Google calendar messages (meeting invites and updates, not meeting reminders) have the header  mailed-by:  calendar-server.bounces.google.com   Alas, filtering by mailedby:calendar-server.bounces.google.com or mailed-by:calendar-server.bounces.google.com does not work.  How can I filter the Google Calendar messages?  At this time I am using   subject:("Invitation:" OR "Canceled Event:" OR "Declined:" OR "Accepted:")   but this does not include meeting updates because their subject sometimes starts with [Update] and Gmail filters cannot handle brackets (even with backslashes &amp;c - believe me, I know regexps).
I'd like to create a Gmail filter to filter all of my Google Calendar notifications into a separate folder.  Google Calendar notifications have headers that look like the following:  Reply-To: Helen Seu &lt;h...@g...com&gt; Sender: Google Calendar &lt;calendar-notification@google.com&gt; Auto-Submitted: auto-generated Date: Thu, 01 May 2014 05:25:28 +0000 Subject: Updated Invitation: Mobile design review @  Fri May 9, 2014 3pm - 4pm From: Helen Seu &lt;h..@g...com&gt; To: "Mike Burton" &lt;m...@g...com&gt;,    In particular, the email Reply-To is the user who created the meeting, and the Sender is calendar-notification@google.com.  For some reason, Gmail does not seem to allow you to filter on the Sender field of an email. Is there another clever way to filter Google Calendar notifications to a separate folder?
How to modify itemize vertical spacing within blocks in beamer? I have noticed that when I have "itemize" environments defined within blocks in beamer, that they introduce more vertical spacing between the top and the bottom of the blocks, than vertical text. Since I have a slide with some text filled blocks and directly after, a slide with only "itemize" holding blocks, I would like the spacing between the itemized text and the block to be the same as for regular text.   On  I have read about "squeeze" option, but it doesn't seem to work...   Additional info:  The problem is in the additional vertical space that is included by the "\itemize" environment. I have found  on how to remove vertical spacing from the itemize list, and I am using this approach. However, it is a brute force approach. What I would like to know if it is possible to have the same vertical space between the end of the itemize list and the bottom of a block, as the one set between the normal text and the bottom of the block.   MWE:   \documentclass[10pt]{beamer} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{exscale} \usepackage{epsfig} \usepackage{fancybox} \usepackage{lmodern} \usepackage[T1]{fontenc} \usepackage{overpic}  \usepackage{multimedia}  \usetheme{default} \usecolortheme{crane}  % -- Modifications that enable the footline and set black color for the fonts  \setbeamertemplate{footline}{%     \begin{beamercolorbox}[wd=\paperwidth,ht=2.25ex,dp=1ex,right, rightskip=1mm, leftskip=1mm]{titlelike}         \textcolor{black}{             %\inserttitle\hfill\insertauthor\hfill\insertframenumber%             Something\hfill Something Else \hfill\insertframenumber%         }     \end{beamercolorbox} }  \setbeamertemplate{itemize items}[square]  \setbeamercolor{item}{fg=black} \setbeamercolor{frametitle}{fg=black} \setbeamercolor{block title}{fg=black} \setbeamercolor{title}{fg=black} \setbeamercolor{block body}{bg=white}  % END Modifications  \begin{document}   \begin{frame}   \frametitle{text blocks}     \begin{block}{block}         block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   blockk     \end{block}{block}      \begin{block}         block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block     \end{block}     \begin{block}{block}         block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block   block     \end{block} \end{frame}  \begin{frame}   \frametitle{itemize blocks}     \begin{block}{item}         \begin{itemize}             \item item              \item item              \item item         \end{itemize}     \end{block}     \begin{block}{item}         \begin{itemize}             \item item              \item item              \item item         \end{itemize}     \end{block}     \begin{block}{item}         \begin{itemize}             \item item              \item item              \item item         \end{itemize}     \end{block}   \end{frame}  \end{document} How do I calculate CRC32 of a string How do I calculate the CRC32 (Cyclic Redundancy Checksum) of a string in .NET? Why do professional print houses use gamut-limiting CMYK? Why does 1+++2 = 3? How does Python evaluate the expression 1+++2?  How many ever + I put in between, it is printing 3 as the answer. Please can anyone explain this behavior  And for 1--2 it is printing 3 and for 1---2 it is printing -1 Why is $\sqrt{-x}*\sqrt{-x}=-x?$ How to count occurrences of one field grouped by values of another? Prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$. $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$ then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)&lt;0$.  Attempt: Let us define a function $h(x) = g(x) g~'(x)$. Then $$h'(x) =  g(x)g''(x) + \left( g'(x) \right)^2 \tag 1$$ If we prove that for some $c \in (-3,3), h~'(c) &lt; 0,$ then $$g(c)g''(c) &lt;0 \tag 2$$ Also, $$\left|f(0)\right| &lt; 1 \implies f'(0) \in (-3,-2\sqrt 2 ) \cup (2\sqrt 2,3) $$ Could someone please advise me how do I move forward from here. Thank you very much for your help in this regard. Calculating $\int_{- \infty}^{\infty} \frac{\sin x \,dx}{x+i} $ I'm having trouble calculating the integral  $$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$  using residue calculus. I've previously encountered expressions of the form   $$\int_{- \infty}^\infty f(x) \sin x \,dx $$  where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$  How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work What measures does Cognito Forms take to ensure Privacy/ Security Level for Completed Forms/ User Data? Some forms may contain confidential information - even something basic like where a person lives.   What measures do Cognito Forms take to ensure that data stored within its site is secure? GRUB does not detect Windows Why can't I use the Delete key to delete in Photoshop CS6? The previous version of Photoshop I used was CS4 and whenever I selected a portion of an image with the magic wand and hit Delete, it simply deleted that portion of the image (rendering it transparent). Easy!  In CS6, this is not the case. Now when I select a portion of an image and hit Delete, it brings up the Fill dialogue.  For example I have this picture (it's a Vauxhall Corsa). I want to use the Magic Wand to render anything that isn't part of the car transparent, so I start by highlighting some clouds with the Magic Wand like so:     Click on images to see larger versions  Now when I hit Delete, instead of that area of the picture being deleted as I would expect based on my experience with CS4, I get this dialogue:    My questions are:   Can this dialogue be used to achieve the functionality I want? If so, how? If not, how do I delete the background in CS6? What is Quantization? What is a suitable word to describe a place where two rivers meet? I've googled for a while and on some sites I've found the word "watershed" as the proposed word. Is it the word that best suits it? I forgot my iPhone 5 passcode when I was changing the code. A window appeared on the device screen saying that iPhone is disabled.  I have Apple ID and password for it. After signing in to iCloud I have erased the device in order to remove the passcode from my iPhone but it still says that device will be erased when device will connect to internet. What would i have to do? How to close entire application when application has more than 2 forms without showing messagebox more than once Find the value of $x,y$ for the equation $12x+5y=7$ using number theory . I tried the solution using other ways like:  $12x+5y=7$  $12x=7-5y$  $x=7-5y/12$  putting the value of $x$  $12(7-5y/12)+5y=7$  both $12$ will cancel out  $7-5y+5y=7$ here comes the problem please help me ? How old should you be when you begin Jedi training? NULL values inside NOT IN clause Linear regression in a loglog-plot with fixed slope Goal: create a plot to compare the empirical convergence rate of a numerical method with the one expected from the theory.  Accordingly, given a loglog-plot, I would like to add the line which has a prescribed slope (in this case, the convergence of the method) and the best intercept to fit the data (e.g., in a least-square sense).  Question: how to do a linear regression with fixed slope in pgfplots?  What I have done so far is the following: (1) compute the regression line (of the logarithm of the quantities), (2) save the intercept, (3) plot the line with the slope expected from the theory and this intercept.  This is a good approximation of what I am looking for, provided the data are in agreement with the theory.  Thanks  MdFB Showing group with $p^2$ elements is Abelian I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.   My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.   I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).   First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?  Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).   Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.