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What makes frictional force the centripetal force of a car turning along a curve? As friction is the opposing force and acts anti-parallel so there is no component of frictional force towards the center,right? Then how can frictional force be centripetal force? Prove $e^x - e^y \leq e |x-y|$ for $x$ belonging to $[0,1]$ I'm not sure how to go about this. Does it involve using MVT?  I got as far as saying $e = \frac{e^x - e^y}{x-y}$. Escaping both forward slash and back slash with sed Can anybody please suggest how I can compute the moment generating function of the inner product of two Gaussian random vectors, each distributed as $\mathcal N(0,\sigma^2)$, independent of each other? Is there some standard result available for this? Any pointer is highly appreciated. grep strange behaviour with single letter words I am removing stop words from a text, roughly using this   I have the following  $ cat file file types extensions  $ cat stopwords i file types   grep -vwFf stopwords file  I am expecting the result: extensions  but I get the ( I think incorrect)  file extensions   It is as if the word file has been skipped in the stopwords file. Now here's the cool bit: if I modify the stopwords file, by changing the single word/letter i on the first line, to any other ascii letter apart from f, i, l, e, then the same grep command gives me a different and correct result of extensions.  What is going on here and how do I fix it?  I'm using grep (BSD grep) 2.5.1-FreeBSD on a Mac OSX GNU bash, version 4.4.12(1) Who is the eponymous Jedi in Return of the Jedi? I've working LDAP authentication with the following setup   AuthName            "whatever"  AuthType            Basic  AuthBasicProvider   ldap  AuthLDAPUrl         "ldap://server/OU=SBSUsers,OU=Users,OU=MyBusiness,DC=company,DC=local?sAMAccountName?sub?(objectClass=*)"  Require ldap-group  CN=MySpecificGroup,OU=Security Groups,OU=MyBusiness,DC=company,DC=local   This works, however I've to put all users I want to authenticate into MySpecificGroup. But on LDAP server I've configured that MySpecificGroup also contains the group MyOtherGroup with another list of users.  But those users in MyOtherGroup are not authenticated, I've to manually add them all to MySpecificGroup and basically can't use the nested grouping. I'm using Windows SBS 2003.  Is there a way to configure Apache LDAP to do this? Or is there a problem with possible infinite recursion and thus not allowed? I have always heard the term used in referring to a single word. When browsing questions on this site, I've seen it used applied to entire phrases, and have suppressed the compulsion to edit them and replace the term with origin. What are some CD Ripping Programs you can use on Ubuntu? What are some CD ripping programs for Ubuntu? Can you list the Pros? What are the Cons? For each program there should be a screenshot, sources to install, and instructions to install and use. Convergence of the series $\sum_n \frac{n^{\log(n)}}{\log(n)^n}$ I have a rough outline of a proof. Is this correct?  $$\sum_{n \geq 2} \frac{n^{\log(n)}}{\log(n)^n} \to a \in \mathbb{R} \iff  \sum_{n \geq 2} 2^n \frac{(2^n)^{\log(2^n)}}{\log(2^n)^{2^n}} \to b \in \mathbb{R}$$ via dyadic criterion (or Cauchy condensation test).  Analyzing the right-hand side:  $$\sum_{n \geq 2} 2^n \frac{(2^n)^{\log(2^n)}}{\log(2^n)^{2^n}} = \sum_{n \geq 2} 2^n \frac{(2^n)^{n\log2}}{(n\log2)^{2^n}} = \sum_{n \geq 2} \frac{(2^n)^{n\log2 + 1}}{(n\log2)^{2^n}} = \sum_{n \geq 2} \frac{2^{n^2\log2 + n}}{(n\log2)^{2^n}} = (\ast). $$  We know for some sufficiently large $N$, $n \log 2 \ge 2$ for all $n\ge N$, so we can bound the series above by:  $$(\ast) \leq C_N + \sum_{n \geq N} \frac{2^{n^2\log2 + n}}{2^{2^n}} = \sum_{n \geq N} 2^{n^2\log2 + n - 2^n} = (\ast\ast),$$  where $C_N$ is the finite sum of the original series until $N$. And once again, for sufficiently large $M$, $m^2\log2+m - 2^m \leq -m^2$ for all $m \ge M$. Hence  $$ (\ast\ast) \leq C_{\max(N,M)} + \sum_{n \geq \max(N,M)} 2^{-n^2} &lt; \infty.$$  We then use the $p$-test to conclude that the sequence does converge. `lmodern` package reduces the size of automatically sized delimiters in math mode Should Stack Exchange have a spam filter? (Does it already and I’m not aware?) Of course spam posts should be flagged, and if they come up in review, rejected.  Over on Mi Yodeya during the past 24 hours I’ve helped delete and reject four spam posts through the review queue, and I can only imagine that others were posted over the course of the night. To clarify, I’m referring to edits and answers that were about 99.274% words that I don’t think any decent person would use in everyday discussions, even if someone got them really angry.   Thing is, if the only thing we can do as a community is delete his posts/reject his edits, and as moderators to ban his account, can’t he just create a new account? Further, some of these posts . Maybe the mods can ban his IP address? I’m not one, so I have no idea what kind of power they wield in that respect. But he can always just get a new computer.   I don’t have the numbers to see what this is like on other sites, but I know this doesn’t happen often on Mi Yodeya. I wonder, then, whether this is the optimal solution, or whether an automatic spam filter be put into place - say, a new review queue that automatically deletes posts or rejects edits that can then be voted on whether to be reposted if it’s a false alarm. Perhaps it can be automatically triggered by certain keywords (like ****, %$#^, or @$+^).   Now, all of these posts that I voted to delete/reject came up in Suggested Edits and First Posts. Maybe such edits would show up in Low Quality in general, and that just wasn’t the case here because of precedence rules. I don’t know. But perhaps the above would be a better solution, if this is even as big of a problem as I’m making it out to be. I have been asked by a client to implement some views for reporting purposes, the views will be accessed via PowerBI, Excel and SSRS. The designated user will only have access to the view and no underlying tables must be available to the user.  The problem I have encountered in that the SQL within the view touches 3 different schemas (All within the same database):   Pupil Provider  Security   The view is:  CREATE VIEW dbo.vTestPermissions AS SELECT a.Column1,        b.Column1,        c.Column1 FROM Pupil.Table1 a JOIN Provider.Table2 b ON a.Column1 = b.Column1 JOIN Security.Table3 c ON a.Column1 = c.Column1   The owners of the Tables/View are as follows:   Pupil.Table1 - Owner Pupil Provider.Table2 - Owner Provider Security.Table3 - Owner Security vTestPermissions - Owner dbo   When I select from the view I get an error:     The SELECT permission was denied on the object 'table3', database   'TEST', schema 'Security'   I have tried giving SELECT permission with and without the GRANT option to the schema, and the tables, but that makes the underlying tables available to the user.  Any help on this will be greatly appreciated. How do you get the root namespace of an assembly? Given an instance of System.Reflection.Assembly. How to understand dot product is the angle's cosine? How can one see that a dot product gives the angle's cosine between two vectors. (assuming they are normalized)  Thinking about how to prove this in the most intuitive way resulted in proving a trigonometric identity: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$.  But even after proving this successfully, the connection between and cosine and dot product does not immediately stick out and instead I rely on remembering that this is valid while taking comfort in the fact that I've seen the proof in the past.  My questions are:   How do you see this connection? How do you extend the notion of dot product vs. angle to higher dimensions - 4 and higher? Firefox 37.0.2 not opening (removing firefox did not help) runing ubuntu 14.04 LTS When running umount /path I get:  umount: /path: device is busy.   The filesystem is huge, so lsof +D /path is not a realistic option.  lsof /path, lsof +f -- /path, and fuser /path all return nothing. fuser -v /path gives:                    USER        PID ACCESS COMMAND /path:            root     kernel mount /path   which is normal for all unused mounted file systems.  umount -l and umount -f is not good enough for my situation.  How do I figure out why the kernel thinks this filesystem is busy? tikz: Ignore one path in bounding box calculation Why can't I use Stack Exchange if I'm under 13 years old, or if I'm under 16 years old and from the EU? Attributing contributions to academic work that occur in Stack Exchange