I bought 2 domains form GoDaddy but with both I am seeing the same problems in that the domain on the www version goes to the old site which is still being hosted.  I have checked the IntoDNS website and in the www record it shows:  Your www.example.com A record is: www.example.com -&gt; example.typepad.com -&gt; cname-cloudflare.typepad.com -&gt;   What can I do to stop this from happening?  Will this eventually be automatically removed and fix itself? Though obviously it's not automatically fixed itself in the long drawn out expiry process...   It's been quite a while for one of them and still hasn't propagated for the www. I'm not having any problems with the normal example.com part of the site.
It's often the case that I want to change name servers because my site is messed up.  Well, it takes a very long time for the name server to propagate.  Is there something in DNS, or registrar or anything I can do to speed that up.  Flushing my dns server using  and it rarely works
All group homomorphism from $ \mathbb{Z} _m $ to $ \mathbb{Z}_n $  How could I find every group homomorphism? Fibonacci combinatorial identity: $F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$ Can someone explain how to prove the following identity involving Fibonacci sequence  $F_n$  $F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$ ? I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges. At first I want to know if my answer is OK.  My try: $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$     $ = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ Now we prove that the last sum converges by the comparison test: $\frac{1}{k}-\log(\frac{k+1}{k}) &lt; \frac{1}{k^2} \Leftrightarrow k&lt;k^2\log(\frac{k+1}{k})+1$ which surely holds for $k\geqslant 1$      As $ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $ \Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ converges and we name this limit $\gamma$ q.e.d Stone-Weierstrass theorem with $p(1/x)$ I am trying to prove that for a continuous $f\colon[1,\infty)\rightarrow\mathbb R$ and $f(x)\to a$ as $x\to\infty$ it could be approximated by $g(x)=p(1/x)$ where $p$ is a polynomial. How to connect to remote X-Server (logged in via ssh) When I'm logged on to another host (e.g via ssh), how do I connect to the XServer of that machine (same user is logged in and is running a desktop (gnome))?  You may ask way I wish to do that: There are commands that don't open an X-Window, e.g. xinput, xhost, etc.. and there are situations where you want to run them from remote. Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? I need to prove  $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?  This is how I prove this:  Prove:$2n ≤ 2^n$, for all integer $n≥1$  Proof: $2+4+6+...+2n=2^n$  $i.)$ Let $P(n)=1         P(1): 2(1)=2^1\implies 2=2$.        Hence, $P(1)$ is true.  $ii.)$  Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$  from the assumption add $2(k+1)$ on both sides so we have   $2+4+6...2k+2(k+1)=2^k+2(k+1)$  I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$.  I feel i'm doing something wrong.   Any help would be appreciated! Ubuntu 14.04 LTS fails to start with a kernel panic What is a substitute for red or white wine in a recipe? Robot tank run amok, its trainer trapped in cave on the Moon Another dim memory from my sci-fi reading binges in the 1980s... this one almost definitely in one of the big hardback collections full of late Golden Age short stories.  The story is told from the perspective of a robot (maybe cyborg?) tank that has stopped responding to control signals and is aggressively "defending" some portion of the Moon, to which it was previously assigned. Maybe a war had ended, or maybe it became confused about an exercise that it thought was real?  There was a person trapped in a cave trying to communicate with the tank. The tank mostly ignores him, so the main purpose of what he says is exposition for the reader. I think this person was an expert on this type of robot/cyborg sent to "talk it down", but the tank had attacked, and this guy was the only survivor. The expert was badly wounded -- I think he mentions his suit had to amputate his leg to seal a hole -- or maybe flood it with water which would freeze to seal the hole but mean that his leg had to be amputated later. He keeps trying to convince the tank to stand down, at one point saying that he knew it "back when it was just a two-wheeler an autocyber in boot camp" or something along that line. (Edit: I must have been reading too fast and/or distracted back then, since per the quote below I seem to have misread "motorcycle" for "autocyber"!)  One detail of the attack stands out -- the tank was low on ammunition, I think, so it used a kind of bomb with an electric/magnetic "catapult" (aka "magnapult") to hit the expert's transport. The tank was proud of itself for figuring out a way to hit it far out of the normal range by 1) using its own top speed to add velocity to the projectile, and 2) building up the catapult's energy past its normal operating range, which caused it pain but made the bomb go much further. It catches the transport by surprise, since they thought they were safely out of range of the armaments it had left.  The premise sounds a lot like a Keith Laumer "Bolo" story, but I'm fairly sure that it's not actually one of them. It had a somewhat grimmer feel to it. I could be wrong, though, since it's been 30 years. Is it necessary to explicitly remove event handlers in C# What is safe to exclude for a full system backup? I want to uninstall VMware Player on my Ubuntu 12.04 LTS system. How should I uninstall it? How to play DVDs in Ubuntu 16.04? When Ron broke his wand, why didn't he apply for Hogwarts financial aid? php String Concatenation, Performance In languages like Java and C#, strings are immutable and it can be computationally expensive to build a string one character at a time.  In said languages, there are library classes to reduce this cost such as C# System.Text.StringBuilder and Java java.lang.StringBuilder.  Does php (4 or 5; I'm interested in both) share this limitation?  If so, are there similar solutions to the problem available? How can I get the amount of available memory portably across distributions? The standard files/tools that report memory seem to have different formats on different Linux distributions. For example, on Arch and Ubuntu.   Arch  $ free               total        used        free      shared  buff/cache   available Mem:        8169312     3870392     2648348       97884     1650572     4110336 Swap:      16777212      389588    16387624   $ head /proc/meminfo  MemTotal:        8169312 kB MemFree:         2625668 kB MemAvailable:    4088520 kB Buffers:          239688 kB Cached:          1224520 kB SwapCached:        17452 kB Active:          4074548 kB Inactive:        1035716 kB Active(anon):    3247948 kB Inactive(anon):   497684 kB  Ubuntu  $ free              total       used       free     shared    buffers     cached Mem:      80642828   69076080   11566748    3063796     150688   58358264 -/+ buffers/cache:   10567128   70075700 Swap:     20971516    5828472   15143044   $ head /proc/meminfo  MemTotal:       80642828 kB MemFree:        11565936 kB Buffers:          150688 kB Cached:         58358264 kB SwapCached:      2173912 kB Active:         27305364 kB Inactive:       40004480 kB Active(anon):    7584320 kB Inactive(anon):  4280400 kB Active(file):   19721044 kB    So, how can I portably (across Linux distros only) and reliably get the amount of memory—excluding swap—that is available for my software to use at a particular time? Presumably that's what's shown as "available" and "MemAvailable" in the output of free and cat /proc/meminfo in Arch but how do I get the same in Ubuntu or another distribution? I want it to be look like this:    I have an HDD and an SSD in my PC, I want to install Ubuntu to SSD, I have windows on HDD. I want to be able to boot ubuntu from the windows 10 uefi boot manager. How do I do it? Help Me Understand My Bicycle's Gears Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? Proving that free modules are flat (without appealing projective modules) Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah &amp; Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.   So my question is,     Is it possible to prove that every free module is flat just by   definitions and without appealing to projective modules?   Thanks!