I'm trying to find a word that encompasses both these ideas. I thought pharisaic would be a bark up the right tree, but it appears to pertain more to hypocrisy and legalism.  Ideally the word should encapsulate the idea that some form of good is being displayed for the primary purpose of self-gain.  I understand that such an action could be hypocritical, but that word does not necessarily imply the act of doing good only for the praise it brings.  The Pharisees are a great example of this (with their ostentatious tithing), but the word named after them doesn't seem to have been applied to that aspect of their character.
There are many examples of this, and I'd like to give a few:   A person who puts a lot of effort to help the community and earns reputation points. But that reputation is the motivation behind helping the community, not because they really want to. A student approaches a professor, and helps them in their teaching and doing research in order to have a good LOR (letter of recommendation). To a SEO, the content of the website must be good (well, this one is a little tricky, but you get the spirit.)   I'm not saying that the reputation/LOR/high-ranked is not deserved. It is. I just want to focus more on the selfish reason(s) behind that.  Any kind of answer can be accepted: single word, phrase, proverb or idiom.
OrderBy and OrderByDescending are stable? Normality of dependent variable = normality of residuals? Express partial derivatives of second order (and the Laplacian) in polar coordinates $z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$  Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$  I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...  $ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question.   Full Question: My intuition tells me it's a trade off between speed and security, but how did the standardisation process select these three seemingly arbitrary key lengths (namely, AES-128, AES-192, AES-256). I know of 2 basic kinds of default web icons for websites:   : place an .ico formatted file at the root of your site named favicon.ico and most desktop browsers will find it and display it in various contexts (bookmarks, tabs). Can also be added in code, which allows you to specify other names and other graphics types to the file, though there are :  &lt;link rel="shortcut icon" href="/favicon.ico" /&gt; &lt;link rel="icon" href="/favicon.ico" /&gt;  : place a png formatted file called either apple-touch-icon.png or apple-touch-icon-precomposed.png at the root of your site and iPhone OS/iOS versions of Safari will pick those up and store them when the site is added to the Home Screen. Can also be added with code like:  &lt;link rel="apple-touch-icon" href="/custom_icon.png" /&gt;    Are there any other default icon graphics for a website a webmaster can include? How to show that the spherical metric satisfies the triangle inequality? For $x,y\in \mathbb R^n$ define  $$d(x,y)={\|x-y\| \over \sqrt{1+\|x\|^2} \sqrt{1+\|y\|^2}}$$ Here $\|x\|$ is the euclidean norm of a vector. How to prove that $d$ (the spherical metric) is indeed a metric?     Progress so far:    $d(x,y)\ge 0$ is obvious. $d(x,y) =0 \iff \|x-y\|=0 \iff x=y$, so the positivity holds. $d(x,y) = d(y,x)$ is clear from the formula, so symmetry holds.    But I am having difficulties with the triangle inequality. Writing it out in coordinates leads to a complicated inequality with square roots all over the place in denominators. Is there a better way? Can't pair apple wireless keyboard Keyboard: Apple Wireless Keyboard (model number MC184LL/B)  System: Macbook Pro Early 2011 running OS X 10.8  The keyboard gets detected and when I try to pair it, the system asks to enter some numbers through the keyboard.   However when I press the key (on the wireless keyboard) nothing happens, the only key that seems to work (i.e. to move to the next key in the code) is 0. I can also press enter to confirm but it always fails (since the only code I can enter is 0000...0).   I've also tried pairing from the bluetooth preference and changing the passcode option. However this won't work either.   Any ideas? The keyboard was paired with another MacBook Pro (I don't have access to it anymore). Is there any method to 'reset' the keyboard to factory settings? How can I retrieve mp3 metadata in Python? Should I put a comma before the last item in a list? Should I put a comma before the last item in a list?     I would like crackers, cheese and some soda.   I would like crackers, cheese, and some soda. Proof a set cannot contain its own powerset - Russell's Paradox NSArray becomes NSCFArray when passed How to restore file copy / transfer status window? $\chi(G)+\chi(G')\leq n+1$ How to prove, that the sum of chromatic numbers of graph and it's complement is smaller then the number of vertices incremented by one?  $\chi(G)+\chi(G')\leq n+1$  The notes from my classes say to color the first graph using sequential algorithm with vertices ordered by their degrees (the ones with lowest appear first). We would get $k_1$ colors, then we color the graph $G'$ in the backwards order and get $k_2$ colors, and the sum $k_1+k_2$ in my notes is magically smaller than $n+1$, but I just don't see why... How can I typeset this accented, dotless i? See the rough sketch above. I am able to typeset the accented i shown in the middle (for which I used the \v macro) and, obviously, the plain and simple i shown on the left also. However, I can't figure out how to typeset the one shown on the right. How can I do that? Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ I simplified to get $6x^4-25x^3+12x^2+25x+6=0$  Or $6 (x^2+1)^2+25x(1-x^2)=0$  Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$  I can't find a way to proceed further. Overriding "+=" in Python? (__iadd__() method) Bootstrapped confidence intervals: different values at every computation? I am running an ANOVA and computing bootstrapped CIs for my effect sizes. I am using for this the Measures of Effect Size Toolbox for Matlab.  To my surprise, if I re-run the same line of code, i.e. compute the same thing for the same data, the lower&amp;upper margins of the CI are slightly different each time.  I know that bootstrapped CIs are built from permutation analyses based on random generations of distributions. Still, I do not remeber reading anywhere this odd property of CIs of being never twice the same! Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$ I met an inequality, I ask, do not mathematical induction to prove that:     Prove \[ \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \] without using induction Possible Beamer bug: Frame number has different color than date in infolines footer I use TeX Live on Mac and I get my weekly updates through the TeX Live Utility. I am working on some slides for an upcoming talk. I am using the same Beamer templates and settings that I always use, so I was surprised to see that the page numbers that I always include at the bottom of my slides suddenly appeared in a different color. Here's an MWE:  \documentclass{beamer} \useoutertheme{infolines} \usecolortheme{beaver} \title{Title} \date{\today} %\setbeamercolor{page number in head/foot}{fg=darkred!60!black} \begin{document} \begin{frame} \end{frame} \end{document}     As you can see in the bottom right corner of the slide the page number appears but its color looks off. Until recently it would look exactly the same color as the date appearing next to it.  By Googling around a bit I managed to figure out how to get this to look normal again using the line commented out in the code above:  \setbeamercolor{page number in head/foot}{fg=darkred!60!black}   By uncommenting this line I got back the desired behavior:    So my question is not so much about how to fix this problem, but about why this is happening in the first place. Is this a bug that was introduced in a recent update, and if so will it be fixed? Is this a deliberate choice on the part of the developers (if so I can't imagine why)? Or am I doing something wrong or is something broken in my TeX setup?  I'd appreciate any insight on this. Thanks in advance! JMustache attribution is incorrectly listed as Butterknife