Derivation of this inequality
$|e^a-e^b| \leq |a-b|$ for complex numbers with non-positive real parts
Moving from SQL 2005 [SQL_Latin1_General_CP1_CI_AS] to 2008 - will I lose any features by using 'backwards compatibility' We're moving from SQL 2005 [Instance and DB have collation of SQL_Latin1_General_CP1_CI_AS]  to SQL 2008 [which defaults to Latin1_General_CI_AS].  I completed a SQL 2008 R2 installation, and used default Latin1_General_CI_AS collation, with the restoration of the database still on SQL_Latin1_General_CP1_CI_AS. The excepted problems occurred - the #temp tables where in  Latin1_General_CI_AS whilst the db was in  SQL_Latin1_General_CP1_CI_AS and this is where I am now - I need advice on the pitfalls now please.  On installation of SQL 2008 R2, I have the option on installation to use 'SQL Collation, used for backwards compatibility' where I have the option to select the same collation as the 2005 database : SQL_Latin1_General_CP1_CI_AS.   This will allow me to not have problems with #temp tables, but are there pitfalls? Would I lose any functionality or features of any kind by not using a "current" collation of SQL 2008? What about when we move (e.g. in 2 years ) from 2008 to SQL 2012? Will I have problems then? Would I at some point be forced to go to Latin1_General_CI_AS? I read that some DBA's script complete the rows of complete databases, and then run the insert script into the database with the new collation - I'm very scared and wary of this - would you recommend doing this? Prove holomorphism of expression I have to prove that given $\gamma$ contour of $G \subset \mathbb{C}$  and $f$ continuous function on $G$. Then the function $g(z) = \int_{\gamma} \frac{f(w)}{w-z}dw$ is holomorphic on $\mathbb{C} - \bar{G}$ My attempt: I started by using that given any parametrization $\alpha :[a,b] \rightarrow \gamma$ we can rewrite $g(z)$ as: $g(z) = \int_a^b \frac{f(\alpha(t))}{\alpha(t)-z}\alpha'(t)dt$ Now using the fundamental theorem I derivate to see where $g'(z)$ is defined and I get: $g'(z) = \frac{f(\alpha(b))}{\alpha(b)-z}\alpha'(b) - \frac{f(\alpha(a))}{\alpha(a)-z}\alpha'(a) $ Here I get stuck, I fail so see how this links back to $g$ being holomorphic at $\mathbb{C} - \bar{G}$. Any hints on how to proceed? Thanks! Can I check iPhone, iPad, Watch battery on Mac? My goal is to have battery status of other Apple device on my Mac (preferably on menubar). I'd like to have a glimpse of the battery status of my iPhone, iPad, Watch, and Airpods.  I'm a user of Bitbar and iStat menu, but I haven't found any widgets can wirelessly check other Apple device's battery status.  What I know is that I can click the Wi-Fi icon on menu bar and it will show my iPhone battery status. To show the percentage, I have to option + click the Wi-Fi icon and hover over my mouse on the iPhone hot spot section. It's pretty clunky, but I guess at least the API is there.  I am also open to third party apps which can let me see battery status across devices. Op Amps: how do they know whether the supply is split? Most op amps can operate in single and split supply configurations. However, I cannot see how this is possible.  If I connect voltages \$V_\text{SY}^+\$ and \$V_\text{SY}^-\$ to the positive and negative supply pins of a op amp, and \$V_+\$ and \$V_-\$ to the non-inverting and inverting inputs, then the output is (if \$A_\text{OL}\$ is the open loop gain of the op amp): $$V_\text{out}=A_\text{OL}(V_+-V_-)+V_\text{SY}^-\tag{Single-supply}~\text{ }~~\text{ }~~\text{ }~~\text{ }~~\text{ }~~\text{ }$$ The above equation is from personal experience using op amps in single supply (i.e. \$V_\text{SY}^- = \text{GND}, V_\text{SY}^+ &gt; \text{GND}\$).  In split supply, we can use the following diagram from MIT's 6.002 introduction to electronics lecture video:  $$V_\text{out}=A_\text{OL}(V_+-V_-)+(V_\text{SY}^+-V_\text{SY}^-)/2\tag{Split-supply}$$  These two equations, however, are mutually inconsistent: If I take an op amp and hook up a single voltage source between its supply pins, how does the op amp know that the resulting voltage is not split between two sources?   It must know - but how does it know? REST Service API executor.executeAsync Call Ends in Failure Function in Sharepoint Hosted App Any Answers Will Help me to Forward and Gain Knowledge in SharePoint Development  While i am Trying to get the list Items Using Rest API Calls in SharePoint Hosted App Using Below Function  function loadingSamples()  {     var JobListName = "RRR Jobs #";     var executor;     // Initialize the RequestExecutor with the app web URL.     executor = new SP.RequestExecutor("/");      //Fetching the choice values from the Job Status Field, to show in the Dropdown field     executor.executeAsync     ({         url: "/_api/web/lists/getbytitle('" + encodeURIComponent(JobListName) + "')/fields/getbytitle('Job Status')?$select=Choices",         method: "GET",         headers: { "Accept": "application/json; odata=verbose" },         success: successHandlerStatus,         error: errorHandler     }); } function successHandlerStatus(data)  {     alert("sucesss") }  function errorHandler(data, errorCode, errorMessage)  {     alert("errorMessage") }   It's always going to error function named errorHandler With  Error Code as -1002 and Error Message as Bad Request and Inside Data -> Body , the Error message is  {     "error":{            "code":"-1,Microsoft.SharePoint.Client.InvalidClientQueryException",         "message":{             "lang":"en-US",             "value":"The expression \"web/lists/getbytitle('DWR Jobs \" is not valid."         }     } }   Note :    This is working code in Sharepoint Hosted App that previously hosted in another Windows server. Now i hosted this App in New windows server,it's successfully hosted but it gives above mentioned Error in Rest Service Call. Cannot upgrade unsupported version of Skype for Linux (v5) I'm currently reading the TeXbook, and I use \show a lot to see how Knuth defined plain-TeX macros in terms of TeX primitives. For instance, to have access to the definition of \centerline, I would write \show\centerline in my input file. When I compile  that file with pdftex in interactive mode from the command line (a Unix terminal, in my case), TeX generates an error message that shows me the definition of \centerline in the terminal and offers me a menu of options (as described in Chapter 6: Running TeX in the TeXbook).  I understand that design choice by Knuth of having TeX generate an error message for each instance of \show: he probably wanted to leave the user some time to read what \show... prints in the terminal before letting TeX resume execution, instead of having to open the log file only after the end of the compilation.  However, instances of \show... in the input file have the undesirable side-effect of raising "errors" (false alarms, really) when pdftex is called from within an IDE such as TeXmaker with option -interaction=nonstopmode. For example,  \show\centerline \bye   when compiled with pdftex -interaction=nonstopmode causes TeXmaker to report     Obviously, those false alarms prove really annoying when debugging. I always have to ask myself: Wait... are those errors due to instances of \show in my input file, or are they bona fide errors? On the other hand, my understanding is that you have to specify  -interaction=nonstopmode   if you want to call pdftex from within an IDE. Therefore, my question is:  Is there any way of preventing TeXmaker from reporting those "false alarms"?  Note: the problem I describe may be present in IDEs other than TeXmaker as well. Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ What is the "rootless" feature in El Capitan, really? I have just learned about the "Rootless" feature in El Capitan, and I am hearing things like "There is no root user", "Nothing can modify /System" and "The world will end because we can't get root".  What is the "Rootless" feature of El Capitan at a technical level? What does it actually mean for the user experience and the developer experience? Will sudo -s still work, and, if so, how will the experience of using a shell as root change? Which types of particles are affected by the wave-particle duality? Circles importing as crossed boxes into Adobe Illustrator Create a variable in .CSS file for use within that .CSS file Possible Duplicate:          We have some "theme colors" that are reused in our CSS sheet.  Is there a way to set a variable and then reuse it?  E.g.  .css OurColor: Blue  H1 {   color:OurColor; } Why do most C programmers name variables like this:  int *myVariable;   rather than like this:  int* myVariable;   Both are valid. It seems to me that the asterisk is a part of the type, not a part of the variable name. Can anyone explain this logic? I've just answered .  After accepting the answer, the OP then changed his question to another problem - which I know I shouldn't have answered (I should have requested as new question) but I updated my answer anyway.  The OP has now edited his question to remove the essence of either the original or new problem, and instead has a statement saying that the problem has been fixed.  What is the correct thing to do here?  Roll the question back so it's in its original state? Roll it back to the "new problem" state? Leave it and let him continue to get downvoted and/or closed by people who didn't see the original question? Why would I use manual camera controls instead of the automatic modes? I'm very new to digital photography, so I still haven't gotten the hang of all this exposure and ISO stuff, etc. I have a Canon EOS T3.   My question is how big of a difference does it make using manual settings rather than auto settings for everything? I've taken some pretty nice photos with all auto settings, and I'm not sure if it could have turned out any better using manual settings. I know that there are some people who have been in the game for years upon years, and to them they can notice the difference between photos taken with manual vs auto settings, but for my situation, do you think it really makes a difference? Is my camera smart enough to determine the best settings for me?  Also, lets say manual settings are better (which I'm sure they are, since they give you greater control). How often are you supposed to adjust your settings? Between every shot, considering they are unique shots? Or does it not matter about the object, but rather the lighting environment? So if I'm in a forest at 2:00 PM on a sunny bright day and I take photos for one hour, should I be constantly adjusting my settings between every unique shot to get the best out of it? Or do I usually just find one setting that works for the environment I'm in? The product of all the elements of a finite abelian group I'm trying to prove the following statements.  Let $G$ be a finite abelian group $G = \{a_{1}, a_{2}, ..., a_{n}\}$.   If there is no element $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = e$.   Since the only element in $G$ that is an inverse of itself is the identity element $e$, for every other element $k$, it must have an inverse $a_{k}^{-1} = a_{j}$ where $k \neq j$. Thus $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$.   If there is exactly one $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = x$.   This is stating that $x$ is not the identity element but is its own inverse. Then every other element $p$ must also have an inverse $a_{p}^{-1} = a_{w}$ where $p \neq w$. Similarly to the first question, a rearrangement can be done: $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot xx^{-1} \cdot \cdot \cdot  a_{n}a_{n}^{-1} = xx^{-1} = e$. And this is where I am stuck since I proved another statement.    Any comments would be appreciated for both problems. I am flying from the US to Edinburgh, UK. My layover is at Heathrow. Will I need to go through customs and check bags again at Heathrow before I take the next flight? Way to list "Contact Us" email address on web site, yet reduce likelihood of spam? How do I programmatically set the value of a select box element using JavaScript? Why is saltatory conduction in myelinated axons faster than continuous conduction in unmyelinated axons?