Linux/bash does not execute the executable that "which" tells me Observation: I have an executable named foo, located in /b/foo. It is compiled against an old header of a dynamic library, causing it to segfault when executed:  $ foo Segmentation fault. // Expected behaviour.   Now, I compile a new version of foo to /a/foo against the new dynamic library that should execute just fine. Directory a/ is in my $PATH before b/, so /a/foo should be selected:  $ which foo /a/foo   When I execute foo, the following happens:  $ foo Segmentation fault.   Therefore, it seems that /b/foo gets executed, whereas "which" tells me that /a/foo should be executed. To make things weirder, when I run the full path $(which /a/foo), things run fine:  $ /a/foo OK!  $ cp /a/foo . $ ./foo OK!     To go yet one step further, if I now delete /a/foo:  $ rm /a/foo   Then /b/foo must surely be chosen, right?  $ which foo /b/foo $ foo bash: /a/foo: No such file or directory $ $(which foo) Segmentation fault. // Expected result.   Nope!  Fix: Source .bash_profile and .bashrc and the problem disappears.  Reproducibility: Every time. Just remove /a/foo, source ~/.bash_profile, create /a/foo, and the above observation re-occurs.  Question: Does anyone have an idea what went wrong here?  Hypothesis: "which" is up-to-date, but the system chooses based on "what was used to be the case". In my example above, /a/foo did not yet exist when the terminal was opened: I had only just created it. Therefore, when /a/foo was created, "which" did detect /a/foo, but the system still chose /b/foo, because it was somehow out of sync? But why is the system out of sync?
Why not use "which"? What to use then? When looking for the path to an executable or checking what would happen if you enter a command name in a Unix shell, there's a plethora of different utilities (which, type, command, whence, where, whereis, whatis, hash, etc).  We often hear that which should be avoided. Why? What should we use instead?
3-4 old computers = general purpose cluster? Non-existence of a bijective analytic function between annulus and punctured disk Let $\mathbb{F}$ be a finite field.  Let $a_1,\dots,a_n \in \mathbb{F}$ be given.  I want to know whether there exists $x_1,\dots,x_n \in \mathbb{F}$ such that  $$a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2 = 1.$$  My question: Can we characterize when this equation has a solution, i.e., by some criterion on $a_1,\dots,a_n$?  Is there a clean characterization, in terms of $a_1,\dots,a_n$ and the properties of $\mathbb{F}$?    Here's what I can see immediately.  First, if $n=1$, there is an easy answer (the equation has a solution iff $a_1$ is a square in $\mathbb{F}$).  Second, if any $a_i$ is a square in $\mathbb{F}$, the equation has a solution (set all other $x_j$'s to $0$), so we can assume that all the $a_i$'s are non-squares and that $n \ge 2$.  Third, if $-1$ is a square in $\mathbb{F}$, then the equation has a solution: we can take $x_2 =  \sqrt{-a_1/a_2} \cdot x_1$.  But what about the case where $-1$ is a non-square in $\mathbb{F}$?  Is it always guaranteed to have a solution in this case? inequality with three variables and condition How to run scripts without typing the full path? I'd like to know if there is a way to run  program/shell script without typing full path:  /opt/idea/bin/idea.sh How can I prove  from first principles that $0!$ is equal to $1$? For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  We need to use induction.   The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:   $P(n_o)$ $( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$   Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.  Suppose,  $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.  $\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$  The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.  $(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$  $\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$  We need to demonstrate that   $(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  holds true for $n \geq 2$  The question is do I just substitute a number that is greater than two after I did the induction? The Frobenius Coin Problem I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:  $$\lim_{n \to \infty} nx_n=0$$  My idea is: proving this fact aims to prove that given $\epsilon &gt;0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n &gt;n_0$ we have $|nx_n|&lt;\epsilon$.  Now, I know that $\sum x_n$ converges, so that given $\epsilon'&gt;0$ there's some $k_0 \in \mathbb{N}$ such that if $k &gt; k_0$ we have:  $$\left|\sum_{i=1}^{k}x_i - S\right|&lt;\epsilon'$$  Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 &gt; \cdots &gt; x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:  $$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$  I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?  Thanks very much in advance for your help. SHA-256 "almost unique"? I have seen numerous references on the internet of people describing SHA-256 as generating an "almost unique" hash.  . there are more.  Is there some mathematical basis to the almost uniqueness? I would have thought that as SHA-256 is an evenly-distibuted (?) hash algorithm which can accept any message (infinitely many) as an input and produces a fixed length (finite) hash - then there will in fact be an infinite number of each possible output - which I would think hardly qualifies as "almost unique".  Is there a basis for this statement? Let's say I have some data, and then I fit the data with a model (a non-linear regression). Then I calculate the R-squared ($R^2$).  When R-squared is negative, what does that mean? Does that mean my model is bad? I know the range of $R^2$ can be [-1,1]. When $R^2$ is 0, what does that mean as well? Does "Root A" follow the fact about Takatsuki and the One-Eyed Owl? In Root A, Takatsuki does seem to be fascinated by Kaneki's name, but,      I didn't at all expect her to be the One-Eyed Owl.   I just looked her up on the  and it says that       she is the One Eyed-Owl as well as the horror novelist.   Is this true only in the manga or is it also true in Root A? I am performing a meta-analysis with 137 study arms. From each arm I have extracted a proportion of a disease. Here are my proportions:  [1] 0.002710027 0.000000000 0.000000000 0.000000000 0.017857143 0.000000000 0.010269576 0.000000000 0.000000000 0.000000000 0.000000000  [12] 0.010000000 0.053691275 0.052631579 0.070351759 0.000000000 0.108333333 0.262626263 0.013043478 0.000000000 0.007000000 0.000000000  [23] 0.000000000 0.000000000 0.240000000 0.000000000 0.033333333 0.000000000 0.000000000 0.000000000 0.000000000 0.013201320 0.000000000  [34] 0.005361930 0.002308403 0.000000000 0.000000000 0.000000000 0.030000000 0.046296296 0.000000000 0.020000000 0.045454545 0.161073826  [45] 0.000000000 0.000000000 0.000000000 0.095238095 0.007194245 0.001472754 0.000000000 0.000000000 0.000000000 0.000000000 0.012048193  [56] 0.000000000 0.048780488 0.000000000 0.086253369 0.033898305 0.011966701 0.004162331 0.062790698 0.005555556 0.004451039 0.287356322  [67] 0.083333333 0.000000000 0.000000000 0.014666667 0.006160164 0.023809524 0.000000000 0.000000000 0.000000000 0.001552795 0.102564103  [78] 0.030555556 0.000000000 0.000000000 0.000000000 0.000000000 0.029411765 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000  [89] 0.003231018 0.000000000 0.037500000 0.040247678 0.000000000 0.000000000 0.007299270 0.014598540 0.000000000 0.015118790 0.000000000 [100] 0.000000000 0.005555556 0.000000000 0.000000000 0.000000000 0.003039514 0.042145594 0.250000000 0.307692308 0.324444444 0.002057613 [111] 0.008333333 0.000000000 0.027522936 0.022598870 0.000000000 0.000000000 0.000000000 0.135278515 0.033333333 0.000000000 0.000000000 [122] 0.003690037 0.003412969 0.000000000 0.017391304 0.004237288 0.033210332 0.000000000 0.008928571 0.038834951 0.030303030 0.142857143 [133] 0.170000000 0.000000000 0.112903226 0.000000000 0.007299270   A majority of the proportions has zero value. I am struggling with whether I should go for logit or arcsine transformation. One source says that arcsine transformation should not be used . On the other hand, I several zero counts. With logit transformation values 0 and 1 are mapped to infinity, and according to another source I should go with arcsine transformation . Furthermore, I found another study close to my topic in which they used arcsine transformation. The default in R with metaprop function is logit.   Which one I should go with my data? I love the  that was recently added onto Stack Exchange. However, this feature seems to be missing in per-site metas.  For instance:  Server Fault Meta:    DBA Meta    The parent websites, do have the edit/improve this post link however. Displaying row/column of point in raster using QGIS? Is it possible in QGIS to display column and row index of pixel in raster?   I mean something similar to value tool. For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $ I proved as follows:  Let $f(x) = \tan x -  x - \frac{x^{3}}{3}.$ Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x &gt; \tan x - x,$ since $\\tan x &gt; x,$ $f''(x) &gt; 0$ so $f'(x)$ is increasing. Since $f '(0) = 0,$ $f '(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$ So $f(x)$ is increasing. since $f(0) = 0,$ $f(x) &gt; 0$ for all $x \in (0, \frac{\pi}{2}).$  But my proof seems complex. Is there any simpler way to prove this? What part of a multi-type damage roll is reduced by a non-type-specific effect? Effects like  (PHB, p. 54) and Spirit Shield (XGtE, p.  10) reduce damage by a roll, irrespective of damage  (PHB 196). However, single damage rolls can have multiple damage types, as confirmed (for wording) by Jeremy Crawford in .   How do you know which part of the damage is reduced?   This could actually make a difference to the amount of damage taken if, for example, the creature taking damage is resistant to one or more of the damage types involved.       Cutting Words: ... When a creature that you can see within 60 feet of you makes an attack roll, an ability check, or a damage roll, you can use your reaction to expend one of your uses of Bardic Inspiration, rolling a Bardic Inspiration die and subtracting the number rolled from the creature's roll.   Using this ability to expand upon the examples found in  and :     Barny, an Ice Devil (MM, p.  75) hits Bobby, rolling 9 slashing damage and 11 cold damage.      Bobby's ally, Vera, is a level 10 bard of the college of Lore, and uses her Cutting Words ability on the attack's damage roll. She rolls a 9, which is subtracted from the Barny's damage roll.      Bobby is wearing his Boots of the Winterlands (DMG, p.  156), giving him resistance to cold.   We know that "resistance and then vulnerability are applied after all other modifiers to damage" (PHB, p.  197) and both slashing damage and cold damage are treated as "one big damage roll" (see the JC tweet), so is all the 9 slashing damage removed by cutting words, and then the cold damage halved for 5 total damage, or is 9 of the cold damage removed by cutting words, leaving 2 to be halved for a total of 1 + 9 = 10 damage taken by Bobby?   Who makes this decision?  When an effect (like cutting words) reduces damage without specifying type from a source containing multiple damage types, what part of the damage is reduced? A. $\lim_{n\to\infty}a_nb_n=?$ given that $\lim_{n\to\infty}a_n=0$ and $ \lim_{n\to\infty}b_n=\infty$  B. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=0$  C. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n= \infty$    These seem to be difficult situations I come across when I have to evaluate limits.   Some examples,    $\lim_{n\to\infty}n^a\sin(1/n^b)$ with $a,b&gt;1$   $\lim_{n\to\infty}{n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty}{1/n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty} nf(x_n)$ [where we only know that $\lim \ f(x_n) = 0]$    My first question is : Are the two situations  A and B  equivalent ? I feel that they are equivalent as $\lim_{n\to\infty}b_n=\infty \iff \lim_{n\to\infty}{1\over b_n}=0$, The problem is that I'm not sure if we can apply the product rule for limits to take the limit inside.  My second question is : What are some ways to solve such problems ? I only know that we can apply stolz- cesaro in some cases. I am non-EU citizen working in Germany. Can I travel to UK without a visa? I am from Non-EU country (from Nepal to be precise) and recently moved to Germany for work. I am applying for BlueCard. I would like to travel to UK later this year, can I go there without visa as I am resident in another EU country?  I can't find this information online. Any links/references will be much appreciated. DNS Resolution for example.com - Can you explain?