Change pivot point to selected point In the attached picture, that the model is in the center of the point. I want to change from pivot point to point, where I want to picture it. In Maya 2017 Student version works like this, when it is possible to change with the D keypad.  Blender is so possible?  Thank you in advance for your help!
Set origin to vertex in edit mode? I want to make the origin in edit mode at a vertex, but it keeps setting it to the last clicked-on face :( So now in edit mode the pivot is on the last clicked-on face ;(
Bold name in bib when having a prefix in last name Speed $v$ of a wave on a tightrope I have the following data structure (a list of lists)  [  ['4', '21', '1', '14', '2008-10-24 15:42:58'],   ['3', '22', '4', '2somename', '2008-10-24 15:22:03'],   ['5', '21', '3', '19', '2008-10-24 15:45:45'],   ['6', '21', '1', '1somename', '2008-10-24 15:45:49'],   ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ]   I would like to be able to   Use a function to reorder the list so that I can group by each item in the list. For example I'd like to be able to group by the second column (so that all the 21's are together)  Use a function to only display certain values from each inner list. For example i'd like to reduce this list to only contain the 4th field value of '2somename'    so the list would look like this   [      ['3', '22', '4', '2somename', '2008-10-24 15:22:03'],       ['7', '22', '3', '2somename', '2008-10-24 15:45:51'] ] Prove $$\int _0^\infty f^2(x) dx \leq \frac{2}{3}\cdot  \max_{x \in \mathbb R^+} f(x) \cdot  \int _0^\infty f(x) dx$$ for $f(x) \geq 0$ and convex.   I know via Holder's we can get without the constant $\frac{2}{3}$ but that should also be true for $f$ not convex. I tried some approaches using convexity but didn't get the desired result.   A hint or reference would also help. Also, it would be interesting to know when equality holds. Are mirror lenses good for wildlife photography in daylight? I've recently seen these Rokinon/Opteka mirror lens with whooping focal length (like 500mm, 600mm, 800mm, 1000mm) at a cheap price. And I have seen some of their same pictures in Flickr and vendor sites. But the only thing is that they usually starts from smaller aperture (f/5.6 and upwards)   Now what I was wondering, as they are cheap with quite good IQ - are they good for any bird photography or wild life photography in proper daylight? Checking for valid floating point number How to handle fragmentation of auto_increment ID column in MySQL User group (company) page that shows summary of all user accounts belonging to the same company/user group How to determine if $n (n \geq 2)$ given numbers can be summed up to a certain number? I'm interested in knowing if there's a theorem for checking if combinations of $n$ given positive numbers can be summed up to a certain number.   For example, if I was given $n = 2$ numbers, which are $k_1$ and $k_2$, and I was asked if I could make a combination like $n_1 * k_1 + n_2 *k_2 = C$, all $\in \mathbb{N}^+$, I could check remainders to see if it is possible.   However, if $n = 3$ or higher, how could I check if combinations of $k_1, k_2, ..., k_n$ can be summed up to a $C$? Thank you for any inputs! Does Mathematica have a "MATLAB Builder" equivalent? I am new to blender. Pressing &#9096; Ctrl&#9095; AltQ opens the quad view. How to maximize any one of the four views of the quad view? Are the rules regarding absolute modifiers too absolute? A common grammar lesson that was taught to me in the US and that I've had to teach abroad in EFL classrooms is that we're not to use adverbs of emphasis with absolute modifiers, just as we're not supposed to use them as comparatives or superlatives.  Classic egregious examples of this mistake include very unique and more perfect, which seem obviously flawed to me. Other instances of further modified absolute modifiers are similarly meaningless or contradictory in nature.  I wonder, though, if the rule that I learned isn't overly broad, and whether it's taught similarly in the UK and elsewhere. I find myself modifying absolute modifiers quite regularly, and in some circumstances I think it's not only logical and correct, but quite meaningful. Phrases like almost exactly or virtually all or practically infinite possess a specific meaning that can't really be replicated using nonabsolutes like very nearly or most or quite long.  In my own speech and writing, I use such phrases exactly as I see fit, so this isn't a request for permission, per se. I'm more curious about whether there's a more appropriate rule regarding absolutes that you rely on (and might be used in a classroom). For symmetric operators one has: $$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$ How to prove this in an unveiling way? Old TV mini-series with a living brain in a lab Lets assume my classes are loaded from a compressed .war file or loaded elsewhere,  how can I discover all the resources in a given package?  Enumerating files will not really work, since this is a war file.  Most likely this will involve using the current classloader?  Is there a library out there that does something like that? Googling revealed only some hacks with listing files. The StackExchange™ MultiCollider SuperDropdown™ says I've got replies, but I've already viewed those. The following question is taken from Royden's Real Analysis $4$th edition, Chapter $3,$ question $28,$ page $67:$     Question: Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and $f$ is finite a.e., that is.       Assume $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise almost everywhere on $E$ to the real-valued function $f$ which is finite almost everywhere. Then for each $\varepsilon&gt;0,$ there is a closed set $F$ contained in $E$ for which    $$\{f_n\}\to f \text{ uniformly on }F \text{ and }m(E\setminus F)&lt;\varepsilon.$$   My attempt:   Let $A = \{ x\in E: f_n\not\to f \text{ pointwise} \}$ and $B=\{ x\in E: |f|=\infty \}.$  By assumption,  $$m(A)=m(B)=0.$$ So $A$ and $B$ are measurable.  Note that  $$m[E\setminus (A\cup B)] = m(E) - m(A\cup B) = m(E).$$ Fix $\varepsilon&gt;0.$ Since $\{f_n\}$ converges to $f$ pointwise on $E\setminus (A\cup B)$ to the real-valued function $f,$ by Egoroff's Theorem, there exists a closed set $F$ contained in $(E\setminus (A\cup B))\subseteq E$ for which  $$f_n\to f \text{ uniformly on } F \text{ and }m[(E\setminus(A\cup B)) \setminus F] &lt; \varepsilon.$$ Since $E$ has finite measure and $F\subseteq E,$ by monotonicity,  $F$ has finite measure. As $F$ is closed, it is also measurable.  Therefore, by Excision property, we have $$m(E\setminus F) = m(E) - m(F) = m[E\setminus(A\cup B)] - m(F) = m[(E\setminus(A\cup B)) \setminus F]&lt;\varepsilon.$$  Is my proof correct? I ask for verification because in this  Kenny's comment about countable union of null sets. I do not use this anywhere in my proof. So I wonder whether my proof miss out something. Drive shape key based on relative camera orientation? How does Apple AppStore rank search results? Superman has been exploited by Poison Ivy in Hush, and he's definitely mad in Injustice. What about Batman? Has Batman ever had to be brought to justice/stopped/killed by his fellow superheroes? (That one time when Superman dreams of Batman snapping Joker's neck doesn't count)  Edit: The Dark Knight Returns doesn't count: Batman wasn't really mad, and he wasn't wrong either.