I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$. I have no Idea where to begin with. Someone in previous post suggested using Taylor, but I can't because we haven't learned it yet
I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$. And it made me think, maybe I can say (about $\arctan(x)$ at $x=0$) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then $2$. Am I right?
How can I make my laptop boot automatically from the hard drive without having to press F10 each time? When should comments be flagged/deleted on per-site metas? We all know , but that guidance is, naturally, geared toward main sites, where we want to keep things clean and focused.  Meta, on the other hand, is a site's "second place", with some different rules and conventions.  For example, questions on main sites can be closed as primarily-opinion-based, but meta is all about opinions.  There's a "discussion" tag and such questions collect many opinions.  Meta is different.  Occasionally I see comment flags on metas and I'm never sure what to do with them.  Setting aside things that are blatantly offensive or spam (that's a no-brainer), is there such a thing as "too chatty" or a comment discussion that should be moved to chat on meta?  I am very conservative in deleting things on per-site metas (except on  posts, which like main should stay focused).  Are there any guidelines specific to per-site metas?  I'm specifying per-site metas and not MSE because MSE seems to have conventions that fall somewhere between "main site" and "per-site meta".  If the answer is the same for all metas then great, but I don't want to assume it'll be the same. iPhone Passcode for Powering Off Noetherian property for exact sequence Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then $M$ is Noetherian is equivalent to $M'$ and $M''$ are Noetherian.  For the ''$\Leftarrow$'' case:  I guess if we let $(L_n)_{n\geq 1}$ be an ascending chain of submodules of $M$, then $(\alpha ^{-1}(L_n))_{n\geq 1}$ is a chain in $M'$, and $(\beta(L_n))_{n\geq 1}$ is a chain in  $M''$. For large $n$ both these chains are stationary. Then why we know the chain $(L_n)$ is stationary?   Thanks! Problem Statement  Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$   Suppose that the following conditions hold:   $f(1)=0$ the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$      $$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$   My attempt  Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$  Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$  $$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$  And so we have $$a = -(p+q+r) = -p^2q^2r^2$$ $$b = pq + pr + qr = p^2 + q^2 + r^2$$ $$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$  Then we get the following relations  $$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$  From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$  However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?  Edit  For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$ How can I add a right-click desktop menu entry to create a text file to Gnome 3 in Ubuntu 16.04.02? Xorg: High CPU Usage (94%) on Moving Mouse (Ubuntu 18.04) whenever I move my mouse enough it causes high cpu usage. Screenshot provided is on a fresh X session.    This also causes some frequent freeze ups if I'm running some other programs e.g gaming.  [1075299.085] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077813.434] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077814.079] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077814.246] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077814.682] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077831.975] (EE) client bug: timer event4 debounce: offset negative (-18ms) [1077831.975] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost. [1077831.975] (II) event4  - Corsair Corsair Vengeance M90 Mouse: WARNING: log rate limit exceeded (5 msgs per 30000ms). Discarding future messages. [1120008.128] (II) event4  - Corsair Corsair Vengeance M90 Mouse: SYN_DROPPED event - some input events have been lost.   Is there any way for the CPU usage not to be this high?  EDIT:  I tried with other mice. The CPU usage appears to increase with the Mouse DPI or to be exact, a generic mouse will cause a CPU usage spike, but not as much as a gaming mouse.  I also checked if I have hardware acceleration enabled.  $ glxinfo | grep "direct rendering" direct rendering: Yes   EDIT 2:  CPU is an AMD FX-6350  GPU is AMD RX Vega 56 Existence of area-preserving diffeomorphism between two sets in $\mathbb{R}^2$ Suppose that we have open sets $A, B\subset \mathbb{R}^2$ such that $A$ and $B$ are simply connected and have smooth boundaries. Furthermore, let us assume that $m(A) = m(B) &gt; 0$, where $m$ is the Lebesgue measure on $\mathbb{R}^2$. My question is as follows: when does there exist an area-preserving diffeomorphism between $A$ and $B$? In this context, an area-preserving diffeomorphism between $A$ and $B$ is a is a $C^{\infty}$-diffeomorphism $\phi : A\to B$ such that $m(S) = m(\phi(S))$ for all measurable $S\subseteq A$. If an area-preserving diffeomorphism doesn't always exist between sets $A$ and $B$ as above, are there any extra conditions we can impose on $A$ and $B$ such that there does exist an area-preserving diffeomorphism?  My thoughts thus far are that we always have a $C^{\infty}$-diffeomorphism between $A$ and $B$: this is a consequence of the . It's also obvious to me that if we can deform $A$ linearly to get $B$, then that deformation will be area-preserving. Furthermore, it seems that aside from this case, there's no reason why an area-preserving diffeomorphism should exist. However, I'm having a hard time proving that there necessarily isn't an area-preserving diffeomorphism between $A$ and $B$ if $B$ is not a linear deformation of $A$. Does anybody have any ideas? In Episode III,  explains to Anakin the power of the Dark Side to save people from death. He mentions  - the Sith who discovered this. Sidious then claims he died, which seems odd, given that kind of power.   Then in Episode VII (The Force Awakens) we find a Sith Lord who has appeared from no where. There seems to be two loose ends that fit nicely together.   My question is: Is there evidence to suggest that Supreme Leader Snoke is Darth Plagueis? Assigning Float Pointers in C How would you prove that the generating function of $\binom{2n}{n}$ is $\frac{1}{\sqrt{1-4x}}$?  More precisely, prove that( for $|x|&lt;\frac{1}{4}$ ):  $$\sum^{\infty}_{n=0}x^n\binom{2n}{n}=\frac{1}{\sqrt{1-4x}}$$  Background: I was trying to solve $$S=\sum^{\infty}_{n=0}\frac{(2n+1)!}{8^n(n!)^2}=\sum^{\infty}_{n=0}\frac{(2n+1)}{8^n}\binom{2n}{n}$$ Which if we let $f(x)$ be the generating function in question, would be simply $$f(x)+2xf'(x)$$ With $x=\frac{1}{8}$. Is there a simple proof of the first identity?  states it without a proper reference ( states it without proof). Is there an easier way of calculating $S$? (which is $\sqrt{8}$, by the way) Supposing we have a table with four columns (a,b,c,d) of the same data type.  Is it possible to select all distinct values within the data in the columns and return them as a single column or do I have to create a function to achieve this? 70s UK science fiction TV show with time travel and telepathic cat I'm trying to identify a science fiction series on UK TV during the 70s, but I can only recall the following:   A party of adults and children. There was a woman with a telepathic bond to her cat: she could look into its eyes to see what it had seen. Time travel was involved. I think it may have been inadvertent or uncontrolled time travel. On one occasion the party encountered an Elizabethan seaman who demanded to know if Elizabeth was still queen. They party replied "err... yes" because they didn't want to have to explain time travel. I have been looking at a few scripts other people wrote (specifically Red Hat), and a lot of their variables are assigned using the following notation VARIABLE1="${VARIABLE1:-some_val}" or some expand other variables VARIABLE2="${VARIABLE2:-`echo $VARIABLE1`}"  What is the point of using this notation instead of just declaring the values directly (e.g., VARIABLE1=some_val)?  Are there benefits to this notation or possible errors that would be prevented?  Does the :- have specific meaning in this context? Word for a smart and stubborn person? I'm trying to find a single adjective that would fit the best to the following person's trait:     One who thinks knows everything/best, and often doesn't listen to orders and instead completes some task his/her own way.    This person, while intelligent, is right about 50% of the time, when doing things its own way.   So far I've thought of independent, idiosyncratic and stubborn, but these words don't imply the "knows everything/best" part of the trait. This was a question my friend found in a book.  A block attached to a spring pulled by a constant horizontal force is kept on a smooth horizontal surface. Initially the spring is in its natural state. Then the maximum work that the applied force F can do is:  (a) $F^2/k$ (b) $2F^2/k$ (c) $4F^2/k$ (d) $F^2/2k$  My friend thought that the answer is (a) since the maximum displacement is $x=F/k$ and $W=Fx=F^2/k$. But he saw that the book's answer is different and he asked this question to me. I at first thought that the answer will be (a), but then I realised that this situation is somewhat similar to a charging capacitor. The final energy of the spring is actually $\frac12kx^2=F^2/2k$. So probably the answer is (d)? But the work done is still (a)? But the book says neither. It's (b). This is the logic in the book:  Method 1: in the book:  $$Fx=\frac12kx^2$$  $$x=2F/k$$  $$W=\frac12kx^2=2F^2/k$$  (since the book is an objective questions book it did not have detailed explanations)  What is wrong in the following methods?  Method 2: by calculating the maximum displacement  $$x=F/k$$  $$W=Fx=F^2/k.$$ Method 3: by calculating the final energy  $$x=F/k$$  $$\frac12kx^2=F^2/2k$$  PS: We both know calculus so don't hesitate to use it in the answers. What's an intuitive way to think about the determinant? In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related? Moving a very large (40 TB) DB from a 2008 server to a 2016 server Zooming to WMS layer after CQL filter in OpenLayers? I created a layer that contains points and I added it to a map as WMS layer in OpenLayers. I did a CQL filter and it works but I'd like to zoom to the selected point when I worked with getextent or getdataextent it doesn't zoom to the right place. It missed with a few Km. Here is my the code of filter:   function filterNom(){             var filterType ="cql";             var filter = document.getElementById('batima').value;             // by default, reset all filters             var filterParams = {cql_filter: null};             if (OpenLayers.String.trim(filter) != "") {                 if (filterType == "cql")                      filterParams["cql_filter"] = "batiment='"+filter+"'";              }             // merge the new filter definitions             wfs_layer.mergeNewParams(filterParams);             var bounds =  new OpenLayers.Bounds             bounds = wfs_layer.getExtent()             map.setCenter(wfs_layer.getExtent(),18);             var center = map.getCenter();             center.transform(new OpenLayers.Projection("EPSG:4326"), new OpenLayers.Projection("EPSG:900913"));             alert(center.toString()); }   and here is the code to add the WMS layer:    wfs_layer = new OpenLayers.Layer.WMS("Patrimoines",         "http:// localhost/geoserver/cite/wms",          {layers: 'cite:patrimoine' , transparent: true },         {isBaseLayer: false, opacity: 1, singleTile: true, visibility: true,          projection: 'EPSG:900913'}); How to edit geometry of multiple layers simultaneously in QGIS 3? I have several line layers (route, pipes, cables) that share the same geometry and would like to edit them simultaneously.   Is there a way to easily snap the geometry of the pipes and cables layers to the route layer so that I can simply change the geometry of the route layer and don't have to repeat all changes on the other two layers as well?  To clarify, I plan a fibreglass route and have three layers. The first one describes the route that I have to take to pass all homes that have to be connected. The second one contains the pipes in which the cables lie and the third layer is the cable layer. So all features on those layers have to have the same geometry. Once I change the pathway of the route layer, I have to apply the same changes to the other two layers to ensure that they are still aligned with the route layer. I'm not sure whether I understand your comments correctly, but I use the basic editing tools in QGIS 3.