Looking for Windows utility to find a pixel of specific color on screen
Get image pixels of prescribed color
Etiquette on sending a thank you e-mail to respondents who gave me helpful information I was rewatching the Breaking Bad episode S04E08 (Hermanos) and I didn't quite understand why exactly Don Eladio killed Gus's partner, but he let Gus live. He says:     The only reason you're alive and he's not, is because I know who you are.   But I still have yet to fully understand. I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:  $u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$  Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$   $\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?   Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$  And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$  We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$  This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?   //not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!   EDIT: corrected notation a little. Same e-mail user, different address? I was wondering why using foo.bar@gmail.com works but not foo.bar@outlook.com.  In Gmail, foo.bar@gmail.com and foobar@gmail.com (and any combination of periods in the username) is equivalent.  Is this a hidden feature in Gmail? How do I change extension of multiple files recursively from the command line? I just designed this MPPT solar charge controller which outputs 4.2 V regulated via MPPT, it uses the SPV1040 IC. This will be used on a satellite with 4 side panels, each with 8 solar cells in parallel outputting 2.6 V at approximately 122 Ma. Each panel will have this exact circuit and all the outputs will be connected in parallel to a 3.7 V lithium ion camera battery.  When active only 1 solar panel will be active and the other MPPT circuits will be offline or providing a very low amount of current. How should I connect the 4.2v to the battery, just directly or using some sort of controller to stop charging at 4.2v? Keep in mind the battery will always have a load and it will charge for around 50m in a 90m cycle due to being eclipsed. I would obviously like the battery to last as long as possible so I am not sure what are the implications of constantly having a battery connected to 4.2 V during 50m. Thanks.   The schematic was made based off a simulation made using the ST Edesignsuite directly from the provider.  Excuse my previous question about this, wasn't properly worded. I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$ By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$  And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\infty$  So I've got an open interval(of convergence) s.t. $|\frac1e 2x|&lt;1\implies |x|&lt;\frac e2$.  Now I need to figure out whether the series is convergent or not at the end points, $x=\frac e2, -\frac e2$.  So there are two series I need to deal with, namely, $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}e^n$ and $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(-1)^ne^n$.  Suppose $(\frac{n}{n+1})^{n^2}$ is increasing. Then $$(\frac{n}{n+1})^{n^2}&lt;(\frac{n+1}{n+2})^{(n+1)^2}\iff 1&lt;(\frac{n+1}{n+2})^{(n+1)^2}(\frac{n+1}{n})^{n^2}$$ $$\iff 1&lt;(1-\frac{1}{n+2})^{2n+1}(1+\frac{1}{n^2+2n})^{n^2}$$  But $(1-\frac{1}{n+2})^{2n+1}\to\frac{1}{e^2}$ and $(1+\frac{1}{n^2+2n})^{n^2}&lt;(1+\frac{1}{n^2})^{n^2}\to e$  So I have $1&lt;\frac1e&lt;1$, which is a contradiction. Hence $(\frac{n}{n+1})^{n^2}$ is not increasing. But I don't know if it's decreasing. (is it true that $(1+\frac{1}{n^2+2n})^{n^2}\to e$?)  If $(1+\frac{1}{n^2+2n})^{n^2}\to e$ is true, then $(\frac{n}{n+1})^{n^2}e^n$ converges to some positive number, so $\sum(\frac{n}{n+1})^{n^2}e^n$ diverges. But I don't know if it's decreasing: can't apply the alternating series convergence test(Leibniz's criterion).  So, can you help me with the two end points? Show that $n$ lines separate the plane into $\frac{n^2+n+2}{2}$ regions Referencing bib files with spaces in the filename We know that the maximum level a dweller can train to on any SPECIAL skill is 10, however this does not consider bonuses given from their outfit.    Is there any additional benefit for going over 10? For example, a dweller with an Intelligence skill of 10 wearing a lab coat that boosts Intelligence by 3?   If it is a hard cap, would it be more useful to use outfits that raise a secondary characteristic instead? From injective map to continuous map Changing content of footline in beamer I think my son was trying to guess the passcode! What do I do?  I can't wait 42 years! Prove that inequality is true for $x>0$: $(e^x-1)\ln(1+x) > x^2$ Nothing shows up in the terminal when I type my password Removing intercept from GLM for multiple factorial predictors only works for first factor in model I am running a binomial logistic regression with a logit link function in R. My response is factorial [0/1] and I have two multilevel factorial predictors - let's call them $a$ and $b$ where $a$ has 4 factor levels  $(a_1,a_2,a_3,a_4)$ and $b$ has 9 factor levels $(b_1,b_2,\dotsc,b_9)$. Therefore:  mod &lt;- glm(y ~ a + b, family = binomial(logit), data=pretend) summary(mod)   The model output would then show all the information about the model as well as the coefficients.  There is a factor level for both a and b missing (a1 and b1) from the summary output. I understand that it is constrained in the "intercept" of the model. I have read that if I want to remove the intercept term and see the estimates for those factor levels I can just add -1 to the model formula, i.e.:  mod2 &lt;- glm(y ~ a + b - 1, family=binomial(logit), data=pretend) summary(mod2)   In the new model (mod2) the intercept term is then gone and variable a's factor-level a1 is given amongst the list of coefficients. But, variable b's factor-level b1 is still missing and given that there is no intercept term anymore, how can I interpret the odds-ratio for that factor level then?  Could someone please explain to me how to get the coefficient for b1 too and why this is happening? Proof that total derivative is the only function that can be added to Lagrangian without changing the EOM So I was reading this:  and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.  So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?   EDIT Ok, I changed my question a little bit:  Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: , equation 14.)  Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&amp;Lifshitz's mechanics book only consider this kind of change in the Lagrangian? In "Seeing like a state" of James C.Scott there is a sentence at the beginning of a paragraph (Acknowledgements xi):     There are a good many scholars whose writings opened up new   perspectives for me or provided outstanding analyses of issues that I   could not have hoped to study so comprehensively on my own.   What worries me is a good many scholars. Is this some type of inversion or ancient English? Is this grammatically right? Why is a here? I thought that appropriate version would be There are many good scholars. Am I wrong? QGIS 3.0 and Qt 5.7 time schedule When will QGIS 3.0 with QT5.7 (possibly) be released as a stable and long term support version? I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$  I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$  I know that $\text{deg}(D)=1+1+1+1-4=0$ and it is clear that $\text{sum}(D)=\infty$  Therefore $D$ is the divisor of a function - we want to find this function  I am given \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}  I am then told that we can express $D$ as follows:  $$D = \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$  However, when I expand this back out, I get the following:  \begin{align}D &amp;= \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\text{div}(y-2x)-\text{div}(x-2)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\left(\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\right)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(0,0)\right]+2\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right] \end{align}  which clearly isn't equivalent to the original $D$ unless $$\left[(2,4)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right]=-4\left[\mathcal{O}\right]$$  Can anyone either explain what I've done wrong or show me how I can obtain the above equation please    This question comes in three parts:   Step 1 - This one Step 2 -  Step 3 -