Trusting p-values when errors are not normally distributed
Confidence interval interpretation in linear regression when errors are not normally distributed
Happened on , fwiw. In particular I voted to reject it, then visited to add comment and vote to close, and wanted to add a  tag, but couldn't because had to wait to see the fate of the edit. How can I prevent my mail from being classified as spam? The situation is somewhat like-  var someVar = some_other_function(); someObj.addEventListener("click", function(){     some_function(someVar); }, false);   The problem is that the value of someVar is not visible inside the listener function of the addEventListener, where it is probably being treated as a new variable. If $|f(z)|\leq |f(z^2)|$ on the unit disk, show that $f$ is a constant function. Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint? Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$. Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.  This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.  Thanks! What happens when a caster targets an object that looks like a creature with a spell that targets only creatures? I have found the command to delete files older than 5 days in a folder  find /path/to/files* -mtime +5 -exec rm {} \;   But how do I also do this for subdirectories in that folder? Multivariate multiple regression validation Android 5.0 (Lollipop) was officially announced on October 15th, 2014.  You can review the high-level changes at the  and the announcement on the official .  Of course, not all devices will get this update right away and others won't get it at all. Each manufacturer and carrier often choose to add their custom modifications which tend to delay the official release.  When will each device get Android 5.0, either officially or unofficially, if ever?  See also: How to delete the files in one folder which are more than 60 days old in UNIX? Even/Odd Binomial Coefficients I was wondering if there's a nice general solution for the following problem:  How many numbers in the $n^\text{th}$ row of Pascal's triangle are even? How many are odd? I've noticed this issue on a handful of other communities, but making the browser window narrow (mine is 1080x960) makes the .post-layout and #sidebar elements overlap. Maybe something to do with the grid CSS? The #mainbar element seems to be the correct width and changing the .post-layout's overflow-x to auto seems to "fix" it  From retrocomputing:    From this post, because we are meta after all:    Specs:   Google Chrome Version 78.0.3904.108 (Official Build) (64-bit) Ubuntu 18.04.3 LTS If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$ I'm trying to prove the following:     If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n&lt;\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2&lt;\infty$, then $\sum_{n=1}^\infty a_n^2 &lt;\infty$.   The context here is functional analysis homework, in the subject of Hilbert spaces.  Here's what I've thought:  Let $f=(a_n)&gt;0$. Then the problem reads: if $\int f\overline{g}&lt;\infty$ for all $g&gt;0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.  I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle &lt;\infty$ for all $g&gt;0,g\in \ell^2$ then $f\in \ell^2$.  I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P  From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$  but it's not useful here. After update to OS X 10.11 I started having these bugs:   Finder sidebar text flashing  Very high CPU load:      I have Macbook Pro Retina 13" Mid 2014  I reset NVRAM.  I deleted com.apple.finder.plist, it didn't help.  Also I did this: , no longer flashes. How to change programmatically exposed filters in views I have a page view with exposed filters. One of the filters is term vocabulary, and that's dropdown list, user can select only one category (term).   I want to set programmatically one term from dropdown list when I came from specific URL. I tried with views_query_alter where $category is tid.    $set_exposed_input  = array();  $set_exposed_input['field_category_tid'] = $category;   if ($set_exposed_input) {     $view-&gt;set_exposed_input($set_exposed_input);  } Prove that the minimum number of cycles is $m-n+1$ The question I have is:   Prove that the minimum number of cycles is $m-n+1$ in a connected graph. Where one cycle is a path that starts that begins and ends at the same vertex. Where $m$ is edges and $n$ is the vertices.   I have no idea where to start. A few hints would be appreciated. Please do not provide me with the answer. I was amused by the expression "Paid a penny and only farted" (related by @FumbleFingers), which suggested a similar Japanese saying: 大山鳴動鼠一匹 - "Find only a small mouse coming out after hearing rumblings and experiencing shaking of the big mountain."  It’s humorous. I chuckled when I read it, and I wrote to the author that I love it.  With that said, the word "fart" reminds me of another Japanese proverb, 百日の説法屁一つ- "A fart ruins 100 days of sermons by the priest", which means that a small mistake (the sound of a fart in front of the congregation) brings all his efforts to naught.  It can’t happen in big Western churches. But it could well happen in a small wooden temple where a priest would preach to 30 – 50 provincial followers. Can you imagine how the proud priest being called 'the venerable' would be embarrased when his loud fart is heard by all his followers during his serious sermon and they start to giggle here and there in the hall?  We also have a similar saying" 九仞の功を一簣に欠く, which literally means "Collapse a 9- meter sand tower with the last pile of sand."  What is the English equivalent of the saying "A fart ruins 100 days of sermons by the priest (bishop): a small mistake depreciates the value of (vitiates) all labor and effort to naught," preferably in such a humorous way? The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied I recently renewed my passport. I also have an expired passport which has my Canada visa (visitor's) valid for another three years. I have already used my Canadian visa which is multiple entry. I am planning to visit Canada this summer.  Since my valid Canadian visa is in my old/expired passport, do I need to renew/transfer my Canada visa or can I simply travel with valid visa in old passport? monitor files (à la tail -f) in an entire directory (even new ones) I normally  watch many logs in a directory doing tail -f directory/*. The problem is that a new log is created after that, it will not show in the screen (because * was expanded already).  Is there a way to monitor every file in a directory, even those that are created after the process has started?