In the book of Linear Algebra by Wernet Greub, at page 206, it is given that     $\phi$ is bounded iff for any $x\in E$, $\exists M$ s.t $$||\phi(x)||  \leq M \cdot ||x||$$   Now let  $\phi : E \to E$ is a bounded linear transformation.Then we denote   $$||\phi|| = sup_{||x|| = 1} ||\phi(x)||,$$ and it follows that    $$||\phi(x)|| \leq ||\phi|| \cdot ||x||, \quad x\in E.$$   However, I couldn't figure out how did he arrived to that conclusion. I mean for $||x|| = 1$, we have $||\phi(x)|| \leq M \cdot ||x|| = M$, hence by defition of supremum, we should have  $$sup_{||x|| = 1} ||\phi(x)|| = ||\phi|| \leq M,$$ hence for an arbitrary $y \in E$ $$||\phi|| \cdot ||y|| \leq M \cdot ||y||,$$ so I couldn't derive the same conclusion.
From the  on the dual of a norm:  $X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number $$\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.$$  At some point in the proof that $L(X,Y)$ is bounded, it seems like they use the inequality $$\|fx\| \leq \|f\|\cdot\|x\|,$$ where $\|x\|\leq 1$.  I can't see why this should holds.   $\|f|\|$ is the supremum of what $|f(x)|$ can be, given that $\|x\|\leq 1$. So, given that $\|x\| \leq 1$, $|fx|$ should be bounded by $\|f\|$. But why should it be bounded by $\|f\|\cdot\|x\|$?
Show group of order $4n + 2$ has a subgroup of index 2. Let $n$ be a positive integer. Show that any group of order $4n + 2$ has a subgroup of index 2. (Hint: Use left regular representation and Cauchy's Theorem to get an odd permutation.)  I can easily observe that $\vert G \vert = 2(2n + 1)$ so $2 \mid \vert G \vert$ and 2 is prime. We have satisfied the hypothesis of Cauchy's Theorem so we can say $G$ contains an element of order 2.  This is where I am stuck  I am confused about how the left regular representation relates to the group. So my understanding at this point is that every group is isomorphic to a subgroup of some symmetric group. My question: is the left regular representation $\varphi : G \to S_G$ an isomorphism? where $G \cong S_G$ or is $S_G$ the same thing as $S_{\vert G \vert}$ and $\varphi$ is only an injection? I'm using Dummit and Foote for definitions.  I saw an argument online that said that since we have an element of order 2, there is a $\sigma \in S_G$ of order 2, but it is a product of $2n + 1$ disjoint 2-cycles. I don't understand how they could claim this and tried working it out on my own but didn't get there. -- They then went on to use the parity mapping $\varepsilon : S_G \to \{\pm1\}$ and since we have an odd permutation $\sigma$, we have $[S_G:\text{ker }\varepsilon] = 2$. I understood their computation but not how that directly shows that $G$ has a subgroup of order 2? unless $G \cong S_G$ because of how left regular representation is defined. (but again, I'm not understanding that concept very well yet.)  So, to be clear about my questions:   What is meant by left regular representation, is it an isomorphism or just an injection? and how would it be used here. If it is an isomorphism, the argument online starts to make more sense, but how can they say that since $\sigma$ is even, it is made up of $2n + 1$ disjoint transpositions? If you have a full, proof, I'd appreciate it, but good hints are just as good!   Thanks you! The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.  I tried the following concept, how to sum 9 using 1-9 only without repetition.  1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,  6:1+2+6 ,7:1+3+5, 8:2+3+4  The answer is 8.  How will it be solved using Binomial theorem. Can I reposition AdSense ads with Javascript (for responsive purposes)? I have a responsive blog where ads are displayed in the sidebar. On mobile devices however, the sidebar is pushed to the bottom below the posts.   So what I'd like to do is move the ads from the sidebar and insert them into the post itself on mobile devices   I do this using javascript:  &lt;div id="main"&gt;     &lt;div id="article"&gt;         &lt;div class="post"&gt;             &lt;p class="intro"&gt;&lt;/p&gt;             &lt;p class="text"&gt;&lt;/p&gt;         &lt;/div&gt;     &lt;/div&gt;     &lt;div id="sidebar"&gt;         &lt;div class="adbox-1"&gt;&lt;/div&gt;     &lt;/div&gt;     &lt;/div&gt;  &lt;script&gt;     jQuery(document).ready(function(){          // AD REPOSITION         if ($(window).width() &lt; 769) {             $('.ad-box1').insertAfter($('.intro'));                    }      }); &lt;/script&gt;   The ads are always loaded in the sidebar, but if the screen width is below 768px, they will be moved somewhere else.  I wonder if this would violate AdSense terms of service? Introduction to Statistical Learning defines the F-statistic as follows: $$\frac{(TSS - RSS) / p}{RSS / (n - p - 1)}$$ I am trying to interpret this formula intuitively - the numerator looks like the $ESS$ per regressor, and the denominator looks like the $RSS$ per observation. This does not seem like an apples to apples comparison; can anyone explain why it makes sense?  They also say that the expectation of the denominator is equal to the variance of the irreducible error if the linear model assumptions hold (I get this, as the denominator is really the Residual Standard Error, which is an unbiased estimator). They also say the numerator is equal to the variance of the irreducible error if the null hypothesis is true. Therefore the F-statistic will be close to 1 if the null hypothesis is true.  But if you let p = 1, that is, apply the F-statistic to single linear regression, it becomes:  $$\frac{TSS - RSS}{RSS / (n-2)}$$  According to the book, if the regressor has no explanatory power, the F-statistic should be close to 1. But if you imagine a data set where the coefficient on X is 0 (i.e. no explanatory power), $TSS$ will be equal to $RSS$, so the numerator and hence the F-statistic should be 0, not 1 as they claim. What is going on?  Furthermore, if you accept that the F-statistic is 1, as they claim, then $(TSS - RSS) = RSS / (n-2)$, which means $ESS = RSS / (n-2)$. If you think of the F-statistics as comparing explained vs. unexplained variation, this does not seem like a fair comparison since this decomposition makes it the ratio of TOTAL explained variation summed across ALL observations vs. the unexplained variation PER observation. Again, what am I missing?  I am just trying to make sense of it in a layman way, apologies if I am missing something obvious. I've taken multivariate calculus and am wondering if I can see a specific function where the order of taking the partial derivative matters. I've been told that there are some exceptions where $ \dfrac{\partial ^2 f}{\partial x \partial y} \ne \dfrac{\partial ^2 f}{\partial y \partial x} $, so I'm curious to see what this looks like.  EDIT: And why would this true? How to debug an unmet dependencies problem? Was unable to find rationale for separate tags accros SE sites.   Tag function According general definition in information systems, a  (such as an Internet bookmark, digital image, database record, or computer file). This kind of metadata helps describe an item and allows it to be found again by browsing or searching. Tags are generally chosen informally and personally by the item's creator or by its viewer, depending on the system, although they may also be chosen from a controlled vocabulary.  On SE - the same:  is a () word or phrase that describes the topic of the question help you identify  or relevant to you.  tag origin and quality  Where from tags are originated and what are tag quality parametres?  problems and contradictions examples related with SE tags  Humanity is facing information overload.  are created for collecting and sharing personal knowledge on questions or problem solving. But tag's descriptions are not matter of personal knowledge - it's just matter of , i.e. vocabulary/dictionary, or taxonomy/ontology or, speaking in general, human lexic conceptual apparatus / framework. So why time has to be spent on fact's duplicating and sharing?  At the same time We are living at the linked data times. IMHO it's the biggest blunder having same tags with different descriptions in different SE sites unlinked. Tags (and so - descriptions) should be unified.  So I'm calling to unify all tags across SE sites as feature request. Maybe automatic linking to Wikipedia and/or some other decent dictionary would be reasonable.  solution examples  This is not a duplicate of Recover harddrive data I have a dell laptop that recently "died" (It would get the blue screen of death upon starting) and the hard drive would make a weird cyclic clicking noises.  I wanted to see if I could use some tools on my linux machine to recover the data, so I plugged it into there.  If I run "fdisk" I get:  Disk /dev/sdb: 20.0 GB, 20003880960 bytes 64 heads, 32 sectors/track, 19077 cylinders Units = cylinders of 2048 * 512 = 1048576 bytes Disk identifier: 0x64651a0a  Disk /dev/sdb doesn't contain a valid partition table   Fine, the partition table is messed up.  However if I run "testdisk" in attempt to fix the table, it freezes at this point, making the same cyclical clicking noises:  Disk /dev/sdb - 20 GB / 18 GiB - CHS 19078 64 32 Analyse cylinder   158/19077: 00%   I don't really care about the hard drive working again, and just the data, so I ran "gpart" to figure out where the partitions used to be.  I got this:  dev(/dev/sdb) mss(512) chs(19077/64/32)(LBA) #s(39069696) size(19077mb)  * Warning: strange partition table magic 0x2A55. Primary partition(1)    type: 222(0xDE)(UNKNOWN)    size: 15mb #s(31429) s(63-31491)    chs:  (0/1/1)-(3/126/63)d (0/1/32)-(15/24/4)r    hex:  00 01 01 00 DE 7E 3F 03 3F 00 00 00 C5 7A 00 00  Primary partition(2)    type: 007(0x07)(OS/2 HPFS, NTFS, QNX or Advanced UNIX) (BOOT)    size: 19021mb #s(38956987) s(31492-38988478)    chs:  (4/0/1)-(895/126/63)d (15/24/5)-(19037/21/31)r    hex:  80 00 01 04 07 7E FF 7F 04 7B 00 00 BB 6F 52 02   So I tried to mount just to the old NTFS partition, but got an error:  sudo mount -o loop,ro,offset=16123904 -t ntfs /dev/sdb /mnt/usb NTFS signature is missing.   Ugh.  Okay.  But then I tried to get a raw data dump by running  dd if=/dev/sdb of=/home/erik/brokenhd skip=31492 count=38956987    But the file got up to 59885568 bytes, and made the same cyclical clicking noises.  Obviously there is a bad sector, but I don't know what to do about it!  The data is still there... if I view that 57MB file in textpad... I can see raw data from files.  How can I get my data back?  Thanks for any suggestions,  Solution:  I was able to recover about 90% of my data:   Froze harddrive in freezer Used Ddrescue to make a copy of the drive Since Ddrescue wasn't able to get enough of my drive to use testdisk to recover my partitions/file system, I ended up using photorec to recover most of my files Export multiple objects to .obj It is possible to export all object from a scene into different obj files (each object in a obj file separately)? Multi-OS Portable TeX system probability of $k$ boxes contain exactly $1$ ball Is it possible to create 3D red/cyan anaglyphs in a hand drawing? There are plenty tutorials online on how to create 3D red/cyan anaglyphs from existing photos in Photoshop or similar software. However, I'm wondering if it is possible to draw a 3D image by hand on for example a whiteboard using markers.  My plan is to tape together a red and blue marker and draw with it using the same rotation/inclination of the marker across the drawing. Will this give a 3D effect when viewed through red/cyan glasses? Or is the 3D anaglyph method more complex than just offsetting red and cyan lines? What happens with the appearance of the folder icons in dark mode of macOS Mojave Finder? In High Sierra I used some custom icons for my folders in the Finder; but with the new update to Mojave and the dark mode, the predefined icons also become darker. However the custom icons don't.  Inspecting an element you find a dark preview, but when you extract the icon this is the clear, the same as there was in High Sierra. E.g.:   Previously visualization of the developer predefined icon. It's dark:  But if you extract that icon, it's clear:    There are no dark icons in the system library, at least not in the same directory as always:  /System/Library/CoreServices/CoreTypes.bundle/Contents/Resources   Are they being actively obscured by the operative system without there being icons with that real appearance? In this case, any idea to apply this method to the custom icons without modifying it into a new coloration? How can I make my laptop boot automatically from the hard drive without having to press F10 each time? I followed the instructions in  to   Once I installed Windows I found that I have to press F10 to select the hard drive every single time to make it boot into Windows. I tried re-enabling Secure Boot but then I am unable to find the hard drive in the bootable devices list.  What should I do?    my labtop is sumsang core i3  i mean that when i wanted to change my windows i can not find cd-rom in the list of F10 i saw the Instructions of this link    i Succeeded to do that and i changed my windows but then i can not go to windows automatically  I had to press F10 to select hard drive and i can go to windows i want To what it was before so I make secure boot (enabled) but I can't find hard drive on the list of F10.  maybe i Erred in doing my new windows  If i changed my windows again i can go to windows automatically or not  please i wait your advice and Instructions 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? I'm exploring the effects of removing the intercept in a logistic regression model.  Assume a model:  $$logit(Y = 1) = \beta_1 x + \beta_2z + 0$$  with $x$ and $z$ being categorical variables with 2 levels each and no intercept.  I understood that having no intercept with categorical predictors produce coefficients that compare the $P(Y = 1)$ in each level of the two predictor against a null case where $P(Y=1) = 0.5$ or $logit(Y=1) = 0$.  I noticed a phenomenon that can understand. Using glm() function in R if you change the order of the variable in the right hand part of the formula, the coefficients change too. But even more oddly, the coefficient of the first variable is always the same.  Here's an R demo:  y &lt;- as.factor(sample(rep(1:2), 30, T)) x &lt;- as.factor(sample(rep(1:2), 30, T)) z &lt;- as.factor(sample(rep(1:2), 30, T))  coef(glm(y ~ x + z - 1, binomial) #        x1         x2         z2  #-0.1764783  0.3260739 -0.1335192  coef(glm(y ~ z + x - 1, binomial)) #        z1         z2         x2  #-0.1764783 -0.3099976  0.5025523    As you can see the first predictors have the same coefficient while the other are different in the two models.  Here is what I expected and instead behave differently than what I though:   Since every level of the two predictors is compared to the same null case, I expected to have the same coefficients in the two models, independently from the order in which I use them. I expected to see the coefficients of every level of every predictor, instead the coefficient for the 1 level of the second predictor is not shown. I therefore assume that only the first variable is compared against the null case, while the second is compared against a reference level; but what is this level? Is it $P(Y = 1 | X = 1 \cap Z = 1)$? Reproducing one of the models WITH the intercept we get:  coef(glm(y ~ x + z - 1, binomial) #        x1         x2         z2  #-0.1764783  0.3260739 -0.1335192  coef(glm(y ~ x + z, binomial)) #(Intercept)         x2          z2  #-0.1764783   0.5025523  -0.1335192    As expected x1 become the intercept, and x2 is likely relative to x1. z1 is missing also in this case and z2 is the same as in the model without intercept.  Thus should I assume that the comparison against the null case $P(Y = 1) = 0.5$ is made only for the first variable in a formula, while the other are compared against the usual intercept? Is this behavior normal? What about the fact that the first coefficient has the same value whichever the order of the predictors in the formula? What if I want to compare all level of each predictor against the null case and have a coefficient for all levels? Or it's theoretically impossible for some reason I don't get? I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived.  For example, when r = 2, the formula is given by: $\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$ according to   Thanks! Metrizability of Hausdorff continuous image of compact metric space Let $f$ be a continuous mapping of a compact metric space $(X, d)$ onto a Hausdorff space $(Y, \tau_1)$. Then $(Y, \tau_1)$ is compact and metrizable.   In one proof the following metric is constructed:  $d_1(y_1, y_2) = inf\{d(a, b) : a \in f^{-1}\{y_1\},\ b \in f^{-1}\{y_2\}\}$, for all $y_1$ and $y_2$ in $Y$.  I'm thinking about how to prove that triangle inequality holds for $d_1$, s.t. $d_1(x, z) \leq d_1(x, y) + d_1(y, z)$  And how about the case when triangle inequality does not hold?    Update: @daniel-schepler suggested to use metric $d_1(x, y) = inf\{d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)\}$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$.  Now we need to prove 3 property of metric: positivity, symmetry and triangle inequality. This is my sketch:   Positivity. Chose two points $x,y \in Y$ s.t. $x \neq y$. Singleton sets $\{x\}$ and $\{y\}$ are disjoint closed sets by Hausdorfness of $Y$ and their preimages are also disjoint closed in $X$ by continuity of $f$. $X$ is compact metric space by hypothesis and therefore a normal space. In normal space every two disjoint closed sets have disjoint open neighborhoods. Let $f^{-1}(x) \subseteq U$ and $f^{-1}(y) \subseteq V$ s.t. $U,V$ are open and $U \cap V = \varnothing$. Therefore there exists an $\varepsilon &gt; 0$, s.t. $x^* \in B_\varepsilon(x^*) \subseteq U$ and $y^* \in B_\varepsilon(y^*) \subseteq U$ and distance between any two closed sets is strictly positive and $d_1$ is positive and zero iff $x = y$. Symmetry. Clearly $d_1$ is symmetric. Triangle inequality. For triangle inequality we can show that in the "worst" case $d_1(x, z) = d_1(x,y) + d_1(y, z)$. What does the colon in Tikz coordinates mean? Altering layer's CRS to correct Layer Misalignment in QGIS I've got a vector layer in my project which isn't aligned with the rest of the project (see attached JPEG).  These are work files so I'm not sure when they were created or how they might have been messed with since they were created.  We moved from ArcGIS to QGIS a couple of years ago.  The infilled blue vector layer is the one I need to move, slightly north and east, so that it correlates to the blue line (also a vector).  All layers (and the project) are currently in EPSG:27700 - OSGB 1936.   If I need to create a new CRS based on OSGB 1936 and give it some new parameters, then how would I go about this?  Further information:  This is on Exmoor in SW England, and the data is uniformly almost exactly 8m out to the SW point to point - i.e I need to shift it 8m NE.