Perhaps I haven't understood the concept of the "Scaling Workflow" and hence my confusion ...  I checked out most of the Q&amp;A for designing large format images on Photoshop (like for huge billboards), and the most popular suggestions were to "scale" the image down and design at high resolution.  Examples -     For instance, if a billboard is to be 20' wide, the vendor may ask you   to send a file 20 inches wide. If the billboard will be printed at   40ppi, then your 20" wide file would need to be set to 480ppi to allow   for the scaling (12X40).   Source:      30 dpi for final output is more than enough for a billboard of that   size. It's not unusual for final output to be 12-15 dpi in this   context. The usual professional billboard workflow in Photoshop is to   build the image at a small scale with high ppi (e.g., 4x6cm @ 300   ppi),   Source:   How does this help?   For example, let's assume a billboard's print spec is 35 meters (W) x 15 meters (H) at a resolution of 50 DPI / PPI in CMYK. And the billboard designing will be done with a raster image using Photoshop.  So how does it help if you design it at 350 cm x 150 cm @ 500 ppi (scaled to 1/10, for example) or at its actual size of 3500 cm x 1500 cm @ 50 ppi?   Calculation -  A 350 cm x 150 cm @ 500 PPI will have ~ 68,898 x 29,530 pixels (350 cm = 137.795 inch x 500 PPI = 68897.5 and 150 cm = 59.06 inch x 500 PPI = 29,530).  A 3500 cm x 1500 cm @ 50 PPI will also have 68,898 x 29,528 pixels (3500 cm ~ 1377.953 inch x 50 PPI = 68897.65 and 1500 cm ~ 590.56 inch x 50 PPI = 29,528).  In both cases, the file size in Photoshop is going to be approximately the same (around 4+ GB!), since the amount of pixels in both the Photoshop file will be the same! When the document size is the same for both cases, how can Photoshop use less resource for the "scaled" down size? How does it become easier to design using a "scaled" down version?    To be clear, I understand the theory of why a lower resolution works with large images for bill boards.    My question pertains to the "scaling" designing workflow that was suggested for editing large raster images (mainly in photoshop). I fail to see what benefit this approach offers, and hence my query ...
(This question is meant to be a resource to direct others to given that this is a common question.)  Given the task of creating an image that will be printed at a certain size, what pixel dimensions and resolution should I be setting my file at?
As the title states, I'm having quite a lot of trouble with the integral:  $$\int_{-\ln 2}^{\ln 2}  \frac{(\sin x+x)^{1/3}}{e^x}dx$$  The problem is that my standard (admittedly sparse) repertoire of tricks seems to have no effect at all. I can't think of any clever substitution that would work, and it looks like a nasty integral all-round.   I was hoping to apply some symmetry arguments (especially considering the limits of integration), but the exponential function is neither even nor odd, so I don't see how this would work.   I'm afraid I have absolutely no idea how to approach this. Two-Weapon Fighting & Bonus Actions Make your own .sty files Sorry to start a subjective question, but want to understand "own.sty" file.  Other than all \usepackage{} what sort of other commands can be added to the makeyourown.sty file.  Can you include \let command, and \newcommands in it.   What sort of precaution / structure newbies should follow. How many MB/sec can I expect copying to USB 3.0 enclosure with SATA drive? If $(a,b)=1$ then prove $(a+b, ab)=1$. Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.   $(a,b)=1$ means $a$ and $b$ have no prime factors in common  $ab$ is simply the product of factors of $a$ and factors of $b$.  Let's say $k\mid a+b$ where $k$ is some factor of $a$.  Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.  So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].  But $(a,b)=1$ so $a$ cannot divide $b$.  We have a similar argument for $b$.  So $a+b$ is not divisible by any factors of $ab$.  Therefore, $(a+b, ab)=1$.  Would this be correct? Am I missing anything? I am working on a proof about Dedekind cuts on the positive rational numbers. I have been stuck for a while on the following point and would appreciate any help.  Given $x\in \mathbb{Q}^+$ such that $x^2&lt;2$, how to find $y\in \mathbb{Q}^+$ such $y&gt;x$ and $y^2&lt;2$? Label legend interval with number of elements in ArcMap? I have made a series of  in ArcMap 10. When I classify the interval the bottom of the dialogue tells how many elements are in the class.  Is it possible to have this number automatically places beside my class labels? When & How often do 4e Monster powers recharge? How often should I roll power recharge rolls and at which point in the monster's turn do I re-roll them?  I'm trying to find the official definition of the recharge mechanic in the online compendium and I'm not seeing it there. iPhone App CrashReport I am trying to use squared loss to do binary classification on a toy data set. I am using mtcars data set, use mile per gallon and weight to predict transmission type. The plot below shows the two types of transmission type data in different colors, and decision boundary generated by different loss function. The squared loss is $\sum_i (y_i-p_i)^2$ where $y_i$ is the ground truth label (0 or 1) and $p_i$ is the predicted probability $p_i=\text{Logit}^{-1}(\beta^Tx_i)$. In other words, I am replace logistic loss with squared loss in classification setting, other parts are the same. For a toy example with mtcars data, in many cases, I got a model &quot;similar&quot; to logistic regression (see following figure, with random seed 0).  But in somethings (if we do set.seed(1)), squared loss seems not working well.  What is happening here? The optimization does not converge? Logistic loss is easier to optimize comparing to squared loss? Any help would be appreciated.  Code d=mtcars[,c(&quot;am&quot;,&quot;mpg&quot;,&quot;wt&quot;)] plot(d$mpg,d$wt,col=factor(d$am)) lg_fit=glm(am~.,d, family = binomial()) abline(-lg_fit$coefficients[1]/lg_fit$coefficients[3],        -lg_fit$coefficients[2]/lg_fit$coefficients[3]) grid()  # sq loss lossSqOnBinary&lt;-function(x,y,w){   p=plogis(x %*% w)   return(sum((y-p)^2)) }  # this random seed is important for reproducing the problem set.seed(0)  x0=runif(3) x=as.matrix(cbind(1,d[,2:3])) y=d$am opt=optim(x0, lossSqOnBinary, method=&quot;BFGS&quot;, x=x,y=y)  abline(-opt$par[1]/opt$par[3],        -opt$par[2]/opt$par[3], lty=2) legend(25,5,c(&quot;logisitc loss&quot;,&quot;squared loss&quot;), lty=c(1,2)) Get index of each capture in a JavaScript regex Bijection between an open and a closed interval Recently, I answered to this problem:     Given $a&lt;b\in \mathbb{R}$, find explicitly a bijection $f(x)$ from   $]a,b[$ to $[a,b]$.   using an "iterative construction" (see below the rule).  My question is: is it possible to solve the problem finding a less exotic function?  I mean: I know such a bijection cannot be monotone, nor globally continuous; but my $f(x)$ has a lot of jumps... Hence, can one do without so many discontinuities?    W.l.o.g. assume $a=-1$ and $b=1$ (the general case can be handled by translation and rescaling). Let:  (1) $X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$, and  (2) $f_0(x):=\begin{cases}     -x-\frac{3}{2} &amp;\text{, if } -1&lt;x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &amp;\text{,     if } \frac{1}{2}\leq     x&lt;1\\ 0 &amp;\text{, otherwise} \end{cases}$,  so that the graph of $f_0(x)$ is made of two segments (parallel to the line $y=x$) and one segment laying on the $x$ axis; then define by induction:  (3) $X_{n+1}:=\frac{1}{2} X_n$, and  (4) $f_{n+1}(x):= \frac{1}{2} f_n(2 x)$  for $n\in \mathbb{N}$ (hence $X_n=\frac{1}{2^n} X_0$ and $f_n=\frac{1}{2^n} f_0(2^n x)$).  Then the function $f:]-1,1[\to \mathbb{R}$:  (5) $f(x):=\sum_{n=0}^{+\infty} f_n(x)$  is a bijection from $]-1,1[$ to $[-1,1]$.  Proof: i. First of all, note that $\{ X_n\}_{n\in \mathbb{N}}$ is a pairwise disjoint covering of $]-1,1[\setminus \{ 0\}$. Moreover the range of each $f_n(x)$ is $f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$.  ii. Let $x\in ]-1,1[$. If $x=0$, then $f(x)=0$ by (5). If $x\neq 0$, then there exists only one $\nu\in \mathbb{N}$ s.t. $x\in X_\nu$, hence $f(x)=f_\nu (x)$. Therefore $f(x)$ is well defined.  iii. By i and ii, $f(x)\lesseqgtr 0$ for $x\lesseqgtr 0$ and the range of $f(x)$ is:  $f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$,  therefore $f(x)$ is surjective.  iv. On the other hand, if $x\neq y \in ]-1,1[$, then: if there exists $\nu \in \mathbb{N}$ s.t. $x,y\in X_\nu$, then $f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$ (for $f_\nu (x)$ restrited to $X_\nu$ is injective); if $x\in X_\nu$ and $y\in X_\mu$, then $f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$ (for the restriction of $f_\nu (x)$ to $X_\nu$ and of $f_\mu(x)$ to $X_\mu$ have disjoint ranges); finally if $x=0\neq y$, then $f(x)=0\neq f(y)$ (because of ii). Therefore $f(x)$ is injective, hence a bijection between $]-1,1[$ and $[-1,1]$. $\square$ Why is triple-DES using three different keys vulnerable to a meet-in-the-middle-attack? Among other sources,  states that triple DES using three seperate keys (k1, k2, k3) is vulnerable to meet-in-the-middle-attacks, while triple DES using only two keys (k1, k2, k1) is not.  Why is this attack possible if all the keys are different, but not if the first and the third are the same? C# Interfaces. Implicit implementation versus Explicit implementation What are the differences in implementing interfaces implicitly and explicitly in C#?  When should you use implicit and when should you use explicit?  Are there any pros and/or cons to one or the other?    Microsoft's official guidelines (from first edition ) states that using explicit implementations are not recommended, since it gives the code unexpected behaviour.  I think this guideline is very valid in a pre-IoC-time, when you don't pass things around as interfaces.  Could anyone touch on that aspect as well? Frequent Write to Non-volatile Memory Why does Unix store timestamps in a signed integer? Why is a signed integer used to represent timestamps? There is a clearly defined start at 1970 that's represented as 0, so why would we need numbers before that? Are negative timestamps used anywhere? I've been trying to figure out how to change the chart from scientific notation to just show the 5 digit numbers. I've tried scale and scale ticks with not success.  How can I change the y-axis and data values to show the decimals not in scientific notation?  \begin{tikzpicture} \centering \begin{axis}[     width=300pt,     height=250pt,     axis lines*=left, % Don't display the top and right lines     ybar, % Display it as a bargraph     enlarge y limits={upper,value=0.2},     legend style={at={(0.5,-0.20)}, % -.20 y-axis     anchor=north,legend columns=-1},     ylabel={Densities},     nodes near coords,     every node near coord/.append style={ anchor=mid west, rotate=90},     symbolic x coords={Class1,Class3,Class5,Class7,Class9,Class11,Class13,Class15,Class17,Class19,Class21,Class23},     xtick=data,     xticklabel style={ inner sep=0pt, anchor=north east, rotate=70 },     nodes near coords align={vertical}, ] \addplot coordinates { (Class1,0.0968) (Class3,0.0887) (Class5,0.1452) (Class7,0.0726)     (Class9,0.0403) (Class11,0.0081) (Class13,0.0161) (Class15,0.0000)     (Class17,0.0161) (Class19,0.0081) (Class21,0.0000) (Class23,0.0081) }; \end{axis} \end{tikzpicture} Create a text file using user input from a batch file Is it possible to write a batch file that will create a popup window for the user to input some specific information, and place that information into a text file? The text file should already have some text in it but I just want a couple lines to have input from the user. This is what the text file should look like:  PARTSN="USER INPUT1" PART=blahblah WORKSTATION=blahblah USERID="USER INPUT2" SIZE=blahblah   Where it says "USER INPUT1,2" is where I want the information from the user placed.  Any help would be great! Is there some trick to burning a cd with the CD/DVD creator?    I'm installing k3b because that's a known entity.  What is this, and where is the "burn" button?  Help just tells me this is Nautilus 3.4.1, but I don't see a .  There's no option to burn in the Files menu, nor does clicking on the whitespace indicate an option to burn anything. How can I get dramatic shallow DOF with a kit lens?