How to read and debug the compressed javascript file?
How do you debug through a compressed JavaScript file?
How to code to skip a shapefile, if it doesn't exist? So if shapefile 'AAA010' does not exist, how can I skip the Merge and Clip and move on to the next part of the script? If possible I would like to keep the current script as it is, but by just adding else or if, Sorry still trying to learn the ins and outs.  import os, fnmatch, arcpy   arcpy.env.workspace = r'C:\Users\sysadmin\Desktop\DataMagement\Comp_3\SA_MGCP3.gdb'   fcMerged = 'MergedFeatureClass' fcClipped = 'fcClipped' fcToClip = r'C:\Users\sysadmin\Desktop\DataMagement\Country.shp'  filesMatched = list()  for root, subfolders, files in os.walk (r'C:\Users\sysadmin\Desktop\DataMagement\MGCP_Raw'):      for f in fnmatch.filer(files, 'AAA010.shp'):              filesMatched.append(os.path.join(root,f))      if filesMatched:         print ' files found - merging'         arcpy.Merge_management (filesMatched, fcMerged)         arcpy.Clip_analysis (fcMerged, fcToClip,fcClipped)      else:          print 'nothing found - skipping' How do I make Units efficiently? I just blew almost all my money on a really cool lookin' ship. How do I regain solvency without spending a lot of time fiddling around in my inventory?  At the moment I've found a sequence of steps that'll turn 50 carbon and 100 Plutonium into ~5000 Units (by creating Suspension Fluid and then turning it into Electron Vapor). Are there better recipes or sequences of recipes I should look for? Or should I just drop the whole thing and go mine a particular mineral? Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals. Finding $\lim\limits_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$ for continuous $f:[0,1]\to[0,\infty)$ Find $$\lim_{n \rightarrow \infty}\left(\int_0^1(f(x))^n\,\mathrm dx\right)^\frac{1}{n}$$if $f:[0,1]\rightarrow(0,\infty)$ is a continuous function.   My attempt:  Say $f(x)$ has a max. value $M$. Then $$\left(\int_0^1(f(x))^ndx\right)^\frac{1}{n}\leq\left(\int_0^1M^ndx\right)^\frac{1}{n} =M$$  I cannot figure out what to do next. Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$  such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...   My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$  Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.   Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon &gt; 0 ,\exists \delta_1 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| &lt; \epsilon  $$  Since this statement must be true for all $\epsilon &gt; 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus  $$\forall \epsilon &gt; 0, \exists \delta_2 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_2 \implies |g(x)-g(x_0)| &lt; \epsilon $$  and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* &gt; 0, \forall \delta&gt;0: \exists x \in  \mathbb{R}, |x-x_0| &lt; \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$  My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon &gt; 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*&gt;0$, so statement 1 and 2 are still true for said $\epsilon$  I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.  Let $|x-x_0|&lt; \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| &lt; - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| &lt;0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation. $\sum x_n^2$ and $\sum y_n^2$ converge, what does $\sum (x_n + y_n)^2$ do? The  tag doesn't appear to be very useful.  It's used in a variety of ways (for reading, writing, showing, transforming, compiling, translating, obfuscating, obtaining, ... source code) on 1747 questions right now, and doesn't appear to be adding any classification value to the site.  Can we get rid of it? In 1735 Euler gave the value of $\gamma$ as $0.577218.$ The constant is generally defined as the limit of the difference between the harmonic series and $\log n:~\gamma= \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k}-\log n.$ Euler apparently relished this sort of calculation and must have taken quite a few terms to get such a good approximation.  My question is whether without a computer one can now prove that  $$\gamma &lt;  \sqrt{1/3}$$   with at least some savings in terms of the type of work Euler apparently expended? I don't think there's any point in raising the bar to $\gamma &lt; \ln 2\sqrt{\frac{23}{29}} $  because it seems to be the same sort of question.       My own thought was to compare $\frac{1}{2}\int_0^{1/3}\frac{dx}{\sqrt{x}}$ to something like $(-1)\cdot\int_0^\infty e^{-u}\ln u~du$ but I expect there's a better way. I know there is a ton of questions and answers about this topic, but none answers my question sufficiently.  Let's say I have a viewport view that I want to render in Cycles. Pressing ALT CTRL Numpad 0 aligns the camera to this view, but it does not solve the whole problem. The thing is, I want PRECISELY this view on the render, preferably in one click, without manually moving the camera back-and-forth, changing sensor sizes etc. fiddling that will remove the unwanted border that appears and blocks the viewport partially. This whole activity wastes a lot of precious time especially if you do this 10 times a day.  What this dream action should do:   match the render resolution (proportions) to the resolution of the current view (can the resolution of current view be read by a Python script?) make the entire view the camera view, without any borders, passe-partout etc. by automatically maneuvering sensor sizes etc. whatever is necessary to make it exactly the same view, the same camera lens angle.   Is it possible to write such a script that would do it all in one click? Is there a pattern to which Star Trek ships get the definite article? Some ships in Star Trek are referred to with definite articles. "The Enterprise", "The Defiant". Other ships are referred to without them. "Voyager", "Enterprise". Is there a pattern to or explanation for this behavior? Converting a DC motor to a generator Can ANY DC motor become a generator if torque is applied to the shaft? example:   And by any, I mean the really cheap ones too.  Or is there a specific type of motor (of which I do not know the name) that can work in both directions (battery to power the shaft OR torque-on-shaft to create a voltage difference between the terminals). Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?   Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.  Or is there a simpler method ? (Maybe not UFD would imply not PID) Is there a number whose absolute value is negative? How do you access the matched groups in a JavaScript regular expression? Fibonacci Recurrence Relations Solve the recurrence relation $f(n) = f(n − 1) + f(n − 2)$ with initial conditions $f(0)=1$, $f(1)=2.$   So I understand that it grows exponentially so $f(n) = r^n$ for some fixed $r$.  This means substituting this $r^n = r^n-1 + r^n-2$ which gives the characteristic equation of $r ^ 2 - r - 1 = 0$. I'm not 100% sure where to move on from here. Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ How to validate phone numbers using regex I'm trying to put together a comprehensive regex to validate phone numbers. Ideally it would handle international formats, but it must handle US formats, including the following:   1-234-567-8901 1-234-567-8901 x1234 1-234-567-8901 ext1234 1 (234) 567-8901 1.234.567.8901 1/234/567/8901 12345678901   I'll answer with my current attempt, but I'm hoping somebody has something better and/or more elegant. Here's an image from another question:    I get what the pull down resistor is there to do, but I don't really understand exactly how it works.  When there is no voltage applied through the base wire, why does any induced current then go through R2, but when a voltage is applied, it goes to the transistor? My understanding is that voltage always follows the path of least resistance, so if applying say, 5v turns the transistor on, that means that the transistor has a lower resistance than R2, so I would also assume any induced current would also flow through the resistor - so why does it then go through R2? Create longer arrows I'm trying to compute the infinite series $\sum_{n=0}^{\infty} ne^{-n}$.  I know the answer is $e/(e-1)^2$, but I don't understand how to find this result.  Thanks for the help!