Is there a non-gendered word for fishermen, that doesn't feel as self-conscious as fisherpersons or as folksy as fisherfolk?
What is the unisex form of a word like fisherman? Do you have to use fisherman and fisherwoman separately, or is fisherperson acceptable? I couldn’t find a dictionary with the word …  In general, what do you do when a word does not have a unisex form?
How to read value of variable, where the name of variable is the value of another variable I think an example is in order. Consider I have these two environment variables setup (by a Continuous Delivery tool): PACKAGE_VER=1.2.3 and USE_VERSION_FROM=PACKAGE_VER.  What I want now is a script that can't predict the name of the variable pointing out 1.2.3, only USE_VERSION_FROM. And so, I'm looking for a way to get my script to know the version to use (1.2.3) without knowing anything other than that USE_VERSION_FROM points out the name of the variable that holds the version number.  My naïve non-working first attempt:  #!/bin/bash VERSION_TO_USE="${${USE_VERSION_FROM}}" Why does a cork ball float to the side of a glass as illustrated in the following GIF? What is the physical phenomenon behind this observation and why does it happen? Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. Sharepoint 2013. Add JS to NewForm.aspx of a list What training do doctors/Healers receive above their school education?   A fresh Hogwarts graduate does not seem to have all the skills required to be a Healer, so s/he would need to be learning those skills somehow/someplace. So how/where do they get those skills?  Also, there is one instance of a doctor being praised for inventing a healing spell. Is there any research system, like funding, trials etc or any other source of innovation? Real valued analytic function defined on a connected set is constant Let $G$ be a connected set and $f : G \rightarrow \mathbb{C}$ a real valued analytic function. Prove that $f$ is constant.    My idea to prove the result is to prove a subset $A \neq \varnothing$ of the connected set $G$ is both open and closed. So $G=A$ Take $f(w) = a$ $$A = \{z\colon z \in G, f(z) = a\}$$ Now I want to show that $A$ is infinite. How to do it? After that it is easy to prove $A=G$. Is Switching the Neutral OK? What’s the worst that can happen if i put a switch in the neutral instead of the phase?   I just found out that 3 light bulbs sockets in my home  are connected this way, it first came to my mind to fix it but then I recalled the flow of current from positive to negative is just a convention since the electrons are charged negatively and they actually flow from N to L so i cant really find an solid excuse to mess with the wiring for the next 2 hrs Is there a robust command line tool for processing csv files? Specify a field in attribute table with raster calculator Consider a common proof of the uncountability of $(0,1]$, as presented  for example:  We assume that the reals can be arranged in a sequence $x_k$, represent every number in $x_k$ by its nonterminating decimal expansion, use the diagonal argument to create a sequence of decimal digits that differs from the nonterminating decimal expansion of each number in $x_k$ and then claim that this sequence of digits so generated must be the decimal expansion of some number not in $x_k$.  My problem is with this last step. Could it not be the case that the new sequence of digits is just the terminating decimal expansion of some number already in $x_k$?  Would it not be more correct to use the sequence $x_k$ to generate a sequence of all (both terminating and non-terminating) decimal expansion of all numbers in $[0,1)$ and then apply the diagonal argument? I added a comment to  but something strange is happening. The end code reads "); but SO decided to change that to ";);.    I did not add a ; in front of the ).  To further explain what's happening here, the following:  "http://example.com/." http://example.com/. "http://example.com." http://example.com. "http://example.com" http://example.com   renders as: change color of pgfplot bar plot I want to plot two bar plots next to each other. This works without a problem. However both use the first color defined in a pre or selfdefined colorseries. I tried to modify the color manually in the plot, but the color and fill definitions have no effect:  \documentclass[border=5mm]{standalone} \usepackage{pgfplots, pgfplotstable}  \begin{document}  \pgfplotstableread[header=false, col sep=comma]{  1, first item 5, second item }\datatable  \begin{tikzpicture}   \begin{axis}[             ,xbar     ,bar width=2ex, y=3ex     ,enlarge y limits={abs=0.75}     ,ytick={data} % (\empty,data,{coordinate list})     ,yticklabels from table={\datatable}{1}   ]   \addplot table [     ,red!20!black,fill=red!80!white,     y expr=-\coordindex, % Use negative coordinate index as y coordinate     x index=0 % Use first column as x coordinate   ] {\datatable};   \end{axis} \end{tikzpicture}  \end{document}   The plot is still in the default blue. Word for something sad and funny at the same time I am trying to understand capitalization rules with Mom and Dad. I believe I have it correct below, but please let me know if I do not.     The one thing I learned from my dad was that it was good to earn the trust of one's children. Dad has shown me how good that a dad can be. One day when I went to his house, and Dad wasn't there, I ran outside....   The above passage is just made-up just so that you would see what I mean about capitalization. Let's put numbers by each one, like so:     The one thing I learned from my dad(1) was that it was good to earn the trust of one's children. Dad(2) has shown me how good that a dad(3) can be. One day when I went to his house, and Dad(4) wasn't there, I ran outside....    This is one where I'm not certain. I believe in this sense I'm showing possession here with "my", and it just didn't seem right saying "my Dad". I don't know what the grammar rule is here, or if I even have it right. This one is obvious—it starts the sentence. In this case, I'm using Dad like a name instead of saying "Larry". Therefore, it should be capitalized. Again, just like 3. Prove that $f(x)$ is irreducible iff its reciprocal polynomial $f^*(x)$ is irreducible. This is what I'm trying to prove:     Let $f(x)\in\mathbb{Q}[x]$ and $\deg(f(x))&gt;1$. Prove that $f(x)$ is irreducible in $\mathbb{Q}[x]$ iff its reciprocal polynomial $f^*(x)$ is irreducible in $\mathbb{Q}[x]$.    Note: The reciprocal polynomial $f^*(x)=x^nf(1/x) \in \mathbb{Q}[x]$, where $n=\deg f$.   So my thoughts are to prove the contrapositive, i.e. $f^*(x)$ is reducible iff $f(x)$ is reducible.   So to prove the ($\Rightarrow$) direction I assume that $f^*(x)=g(x)h(x)$, so I get that $g(x)h(x)=x^nf(1/x)$ which implies $f(1/x)=(1/x^n)g(x)h(x)$. I want to say this somehow makes $f(x)$ reducible but I am unable to proceed as $f(1/x)\notin \mathbb{Q}[x] $. I thought about substituting $y=1/x$, which gives $f(y)=y^ng(1/y)h(1/y)$, but I am unable to show that this is irreducible in $\mathbb{Q}[y]$. If I could, then it would be irreducible in $\mathbb{Q}[x]$ since $\mathbb{Q}[y]\cong \mathbb{Q}[x]$   Any suggestions or hints will be appreciated. Can the Hubble constant be measured locally? The Hubble constant, which roughly gauges the extent to which space is being stretched, can be determined from astronomical measurements of galactic velocities (via redshifts) and positions (via standard candles) relative to us. Recently a value of 67.80 ± 0.77 (km/s)/Mpc was published. On the scale of 1 A.U. the value is small, but not infinitesimal by any means (I did the calculation a few months ago, and I think it came out to about 10 meters / year / A.U.). So, can you conceive of a measurement of the Hubble constant that does not rely on any extra-galactic observations?  I ask because, whatever the nature of the expansion described by the Hubble constant, it seems to be completely absent from sub-galactic scales. It is as though the energy of gravitational binding (planets), or for that matter electromagnetic binding (atoms) makes matter completely immune from the expansion of space. The basis for this claim is that if space were also pulling atoms apart, I would naively assume we should be able to measure this effect through modern spectroscopy. Given that we are told the majority of the universe is dark energy, responsible for accelerating the expansion, I wonder, how does this expansion manifest itself locally?   Any thoughts would be appreciated. How do I remove gapps from CyanogenMod I installed CyanogenMod 11 on my HTC One with CyanogenMod Installer. So I got also the google Apps (gapps) on my device. But I need a CM11 without gapps. I did then the following without success:   copy cm-11-20140308-SNAPSHOT-M4-m7.zip to /sdcard/ boot to recovery mode -> recovery -> ClockworkMod Recovery v6.0.4.6: "wipe data/factory reset" "install zip" choosing the zip-file "reboot system now"  So I got a new CM11 but with gapps still installed.  What should I do in addition to "wipe data/factory reset" to get rid of this apps? Covariant derivative ambiguity I'm studying general relativity and am running into an ambiguity with the covariant derivative. The covariant derivative acting on a scalar is, in a co-ordinate basis, simply $$\nabla_X f = X^a \nabla_a f = X^a \partial_a f $$ whilst when acting on vectors or tensors, the covariant derivative includes extra terms involving the connection coefficients $\Gamma$. My problem is that I'm not sure what happens when the covariant derivative acts on the components of a vector. So the $X^a$ in the above expression are each scalar functions, and so when we write $\nabla_Y X^a$, do we assume the covariant derivative acts on $X^a$ as scalars, or do we include connection coefficient terms?  To clarify, consider $$ \nabla_X Y = X^a \nabla_a(Y^b e_b) = X^a (\nabla_a Y^b) e_b + X^a Y^b (\nabla_a e_b) = X^a (\partial_a Y^b) e_b + X^a Y^b \Gamma^c{}_{ba} e_c $$ As you can see, when the covariant derivative acts on the $Y^b$, it acts as a simple partial derivative, since the $Y^b$ are here considered functions premultiplying the vectors $e_b$.   However, in this equation from  page: $$(\nabla_X \nabla_Y Z)^a = X^c \nabla_c Y^b \nabla_b Z^a = X^c Y^b \nabla_c \nabla_b Z^a + (X^c \nabla_c Y^b) \nabla_b Z^a = (\nabla^2_{X,Y} Z)^a + (\nabla_{\nabla_X Y} Z)^a$$ When the covariant derivative acts on the $Y^b$ term it treats it like a vector.  What's going on here? Thanks. What is potential energy truly? I have a problematic question for which I have been unable to attain a satisfactory answer. What is  truly?   -  I have read about how potential energy can be seen as the "highering" of an object within a field, be it gravitational, electromagnetic, etc.   In a way, an object is lifted to a higher position, and when it eventually falls down again energy is released back.  The object does not even have to be "lifted" beforehand, as with the case of asteroids.   However, since energy cannot be destroyed, it must somehow be stored, yet how does energy truly store itself within a field? It seems hard for me to grasp that energy can physically be stored within the positioning of an object?  I can't really get my head around how a field conserves energy within it. Kinetic energy on the other hand, can be, for me, intuitively understood, as the object really is "moving" and energy manifests itself through motion.  Can anyone elaborate on this, perhaps give reasoning as to why a field can in fact "store" energy? Please point out any fallacious or naive views I seem to hold. Proof that the coefficients in an OLS model follow a t-distribution with (n-k) degrees of freedom