I made something with the mirroring tool and now when I'm sculpting there's a gap between the two sides? How can I fix that?
When I add a mirror modifier it sometimes doesn't mirror correctly, giving me an overlapping mesh. How can I fix this?
How would one prove that $$\lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}=\varphi$$  where $F_n$ is the nth Fibonacci number and $\varphi$ is the Golden Ratio? Can I use a higher rank tensor field as the metric field? What a metric should necessarily do is it should give me a way to associate a frame invariant number with a given pair of spacetime events. Now, if I use a higher rank tensor field (say, for example, a tensor field of rank 3: $g_{\mu \nu \rho}$) then also I can certainly produce a frame invariant scalar out of a given displacement $\vec{A}$ in this trivial way: $ I := g_{\mu \nu \rho} A^{\mu}A^{\nu}A^{\rho}$. Since the components of the higher rank tensor and that of the displacement vector are going to transform in a covariant and a contravariant manner respectively, the cooked up quantity is certainly a scalar.   Another crucial property, which I think has more direct physical content than the previous one, is that between two inertial frames, at least one such transformation should exist that leaves at least one metric invariant. i.e., There should exist at least one combination of transformation matrix $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}$ and metric $g_{\alpha \beta \gamma}$ that satisfies the following equation:      $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}\displaystyle\frac{\partial{x^{\beta}}}{\partial{x^{\nu '}}}\displaystyle\frac{\partial{x^{\gamma}}}{\partial{x^{\rho '}}} g_{\alpha \beta \gamma} - \delta_{\mu '}^{\alpha}\delta_{\nu '}^{\beta}\delta_{\rho '}^{\gamma} g_{\alpha \beta \gamma} =0$    The last thing I can think of that can put a restriction on the choice of a tensor as a metric is the existence of a possibility of finding a metric compatible symmetric connection field. Following the usual procedure of finding the expression for a metric compatible symmetric connection field, I reached following condition (unlike the case of the usual two rank metric where we get a full-fledged expression) for the connection in the terms of the metric:      $g_{\nu \rho k_1} \Gamma^{k_1}_{\mu \lambda} - g_{\lambda \mu k_2} \Gamma^{k_2}_{\rho \nu} = \displaystyle\frac{1}{2} (\partial_{\nu}g_{\rho \lambda \mu} + \partial_{\rho}g_{\lambda \mu \nu}  - \partial_{\lambda}g_{\mu \nu \rho}  - \partial_{\mu}g_{\nu \rho \lambda}) $   My question is that if it is possible to satisfy the two highlighted conditions then can we use such a 3 rank (or even higher rank tensors with similarly produced conditions) tensor fields as metric fields?   PS: This is NOT a proposal for a new home-production theory of gravity (or that of anything for that matter) but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric. Thank you. Is there offside when the ball deflects off an opponent? I was refereeing a 10 year old's junior game today and this situation happened. The home team's defender attempted to  pass from their own half to their forward who was in an offside position beyond the halfway line.  I was going to blow the whistle but the ball deflected off an away defender and then went to the originally offside attacker who put the ball wide.  I viewed this as onside at the time because the ball struck the away team player, however now I think it was offside because at the time of pass, the home player was offside.  Is this offside or not? What is the difference between a "link function" and a "canonical link function" for GLM What's the difference between terms 'link function' and 'canonical link function'? Also, are there any (theoretical) advantages of using one over the other?  For example, a binary response variable can be modeled using many link functions such as logit, probit, etc. But, logit here is considered the "canonical" link function. Wifi doesnt work on fresh ubuntu 16.04 Usage and origin of "sister" in expressions like "sister company, sister ship, sister site" etc The term  is often  used figuratively  to refer, for instance,  to a “sister company” for a company  within the same group, or to a “sister  site” for  sites that belong to the same family. This connotation as explained by the   means:         belonging to a pair or group of similar and related things, such as businesses, usually owned or operated by the same person or organization:         our sister company in Australia.          also, from :        You can use sister to describe something that is of the same type or is connected in some way to another thing you have mentioned.    ⇒ ...the International Monetary Fund and its sister organisation, the World Bank.   ⇒ ...Voyager 2 and its sister ship, Voyager 1.       This usage appears to be common with things that are  regarded as feminine and are associated, as if by kinship,  with other similar things that belong to the same group. An early  usage example of this is the "sister" referred to  ships:         the US battleship Missouri and her sister ship, the Wisconsin.      In other instances,  the "feminine" issue is less explicit as in the case of internet sites, so I guess this usage has to do with the fact that English is less gender specific when it comes to things or abstract entities.   Questions:    Does “sister” apply to whatever entity that belongs to the same group irrespective of its real or perceived gender? or could "brother" be used instead?  Where does this usage come from? How to monitor disk activity? How do I view my Disk Activity on Ubuntu? For example, How can I monitor what is being written and read from disk as well as the percentage of disk usage?  I am looking for an application similar to the Resource Monitor in Windows. If $q$ is coprime to $a$ then $a\mid nq-1,\,$ so $q$ is invertible mod $a$ I think the following is true:  If $q$ is coprime to $a$ then $a\mid (nq-1)$ for some $n\in\mathbb Z$.  I have started to first reducing $n\in\mathbb Z$ to $n\in\{1,\ldots,a-1\}$ and then considering all the cases $q = ka + j$ for $j=1,2,3,\ldots,a-1$. I stopped at $j=3$ and found that this is kinda messy. Does anyone have an elegant proof for this (maybe using some results from number theory)?  [Note: question has been generalized from prime $q\nmid a$ to $\,q$ coprime to $a$] How do I design a user friendly survey with more than 150 questions? I am designing UI/UX for a survey webapp which has more than 150 questions. Right now they are using 5 radio buttons for each question. Similar to the image below:      So the question is, how to make it super user friendly, easy and fun to do?   Further I am looking for a solution from responsive web design perspective too.  I hate designing it like the following image below but what are some other options. I have a valid B1/B2 US visa expiring about a year from now. The passport it's in expired a few years ago, and I recently got a new one.   The passport number does not match, but the national ID number, mentioned in both, does (it used to be the same as your national ID, but not anymore)  Can I travel to the US carrying both passports? Or is it asking for trouble? (I read a few years ago that it was possible, but that might have changed)  Should I just get a new visa to avoid potential complications? How would I be able to get N results for several groups in an oracle query.  For example, given the following table:  |--------+------------+------------| | emp_id | name       | occupation | |--------+------------+------------| |      1 | John Smith | Accountant | |      2 | Jane Doe   | Engineer   | |      3 | Jack Black | Funnyman   | |--------+------------+------------|   There are many more rows with more occupations.  I would like to get three employees (lets say) from each occupation.  Is there a way to do this without using a subquery? Why do some questions not start with an auxiliary verb? When I learned English, my teachers told me that all questions must have an auxiliary verb at the beginning, just like Are you mad? or Is she playing? do.  But when watching some movies or talking with people who speak English, they just ask using things like You mad? and She’s playing?   Of course, the kind of intonation let me know they are asking a question, but why this happen? I have two LED lamps in my corridor, and when I switch them off, they keep emitting light (not very strong, but it's annoying at night).  What can it be? How can I fix it? How do I mount shared folders in Ubuntu using VMware tools? I'm very confused about tenses.  I have examples in two different situations here. Situation 1: I went to New York two months ago and was talking to someone about it.  A. This is the first time I've been to New York. B. That was the first time I'd been to New York.  Which is correct in a daily conversation?  Situation 2: I'm going to New York next month and talking to someone.  A. This will be the first time I've been to New York. B. This is the first time I've been to New York. C. This will be the first time I go to New York.  Which sentence is appropriate?  According to a grammar article I've read before, I think it says that I can use this sentence in both cases:  This is the first time I've been to New York.  But I'm not sure if this is really so. So I'd like to know which tense I should use in each case. Could you explain it? Create a shortcut for URL? Possible Duplicate:             I faced this issue today when I was replying to a comment posted which used @Transaction annotation.     @Trevor: @Transaction and ... @Transaction   When typing this comment, I got a warning. Multiplicative group $(\mathbb R^*, ×)$ is group but $(\mathbb R, ×)$ is not group, why? $\mathbb R^*$ refers to $\mathbb R$ without zero.  Please explain the statement "$\mathbb R$  is not a group under multiplication, it is a group under addition." in the . Basically:   Why is multiplicative group $(\mathbb R^*, ×)$ a group? Why is $(\mathbb R, ×)$ not group? Why are $(\mathbb R^*, +)$ and $(\mathbb R, +)$ groups? Computational complexity of Fibonacci Sequence In the movie Interstellar,  Cooper (played by ) enters a black hole and sends a message to Murph (his daughter). Later, he is found by rangers, but I don't understand how he gets out of the black hole?  As far as I know, nothing can get out of a black hole, as it has extremely high gravity.