I found the following strange behavior. Adding some floats result in "random" accuracy.  so first I run  go version go1.12 darwin/amd64 on macOS Mojave (10.14.3) with Intel i7 2,6 Ghz  the behavior occur in the following Example:  func TestFloatingAddition(t *testing.T) {     f1 := float64(5)     f2 := float64(12.1)     f5 := float64(-12.1)     f3 := f1 + f2 // 17.1     f4 := f3 + f5 // 5.000000000000002     if f4 != f1 {         t.Fatal("addition is not reversable")     } }   Can someone explain to me why f4 takes on this strange value, and what can I do to fix it?
Consider the following code:   0.1 + 0.2 == 0.3  -&gt;  false   0.1 + 0.2         -&gt;  0.30000000000000004   Why do these inaccuracies happen?
Upgrade to the latest PHP version in Ubuntu Bootstrap 3 - Why is row class is wider than its container? I know that it holds for $n=5$, so the first step is done.  For the second step, my IH is: $4^n &gt; n^4$, and I must show that $4^{n+1} &gt; (n+1)^4$. I did as follows:  $4^{n+1} = 4*4^n &gt; 4n^4$, and if I show that $4n^4 &gt; (n+1)^4$ holds for $n\geq 5$ I'm done, so I expand the right hand side and arrive to:  $$4n^4 &gt; n^4 + 4n^3 + 6n^2 + 4n + 1$$  And this holds iff: $$3n^4 - 4n^3 - 6n^2 - 4n - 1 &gt;0$$ And here I'm stuck, I don't really know how to show that the given inequality holds. Any hints? When I use the listings package with frame in my code the frame "breaks" when the page ends and it is open in the end of the page. I want the frame to be complete with four sides and continue in the next page again with all four frame sides complete. How can I do this? Why does the frame "breaks"? How should I detect unnecessary #include files in a large C++ project? As in the title I have to evaluate this triple integral: $$\iiint_{[0,1]^3}\frac{dx\,dy\,dz}{(1+x^2+y^2+z^2)^2}$$ I've been trying to solve this since a week ago. The first thing I've done was understand the meaning of the integral. I think this integral represents the mass (as an example) of a unitary cube which contains materials of different density. The values of the materials density are, point for point, the inverse of the square of spheres centered in the origin plus one. The max value of the density is $1$ in the origin of the cube and the min value is $\frac{1}{16}$ on the opposite vertex. I suppose that the value of the integral is $\frac{\pi^2}{32}$ I've tried to use simple substitutions without any results, so I tried to change the coordinates with spherical and cylindrical systems. The spherical coordinates give me an incredibly long sum of integrals and I doubt that they're all integrable as elementary functions. The cylindrical gives me the following result $$\frac{\pi^2}{16}-\int_0^\frac{\sqrt2}{2}{\frac{\arctan{\sqrt{\frac{u^2-1}{u^2-2}}}}{\sqrt{2-u^2}}du},$$ which I'm not able to solve. My instinct tells me that there is a trick in some steps where I can observe that a difficult integral actually is exactly half of another one simpler but I can't figure out where. I'll appreciate any kind of suggestions. Mount a VirtualBox drive image (vdi)? Is it possible to mount a VirtualBox drive image (.vdi) so the contents can be viewed in Nautilus etc.?  I have a windows 2000 .vdi which won't boot ("inaccessible boot device") after upgrading from VirtualBox 2.x to 3.1.6. I believe the IDE drive details have changed and that all I need to do is access the internal drive image and edit the Windows boot.ini to point to the new location. Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$ I'm looking for an algorithm which when given a latitude and longitude pair and a vector translation in meters in Cartesian coordinates (x,y) would give me a new coordinate. Sort of like a reverse Haversine. I could also work with a distance and a heading transformation, but this would probably be slower and not as accurate. Ideally, the algorithm should be fast as I'm working on an embedded system. Accuracy is not critical, within 10 meters would be good. $x^3+x+1$ irreducible over $\mathbb{F}_2$, $\theta$ root, compute powers of $\theta$ Problem statement:     Show that $x^3+x+1$ is irreducible over $\mathbb{F}_2$ and let $\theta$ be a root. Compute the powers of $\theta$ in $\mathbb{F}_2$.   I am having trouble computing the powers of $\theta$. $x^3+x+1$ is irreducible, since any cubic polynomial over a field is reducible if and only if it has a root in the field. Since $\mathbb{F}_2=\{0,1\}$, we see that none of these is a root of $x^3+x+1$.  To compute the powers of $\theta$, note that $\theta^3+\theta+1=0$. Hence $\theta^3=-(\theta+1)$. Is this a good enough answer for $\theta^3$?  To compute $\theta^2$, note that $\theta = -(\theta^3+1)$. Therefore, $$\theta^2=(\theta^3+1)^2 = \theta^6 + 2\theta^3 + 1 = (-\theta-1)^2-2\theta-2+1,$$ which just reduces to $\theta^2$. Any ideas for computing these powers?  Thanks! What is the best word describing the nature of expiring in a short amount of time, connoting "losing public attention"? Is there an adjective that describes the nature of expiring, connoting a sense of losing public attention, in a very short amount of time? For example,  In a fast-paced era, news articles are ____. During the 1990s, compared with traditional technologies, computer  technologies are ___; for example, a new generation of CPU technologies usually comes out every one to two years.  (I thought about &quot;perishable,&quot; but it does not seem connoting &quot;losing public attention.&quot;) How do I check the status of the CMOS battery? Blender Freestyle : Vertical lines On Blender , a picture is shown with two Suzannes made of vertical/horizontal lines.    Click to enlarge  How to reproduce this effect with Blender Freestyle? Using modifiers? Custom shaders? ERROR running force:source:pull: Entity of type 'CustomApplication' named 'Contact' cannot be found Do strong and weak interactions have classical force fields as their limits? Electromagnetic interaction has classical electromagnetism as its classical limit. Is it possible to similarly describe strong and weak interactions classically? I really tried hard to do this on my own, but I keep failing ... Using multi-column and multi-row commands confused me. I know some of you might frown upon such a request but I really need help getting this table implemented .. So I'm sorry if I upset anyone.  My preamble has :   \documentclass[12pt,a4paper]{report} \usepackage[lmargin=3.81cm,tmargin=2.54cm,rmargin=2.54cm,bmargin=2.52cm]{geometry} \linespread{1.5}     Is there a way I can convert this from excel to LaTeX form ? I tried an add-in before but it didn't work for me. Also online generators cannot combine cells the way I want them (unless I didn't find a better table generator ? ) Automatically put PGFPlots xtick scale label in x axis label I'm trying to put xtick scale label (defined by pgfkey /pgfplots/xtick scale label code/.code) in x axis label, between variable name and unit.  I've tried executing using the option xlabel = {$ x $ (\pgfkeysvalueof{/pgfplots/xtick scale label code/.code} m)} to put in but this doesn't work.  I've also tried the following xtick scale label code/.code={\pgfkeyssetvalue{/pgfplots/xlabel}{#1}} to force PGFPlots to change my xlabel but it doesn't work as well.  I know I can do this by hand, but I'm interested in automatic approach. Is there a workaround or a solution to this?  My MWE is:  \documentclass{article} \usepackage{pgfplots} \pgfplotsset{compat=1.14}  \begin{document}      \begin{tikzpicture}     \begin{axis}[         scaled x ticks=true,         scaled y ticks = false,         y tick label style={/pgf/number format/fixed},         %xtick scale label code/.code={\pgfkeyssetvalue{/pgfplots/xlabel}{#1}},         xlabel = {$ x $ (\pgfkeysvalueof{/pgfplots/xtick scale label code/.code} m)},         ylabel = {$ y $},     ]      \addplot[line width=3pt] coordinates {(0.0,0.0)(100,1)(400,2)(10000,0)};      \end{axis}     \end{tikzpicture}  \end{document}   Output from this code is: Are these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ correct? Mirror cells so that a change in either cell updates the other I have 2 cells containing the same picklist values on separate sheets. I want the cells to mirror each other so that whenever I choose a new value in either sheet 1 or sheet 2, the value on the other sheet is updated to match the newly selected value. Is this possible?  UPDATE   Worksheet 1:   Worksheet Name: 'Live Forecast Tool' Cell containing picklist: C3:D3 (merged) Picklist values within range: ='SF MRF'!H5:Y5  Worksheet 2:   Worksheet Name: 'AH' Cell containing picklist: C3:D3 (merged) Picklist values within range: ='SF MRF'!H5:Y5    Cell C3:D3 needs to update to show the value that has been changed on the other sheet. Homemade coconut milk separates into different layers after a day in the fridge