I set a bevel on a curve, but supposedly there is a way to scale vertices of a curve so that they also scale the radius of the corresponding bevel. How can this be accomplished?
Is there a way to scale up the bevel object on one end of a curve, a little bit over each point?  I notice the vertex "weight" slider will tweak how much influence each vertex has on the others, is there something like this that can scale the bevel object up a little bit?  Preferably with the option to constrain X,Y,Z independently
How to Truncate a string in PHP to the word closest to a certain number of characters? I have a code snippet written in PHP that pulls a block of text from a database and sends it out to a widget on a webpage.  The original block of text can be a lengthy article or a short sentence or two; but for this widget I can't display more than, say, 200 characters.  I could use substr() to chop off the text at 200 chars, but the result would be cutting off in the middle of words-- what I really want is to chop the text at the end of the last word before 200 chars. UnboundLocalError: local variable 'url_request' referenced before assignment Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. Any tips or hints would be greatly appreciated. Installing Guest Additions causing problems I am trying to install Guest Additions in VirtualBox with Windows XP host and Ubuntu 12.04 guest. I followed the above steps:  Devices-&gt; Insert Guest Additions CD Image. After that I locate the folder /media/VBoxAdditions4.3.4_91027/ and run sudo ./VBoxLinuxAdditions.run   I want to be able to create shared folders and adjust the resolution of the screen. I achieved to do the first issue, but I couldn't do the second one. I didn't get any message, however it is not possible to change the resolution of the screen and I cannot switch to fullscreen. What am I doing wrong?  I found in the instal log the following message: unknown version of the X Window System. I've tried the solution of the proposed link but I got the following message:  Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies: virtualbox-guest-x11 : Depends: xorg-video-abi-11                     Depends: xserver-xorg-core (&gt;= 2:1.10.99.901) E: Unable to correct problems, you have held broken packages.   I tried also to install with the command sudo apt-get install virtualbox-guest-dkms but nothing is changing. During the installation of dkms I got the message: Error! Bad return status for module build on kernel 3.13.0-32-generic(i686). I am having some trouble with the following problem:     Let $K\subseteq L$ be fields. Suppose $\alpha,\beta\in L$ are algebraic elements over $K$ of degrees $p,q$ respectively, where $p$ and $q$ are distinct primes. Show that $\alpha+\beta$ is algebraic over $K$ with degree $pq$.   I have shown that $\alpha+\beta$ is algebraic over $K$, but I am having trouble showing that the degree is $pq$. I have previously shown that $[K(\alpha,\beta):K]=pq$, so I have been trying to use this result. I thought that if I could show that $K(\alpha,\beta)=K(\alpha+\beta)$ then I'd be done. To show this I first noted that clearly $K(\alpha+\beta)\subseteq K(\alpha,\beta)$. Then we have $$pq=[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha+\beta)][K(\alpha+\beta):K]$$ so that $[K(\alpha+\beta):K]=1,p,q,pq$. I tried looking at the cases when it is equal to $1,p,q$ and deriving a contradiction, but have been unsuccessful.   Another way I thought of solving this is to show that $\alpha,\beta\in K(\alpha+\beta)$, which would allow me to conclude that $K(\alpha+\beta)=K(\alpha,\beta)$, but I am not sure how to complete this either.    I am looking for some assistance to show that $[K(\alpha+\beta):K]\neq 1,p,q$, or that $\alpha,\beta\in K(\alpha,\beta)$. If you have any other solutions to this problem I would like to see those too. proof of functional completeness of logical operators I am looking for some (may be rarely know to nonnative English speakers) synonyms for  "immensely valuable" or "extremely valuable". What harm can be done with a copy of one's passport? In one of the questions on this site, it was suggested the use of  to save a backup copy of your passport, in case you need to proof who you are if your original document is lost.   I like the idea, but what if someone gets access to your Dropbox folders and is able to acquire the copy of your passport.  The same goes for when you leave a copy of your passport at a hotel receptions desk. Once someone has the info printed on that copy, would that make it easier to steal your identity? How do you disable an application specific hotkey, for example, "Command-P"? In many applications, Command-P is used to print.  My fingers are accustomed to emacs settings and reflexively press Command-P not for printing but for browsing history. It's annoying to see the print pop-up window every time when I press Command-P. I don't need a shortcut to print. How do you block a shortcut globally? Possible Duplicate:    I am using 10.04 LTS - Lucid Lynx. I shutdown the computer while a software update was running. Then after I restarted my computer and while the update manager was running, I got this error:  Could not download all repository indexes The repository may no longer be available or could not be contacted because of network problems. If available an older version of the failed index will be used. Otherwise the repository will be ignored. Check your network connection and ensure the repository address in the preferences is correct. GPG error: http://toolbelt.herokuapp.com ./ Release: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY C927EBE00F1B0520 Failed to fetch  http://ppa.launchpad.net/rabbitvcs/ppa/ubuntu/dists/**DISTRIBUTION**/main/binary-i386/Packages.gz 404  Not Found   I went through the answers to  and I tried this $ sudo launchpad-getkeys  Please wait... launchpad-getkeys is running an update so it can detect the missing GPG keys Trying to import all the missing keys gpg: requesting key 0F1B0520 from hkp server keyserver.ubuntu.com gpgkeys: key C927EBE00F1B0520 not found on keyserver gpg: no valid OpenPGP data found. gpg: Total number processed: 0 launchpad-getkeys has finished importing all missing GPG keys.  Try running sudo apt-get update - you shouldn't see any key  errors anymore  But when I run sudo apt-get update again, it again shows that the key was not found on the keyserver. How to solve this problem? Proof that every finite dimensional normed vector space is complete Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough. Example of non-isomorphic structures which are elementarily equivalent What is the right way to indent your source code? I am a economics major undergraduate student who will apply for grad schools in the coming semester. I think I have quite a strong research experience in mathematical statistics (asymptotic statistics to be specific), but, unfortunately, not much in economics itself. So I am curious if economics departments in graduate schools will care about my research experience in statistics?  Maybe a broader question is if grad schools care about applicants' research experience in a different (maybe even unrelated) field at all? I guess in my case statistics and economics are still quite correlated, but what about other more distant combinations of target departments and previous research experience? I have noticed that some users have taken it upon themselves to create  that allow you to see a user's rank (percentile) among SO developers.  I like this.  I was wondering if the owners of the website are planning something like this in the future?  If so, are you planning on making this by tag?  (I care less about where I rank when compared to developers of a different language).  What I'm looking for is something like this:  C#:      User is ranked #XX (Top XX %) ASP.Net: User is ranked #YY (Top YY %) Removing duplicate strings from an array? Is tipping mandatory at Restaurants and Bars in Germany? Conversion Error setting value for 'null Converter' - Why do I need a Converter in JSF? Necromancy In Harry Potter How is Necromancy accomplished in the Harry Potter universe?  I thought that people cannot be raised from the dead in Harry Potter.  Were bodies just use as puppets that can only do simple tasks, for instance when Dumbledore and Harry were getting the locket, the  were there to guard the locket.  Also, besides Inferi were there any Zombies in Harry Potter? Travel to Australia with new passport (with visa in old passport) I was granted a visa to Australia few months ago. I’d like to travel to Australia in few days. However, I’ve just renewed my passport and I don’t have enough time to inform them about my passport renewal. Can I travel to Australia by carrying both my old and my new passports?