C++ Undefined reference to 'LinkedList::append(int)'
Why can templates only be implemented in the header file?
How to show that any separable space is CCC How was this character killed without Avada Kedavra? "Disappear" as a transitive verb Transparent text cut out of background Why do only Hobbits use a combination of first- and last names? While all the Hobbits I can think of have a combination of first- and lastname (e.g., Bilbo Baggins, Peregrin Took, the Sackville-Baggins's,...), I can't think of any character* of another race that has a lastname:   Men seem to like to add a reference to their ancestors (e.g., Aragorn, Arathorn's son or Isildur's heir) Dwarves seem to like descriptive or honorary names (e.g., Thorin Oakenshield, Dain Ironfoot). They are not lastnames though, since they do not get passed on to descendants. Elves don't have a need for lastnames, since elvish names are unique. Some nevertheless seem to have an additive (e.g., Legolas Greenleaf). Likewise, I can't find any usage of lastnames among Ents, Eagles, Istari, Orcs, trolls,...   But why would a society that is quite rural and not a leader of modernization be the only one to use a modern naming system with first- and lastnames?  *Except for Tom Bombandil. But he is, well, "special". Prove by induction that $\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$ As the title says I need to prove the following by induction: $$\sum_{i=1}^{2^n} \frac{1}{i} \ge 1+\frac{n}{2}, \forall n \in \mathbb N$$   When trying to prove that P(n+1) is true if P(n) is, then I get that:  $$\sum_{i=1}^{2^{n+1}} \frac{1}{i} = (\sum_{i=1}^{2^n} \frac{1}{i} ) +(\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} )$$ using the inductive hypothesis. And for P(n+1) to be true it would be sufficient to prove that  $$1+\frac{n}{2} + (\sum_{i=1}^{2^n} \frac{1}{2^n+i} ) \ge 1+\frac{n+1}{2}$$  or rewriting $$\sum_{i=1}^{2^n} \frac{1}{2^n+i} \ge \frac{1}{2} $$ and here I'm a bit stuck  Thank you for your help! Computing the best constant in classical Hardy's inequality Classical Hardy's inequality (cfr. Hardy-Littlewood-Polya Inequalities, Theorem 327)  If $p&gt;1$, $f(x) \ge 0$ and $F(x)=\int_0^xf(y)\, dy$ then   $$\tag{H} \int_0^\infty \left(\frac{F(x)}{x}\right)^p\, dx &lt; C\int_0^\infty (f(x))^p\, dx $$  unless $f \equiv 0$. The best possibile constant is $C=\left(\frac{p}{p-1}\right)^p$.   I would like to prove the statement in italic regarding the best constant. As already noted by Will Jagy , the book suggests stress-testing the inequality with   $$f(x)=\begin{cases} 0 &amp; 0\le x &lt;1 \\ x^{-\alpha} &amp; 1\le x \end{cases}$$   with $1/p&lt; \alpha &lt; 1$, then have $\alpha \to 1/p$. If I do so I get for $C$ the lower bound  $$\operatorname{lim sup}_{\alpha \to 1/p}\frac{\alpha p -1}{(1-\alpha)^p}\int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx\le C$$  but now I find myself in trouble in computing that lim sup. Can someone lend me a hand, please?    UPDATE: A first attempt, based on an idea by Davide Giraudo, unfortunately failed. Davide pointed out that the claim would easily follow from   $$\tag{!!} \left\lvert \int_1^\infty (x^{-\alpha}-x^{-1})^p\, dx - \int_1^\infty x^{-\alpha p }\, dx\right\rvert \to 0\quad \text{as}\ \alpha \to 1/p. $$  But this is false in general: for example if $p=2$ we get   $$\int_1^\infty (x^{-2\alpha} -x^{-2\alpha} + 2x^{-\alpha-1}-x^{-2})\, dx \to \int_1^\infty(2x^{-3/2}-x^{-2})\, dx \ne 0.$$ Reinstall software-center in Ubuntu 16.04 How to combine two jQuery results How do you combine two jQuery search results? eg:  var $allFoos = $('.foo'),     $allBars = $('.bar')     $allFoosAndBars = $allFoos + $allBars;   Obviously, I just made up that last line, but I hope it makes it sorta clear what I mean. To be clear, the example is greatly simplified, and it could be any arbitrary sets i'm talking about, so $('.foo, .bar') is not what I'm after. Verify proof that sequence converges How to limit zoom level in OpenLayers 2.x I would like to have only 3 levels of zoom available (from level 6 to level 8).  demolayer2 = new OpenLayers.Layer.WMS( "abc...","http://localhost:8080/geoserver/gwc/service/wms", {layers: 'def...', transparent:"true", format: 'image/png', numZoomLevels: 3, minZoomLevel: 6 },{isBaseLayer:false},  { tileSize: new OpenLayers.Size(256,256)});  map.addLayer(demolayer2);   map.addControl(new OpenLayers.Control.LayerSwitcher({'div':OpenLayers.Util.getElement('layerswitcher')}));  map.zoomToExtent(new OpenLayers.Bounds(-4.601615515076983,39.8769407866263,-3.0527184873294764,41.16710732525929)); MWE:  \documentclass{article} \begin{document} $\wedge{\small\wedge}{\tiny\wedge}$ \end{document}   All the \wedges have the same size... Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} &gt; 1$$  adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is true by the inductive hypothesis, the 1 is already less than all those values? How to find and replace a string by increasing its numerical part? Explaining to laypeople why bootstrapping works I cite elements from my bibfile using \citet{key}. The reference appears as "Name1, and Name2 (year)" in text where year is a link leading to the bibliography. Now, I want to cite two papers from the same authors from different years. Instead of     Model XYZ is analyzed by Name1, and Name2 (2019) and Name1, and Name2 (2020).   I'd prefer to have something like     Model XYZ is analyzed by Name1, and Name2 (2019, 2020).   Of course, now 2019 should be a link to the 2019 paper in the bibliography and 2020 a link to the 2020 paper in the bibliography. Any ideas of I could adjust the \citet command?    I use biblatex as follows.  \usepackage[style=ext-authoryear,              sorting=nyt,             backend=biber,              natbib=true,                          uniquename = false,              uniquelist = false,             firstinits=false,              hyperref=true,             dashed=false,             backref=false,              maxbibnames=99,              maxcitenames=3,             isbn=false,              doi=false]{biblatex} Prove that $A^*A + I$ is invertible Let $A$ be an $m\times n$ matrix. Prove that $A^*A + I$ is invertible.   I'm not sure what to do because everything I try ends up at $A^*A$ again and not $AA^*$. Do I have to enter the UK on my UK passport? How can I disable a USB port in Windows XP to make it available only for smart phone battery charging?   I don’t want the users to be able to see the USB drive in My Computer, and the USB device should stay disabled in Device Manager.   Is there any way to accomplish this? Convert Month Number to Month Name Function in SQL I have months stored in SQL Server as 1,2,3,4,...12. I would like to display them as January,February etc. Is there a function in SQL Server like MonthName(1) = January? I am trying to avoid a CASE statement, if possible.