I build an SharePoint 2013 Sandbox Event Receiver which works in my local on-prem dev environment.  However, when pushing it up to my SharePoint Online tenant it does not work nor even fire at all.  I can see the event receiver attaching and detaching from my list using the SharePoint  with each activation/deactivation so I know it is attaching properly.   Are there any special configurations necessary for it to work on SPOnline?  Thanks
using System; using System.Security.Permissions; using Microsoft.SharePoint; using Microsoft.SharePoint.Security; using Microsoft.SharePoint.Utilities; using Microsoft.SharePoint.Workflow;  namespace Change_Title_Event.Change_Title {     /// &lt;summary&gt;     /// List Item Events     /// &lt;/summary&gt;     public class Change_Title : SPItemEventReceiver     {        /// &lt;summary&gt;        /// An item is being added.        /// &lt;/summary&gt;        public override void ItemAdding(SPItemEventProperties properties)        {            properties.AfterProperties["Title"] = "Kevin: Changed by event receiver at" + DateTime.Now.TimeOfDay.ToString();            base.ItemAdding(properties);        }       } }   Where simple example, which works perfect on my local Sharepoint 2010. It is sandboxed.  But when I package this solution, it do not work on Sharepoint Online. I does not give any Errors either.  I think I am doing something wrong, something simple, but I can't simply find anything on Google about it.  Here is another who asks on the MSDN forum, but gets a very wierd reply, that I do not understand.
$$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$ Why didn't Harry himself drink Polyjuice Potion instead of all others in Deathly Hallows Part 1? In Harry Potter and the Deathly Hallows Part 1, it is shown that members of the Order of Phoenix come to rescue Harry from his house at Privet Drive. There, they all drink Polyjuice Potion and assume the form of Harry to confuse death eaters in identifying who is the real Harry.  An obvious, more safe, better choice would be to change Harry into someone else. Maybe a muggle, like his cousin Dudley. And then, safely take him to the Weasleys, maybe apparating. That way, death eaters won't ever find out. Isn't that correct?  Why didn't they take this path, and follow a rather complicated and risky one? How can $\pi$ be irrational if it can be calculated using an infinite series of rational numbers? Exception messages in English? Euclidean Algorithm for GCD of polynomials I am struggling to use the Euclidean algorithm for polynomials.  Given something like $$GCD(x^5+1, x^3+1)$$  I can easily use it as follows:   $$x^5+1 = x^2(x^3+1) -x^2 +1 \\ x^3+1 = -x(-x^2+1) + x +1 \\ -x^2+1 = (x+1)(-x+1)$$  GCD = x+1  But for something like $$GCD(2x^2+6x+3, 2x+1)$$   I cannot figure out how to do it using the same method. I start like this:  $$2x^2+6x+3 = x(2x+1) + 5x+3\\ 2x+1 = \frac{2}{5}(5x+3) - \frac{1}{5}\\$$  and I am not sure how to continue. Any help would be appreciated, sorry for my poor attempt at using LaTeX. Rooting Stock Firmware - GSII with NFC Background reflected in glossy glass, but I didn't set it I've made a roughness 0 glass pane, when I'm in lookdev mode it's reflecting a background that I didn't put there. I know I could just ignore it, but I'd rather be able to see what it'll mostly look like when I'm in the later stages. I also just want to know how on Earth I managed to set this background without realising it  Reflection completely ignores any objects I put in front of it. If I placed this glass inside my house, it would still just reflect trees (This is the point where I felt I couldn't ignore it - reflections are going to play a big part in my project and I don't want to be in render mode all the time)  The nodes are the default glass BSDF, except I set roughness to 0  The glass is set to 20mm thick, if that makes a difference  The background moves with my perspective and moves when I relocate the glass  I have a few HDRI's on my PC, but none look remotely like this   Also, I'm using cycles as I find Evee buggy as hell I'm quite confused about the following regular expressions I found in a shell script:  ${0##*/} ${0%/*}   How do they work? What is the closure of an open ball $B_X(\mathbf{a},r)$ in $X=\mathbb{R}^n$? Suppose we have the open ball $B_{X}(\mathbf{a},r)$ and the closed ball $\bar{B}_{X}(\mathbf{a},r)$ of radius $r$ about $\mathbf{a}\in\mathbb{R}^n=X$ with the Euclidean metric $d_2$.  What is the closure of an open ball $B_{X}(\mathbf{a},r)$ in $(\mathbb{R}^n,d_2)$?  First, I have defined the sets like this:  $$B_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})&lt;r\}$$ $$\bar{B}_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})\le r\}$$  Then, let $U=B_X(\mathbf{a},r)$ and claim the closure is $$\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}=\bar{B}_X(\mathbf{a},r).$$  Since closed balls are closed then we have $$\bar{U}\subset\bar{B}_X(\mathbf{a},r).$$  But, how can I prove the other direction $\bar{B}_X(\mathbf{a},r)\subset\bar{U}$?  Since it's a containment argument I know that I have to take any element from $\bar{B}_X(\mathbf{a},r)$ and show it is in $\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}$.  But I do not know how to get an expression for $\mathbf{x}_n$.  I would appreciate any help. Is there a way to start over?  No spoilers here, but after a certain point in the game I wish I had done something different, so I'd like to perhaps start over and make some different decisions based on my current game play experience. Distributing distinct apples and pears to three people so that each person receives at least one pear How many ways are there to distribute seven distinct apples and six distinct pears to three distinct people such that each person has at least one pear.   Note: The first part of the problem was the same but the word distinct was replaced with identical apples and pears. That answer is $360$ ways.   What has to be done differently when the word identical is replaced with distinct? I want to add  on my blog hosted on Blogger. The only solution I found is , but it requires a hosting server. Any other option? Maximal order of an element in a symmetric group If we let $S_n$ denote the symmetric group on $n$ letters, then any element in $S_n$ can be written as the product of disjoint cycles, and for $k$ disjoint cycles, $\sigma_1,\sigma_2,\ldots,\sigma_k$, we have that $|\sigma_1\sigma_2\ldots\sigma_k|=\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)$.  So to find the maximum order of an element in $S_n$, we need to maximize $\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)$ given that $\sum_{i=1}^k{|\sigma_i|}=n$. So my question:      How can we determine $|\sigma_1|,|\sigma_2|,\ldots,|\sigma_k|$ such that $\sum_{i=1}^k{|\sigma_i|}=n$ and $\operatorname{lcm}(\sigma_1,\sigma_2,\ldots,\sigma_k)$ is at a maximum?   Example  For $S_{10}$ we have that the maximal order of an element consists of 3 cycles of length 2,3, and 5 (or so I think) resulting in an element order of $\operatorname{lcm}(2,3,5)=30$.  I'm certain that the all of the magnitudes will have to be relatively prime to achieve the greatest lcm, but other than this, I don't know how to proceed. Any thoughts or references? Thanks so much. How do I remove the limit on PC keyboard button presses? Whenever I play a PC game with the directional keys, I can only press two buttons at a time.  For instance if I am playing a shooter like R-Type, I cannot hold down the fire button while moving diagonally at the same time.  The computer simply ignores the extra button press.  How do I remove this limit so I can shoot while moving diagonally?  This limitation seems to apply for every game.  I am running Windows 7 if it makes any difference.  I noticed the same problem back when I used Windows XP. Why does this not work with an imaginary exponent? Could someone please tell me why this exponentiation rule does not work when the exponent is imaginary?  The following seems to be true but is only true if $x$ is an integer.  $$e^{2\pi ix} = \left( e^{2\pi i} \right)^x = 1^x = 1$$ Deleting photos on the phone - does it affect them on iCloud? example of a non differentiable manifold Take the manifold: The graph of $|x|$ on $(-1,1)$, with the induced topology from $\mathbb{R}^2$. This is a topological manifold, which is homeomorphic to $(-1,1)$ by projection. Is it a differentiable manifold? I believe it is, because we can take an atlas with only the projection, and then we will have only one transition map which is the identity, and therefore differentiable. But I'm not sure I get the definition of a differentiable manifold right...  Thanks! Auto.arima with daily data: how to capture seasonality/periodicity? I am fitting an ARIMA model on a daily time series. Data are collected daily from 02-01-2010 to 30-07-2011 and are about newspaper sales. Since a weekly pattern in sales can be found (the daily average amount of copies sold is usually the same from Monday to Friday, then increases on Saturday and Sunday), I am trying to capture this "seasonality". Given the sales data "data", I create the time series as follows:  salests&lt;-ts(data,start=c(2010,1),frequency=365)   and then I use the auto.arima(.) function to select the best ARIMA model via AIC criterion. The result is always a non-seasonal ARIMA model, but if I try some SARIMAs model with the following syntax as example:  sarima1&lt;-arima(salests, order = c(2,1,2), seasonal = list(order = c(1, 0, 1), period = 7))   I can obtain better results. Is there anything wrongs in the ts command / arima specification? The weekly pattern is very strong so I would not expect so many difficulties in capturing it.  Any help would be very useful. Thank you, Giulia Deppieri  Update:  I have already changed some arguments. More precisely, the procedure selects ARIMA(4,1,3) as the best model when I set D=7, but AIC and the others good of fit indexes and forecasts as well) do not improve at all. I guess there's some mistakes due to confusion between seasonality and periodicity..?!   Auto.arima call used and output obtained:  modArima&lt;-auto.arima(salests,D=7,max.P = 5, max.Q = 5)     ARIMA(2,1,2) with drift         : 1e+20  ARIMA(0,1,0) with drift         : 5265.543  ARIMA(1,1,0) with drift         : 5182.772  ARIMA(0,1,1) with drift         : 1e+20  ARIMA(2,1,0) with drift         : 5137.279  ARIMA(2,1,1) with drift         : 1e+20  ARIMA(3,1,1) with drift         : 1e+20  ARIMA(2,1,0)                    : 5135.382  ARIMA(1,1,0)                    : 5180.817  ARIMA(3,1,0)                    : 5117.714  ARIMA(3,1,1)                    : 1e+20  ARIMA(4,1,1)                    : 5045.236  ARIMA(4,1,1) with drift         : 5040.53  ARIMA(5,1,1) with drift         : 1e+20  ARIMA(4,1,0) with drift         : 5112.614  ARIMA(4,1,2) with drift         : 4953.417  ARIMA(5,1,3) with drift         : 1e+20  ARIMA(4,1,2)                    : 4960.516  ARIMA(3,1,2) with drift         : 1e+20  ARIMA(5,1,2) with drift         : 1e+20  ARIMA(4,1,3) with drift         : 4868.669  ARIMA(5,1,4) with drift         : 1e+20  ARIMA(4,1,3)                    : 4870.92  ARIMA(3,1,3) with drift         : 1e+20  ARIMA(4,1,4) with drift         : 4874.095   Best model: ARIMA(4,1,3) with drift           So I assume the arima function should be used as:  bestOrder &lt;- cbind(modArima$arma[1],modArima$arma[5],modArima$arma[2]) sarima1&lt;-arima(salests, order = c(4,1,3))   with no seasonal component parameters and period specifications. Data and exploratory analysis show that the same weekly pattern can be approximatively considered for each week, with the only exception of August 2010 (when a consistent increase in sales is registered). Unfortunately I have no expertise in timeseries modeling at all, in fact I am trying this approach in order to find an alternative solution to other parametric e non-parametric models I have tried to fit for these problematic data. I have also many dependent numeric variables but they have shown low power in explaining the response variable: undoubtedly, the most difficult part to model is the time component. Moreover, the construction of dummy variables to represent months and weekdays turned out not to be a robust solution. I am looking to replicate black &amp; white maps in QGis, using hatching patterns and greyscales to indicate different categories of a continuous variable (example below).  It is possible/straightforward to do so in QGis? At first sight I only see color gradients and symbol fills.   ). Restore the state of std::cout after manipulating it