Is this T or F? and most importantly, why?  I'll be using any answers for a basis or completely my other questions, since my understanding  is still a little poor.
I am trying to prove:     $a\mid b \land b\mid c \Rightarrow a\mid c $   $a\mid b$ means that a divides b if there is an integer k, that $b=k\cdot a$  Please give me a hint on how to start, because I have no idea.
How can I get rid of those little white dots? I am using Blender 2.70a. I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) I stumbled upon  on last weekend, it introduces an integral trick that exploits . On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral $$ \int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2} $$ using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.     Here are what I have tried so far:   A first instinct would be do something like:$$ I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx $$ for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode: $$ \frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)} $$ which gives: $$ I(b) = \frac{C}{\sqrt{f(b)}}. $$ However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework. A second try involves an exercise on the same page: $$ I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx. $$ Taking derivative and rescaling the integral using change of variable we have: $$ I'(b) = -2I(b). $$ This gives us another impossible to solve constant $C$ in: $$ I(b) = C e^{-2b} $$ without leaving this framework yet again. The third try is trying modify Américo Tavares's answer in this : $$ I(b) = \int^\infty_0 be^{-b^2x^2}\,dx. $$ It is easy to show that: $$ I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0 $$ by an integration by parts identity: $$ \int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx . $$ Then $I(b) = C$, ouch, stuck again at this constant.     Notice in that  question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.  Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"? My question is how I can intergrate Salesforce with different FTP servers in real time. I have read some articles about it, but to be absolutely sure I thought it would be a good idea to post a question here.  I want to be able to send and receive data from Salesforce as a csv file to a ftp server. How I'm thinking about doing this is listed below.  Send the data:  Create a apex class which query the necessary data in Salesforce and create a csv file. I guess this is trivial? Then send the file to a web service created in Salesforce, just by calling it in the apex code. Create a web service in Salesforce which receives the data from the apex code. This web service should in some way send the data to the appropriate ftp server. I guess I can simply modify some of the code provided by Salesforce to make this work? The file is sent. Receive the data  This one is a bit trickier. I guess I have to pull the data from the FTP server by using my web service, which then send the file collected to the appropriate apex class. Or, the FTP server could send the data to the SFDC web service, I don't know if this is possible? The web services on the other side that sends the data to the ftp server, could send it directly to the SFDC web service, which then again forward it to the apex class. I guess the web service will use the remote site url to connect to the SFDC web service? Is Myopia Reversible/Curable? Why are vias placed this way on a PCB? I used to check complex commercial PCBs specially those of graphics cards to see how professional PCB designers do their layout and learn from their techniques.  When I checked the card shown below I noticed two things regarding the placement of vias:  (A higher resolution image is ).   The PCB is surrounded by stitching vias all around the edges. What's the role of all of these? I think they're connected to ground to act as a shield, if that's true, I can't understand technically how by this placement they achieve this shield? By looking closer to the mounting holes, I noticed they added vias all around the pad, why? I added a user account to the admin group and discovered I added wrong user. So I now have to remove the account from the admin group.   How can I remove an account from a group without deleting the user? Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.  Let $a, b \in \mathbb R$. Find the limit $$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$ In my experience, input type="text" onchange event usually occurs only after you leave (blur) the control.  Is there a way to force browser to trigger onchange every time textfield content changes? If not, what is the most elegant way to track this “manually”?  Using onkey* events is not reliable, since you can right-click the field and choose Paste, and this will change the field without any keyboard input.  Is setTimeout the only way?.. Ugly :-) No such file or directory but I can see it! I'm trying to run a python script, on a headless Raspberry PI using winSCP and get the following error message:  Command '"./areadetect_movie_21.py"' failed with return code 127 and error message /usr/bin/env: python : No such file or directory.   When I try and run from terminal, I get:  : No such file or directory.   I try a similar python script, in the same directory, with the same python shebang, the same permissions and using the same user pi, and it works.  I also do a ls and I can see the file, so I don't know why it will not run. Sieve of eratosthenes infinite primes I don't understand why the sieve of eratosthenes doesn't prove an infinite number of primes.  For example the number 2 eliminates 1/2 of an infinite number of numbers.  The next prime, 3, eliminates 1/3 of the remainder, not 1/3 of all the numbers and so on.  5 would eliminate 1/5 of the remaining numbers. Why doesn't that prove an infinite number of primes? Thanks 1) Up until now, during practical work sessions, I always used these formulas for uncertainty propagation:  if $C = A+B$  or  $C = A-B$ $$\Delta C = \Delta A + \Delta B$$ if $C = AB$ or $C = \frac{A}{B}$ $$\frac{\Delta C}{C} = \frac{\Delta A}{A}  + \frac{\Delta B}{B}$$ if $C = A^m$ $$\frac{\Delta C}{C} = |m|\frac{\Delta A}{A}$$   These formulas are derived from the expression for the differential of a function of multiple variables:  if $C = f(A,B)$ $$ dC = \frac{\partial C}{\partial A}dA + \frac{\partial C}{\partial B}dB \Rightarrow \Delta C = \frac{\partial C}{\partial A}\Delta A + \frac{\partial C}{\partial B}\Delta B$$  and I think that makes sense because the value C is just a function of two variables that happen to be measurements.  2) But this morning, I was told that this is wrong and I should actually use these instead:  if $C = A+B$  or  $C = A-B$ $$(\Delta C)^2 = (\Delta A)^2 + (\Delta B)^2$$ if $C = AB$ or $C = \frac{A}{B}$ $$\left(\frac{\Delta C}{C}\right)^2 = \left(\frac{\Delta A}{A}\right)^2  + \left(\frac{\Delta B}{B}\right)^2$$  which are apparently derived from a general formula:  $$(\Delta C)^2  = \left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...$$  or   $$\Delta C  = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 +   \hspace{0.3cm}...}$$  3) So:   Why is this formula better?  Where does it come from? What does it actually represent? (Do I recognize the shape of a norm in that last formula?)  What is wrong with the formula of the differential?   I'm asking a lot, but the propagation of uncertainty always confused the ship out of me. Converting MultiLinestring to Linestring with PostGIS I have a multiline string and want to convert it to a linestring in order to use some functions that only work with linestrings. I used ST_Dump() and got a collection of my geometries. If I merge them back together, I still get the multilinestring. Should I get the Npoints and combine them together or something, in order to end up with a simple linestring? ST_LineMerge() also does not work on my multilinestring, it returns the same. my example,  MULTILINESTRING((-3.16420835153456 55.9269166007097,-3.164222 55.926918),(-3.1642070167833 55.9269296196706,-3.16421351659546 55.9268662214904),(-3.16421351659546 55.9268662214904,-3.16421636372824 55.9268384509897),(-3.16421636372824 55.9268384509897,-3.16422182573761 55.9267851753802),(-3.16422182573761 55.9267851753802,-3.16422870102352 55.926718114886),(-3.16422870102352 55.926718114886,-3.16423309121073 55.926675293667),(-3.16423309121073 55.926675293667,-3.16423565148822 55.9266503211093),(-3.16423565148822 55.9266503211093,-3.16424103159897 55.9265978443265),(-3.16424103159897 55.9265978443265,-3.16424680776317 55.9265415044985),(-3.16424680776317 55.9265415044985,-3.16425267254583 55.9264843002995),(-3.16425267254583 55.9264843002995,-3.16425541048045 55.9264575949012),(-3.16425541048045 55.9264575949012,-3.16426111146586 55.9264019883556),(-3.16426111146586 55.9264019883556,-3.1642667032531 55.9263474469124),(-3.1642667032531 55.9263474469124,-3.16426957768543 55.9263194101362),(-3.16426957768543 55.9263194101362,-3.16427488261739 55.9262676666359),(-3.16427488261739 55.9262676666359,-3.16428009893088 55.9262167875066),(-3.16428009893088 55.9262167875066,-3.164282741107 55.9261910161221),(-3.1642875546472 55.9261440655823,-3.164282741107 55.9261910161221),(-3.1642875546472 55.9261440655823,-3.16429466890915 55.9260746741522),(-3.16429466890915 55.9260746741522,-3.16430092974527 55.9260136069079),(-3.16430092974527 55.9260136069079,-3.16430822838418 55.9259424170929),(-3.16430822838418 55.9259424170929,-3.16431547242401 55.925871759829),(-3.16431547242401 55.925871759829,-3.16431448732505 55.9258328901507),(-3.16431770120536 55.9257327846001,-3.16431547242401 55.925871759829),(-3.164339 55.925777,-3.16431770120536 55.9257327846001)) Indicator Function Here is  an excerpt from "All of Statistics" by Larry Wasserman (page 23):     Given an event A, define the indicator function of A by      $$ I_{A}(\omega)= \color{Red}{I(\omega \in A)} =\begin{cases}  1 &amp;  \text{ if } \omega \in A \\   0 &amp; \text{ if } \omega \notin A   \end{cases} $$   I am not convinced, referring to term colored red, with the notation here. Since this is a widely referred text, I would like to know that whether this is a typo or I am missing something here. Mirror Modifier making ugly smoothing on mirrored faces The model is vivisected right down the middle along the z axis. The right leg is rendering funny and I assume if I can get that fixed everything else will fall into place. I tried flipping normals, changing shading types - can't think of anything else to do. Why are RSA key sizes almost always a power of two? I know that other bit sizes are possible, e.g. this HTTPS server seems to have a 9000 bit key , but it's very rare that one sees a key not of size 1024, 2048, 4096, etc. bits in common usage - what is the reason for this? Is it just customary, or is there a cryptographic advantage? What happens to objects that enter a Stargate from the destination side? We know that matter can only travel one way through a Stargate, ie. from the dialing gate to the destination. What happens to things that enter the wormhole from the destination side?   For example, if someone dials a gate that has tipped on its side (as in the SG-1 episode Hundred Days) and walks through, what happens when he reaches the other side and gravity causes him to immediately fall back into the gate?  Does the matter disappear, or does it pass through to the other side of the destination gate, as if the wormhole were not there?  Related: How can I display "recent folders" in the sidebar of "Save as" dialogue? I was using 14.04 and upgraded to 16.04 yesterday.   In 14.04 I had recent folders on top of the sidebar of the Save as dialogue when I saved something from the net (Firefox if that helps). But that feature disappeared in 16.04 and I really want to have it back.   Here is a picture of what I'm trying to explain:  (the save as dialogue in 16.04 without the recent folders on top) Enter from password field doesn't submit SE login form On  page for a site where I don't yet have an account, if I select the "Log in using Stack Exchange" box, when I hit enter from the password form, nothing happens.  Minor, but... What's the origin of "I'm down with it"? I understand it's an expression of agreement. What exactly does it mean and where did it originate from?