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Is there an equivalent to Google Sketchup? Possible to have 2 unity launchers? If nonnegative $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$, then $\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx >4$ Sum of nth roots of unity Question: If $c\neq 1$ is an $n^{th}$ root of unity then, $1+c+...+c^{n-1} = 0$  Attempt: So I have established that I need to show that $$\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}}=\frac{1-e^{\frac{ik2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}} =0$$ by use of the sum of geometric series'. My issue is proceeding further to show that this is indeed true. Why does $F=ma$? Is there a straightforward reason? Why is force = mass $\times$ acceleration? I have searched in many sites but didn't actually get at it. Simply I want to know that if a mass in space moves (gains velocity thus further accelerates), how can I think, postulate and further believe that force = multiplication  of mass and its acceleration ? RegEx: Grabbing values between quotation marks Is zero odd or even? LCM is associative: $\ \text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$ Definition: The least common multiple of $a.b\in\Bbb{Z}$ is the smallest  $k\in\Bbb{N}$ that is divisible by both a and b.         Proposition: $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))\tag{2}$    Proof: We want to show $\text{lcm}(\text{lcm}(a,b),c)\mid\text{lcm}(a,\text{lcm}(b,c))$ and $\text{lcm}(a,\text{lcm}(b,c))\mid \text{lcm}(\text{lcm}(a,b),c)$ such that $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$.  Let $k=\text{lcm}(\text{lcm}(a,b),c)$, then $c\mid k$ and $\text{lcm}(a,b)\mid k$, but if we let $l=\text{lcm}(a,b)$, then $a\mid l\,\,\text{and}\,\, b\mid l$. Therefore $c\mid k\,\,,a\mid k\,\,\text{and}\,\, b\mid k$, but then $\text{lcm}(b,c)\mid k$. Thus $\text{lcm}(a,\text{lcm}(b,c))\mid k$      Question: Does this makes sense, ifso can I do the same for $\text{lcm}(\text{lcm}(a,b),c)\mid \text{lcm}(a,\text{lcm}(b,c))$? I really get confused when to use say and when to use tell. Which is appropriate in the following:   What did he tell? / What did he say? What are you saying? / What are you telling? I have a PHP script that can encode a PNG image to a Base64 string.  I'd like to do the same thing using JavaScript. I know how to open files, but I'm not sure how to do the encoding. I'm not used to working with binary data. I have a large mesh with > 1.5 million faces. I would like to uniformly add a texture to the object without UV mapping (as blender freezes every time). The texture is quite simple and can be applied evenly throughout the object. I have a project where I have a few layers (State of Texas, cities, roads, etc.). They are all resaved and projected to NAD83 / UTM zone 14N EPSG: 26914. The project CRS is also NAD83 / UTM zone 14N EPSG: 26914 and on the fly projection is enabled. I added a text delimited file and cannot get them to line up. I checked the attribute table of the delimited file and the coordinates are correct. I cross-checked this with a point close to where a few of the points from the delimited file are and they are very close so in theory they should be lining up. I tried to save the delimited file and save it with the same projection and add the shp file back in and that did not work. Anyone have any ideas? I can't think of anything else to try. Catalina not allowing to change read/write permissions despite having SIP disabled Currently Running: macOS Catalina 10.15.1 Issue: Ever since I upgraded to Catalina, macOS is not allowing me to change read and write permissions on system protected files, hence, I am not able to tweak the system as per my likes and dislikes. Steps Tried To Solve The Issue: SIP, always remains . So, no question arises regarding any conflict with system integrity protection. I have also tried to force change the permissions on the system files I want to edit using certain brute force commands, which I can't recall the name of, right now. But, I have tried. As of now, I am still unable to change read and write permissions on system protected files and resources. I have 2 monitors connected to a NVIDIA GeForce GTX 660:   Left is 1920x1080 via DVI (Samsung SyncMaster 2443BW) Right is 2560x1440 via DisplayPort and is set to Primary (Samsung SyncMaster SA850)   I often leave my computer running overnight (working on long render job, doing backups, etc) so I only power off my monitors manually. However, if I power off the primary monitor, it is disabled completely as if it were unplugged from the graphics card (i.e. icons are moved, open apps are moved and resized to fit on the new primary monitor). If I switch from DP to a DVI cable, this does not occur. It seems to be an issue when using DisplayPort.  Is there a way to disable the monitor from being disabled completely in Windows when I power it off physically?   I have found no relevant settings in either Windows Display settings, nor the NVIDIA control panel software.  Edit #1: I'm using an .  Edit #2: Here is a screenshot of my NVIDIA Control Panel v7.2.710.0:   Edit #3: Here are screenshots of my Monitor configuration:    Clicking on Advanced settings reveals a Generic PnP monitor running at 59 Hz? 60 Hz is the only other option available.    Here is my secondary monitor which is connected via DVI:    Edit #4: If I open the "Screen Resolution" configuration panel and power the monitor off (via the power switch), this is what is shown:    It looks like I disconnected the monitor from the video card. jQuery get the rendered height of an element? How do you get the rendered height of an element?  Let's say you have a &lt;div&gt; element with some content inside. This content inside is going to stretch the height of the &lt;div&gt;. How do you get the "rendered" height when you haven't explicitly set the height. Obviously, I tried:    var h = document.getElementById('someDiv').style.height;   Is there a trick for doing this? I am using jQuery if that helps. Arbitrary precision spline interpolation If the laws of random variables weakly converge to a point mass at $c$, then the random variables converge in probability to $c$. This is question 10.3.1 from Rosenthal's First Look at Rigorous Probability.  Suppose $\mathcal{L}(X_n) \Rightarrow \delta_c$ for some $c \in \mathbb{R}$. Prove that $\{X_n\}$ converges to $c$ in probability.  Here $\mathcal{L}(X_n)$ refers to the distribution of $X_n$.  Is the following proof correct?  Is a simpler proof available?  Is there a simple proof using Skorohod's theorem?  I attempted to prove the contrapositive. Suppose $X_n \not \to c$ in probability.  This means that there is an $\varepsilon &gt; 0$ so that for all $N$, there exists an $m &gt; N$ such that  \begin{equation}   P(|X_m - c| \geq \varepsilon) &gt; \eta, \text{ for some $\eta &gt; 0$}. \end{equation}  Consider  \begin{equation}   B := \{x \in \mathbb{R} : |x - c| \leq \varepsilon\} \in \mathcal{B} \text{ the Borel sets on $\mathbb{R}$}. \end{equation}  Then $B^c \in \mathcal{B}$ and denoting the distribution of $X_n$ as $\mu_n$ we have  \begin{equation}   \mu_n(B^c) = P(X_n \in B^c) = P(|X_n - c| \geq \varepsilon) &gt; \eta. \end{equation}  For all $N$, there exists an $m &gt; N$ such that the cumulative distribution function of $X_m$ is not the same as the cumulative distribution function of $\delta_c$; in particular it's mass changes somewhere at least $\varepsilon$ distance from $c$. Finally because $\delta_c(\{x\}) = 0$ for all $x \neq c$ this completes the proof. Breaking of diffeomorphism invariance after fixing a background metric Prove: If $A$ is a finite set and $B$ is an infinite set, then $B \setminus A$ is an infinite set.   I've tried proving the contrapositive, that if $B \setminus A$ is a finite set then $A$ is infinite or $B$ is finite, but I don't think this is the way to go. What does closed form solution usually mean? This is motivated by  and the fact that I have no access to Timothy Chow's paper  indicated there by Qiaochu Yuan.  If an equation $f(x)=0$ has no closed form solution, what does it normally mean? Added: $f$ may depend (and normally does) on parameters.   To me this is equivalent to say that one cannot solve it for $x$ in the sense that there is no elementary expression $g(c_{1},c_{2},\ldots ,c_{p})$ consisting only of a finite number of polynomials, rational functions, roots, exponentials, logarithmic and trigonometric functions, absolute values, integer and fractional parts, such that   $f(g(c_{1},c_{2},\ldots ,c_{p}))=0$.