Consider the sequence of complex-valued functions $f_n(z)=z^n$, for $n=1,2,3,\dots$. How to show that this sequence does not converge uniformly to $f(z)=0$ on all of the unit disc. However it converges uniformly in every compact subset of the unit disc to $f(z)=0$?
Show that $f_n(z)=z^n$ converges locally uniformly on $|z|&lt;1$ but not uniformly.   I am really confused. I saw a seemingly reliable reference stating that $z^n$ does not converge uniformly to $0$ on $\{|z|&lt;1\}$ but it does converge uniformly on $\{|z|\le r,r&lt;1\}$. What is the nuance between the two former sets? With $r_n&lt;1,r_n\to 1$, shouldn't one conclude that $f_n(z)$ uniformly converges on $\{|z|&lt;1\}$? And besides, this source tells me what I have to prove for Homework is disproved, which really makes it more frustrating.  I would appreciate any help.
How would Voldemort restore himself from a horcrux? If Voldemort's body dies, what process would he use to restore himself from a horcrux? His plans seem sloppy for the pains he had taken. For example, why create horcruxes from pieces of jewellery when they dont have the agency to seek out a new body by themselves?  He needed Quirrel, a person with unlikely ambitions to seek him out to gain an useful body. When that body was gone, he needed Barty Crouch, someone with questionable agency (he could be imprisoned) to transport Harry and draw his blood.  Nagini and the diary seems the only horcruxes with agency, the others are mostly inert.  What plans were ever taken to restore Voldemort from his horcruxes? Linear Transformations: Proving 1 dimensional subspace goes to 1 dimensional I am having trouble understanding this whole question, and how to prove it.  Let $F:\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation. Prove that if $L$ is a $1$-dimensional subspace of $\mathbb{R}^n$, then  $F(L)=\{y\in\mathbb{R}^m: y=F(x)\text{ for some }x\in L\}$  the image of $L$ under $F$ is a subspace of $\mathbb{R}^m$ with dimension $1$ or $0$. Hence prove that if the kernel of $F$ is $\{0\}$, then the image of $L$ is a $1$-dimensional subspace of $\mathbb{R}^m$ (and so, in this case, $F$ takes lines in $\mathbb{R}^n$ to lines in $\mathbb{R}^m$). When is the present perfect tense used instead of the past tense? When is the present perfect tense used instead of the past tense?    I know that the present perfect tense is used when some adverbs (e.g., never, ever) are present in the sentence; the same is true for sentences like the following one.     When you returned, I have been at home since 3:00 PM.   In which other cases should I use the present perfect? Do the following sentences require it?     I have walked downtown everyday for a year.        I have been at home since 3:00 PM. How can I start automatically a command on Ubuntu 18.04? Why are airline passengers asked to lift up window shades during takeoff and landing? I have always wondered why the windows shades have to be raised during take off and landing. I can't think of any safety issues by having a closed window. I'm using TikZ to plotting some functions for a document. For example, I want to plot 1/(x+1). So I did something like this:  \begin{tikzpicture}[&gt;=stealth,x=3mm,y=3mm] \draw (0,5) node[above]{\fimg $y=f(x+1)=\dfrac{1}{x+1}$}; \clip (-5,-5) rectangle (5,5); \ejes[1]{-4}{4}{-4}{4} \draw[samples=100] plot(\x,{1/(\x+1)}); \end{tikzpicture}   where \ejes[style]{xmin}{xmax}{ymin}{ymax} is command, that I have defined before to draw the axes (join with \fimg). My expected output for the code below was just the function, but I get this:    So, TikZ automagically put the vertical asymptote at x=-1. That isn't a bad thing, but now I want to have some control over it. For example I want dotted lines for vertical asymptotes. How can I do that? I'm working on a web application that needs to prints silently -- that is without user involvement.  What's the best way to accomplish this?  It doesn't like it can be done with strictly with Javascript, nor Flash and/or AIR.  The closest I've seen involves a Java applet.  I can understand why it would a Bad Idea for just any website to be able to do this.  This specific instance is for an internal application, and it's perfectly acceptable if the user needs to add the URL to a trusted site list, install an addon, etc. How do I choose between "as to", "on", "of", and "about"? How to Fix:Though favicons are added, it doesn't show in Google Search Why has my database connection failed? Minimal polynomials and cyclic subspaces I'm trying to make my way through two problems in Curtis's Linear Algebra, chapter 25.  One of the two problems is this one, #5:     Prove that $V$ is cyclic relative to a linear transformation $T \in \mathcal{L}(V)$ if and only if the minimal polynomial of $T$ is equal to the characteristic polynomial.   Part of the issue is that I'm not really sure what it's asking. The statement that $V$ is cyclic seems to suggest that the entire vector space (say it has dimension $n$) is generated by some vector $v \in V$, along with $Tv, T^2v, \ldots T^{n-1}v$; Curtis's definition of a cyclic subspace (as opposed to a space, though I can't see why there should be a difference) seems to reinforce this:     A $T$-invariant subspace $V_1$ of $V$ is called cyclic relative to $T$ if $V_1 \neq 0$ and there exists a vector $v_1 \in V_1$ such that $V_1$ is generated by $\{ v_1, Tv_1, T^2v_1, \ldots T^kv_1 \}$ for some $k$.   However, I've seen several examples that seem to directly contradict this interpretation. For example, we recently had a take-home midterm that required we find elementary divisors, rational canonical form, etc., of a $6\times 6$ matrix; a classmate who received a perfect score on her test found that her matrix's characteristic polynomial was equal to its minimal polynomial &#151; it was $(x+8)^2 (x+14)^2 (x^2+61x+80)$ &#151; and yet the vector space could be written as $\left&lt; v_6 \right&gt; \oplus \left&lt; v_4 \right&gt; \oplus \left&lt; -v_2+v_3+v_6 \right&gt;$ &#151; i.e., as the direct sum of three cyclic subspaces, not as one cyclic subspace.  So that suggests that "$V$ is cyclic" might mean the direct sum of cyclic subspaces. However, while working on the exercise that follows (#6), I came across this:  $$S= \left(\begin{array}{ccc} -1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 2 \end{array}\right)$$  Here, the characteristic polynomial is not equal to the minimal polynomial, which is $m(x) = (x-1)(x+1)(x+2)$. The eigenspace associated with $\lambda = 1$ has dimension $2$. And so it seems as though, if $V$ has basis $\{ v_1, v_2, v_3, v_4 \}$, we can write $V = \left&lt; v_1 \right&gt; \oplus \left&lt; v_2 \right&gt; \oplus \left&lt; v_3 \right&gt; \oplus \left&lt; v_4 \right&gt; $. This would seem to contradict my second interpretation of the original question.  Obviously I'm missing something. Could someone fill me in on what it might be? (I've tried to be thorough, but I'm also trying not to make this ridiculously long; if I've left out necessary information, I'll be happy to append it.) Does anyone know the (specific) reasons that it is recommended that old LTS server users wait until the point release to upgrade?  Is it just a matter of bug fixes and waiting for improved stability for production servers, or is there something more fundamental than that? How to place the intercept of x and y axes at (0 , 0) and extend the x and y axes to the edge of the plot Maximum likelihood when usual procedure doesn't work I am trying to get the maximum likelihood estimate for the parameter $p$. The distribution is the following:   $$ f(x\mid p) = \begin{cases} \frac{p}{x^2} &amp;\text{for} \ p\leq x &lt; \infty  \ \\ 0 &amp;\text{if not} \end{cases} $$  The sample has size $n$.  The problem is, when I try to estimate it by the procedure I know, I would have to estimate the likelihood function and obtain the derivative of the log-likelihood. We'd have:   $$ L(p; x) = \frac{p^n}{\prod_{i=1}^{n} x^2_i}$$  $$ \ln L(p;x) = n \ln(p) - \sum_{i=1}^{n} \ln(x_i^2) $$   For the derivative:  $$ l'(p;x) = \frac{n}{p} = 0$$   And I am stuck because it has no solution for $p$. How do I evaluate this?  Thanks!   EDIT:  So in this case I can use the indicator variable to write:   $$ L(p;x) = \frac{p^n}{\prod_{i=1}^{n}x_i^2} I_{(x_i \geq p)}$$   for $i = 1,2, \ldots n$ in the indicator variable. So the "closest" non-null value of $x \in X$ to $p$ is $min(X_1, \ldots X_n)$. Is that the point? What does "had had" mean? How does this differ from "had"? For example, what is the difference between the following two sentences:     I had a bad day      I had had a bad day If $\Sigma \vdash \varphi$ then $\Sigma \vdash \forall x \varphi$. Why when $(\varphi \implies \forall x \varphi)$ is not true? In the book of first order logic I'm reading they say that if $(\varphi_1,...,\varphi_n)$ is a proof from a set of formulae $\Sigma$, then so is $(\varphi_1,...,\varphi_n,\forall x \varphi_j)$ for all $1\leq j \leq n$. They assume that as an axiom but it doesn't make sense to me. Why if $\varphi$ is true then so is $\forall x \varphi$? For example if $\varphi$ is $x=1$ that wouldn't mean that $\forall x : x=1$ , or would it?  Maybe what I don't understand is: why $(\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)$ is true but $(\varphi \implies \forall x \varphi)$ is not true?  The book is this: How might a corrupt partition in TempDB result in DBCC CHECKDB reporting no issue? CSS to make HTML page footer stay at bottom of the page with a minimum height, but not overlap the page I have the following page (deadlink: http://www.workingstorage.com/Sample.htm ) that has a footer which I can't make sit at the bottom of the page. I want the footer to  stick to the window bottom when the page is short and the screen is not filled, and stay at the document end and move down as normal when there is more than a screenful of content (instead of overlapping the content).  The CSS is inherited and befuddles me. I can't seem to change it properly to put a minimum height on the content or make the footer go to the bottom. How do I install make? Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous map such that $\overline{f(\mathbb{N})} = \mathbb{R}$. Prove that $f$ is surjective.  I tried to use the definition of the continuity and $\overline{f(\mathbb{N})} = \mathbb{R}$ to show that $f$ is surjective but it didn't work. Since $f$ is continuous, the preimage of every open set is open. Also it results in $f(\overline{A})\subseteq \overline{f(A)}$ for every set $A$. How this couples with $\overline{f(\mathbb{N})} = \mathbb{R}$ to show surjectivity?