Is it possible to rotate an HDRI background without rotating the lighting it gives off? I know that you can go into shading, world, and then add a mapping node and that rotates it, but that also rotates the light, I wanna keep the light in the same place but rotate the background. Is there any way to do this? Thanks!
I figured out how to use a normal image as the background while using an HDRI for environmental lighting. But I can't figure out how to use a different HDRI as the background.  Please note I already read this answer:  but this appears to be different for Eevee in 2.8.  Thanks!
Generate all combinations from multiple lists The topic which haunted me for two years until I gave up on it. But now I am doing engineering and this topic suddenly popped out of my textbook from nowhere. I seriously need to understand this topic because I wasn't able to do so in the past. I have read many book on 'energy' and it came to nothing. Maybe just because the many books I have read all said 'energy' is something which doesn't 'exist'! It is 'abstract'! It is just a number which represents the state or orientation of a system. But then I see so many examples that 'uses' energy to do 'work'. But the question here is that if anything doesn't exist in this universe, then how can the same thing be used to do something that exist?  My problem is that I didn't understood the topic 'energy' and all the other topics related to it (work, power etc.) Why did my TeX-related GUI program stop working in Mac OS X Yosemite? What lifetime can be expected of the typical hard disk? Or are there big differences between different types? And does it make a difference if it is used heavily instead of never being connected to system (for example serving as a backup medium)? How do I show that $a^p + b^p > (a + b)^p$ for $a,b>0$ and $0<p<1$? A question on my math homework asks us to show that if $0 &lt; p &lt; 1$ and $a, b &gt; 0$, then $a^p + b^p &gt; (a + b)^p$. I have no idea how to do this, any pointers? How can I make a .jar file executable? I'm trying to run a jar application under Ubuntu, so I installed OpenJDK Java 7 Runtime, but when I open this application I got this message :     The file '/home/aimad/Programms/jMerise/JMerise.jar' is not marked as   executable.  If this was downloaded or copied from an untrusted   source, it may be dangerous to run.  For more details, read about the   executable bit. why does OSX add extra '._' when I tar a directory? I'd just like to know why if I "tar -czf" a file/directory, osx adds a:  ._  for each file? I see these when I untar in linux. Or when I work with the uncompressed project in eclipse as it doesn't like them at all. I use 10.7.5. Is there a way to mix LIKE operator with IN ? Something Like  SELECT Id FROM Account WHERE Name LIKE IN:nameList   Where nameList is list of String? Is this is possible ? (I can make a dynamic query by doing OR with all the elements, but was looking for a more elegant way to do this) Rejecting a PhD offer after having accepted it. Advice? I have accepted a PhD offer from a very good university to do a project very close to my interests. I have told the project supervisor I accept it. In the meanwhile I have managed to secure a PhD offer from another 2 universities much better in world rankings and with better research groups and better faculty. Additionally these 2 universities do research exactly on what I want to devote my PhD research.  My only commitment to the first university is a mail confirming I accept. Nothing else. I understand declining the offer after already having accepted it is quite dishonest but it is a huge decision that will affect my future in a very significant level.   I would like your opinion and I would also like to ask for possible consequences of such an action. As far a postdocs concerned, in the first university there is no chance I ever get a postdoc since they do not actually do research on my area (except for the prospective supervisor and even he not as much).  UPDATE I have decided to do as I was afraid to. I have decided to choose a different university than the first one that I have already accepted the offer. My offer acceptance included:    An online "I accept" statement. An e-mail I sent stating "I accept the scholarship and this University is my first option"   I do not have any legal commitments since I have not signed anything. The supervisor of the first university though is really nice to me and continuously was sending me mails. How do I very nicely tell her I will not be joining in the end? Additionally the place I will go in the end will be for research in Mathematics and not Physics. Can I use this as a part of a sensical excuse?  WHAT I HAVE CHOSEN AND WHAT THE CONSEQUENCES WERE  I think that I should let people, especially other beginning grad students know what happened with my case. I have chosen to nicely tell the first university I got accepted that I will not be attending and will not be pursuing a PhD there since I decided I have different research interests (which is true). I had applied there in the first place since I was not sure what I wanted to do and because it was a nice program. Despite that, I got an offer from a World top university to work on 100% my research interests. I explained this to them, the first uni, and they happily let me go and wished me good luck, especially the supervisor who faught a lot secure me funding. I was lucky. Does Blender support dual-GPU rendering if the two GPUs aren't the same? I have a GTX 1060 6GB in my current build, and I also have a GTX 960 2GB in storage. I was wondering if I got , and got both of my GPUs in one build (not SLI or CrossFire, just in the same system) would I be able to use both GPUs for rendering simultaneously? I don't care if this means having them both render separate tiles in one instance of Blender, or rendering separate frames in two instances. I'm using Cycles for everything I do. Limit $\frac{x^2y}{x^4+y^2}$ is found using polar coordinates but it is not supposed to exist. Consider the following 2-variable function:     $$f(x,y) = \frac{x^2y}{x^4+y^2}$$   I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$.   I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:  $$ x = r \cos \theta, \qquad y = r\sin\theta $$   Hence,  $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$  This simplifies to,  $$ \lim_{r \to 0} \frac{r^3 \cos^2\theta\sin\theta}{r^2(r^2\cos^4\theta + \sin^2\theta)}$$  Simplifying $r^3/r^2$, we finally get;  $$\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$  Now from the above, we find that as $r \to 0$ the limit is $0$.   I wanted to verify this answer so I checked on Wolfram Alpha. Explicitly without changing to polar coordinates, it said that the limit does not exist at $(0,0)$ and rightly so. Then how is it that with polar coordinates, the limit exists and is $0$? Am I doing something wrong in this method?  Also, what should I do in this situation, and when should I NOT use polar coordinates to find limits of multi-variable functions? Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means Prove that if $\lim_{n \to \infty}z_{n}=A$ then:  $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|&lt;\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps.... Is “fortnight” an outdated word? I'm studying English vocabulary books, I find this word in each unit describing time words (days, weeks, months...).      There are 2 weeks in a fortnight.   Almost I haven't seen this word before in social websites, emails or movies. the only place I heard this word is in historical movies.  So, is it a outdated word or native speakers still use it? Error: Declaration of MyClass::start_lvl() should be compatible with that of Walker_Nav_Menu::start_lvl() I'm currently working on a WordPress site and whenever in the admin I go the Appearance > Menus page I get the following error:     ErrorException: Runtime Notice: Declaration of Walker_Nav_Menu_Edit::start_lvl() should be compatible with that of Walker_Nav_Menu::start_lvl() in wp-admin/includes/nav-menu.php line 203   It seems like some sort of PHP compatibility issue. I'm running PHP 5.3.10 on my local dev machine.  What would be the best way to remedy this problem? Show that $\mathbb{E}(T) = \sum\limits^\infty_{k=1}\mathbb{P}(T \geq k)$ for $T$ nonnegative integer valued and $E[T] < \infty$ Let T be a non-negative integer-valued random variable with $\mathbb{E}(T) &lt; \infty $. Prove that $\mathbb{E}(T) = \sum^\infty_{k=1}\mathbb{P}(T \geq k)$.  Had a few attempts, haven't really got anywhere. I'm wondering as I'm typing this if proof by induction is a good way to go.  Edit: One major thing I forgot to add, am I correct in thinking that also, $\mathbb{E}(T) = \sum^\infty_{k=1}k\mathbb{P}(T = k)$? Using dpkg to install upgrade and dist-upgrade packages Can it happen that the image of a functor is not a category? Full Wi-fi/ethernet signal but no internet Invariance of a maximally entangled state under unitary operation $U \otimes U^\dagger$ Apparently the (d-dimensional) maximally entangled state, $|E \rangle = \sum_{i} |ii\rangle /\sqrt{d}$ is invariant under operations of the form $U \otimes U^\dagger$. I want to prove this result, which amounts to showing that  $$\left(\sum_i U | i \rangle \otimes V |i \rangle = \sum_i | ii \rangle \right)\Rightarrow (V = U^\dagger )$$  I have no idea how to even start. I suppose it's some simple linear algebra result, but I don't see it. A hint would be appreciated. I'm having trouble writing a regular expression that matches valid IPv6 addresses, including those in their compressed form (with :: or leading zeros omitted from each byte pair).    Can someone suggest a regular expression that would fulfill the requirement?  I'm considering expanding each byte pair and matching the result with a simpler regex.