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Why does the introduced part-title in letter appear twice?
Bound on the difference of two determinants Let $A$ and $B$ be two real, $n\times n$ matrices. Using Hadamard's inequality, it is not hard to show that $$ \left|\det A - \det B \right| \leq \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. $$ Where $\|A\|_2=\sqrt{\sum_{i,j}a_{ij}^2}$.  From this, I can derive a sup bound, for example  $$ \left|\det A - \det B \right| \leq n^{n+1} \|A-B\|_{\infty} \max (\|A\|_{\infty}^{n-1},\|B\|_{\infty}^{n-1}). $$ Where $\|A\|_\infty=\sup_{i,j}|a_{ij}|$.   The constant $n^{n+1}$ is not the best bound possible : any reference (or proof) for a better (or the best) one? I show below that one can obtain $n^2(n-1)^{n-1}$, but that isn't much better. I just tried  $10^5$ random matrices on Maple and obtained a maximal constant (much) smaller  than one : this is not a proof, but it looks like there is room for improvement nevertheless.    Just for completeness (and in case someone sees a factor I missed), to get the first bound, writing $A=[A_1,\ldots,A_n]$ in terms of its column vectors, an expansion shows \begin{eqnarray*} \det A &amp;=&amp; \det (A_1 -B_1,A_2,\ldots,A_n) + \det (B_1,A_2,\ldots,A_n) \\   &amp;=&amp; \sum_{j=1}^n \det (B_1,\ldots, B_{j-1}, A_j -B_j,A_{j+1},\ldots,A_n) \\ &amp;&amp; + \det B,  \end{eqnarray*} Thus by Hadamard's inequality, \begin{eqnarray*} \det A -\det B &amp;\leq&amp;  \sum_{j=1}^n \prod_{i=1}^{j-1} \|B_i\|_{2}\prod_{i=j+1}^{n} \|A_i\|_{2} \|A_j-B_j\|_{2} \\ &amp;\leq&amp;  \|A-B\|_{2} \sum_{j=1}^n \|B\|^{j-1}_{2}\|A\|^{n-j}_{2} \\  &amp;=&amp; \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. \end{eqnarray*}  The second bound is just that $x^n -y^n\leq n \max(|x|^{n-1},|y|^{n-1}) |x-y|$ and $\|A\|_2 \leq n\|A\|_\infty$.    Another approach is calculus, namely, to write that $\det B - \det A = f(1)-f(0)$, with $f(t)=\det(A + t(B-A))$.   By the mean value theorem $f(1)-f(0)\leq \max |f^\prime (t)|$.   We can compute that $$f^\prime(t) =  {\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)$$  (if I did not mess up, using the formula for the differential of a determinant, where Cofm means the matrix of Cofactors).   Then, it should deliver something better, if there is a nice way to bound it. The simplest thing is to use Cauchy-Schwarz, namely $$ \left|{\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)\right|\leq \|B-A\|_{2} \|{\rm Cofm}(A +t(B-A))\|_{2} $$ and then, by lack of a better idea,  $$ \|{\rm Cofm}(A +t(B-A))\|_{2}\leq n \max_{ij} |{\rm Cof}_{i,j}(A +t(B-A))|, $$ and brutally, $|{\rm Cof}_{i,j}(A +t(B-A))|\leq ((n-1) \max(\|A\|_\infty,\|B\|_\infty))^{n-1}$ gives a slightly better constant, namely $$ n^2(n-1)^{n-1} &lt;n^{n+1} $$ but that still seems a very rough way to bound a determinant, as it is never sharp, since to attain this bound all coefficients should be equal, and therefore the cofactor would be zero. They are of the same order in $n$, and I suspect this order is wrong. While browsing Stack Overflow, I noticed that I lost a lot of reputation:        Is this serial downvoting? Will it be corrected by a rep-recalculation? How often are those done?    Here's a second instance: A while back, I read a story in a collection, but I don't remember very much. I know that it was about some missionaries on Mars. Eventually they found a bunch of spheres that apparently have escaped sin by leaving the material world. I don't remember anything else about it. Someone told me that it was in "The Martian Chronicles" by Ray Bradbury, but I checked my copy and couldn't find. I'm pretty sure that it was by Ray Bradbury though. Help would be greatly appreciated. Keyboard Layout indicator doesn't display current layout in Ubuntu 18.04 (empty drop down box) I see an empty drop down box where the language code of the  should be.  To reproduce the bug: Lock screen or log out and log in again.    I also get the same behavior on the lock screen at the time when I try to log in with my password:    After I tried the workaround (below) the language code on the login screen disappeared as soon as I typed the first character of my password.  Workaround (not permanent): Switching with keyboard (Super+space) or mouse.  System info: Ubuntu 18.04.1 LTS, GNOME 3.28.2.    Please  if also affects you. Is there a reason behind the ordering of letters in the English alphabet? Is there a reason behind the ordering of letters in the English alphabet? i.e. why are we taught “A,B,C,D,E,F,...,Z”? Why not “L,A,S,U,I,Z,...,C”?  I am asking this because, in some of the languages I know, I am told that the ordering of the letters in the alphabet is based on the ease with which they can be pronounced or the frequency with which they are used or depending on the part of the vocal cord that needs to be stressed to pronounce the letter.  Is there a similar rationale?  EDIT  You can see a couple of references for arrangement of letters in Sanskrit  and . Nexus 7 2013 error upgrading to 5.0.2 OTA I updated to 5.0.1 via OTA sometime ago, and it worked fine. Now I got a notification that 5.0.2 is ready but when Android shuts down to install update it returns error (with no other message. My Nexus 7 is rooted but has the stock recovery.  Is there an easy fix that doesn't require factory reset or installing recovery? Virtual vs Real image How to make my own Dropbox / Ubuntu One server at home? Does anybody know of any resources that can show me how to make my own "Dropbox, Ubuntu One" server at home?  I really like the idea of these services, but I don't want to put my 'stuff' in the clouds. Ideally, it should have a client that runs on Linux and Windows.  I tried to setup iFolder on my Ubuntu 10.04, but without any success so far. I have an Ubuntu 14.04 server that has no external monitor connected.  I use NoMachine to remote control the machine.  When I do so, the Unity/Gnome interface doesn't see any monitors connected, so I can only use 800x600 when connecting with NoMachine.  If I plug in a monitor, I can set the resolution to whatever size the NoMachine window is on the remote computer.  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For the return trip there is only one option to get the low price.  Selecting the flights this is what I get:      I'm not sure how to use the fare codes the ITA website provides to purchase this itinerary.  These are some of the things I have tried to book the complex multi-carrier flight from ITA:  My first method was to try coding into HIPMUNK, using the multi-city option, my departure, connections, airlines flights number and dates. Following this method:     And the results page:    Here's what my code into HIPMUNK looks like:  ORD::LH9151 FRA LH756 for the first journey and   BOM::LX155 ZRH SN5104 BRU SN8803 for the return trip.  Don't think I'm making any mistakes in coding it, but there are two adjustments I also made which failed:   I used the dates of the first leg of the flight then tried using the other dates as well. So for example Aug 9th and Aug 10th. I used the codes for the operating carriers and when it didn't work, the marketing carrier leaving the flight number unchanged.   This always returned no results. Breaking up the itinerary to just the oneway from ORD to BOM worked but gave me a price of $624. For some reason the second return trip was creating problems. I tried breaking he return trip down into different segments but no luck. Anyone knows why the return trip is creating difficulties? Coding issues/ hipmunk doesn't have such a large database?   Then I tried method 2: Calling up the airlines  United was able to find all the segments, even price it correctly sometimes (sometimes not) and gave me a confirmation number. I was to call them in 12hrs to check if they were able to procure the seats but when I did call they were not able to get the seats for the SWISS flight from BOM to ZRH and/or from ZRH to BRU. I've tried this at least 5 times changing small things everytime, but the same result. It seems Lufthansa waits for SWISS to confirm the seats but SWISS doesn't respond in time and I'm not able to book.  I tried calling Lufthansa but they were not even able to find the segments and instead told me to look at booking online where I will be able to find it cheaper.  Any insights into the methods that failed or suggestions on how to book this itinerary at the $900 price will be much appreciated. Thanks in advance. number of subsets of even and odd Render not showing UV texture I have added a texture to a cube in the UV editor and it shows up fine in textured view  But when I render it there is no texture  What have I done wrong? What's wrong with this query: INSERT INTO Users( weight, desiredWeight ) VALUES ( 160, 145 ) WHERE id = 1;  It works without the WHERE clause. I've seemed to have forgot my SQL. Why upload rates are lower than download rates in normal internet connections? How to close aberrant tooltips in Windows So, here I am, trying to watch The Cleveland Show (TM).      And, as happens from time to time, some program or another has left a modal tooltip.  I don't know which program; how can I lose the tooltip, short of logoff/reboot?  Currently happening on Windows 10, but I have seen it on other versions. Blender Cycles Bake Not Looking Like Object in Rendered View I have a couch in Blender that looks like this in rendered view.    I baked it with a combined texture and this is what it came out to be.    This is what it looks like added to the mesh.    Why does the view with the baked image not look exactly like the rendered view?  Does it have something to do with my mesh and faces?  This is my mesh.    This has never happened to me before as I baked a couple things before this which came out perfectly. Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.     What is the probability that he wears all three shirts after 6 days?   My try:   $$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$  First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.  Thanks in advance. I noticed the other day when serving the public that when asking for something, people were saying "Can I get an xyz, please". The previous time I had such a job it was "Could I have an xyz", or "May I have..." if they were being very polite.  Does anyone know when this trend started and what caused it?