Forcing arbitrary figure location
Force figure placement in text
I own a 2016 Macbook (with the USB-C charging) - at the time and in response to all of the reports of poor cables and poor chargers, I bought a meter so I could see what the throughput of my cables/chargers was. However, having run some tests I'm now not sure what to consider normal / what is harmless to the battery.   For all tests I used an original Apple C-C cable - though was able to replicate with two third party cables I own.  I have the original Apple block charging at 14.3 V / 1.99 A and a third party block that charges the same.  Another third party block which advertises USB PD and 3 A is currently charging at 20 V / 1.43 A  I've done some research online but I can't find any articles that indicate what a normal USB PD output is relative to the devices that connect to it and would appreciate any guidance / assistance.   TL;DR: Have Apple charger which charges at one voltage/amperage. Bought several others advertised at PD which are higher voltage, lower amperage. Not sure what is 'safe', or 'fast', or does least overall damage to battery. Water Wars: Time Travel TV Show? Uses of the definite article (the) in generic noun phrases How could Luke hide from Vader during the duel on Death Star II? In Return of the Jedi, Vader throws his lightsaber at Luke causing the catwalk come crashing down. Luke uses this in his advantage and hides from him under the throne room platform. Unable to locate his son, Vader tried to lure Luke out of hiding.  But how could Luke conceal himself from Vader in the first place? Did he use a Force power of some kind to blur Vader's sensors? I mean, the underneath of the platform was pretty much an open space with very little possibilities for hiding. The only places I see in the movie are the openings on the both sides of the floor (pointed by the yellow arrows):    The background of a close-up to Luke doesn't give much of a clue (although there is a round shape which could be the bottom of a window frame):    But as Luke reveals himself, he appears to be beneath of the staircase to the dais, right in the middle of it:    There's no way he could have hidden there all the time, because just a couple of seconds before, Vader walked towards and stared at the very same spot:    My first ideas were that Luke used the Force to make himself invisible, or to teleport him to the position. But after taking a look at one of the cross sections of the Emperor's throne room, I assume Luke could have slid almost straight under the floor of the lower level (yellow arrow) and then later  back to the staircase (although we don't see it in the movie, and I'm not sure if Luke knew the ability):    Does someone know better? How to Check If A Plugin Is Enabled Through API? How to capture Control+D signal? I was playing around with some integrals and noticed that some integrals of the form:  $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form. I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.  $$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$ Let's start by letting $x-1=t\,$ thus: $$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$ With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$ $$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$ $$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$ $${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$ $$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$ Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue. $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$ Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$ There are more examples that I found by checking and try such as:  $$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$ $$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$ $$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$ And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.   The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$? pgfmathanglebetweenpoints always returns 90 What is the difference between 'tell' and 'say' How to have a figure with caption in multicol environment? I am having the following problem with adding a figure inside the multicol environment:  \documentclass[10pt, twoside]{book} \usepackage{graphicx} \usepackage{subfig}  \begin{document}   \begin{multicols}{2}  a test file for picture inclusion along with the captions and labeling \begin{figure}  \centering  \includegraphics[width=5cma]{Cauchy_sequence_illustration-png}\\  \caption{Lets see}\label{pinki} \end{figure} \end{multicols}  \begin{figure}  \centering   \includegraphics[width=5cm]{Cauchy_sequence_illustration-png}\\  \caption{Lets see}\label{pinki} \end{figure}    \end{document}    After this while I am pressing Ctrl + Shift + X (usually I compile like this in DVI format and then clicking "dvi--->pdf" I convert it into PDF document.) the picture in multicol environment is not shown in YAP Viewer but the second one (which is outside of multicol environment) is shown along with the caption.  Kindly help me how to handle this problem. Please note that I am using the figure in .ps format. And I am not allowed to use any other format according to the instruction. What good are free groups? In , one learns two definitions of free groups associating with sets.        Let $A$ be a set, the free group of $A$, $F(A)$ is the initial object in the category $\mathcal{C}$, where \begin{equation} \operatorname{Obj}(\mathcal{C})=\{A\xrightarrow{g}G\}, \end{equation}where the codomain $G$ are groups, and \begin{equation} \operatorname{Hom}(\mathcal{C})=\{\text{Commutative Diagrams } A\xrightarrow{g_1}G_1\xrightarrow{\phi}G_2\xleftarrow{g_2}A\}  \end{equation} where $\phi$ are group homomorphisms.   Also $F(A)$ has the concrete construction with elements being non-redundant words with alphabet $A$, and the group multiplication being juxtaposition and reduction.      But what good are free groups? Why are they useful?  says they are useful in topology, but does not explain why explicity.   Can someone give some examples that an undergraduate can understand? I am asking about examples that can show the usefulness of this abstract construction. So I guess the identification of $\mathbb{Z}=F({a})$ does not really count. How can I resize NTFS partition in GParted? How a Gmail username can be taken but not recognised simultaneously? Being at  I enter a desired username (without @gmail.com which is supposed), and I get error:     "That username is taken. Try another."   But when I try to login into Google with that xxx@gmail.com at  it fails with:     "Sorry, Google doesn't recognise that email."   How that can be?  P.S.: I want that username still. I installed CCSM to enable window previews in Unity. But when I went into extras, there was no 'window preview' option.  So I tried experimenting with other options in Compiz and now there are no borders in the dialogue boxes (close, minimize, restore buttons are not there, I had enabled some 'flat....' option in Compiz.  Later, I uninstalled Compiz and then reinstalled it. Now it says      "Failed to load session Ubuntu"   I can't even access through the guest account, I am new to Ubuntu.(please explain in detail).  All I have access to is the login screen (it still shows my custom wallpaper). Please help, I don't want to lose my stuff. What recourse do I have if I believe a moderator has abused their privileges? Recently, the TeX.SE question  got onto the Hot Network Questions list.  Like many questions on TeX.SE, it contains some TeX markup in its title.  Unfortunately, on , this markup gets parsed and rendered as math:    I propose that either:   the entire Hot Network Questions sidebar should be excluded from MathJax parsing by styling it with the tex2jax_ignore class, or at least, any hot questions from TeX.SE should be so excluded from MathJax parsing.   Or I suppose we could always implement , which asks for all questions with TeX in the title to be excluded from Hot Network Questions entirely.  That might be throwing out the baby with the bathwater, though.    Ps. This issue is partially masked by , which causes MathJax not to be rendered in many elements that are hidden by default and revealed using jQuery .show(), such as all but the first five questions in the Hot Network Questions sidebar.  The  user script contains a generic fix for that bug, so users who have it installed will see this issue more frequently, but it occurs even without SOUP whenever a question with TeX markup in the title happens to show up among the top five hot questions that are always visible by default. Is Ghost Adventures real or fake? Calculating MEAN and STANDARD DEVIATION confussed Magic 8 Ball Problem This problem is probably simple enough to have an analogous problem, I just don't know the name so I'm going to describe it and hopefully somebody can point me in the right direction.  The problem is this: estimating the number of sides of a .   Let's say you perform the following process:   Shake the "Magic 8 Ball" and mark down the result If the result has not previously been seen, add it to the set of possible results and label the trial "N" (for "new") If the result has been previously seen, label the trial "O" (for "old") Repeat all steps   I imagine the following kind of sequence occurring: NNNNONNOONOONOOONOOOONONNOONOOONNOONOOOOOONOOONOOOOOOONOOOOOOOONOOOOOOOO...  Now imagine we don't know there are twenty sides to a Magic 8 Ball. Or imagine that you have a Magic 8 Ball with 1000 sides. As the number of "new" trials approaches the actual number of sides, we'll get increasingly more "old" trials showing up in the mix. Once we've seen all of the possible results, we'll always get "old" results. But we're never 100% sure we've seen every single possible result.  So here are the questions I'm interested in:   As we proceed with trials, can we estimate the total number of "sides" of the Magic 8 ball based on the number of "new" and "old" trials up to this point? Can we calculate a probability that our current estimate is correct, or a probability that some bounded estimate is correct? Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$  I rearranged it   $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$  My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?  Thanks