cursor on unity launcher "refreshes" desktop in 16.04LTS
Ubuntu 16.04 Jan 3 2018 update causing screen/access issues
I remember watching this one show when I was a kid in the 80's/early 90's in Canada. It was a live-action show about these two kids and had some puppet/muppet-like characters, but it kind of had a dark/mysterious element to it. I can't remember too much about it except portions of some episodes, but the main characters were kids, and they had some mysterious artifact (kind of like a compass?) that took them to this other world where there were giants and other mysterious beings that we rarely saw. The two kids weren't the only one's going to this place either, they find out kids from all around the world that also had this same artifact (and they all fit together like a round puzzle?). I seem to recall that the giants in this other world were the ones who made the artifacts, and were dying/or in desperate need of the help of the kids.  A couple episodes I remember:   There was one where the leaves were falling, and the world was doomed to end as soon as the last leaf touched the ground, so they're trying to figure out how to prevent all the leaves from falling, but after a long and suspenseful episode (for a kid) the last leaf dramatically falls, but out of no where the main girl catches the leaf and saves the world (it was very exciting for my young self, and I remember being very impressed with the pretty girl for thinking to catch the leaf).     There was another episode where the kids meet these two old guys with long white hair (muppets) sullenly working in a mine/tunnel, they're in tattered old clothes working away with their pick axes or hammers and they are connected to each other at the beard. This long old white beard grows out of one of their faces right into the other's. The kids save the day by cutting the old guy's beard in the middle, freeing them from each other. They celebrate being free from their bonds and excitedly thank the kids, but after the kids leave they continue to work away, this time in much brighter spirits. Seemingly unbeknownst to them though, as they work, their beards slowly and subtly grow back together.   Anyone else remember this show? I can type:  dirname ~/home/blah/file.zip   for instance, and this works fine alone, but when I use this syntax as the parameter for some command it always reads 'dirname' as the desired directory. i.e.:  cd dirname ~/home/blah/file.zip bash: cd: dirname: No such file or directory   Basically, how do I get 'dirname file' to be read as one entity? What is the name of this game involving racing to the other side of the board and putting down barriers? Does anyone recognize this game, where players take turns in a square board, each turn a player can move his pawn or place a wall and the objective is to reach the other side of the board before the other player.  Here is a picture of a near end game. Evaluating limit $\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}$ Did Anakin and Obi-Wan use the Force to protect themselves from the heat on Mustafar? While duelling that close to lava, I would suspect that they would get dangerously hot and dehydrated, and possibly burned. Did they use the Force to protect themselves in some way, for example lower their body temperatures and cool the air around them while duelling? Is it possible to write any bounded continuous function as a uniform limit of smooth functions Is $C^\infty(\mathbb{R})\subset C_b(\mathbb{R})$ dense? I.e. is any continuous bounded function $f:\mathbb{R}\to\mathbb{R}$ the uniform limit of smooth functions?   On any bounded interval this is true since by Stone-Weierstrass polynomials are smooth and dense in the continuous functions. I guess the result is still true for $\mathbb{R}$ but I don't know how to prove it. Since I don't have any integrability conditions on $f$ the technique of convoluting with mollifiers is probably not applicable. What's a big-vocabulary word for someone with a big vocabulary? What's a big-vocabulary word for someone with a big vocabulary?  I'd like to say I'm looking for a "_____". How can I get a new browser session when opening a new tab or window on Firefox/Chrome? When opening new Internet Explorer window, a new browser session is issued. Therefore you can login with two different accounts on the same service (let's say ). The same does not apply when opening a new tab in the same window in Internet&nbsp;Explorer.  How can I get a new browser session when opening a new tab or window on Firefox?  How about Google Chrome? Residue theorem with $\frac{1}{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) d\theta$. The following integral should be doable using the residue theorum:     $$\frac1{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) \,d\theta$$ Prove by mathematical induction that: $\forall n \in \mathbb{N}: 3^{n} > n^{3}$ Prove by mathematical  induction that:   $$\forall n \in \mathbb{N}: 3^{n} &gt; n^{3}$$  Step 1: Show that the statement is true for $n = 1$:  $$3^{1} &gt; 1^{3} \Rightarrow 3 &gt; 1$$  Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$  The general idea I had was to start with $(p+1)^{3}$ and during the process substitute in $3^{p}$ for $p^{3}$ as an inequality.  $$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 &lt; 3^{p} + 3p^{2} + 3p + 1$$  Now, if it can be shown that:  $$\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}$$  ...the proof is complete. This is because $3^{p+1} = 3 \cdot 3^{p}$ and one of those three have already been used.   We do this by mathematical induction. First, the base case of $n = 1$:  $$3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}$$  ..which turns out to be false.  What are some more productive approaches to this step? Scale camera to fit screen size unity My background and gameobjects are 2d sprites not UI images ,I want to know how to make my background scale with screen size and positions of 2d sprits on it don't change after building game (webGl) like scale canvas with screen size and anchor points in UI elements .  Before building the game   After building Seeking Symbology for Interactive Mapping? When a user clicks on a map what should they expect to see?   I would say a symbol for where they clicked.  A pushpin seems to be the most accepted icon for generic data. Google has pushed the upside down tear drop.   What are other ideas/options and is there a place for finding professional gis oriented icon packs?  I ended up creating this,     This just popped up from the guys at map box. Clean This is a 2015 IMO problem.  It seems difficult to solve.     Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.   Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$. Why is gradient the direction of steepest ascent? $$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$ The definition of the gradient is  $$ \frac{\partial f}{\partial x_1}\hat{e}_1 +\  ... +\frac{\partial f}{\partial x_n}\hat{e}_n$$  which is a vector.  Reading this definition makes me consider that each component of the gradient corresponds to the rate of change with respect to my objective function if I go along with the direction $\hat{e}_i$.  But I can't see why this vector (defined by the definition of the gradient) has anything to do with the steepest descent.  Why do I get maximal value again if I move along with the direction of gradient? Why is the wild card character * so different between the zip and rm commands? I put together a script to do some file operations for me. I am using the wild card operator * to apply functions to all files of a type, but there is one thing I don't get. I can unzip all files in a folder like this  unzip "*".zip   However, to remove all zip files afterward, I need to do   rm *.zip   That is, it does not want the quotation marks. The unzip, on the other hand, does not work if I just give it the * (gives me a warning that "files were not matched").   Why is this different? To me, this seems like the exact same operation. Or am I using the wild card incorrectly?   in Unix do not really go into this, and I could not locate anything in the rm or zip docs.  I am using the terminal on a Mac (Yosemite). I recentely tried to log-in / sign-up for  using my SE account (which works on ,  and sign-up here) but it showed me an OpenID related error. (See below)  I could reproduce this error across multiple days and browser refreshes and using my other OpenID provider.  Error message (shown in red, obfuscated my e-mail address):  Unable to log in with your OpenID provider:  The openid.return_to parameter (http://area51.stackexchange.com/users/authenticate/?s=bd394805-cf55-4e8e-ad73-21485a09a989&amp;dnoa.userSuppliedIdentifier=https%3A%2F%2Fopenid.stackexchange.com%2Fuser%2F642b1038-5c91-40d4-83ab-eb92db46c3d1) does not match the actual URL (https://area51.stackexchange.com/users/authenticate/?s=bd394805-cf55-4e8e-ad73-21485a09a989&amp;dnoa.userSuppliedIdentifier=https%3A%2F%2Fopenid.stackexchange.com%2Fuser%2F642b1038-5c91-40d4-83ab-eb92db46c3d1&amp;openid.claimed_id=https%3A%2F%2Fopenid.stackexchange.com%2Fuser%2F642b1038-5c91-40d4-83ab-eb92db46c3d1&amp;openid.identity=https%3A%2F%2Fopenid.stackexchange.com%2Fuser%2F642b1038-5c91-40d4-83ab-eb92db46c3d1&amp;openid.sig=4emyzuA31ChHU0LabPGJXldbN8GeKN%2FFBJXJy69yD10%3D&amp;openid.signed=claimed_id%2Cidentity%2Cassoc_handle%2Cop_endpoint%2Creturn_to%2Cresponse_nonce%2Cns.alias3%2Calias3.mode%2Calias3.type.alias1%2Calias3.value.alias1%2Calias3.type.alias2%2Calias3.value.alias2&amp;openid.assoc_handle=hrQL%21IAAAAGzBtFeNGBLU2o04XSWr7ISRAbO-B2tZCK3L61IzAsuQQQAAAAEdoTN6rf9fqaP1oekJGXGXJeB8fNSVuByptp5yx1VpQbAzQwF6HJzh6a0DwedydNJ-wwpJQoN5dOLEFmndZt-L&amp;openid.op_endpoint=https%3A%2F%2Fopenid.stackexchange.com%2Fopenid%2Fprovider&amp;openid.return_to=http%3A%2F%2Farea51.stackexchange.com%2Fusers%2Fauthenticate%2F%3Fs%3Dbd394805-cf55-4e8e-ad73-21485a09a989%26dnoa.userSuppliedIdentifier%3Dhttps%253A%252F%252Fopenid.stackexchange.com%252Fuser%252F642b1038-5c91-40d4-83ab-eb92db46c3d1&amp;openid.response_nonce=2015-06-12T12%3A30%3A42Z7tADDeYZ&amp;openid.mode=id_res&amp;openid.ns=http%3A%2F%2Fspecs.openid.net%2Fauth%2F2.0&amp;openid.ns.alias3=http%3A%2F%2Fopenid.net%2Fsrv%2Fax%2F1.0&amp;openid.alias3.mode=fetch_response&amp;openid.alias3.type.alias1=http%3A%2F%2Faxschema.org%2Fcontact%2Femail&amp;openid.alias3.value.alias1=mycorrectaddress%40web.de&amp;openid.alias3.type.alias2=http%3A%2F%2Faxschema.org%2FnamePerson&amp;openid.alias3.value.alias2=SOJPM) the request was made with.    As noted in the comments I'm not the only one, so I'd guess it may be not something I can fix. So will this be fixed?  Repro steps (using SE OpenID):   Get an Stackexchange OpenID account (like I have) (optional?) Add other OpenID providers and check they're working (StartSSL in my case) Please note: "You" don't have an "account" for the Area 51 subpage yet and still need to do the initial sign-in. Go to the Area 51 SE page Hit "Log-in" in the top-bar Hit the "log-in using stackexchange" button Enter your e-mail and password and submit Get the above error   Repro steps (using other OpenID)   Perform as above (until step 4), but this time step 2 is mandatory Hit "more OpenID options" Provide your OpenID link XXX.provider.com (works on crypto, server fault, ...) Authenticate against your provider Receive a variation of the above error (can be included if needed) states that if $a, b \geq 0$, $p, q &gt; 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$ab\leq \frac{a^p}{p} + \frac{b^q}{q}$$ (with equality only when $a^p = b^q$).  Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn't figure it out.  I kept trying to manipulate the expressions algebraically, and I couldn't get anywhere.  But every proof that I've seen since uses calculus in some way to prove this.  For example, a common proof is based on this  and integration.  The proof on Wikipedia uses the fact that $\log$ is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I'm wrong).  Can this be proven using just algebraic manipulations?  I know that that is a somewhat vague question, because "algebraic" is not well-defined, but I'm not sure how to make it more rigorous. But for example, the proof when $p = q = 2$ is something I would consider to be "purely algebraic":  $$0 \leq (a - b)^2 = a^2 + b^2 - 2ab,$$ so $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$ Why does the odd part of the function integrate to $0$? When evaluating the integral $$\int^1_{-1}\frac{\cos(x)}{1+e^{1/x}}dx,$$ the author says that the odd part of the function integrates to $0$ over $-1$ to $1$. I have calculated the even part of the function ($\frac{1}{2}\cos(x)$), so the integral is equal to $\sin(1)$. But why exactly does the odd part integrate to $0$? Why square the difference instead of taking the absolute value in standard deviation? In the definition of standard deviation, why do we have to square the difference from the mean to get the mean (E) and take the square root back at the end? Can't we just simply take the absolute value of the difference instead and get the expected value (mean) of those, and wouldn't that also show the variation of the data? The number is going to be different from square method (the absolute-value method will be smaller), but it should still show the spread of data. Anybody know why we take this square approach as a standard?  The definition of standard deviation:  $\sigma = \sqrt{E\left[\left(X - \mu\right)^2\right]}.$   Can't we just take the absolute value instead and still be a good measurement?  $\sigma = E\left[|X - \mu|\right]$ Another point of view that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is cyclic. I was thinking that the product of groups $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is not cyclic, but $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is cyclic if p is an odd prime. Then I found logic that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is certainly cyclic if $m$ and $n$ are coprimes.  I searched on google a proof of that, to compare with mine and the answer is that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/mn\mathbb{Z}$ (which is clearly cyclic) $\iff$ $m$ and $n$ are coprimes.  But I got another proof and I want to know what you think about it :     I claim that $([1]_m,[1]_n)$ generates $\mathbb{Z}/m\mathbb{Z}$   $\times\mathbb{Z}/n\mathbb{Z}$, where $[a]_d$ is the canonical   projection of $a\in \mathbb{Z}$ into $\mathbb{Z}/d\mathbb{Z}$.       Let's take multiples of $([1]_m,[1]_n)$ until we get $m \cdot$   $([1]_m,[1]_n) = ([0]_m,[m]_n)$ (where I suppose $m=\min(m,n)$...   otherwise, consider $\mathbb{Z}/n\mathbb{Z}$   $\times\mathbb{Z}/m\mathbb{Z}$ and $\min(m,n)=n$).       The next step is $([1]_m,[m+1]_n)$. This pair is not equal to   $([1]_m,[1]_n)$, since $m+1 \not\equiv1$ mod $n$ ($m$ and $n$ are   supposed to be coprimes). We can now continue adding $([1]_m,[1]_n)$   to the results.      The last step is precisely reached when you got   $([mn]_m,[mn]_n)=([0]_m,[0]_n)$. At this point, the fact that $m$ and $n$ are coprimes is absolutely necessary because if they are not coprimes, $lcm(m,n)\ne m\cdot n$ and   $([lcm(m,n)]_m,[lcm(m,n)]_n)=([0]_m,[0]_n)$ but all the $m \cdot n$ possible pairs are not reached, only the $lcm(m,n)$ are !