y)/\sqrt2$ are i.i.d., only that $X$ and $Y$ are. |
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Jun 13 |
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Absolute and conditional convergence of $\sum_{n=1}^\infty\frac{n(x-1)^n}{2^n(3n-1)}$ and $\sum_{n=1}^\infty\frac{1}{2n -1}(\frac{x+2}{x-1})^n$ Tip for a): Determine the radius of convergence of the series $\sum\limits_n\frac{n}{3n-1}z^n$. Tip for b): Determine the radius of convergence of the series $\sum\limits_n\frac1{2n-1}z^n$. |
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Jun 13 |
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Solving two greatest integer function equations Or, more simply, $n\leqslant x\lt n+1$ and $m\leqslant y\lt m+1$ hence $n^2\leqslant39\lt n^2+n$ and $m^2\leqslant68\lt m^2+m$, hence... |
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Jun 13 |
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Solving two greatest integer function equations Insight: $x=n+u$ and $y=m+v$ with $n$ and $m$ integers and $u$ and $v$ in $(0,1)$. Since $xn=n^2+un$ and $ym=m^2+vm$ are integers, $u=i/n$ and $v=j/m$ for some integers $i$ and $j$. Hence one looks for integers $(n,m,i,j)$ with $0\leqslant i\lt n$, $0\leqslant j\lt m$, $n^2=39-i$, $m^2=68-j$. Can you continue? |
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Jun 13 |
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Maximum likelihood estimation and efficiency leonbloy: Hmmm... I am stupid and you are absolutely right. Thanks for the explanation. |
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Jun 13 |
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How to prove that the following iteration process converges? Your expression for $p_{n+1}-p$ is incorrect. |
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Jun 13 |
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Maximum likelihood estimation and efficiency @fgp Re your last comment to your answer, let me quote myself: starting from l'(p). |
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Jun 13 |
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Maximum likelihood estimation and efficiency Right. Note that $\hat p$ is unbiased, for every $n\geqslant1$. |
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Jun 13 |
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Likelihood Function for the Uniform Density. Should one relate this nearly auto-replicating phenomenon to the nearly instantaneous acceptation of answers? |
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Jun 13 |
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How to prove that the following iteration process converges? Thus, the result holds for every $p_0$ close enough to $p$, right? For other starting points, I suggest to write down $p_{n+1}-p$ as a function of $p_n-p$. |
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Jun 13 |
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How to prove that the following iteration process converges? To apply the theorem, you may start by identifying $f$. |
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Jun 13 |
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Maximum likelihood estimation and efficiency leonbloy: The formula at the end is wrong (but the first error is when $p_0$ appears, hence you should trace it easily). |
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Jun 13 |
awarded | complex-numbers |
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Jun 13 |
awarded | homework |
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Jun 13 |
awarded | statistics |
