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Show $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$. Let $X=(0,\infty)$. Define two metrics on $X$ by   $d_{1}(x,y)=|x-y|$ and   $d_{2}(x,y)=|x-y|+|\frac{1}{x}-\frac{1}{y}|$   for all $x, y \in{X}$.   Let $(x_{n})$ be a sequence in $X$ and $x\in{X}$.   Show  $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$.  Note: $d_{1}$ and $d_{2}$ are then said to be equivalent metrics on $X$.  Since $d_{1}$ and $d_{2}$ are equivalent metrics, will it be sufficient to show $c*d_{2}(x,y)\leq{d_{1}(x,y)}\leq{C*d_{2}(x,y)}$?  So far I have the following: $d_{1}(x,y)=|x-y|\leq{|x-y|+|\frac{1}{x}-\frac{1}{y}|}=d_{2}(x,y)$. Therefore, $d_{1}(x,y)\leq{C*d_{2}(x,y)}$ where $C=1$. Is there a $c$ that would satisfy the left-hand side of the inequality?  Any help would be appreciated. (this is not homework) $$ A $$  Light is travelling away from the absolute center of the Schwarzchild black hole (i.e. none of that spinning/stretching field) at $\ C $, and so shouldn't $\ \Delta C$ need to be greater than $\ C $ in order for the light to collapse into the black hole, therefore violating the relativistic limit of $\ C $?    It seems like light would just 'hover' right in the center of a black hole then. Because if it goes away from it, the event horizon pulls it back, and if it goes into it, the velocity change must be greater than $\ C $ which violates general relativity; so technically there's no way for the light to move away, and so it would just be still.  The thing is that the light must move at the speed of light or general relativity doesn't work. It's massless... But in the center of a black hole, if the above is true (it can't move or escape), then it is not travelling at the speed of light. So it cannot be still.  It cannot be spinning in a constant orbit either, because that would require velocity in an outward angle. And it cannot move in an outwardly fashion, only in an inwardly one.  Is it spinning in an infinitely decaying spiral towards the center?  If so what happens to it if and when it reaches the direct and complete center, i.e. The case of it being 'shot' from the center.    $$ B $$  If you shot light at an angle towards a black hole, but not directly to the center of it  then that light from the outside will eventually become the same as the light that I said was shot from the center to the outside; because in both cases, they end up in an infinitely decaying spiral.    If so it would end up exactly like the light in the above paragraph, so then we need to figure out what happens to that light to figure this one out, or vice versa.    $$ C $$  If you shot light directly towards the center of a black hole it would have no spiral to decay into, so therefore it must stop at some point in the center of a black hole. If it stops though, it loses it's momentum and violates general relativity again because it's energy "disappears".    If so then it violates conservation laws. So I'm assuming this simply is not true.    Is the light travelling into a black hole somehow still travelling at the speed of light, in every direction possible where that direction keeps it facing towards the center?  If so, why is there not a point at which the only thing left for the light to do is stay still, since all other paths would increase the distance between it and the black hole's center, and therefore become Case A. And if there isn't, what would happen if there was?  Also, if it's a time and observation answer where it supposedly takes and infinite amount of time for C to hit the center, wouldn't it be the case that the black hole is 'eating' all of the available light in the universe and keeping it in stasis?  Also, what happens in a Kerr black hole then? Does it just rip the field of time/gravity at an angle such that the light would never get closer than the center? Not booting from USB or CD (SYSLINUX Message) What is Ubuntu's status on the Meltdown and Spectre vulnerabilities? What are the strengths and weaknesses of each follower? SF short story ID, written 40s—60s, primitive aliens who are much smarter than us I'm looking for a story that was probably written somewhere from the 1940s to the 1960s. I think I read it in an anthology, possibly in the 1970s.  A group of human explorers landed on a planet with primitive aliens, living in hunter-gatherer villages. They hoped to make a profit from trading our goods for—well, I'm not clear on what they wanted from the aliens.  The native people learned the human language before the explorers could learn theirs. Then the natives suddenly started producing technological innovations far beyond what they originally used. I remember that wooden sailing ships suddenly replaced their canoes. All the natives had to do was hear a story once from the explorers before copying the technology in the story.  Finally, the explorers realized that the natives were much, much smarter than us and that Earth faced a serious problem.  I'm fairly certain that it wasn't a Heinlein story, but I remember that it had some qualities similar to his work. Allow new users to comment for X times in a day While reviewing "First Posts" on Stack Overflow, I often come across new users posting comments as answers. It seems (at least to me) they are not doing this because they are unaware of how Stack Overflow works but because they can't comment on any question they are forced to put it as answer.  I have found duplicate questions for my request like:   ,  and    but the best explanation I have found is an  to .     I think the main reason this isn't supported is there's no way to   downvote comments, which would make them attractive to spammers. If   somebody comments on the top answer in a question the only way to get   rid of it is for six people to flag it. Comment flags don't show up on   10k tools, so it needs to be six people that happen to stumble across   the spam comment, or a moderator that checks the list manually. People   can comment every 15 seconds, so that can get out of hand pretty   quickly. Comments also aren't listed on the user's profile page, so   it's hard to see at a glance if a user has been spamming   So I am suggesting allowing new users (with less than 50 reputation points) to comment only 5 times every day or only 25 times times until they reach 50 reputation points or any other variation of it. Difference between "in time" and "on time" I have an appointment at 8 and I arrive there at 7:55, is it "on time" or "in time"? What about "the nick of time"? Which Content Management System (CMS)/Wiki should I use? This is a general, community wiki  question to address non-specific "I need a CMS or Wiki that does x, y, and z..." questions.  If your question was closed as a duplicate of this question and you feel that the information provided here does not provide a sufficient answer, please open a discussion on .    I have a list of features that I want for my website's Content Management System (CMS) - where can I find a [free] script that includes all of them? End of javascript sidebar workarounds? (Salesforce Classic / Aloha) Calculating proportional area of polygon within another layer's polygon using QGIS? A function $f : I \rightarrow \mathbb R$ on an interval $I$ is convex   if $f((1-t)x+ty)\le (1-t)f(x)+tf(y), \forall x,y \in I, t \in [0,1]$      Assume now that $I$ is an open interval.  Show that if $f$ is convex   then for each $c \in I$ there exists $m \in \mathbb R$ such that $m(x  − c) + f(c) \le f(x)$ for all $x \in I$ , and if in addition $f$ is   differentiable at $c$ then $f'(c)$ is the unique $m$ that works. In   general, must m be unique?   (we have previously shown that convex functions are continuous)  I think I'm missing the point of the question.  Surely, by the mean value theorem we can find some $y\in (x,c)$ such that $f'(y)=\frac{f(x)-f(c)}{x-c}$ setting $m=f'(y)$ we get $m=\frac{f(x)-f(c)}{x-c}, m(x-c)+f(c)=f(x)$ such an $m$ fits the inequality we're asked to show.  I also cannot see why $f'(c)$ would be the unique solution or why we would ever get a unique solution of the inequality at all for that matter.  Clearly I'm misunderstanding the question, so if someone could point out where I would appreciate that.  Thank you I would like to convert tab to spaces in gVim. I added the following line to my _vimrc:  set tabstop=2   It works to stop at two spaces but it still looks like one tab key is inserted (I tried to use the h key to count spaces afterwards).  I'm not sure what should I do to make gVim convert tabs to spaces? What is the result of $ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $ and why? NVMe PCIe x4 SSD on M.2 PCIe x2 slot? i'm about to get a new SSD for my PC, and before spending the money i have some doubts about my decision i would like to clarify First, my mobo is a B450 Aorus Pro Wifi, it has 2 M.2 slots, one of them full length (22110) compatible with PCIe 3.0 x4 and the other one standard length (2280) and support only for PCIe 3.0 x2 About the drive, i'm aiming for a 1tb XPG SX8200 Pro , which is a PCIe 3.0x4 SSD My first M.2 slot is already taken by the main SSD, which i use for the OS and other stuff, i can't really use it. Will a PCIe x4 SSD work on a PCIe x2 slot? How much performance will be lost if it does? Will it be worth it? An Open Interval and a Half-open Interval are not Homeomorphic How do I search for questions with specific tags?       For more information about searching, see &quot;&quot; in the . I recently saw a lecturer prove the following theorem (assuming the result that every analytic function is locally 1-1 whenever its derivative is nonzero): Let $\Omega \subset \mathbb{C}$ be open, and let $f : \Omega \to \mathbb{C}$ be 1-1 and analytic on $\Omega$.  Then $f&#39;(z_0) \not = 0$ for every $z_0 \in \Omega$.  I got the basic idea behind the proof: we assume for contradiction that $f&#39;(z_0) = 0$, and, assuming without loss of generality that $z_0 = f(z_0) =0$, we have (from the power-series expansion) that $f(z) = z^kg(z)$ for some analytic $g$ in some disk at the origin (i.e., $z_0$) and some $k \ge 2$. Since $z^k$ is not 1-1 in any such disk (because there are multiple roots of unity), then $f$ isn't either.  However, the proof he gave was rather awkward and technical- it involved defining three different axillary functions, even though the idea was simple, and I've since forgotten how it exactly worked.  In any case, I'm convinced there's a better way.  The problem is that I'm having trouble turning the idea into a real proof- I know that it obviously follows if $g$ is 1-1, but I'm also pretty sure that that is too strong an assumption.  Am I missing something, or does the argument just have to be more complicated? jQuery not working when a new row is added to table Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$ Applying the , I found: $$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where $\Psi^{(1)}(k)$ is the polygamma function. Is it known any sharper bound for the sum $S$? Thanks.