Mostly, if an airport, however minor, has scheduled service, it will be to another airport with more services and so one can connect through to any airport in the world. E.g. one can fly from  to  to LHR to anywhere else.  But is there a pair or set of airports where this doesn't hold, creating an "island" of unconnected air transport?  (Fixed wing or I guess helicopter).
A commercial airport being any airport that has regular scheduled passenger flights.  The mathematics of  got me thinking: is the network of scheduled flights a connected graph?  That is, is there a pair (or triplet or even larger set) of airports such that they only have scheduled flights with each other?  Or can you get from any airport to any other?  Ignore vagaries like cargo and general aviation. I'm looking for two airports that I can shuttle between, but would have to drive (or sail) to get to one or the other of them.
if $f^{(n)}(x_0),g^{(n)}(x_0)$ exist and $\lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n}=0$ then $f^{(r)}(x_0)=g^{(r)}(x_0 )$ for all $0\leq r\leq n.$ Show:   if $f^{(n)}(x_0),g^{(n)}(x_0)$ exist and   $$\lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n}=0$$   then $f^{(r)}(x_0)=g^{(r)}(x_0 )$ for all $0\leq r\leq n.$   My attempt:  let n = 1,  since $f'(x_0),g'(x_0)$ exists then The taylor polynomial exists: $$T_{f,1}(x)=f(x_0)+f'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{(x-x_0)}=0$$  $$T_{g,1}(x)=g(x_0)+g'(x_0)(x-x_0), \lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{(x-x_0)}=0$$  Then, \begin{align} \lim_{x\to x_0} \dfrac{f(x)-g(x)}{(x-x_0)^n} &amp; =\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-g(x)+T_{f,1}(x)-T_{f,1}(x)+T_{g,1}(x)-T_{g,1}(x)}{(x-x_0)^n}\\ &amp;=\lim_{x\to x_0} \dfrac{f(x)-T_{f,1}(x)}{x-x_0}-\lim_{x\to x_0} \dfrac{g(x)-T_{g,1}(x)}{x-x_0}+\lim_{x\to x_0} \dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=\lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0}\\ &amp;=0 \end{align}  Then, \begin{align} \lim_{x\to x_0}\dfrac{T_{f,1}(x)-T_{g,1}(x)}{x-x_0} &amp;=\lim_{x\to x_0}\dfrac{f(x_0)+f'(x_0)(x-x_0)-g(x_0)-g'(x_0)(x-x_0)}{x-x_0}\\ &amp;=\left(f(x_0)-g(x_0)\right)\lim_{x\to x_0}\dfrac{1}{x-x_0}+f'(x_0)-g'(x_0) \\&amp;=0 \end{align} I am confused because  $\lim_{x\to x_0}\dfrac{1}{x-x_0}$ does not exist. I am using NCR18650 batteries with the tp4056 charging module.  When I connect the charger at 2.8&nbsp;V (fully discharged), the tp4056 IC is getting hot at a 1&nbsp;A charge current. Is it OK to reduce this charge current by changing the charge current selecting resistor? So the tp4056 IC heat would be reduced, wouldn't it?  For my application I don't need a quick charge. With a low current, charging time will be increased, but it does not matter for my application.  Will reducing the charge current affect the battery's lifetime? Or will it damage the battery?  I am trying to reduce the heat of the tp4056 IC by reducing the charging current, maybe from 1&nbsp;A to 500&nbsp;mA or less. The 18650 battery never heats for a 1&nbsp;A charge current. On thanksgiving day a woman posted in a facebook group that her father had just passed away (left her money) and in good deed because it was thanksgiving she was offering to pay people’s debts. Well I inboxed her and she pretty much wants me to open up a bank account with First Tech Credit Union and give her my log in and password info. Ofcourse its sketchy!   So I questioned her and asked why she wanted that instead of her sending me the money directly from her bank to mine. She then responded because she wants to keep records of my debts. (She asked for proof and I sent her screenshots ONLY of my outstanding balances) I heard about the depositing checks scam so I questioned her about how she would deposit the funds and she said cash.  Her facebook page is 100% legit, I screenshotted her pics and family profiles etc. We came to an agreement that I would only accept her help if she sends me a picture of her ID before me sending this info.   She literally just wants to deposit the money to me, She isn’t asking me to send nobody money from that account or anything. Of course the account will be empty when I provide it to her. When she does deposit the money I am planning to immediately change the password and obviously stay watching my account activity while she deposits the funds.   She actually also provided pics of other accounts she funded. My question is, is this a scam? And if so how could she possibly scam me? Specially if i have her information and A pic of her ID. She is a christian lady and looks like she has a good heart, We talked about my struggles and she actually looked into why I needed to be assisted with the money. By no means was she pushy or anything. I want to believe that she just wants to help me but I am also too skeptical. Mclaurins with $e^{\sin(x)}$ To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:  $e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$  $\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$  $\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$  $\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$  In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$  However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$.  What did I do wrong? I can't seem to find what or where. How do I defeat the Kayran in the first chapter of the Witcher 2? In the first chapter you are supposed to slay the Kayran, this enormous beast:    I'm having a hard time with it. I drank the Mongoose potion for poison resistance and I have the Kayran trap, but it just hits me too often with its tentacles. Especially when the Kayran swings a tentacle sideways, not from above, I have a very hard time evading it.  Is there any trick besides not getting hit by the tentacles?  With a lot of luck I'm getting as far as the point where it's throwing stones. Any idea what I'm supposed to do then? How to group distinct values from a list using odata rest api Hi I have a list with two fields "Regions" &amp; "Countries" I am using a rest api to retrieve the list of just regions like so: http:mysitename/_api/web/lists/getByTitle('TaxTool')/items?$select=Region  In the results there are lots of duplicates and I was wondering how to return unique values. Can I insert elements to the beginning of an element using .appendChild()? If the user selects an option in a dropdown box, there must be added a label and a textbox. Using appendChild, these elements get added to the end of the container.  var newFreeformLabel = document.createElement('label'); newFreeformLabel.innerHTML = 'Omschrijving:';  var newFreeformField = document.createElement('input'); newFreeformField.className = 'textfield'; newFreeformField.name = 'factuur_orderregel[]';  var newFreeformSpacer = document.createElement('div'); newFreeformSpacer.className = 'spacer';  container.appendChild(newFreeformLabel); container.appendChild(newFreeformField); container.appendChild(newFreeformSpacer);   The issue is that these elements should be inserted at the beginning of container, not at the end.  Is there a solution to this in PrototypeJS? Linear regression with fixed point I have a data set to which I want to add a trend line. For my system, there will always be a fixed point on (2, 0). I have the line representing this system, but currently I do not account for the fixed point. Is there functionality to achieve this?  \begin{figure} \begin{tikzpicture} \begin{axis} [     ylabel={Some Time [ms]},     ]     \addplot[only marks, color=red, error bars/.cd,                        y dir=both, y explicit]         table [x=num_of_nodes, y=MyData, y error=stdev_delay_var] {MyData.dat};     \label{plot_one}     \addplot +[mark=none, dashed]          table[y={create col/linear regression={y=MyData}}]         {MyData.dat}; \end{axis} \end{tikzpicture} \caption{Test Results} \label{fig:test_results} \end{figure} Change the height of a function being plotted How do I fix inconsistent Textarea bottom margin in Firefox and Chrome? How to validate phone numbers using regex I'm trying to put together a comprehensive regex to validate phone numbers. Ideally it would handle international formats, but it must handle US formats, including the following:   1-234-567-8901 1-234-567-8901 x1234 1-234-567-8901 ext1234 1 (234) 567-8901 1.234.567.8901 1/234/567/8901 12345678901   I'll answer with my current attempt, but I'm hoping somebody has something better and/or more elegant. I have Xubuntu 18.04.3, and whenever I try to install a package through apt-get I get the following message.    Removing linux-image-5.0.0-31-generic (5.0.0-31.33~18.04.1) ... /etc/kernel/postrm.d/initramfs-tools: update-initramfs: Deleting /boot/initrd.img-5.0.0-31-generic /etc/kernel/postrm.d/zz-update-grub: Sourcing file `/etc/default/grub' /usr/sbin/grub-mkconfig: 35: /etc/default/grub: FROM:: not found run-parts: /etc/kernel/postrm.d/zz-update-grub exited with return code 127 dpkg: error processing package linux-image-5.0.0-31-generic (--remove):  installed linux-image-5.0.0-31-generic package post-removal script subprocess returned error exit status 1 Errors were encountered while processing:  linux-image-5.0.0-31-generic libdvd-pkg: Package libdvdcss2-1.4.2-1 was removed, stop processing... E: Sub-process /usr/bin/dpkg returned an error code (1)   Contents of /etc/default/grub:  # If you change this file, run 'update-grub' afterwards to update # /boot/grub/grub.cfg. # For full documentation of the options in this file, see: #   info -f grub -n 'Simple configuration'  GRUB_DEFAULT=0 GRUB_TIMEOUT_STYLE=hidden GRUB_TIMEOUT=10 GRUB_DISTRIBUTOR=`lsb_release -i -s 2&gt; /dev/null || echo Debian` GRUB_CMDLINE_LINUX_DEFAULT="quiet splash" GRUB_CMDLINE_LINUX="persistent"  # Uncomment to enable BadRAM filtering, modify to suit your needs # This works with Linux (no patch required) and with any kernel that obtains # the memory map information from GRUB (GNU Mach, kernel of FreeBSD ...) #GRUB_BADRAM="0x01234567,0xfefefefe,0x89abcdef,0xefefefef"  # Uncomment to disable graphical terminal (grub-pc only) #GRUB_TERMINAL=console  # The resolution used on graphical terminal # note that you can use only modes which your graphic card supports via VBE # you can see them in real GRUB with the command `vbeinfo' #GRUB_GFXMODE=640x480  # Uncomment if you don't want GRUB to pass "root=UUID=xxx" parameter to Linux #GRUB_DISABLE_LINUX_UUID=true  # Uncomment to disable generation of recovery mode menu entries #GRUB_DISABLE_RECOVERY="true"  # Uncomment to get a beep at grub start #GRUB_INIT_TUNE="480 440 1"  FROM: GRUB_CMDLINE_LINUX_DEFAULT="" GRUB_CMDLINE_LINUX="" TO: GRUB_CMDLINE_LINUX_DEFAULT="ipv6.disable=1" GRUB_CMDLINE_LINUX="ipv6.disable=1" Can I clear an area of my prison for everyone BUT max security prisoners? How can I make sure that only low and medium security inmates are in a specific area? Is it possible to reverse the overlay you can set? It seems like you are only able to clear areas for low OR medium security prisoners, but sharing is not possible. The thing I want here is to make sure that max security prisoners are not in certain areas. How many points in RSA, such that $m^e = m \bmod n$ For every RSA-cryptosystem, there exist some messages $m$, for which it holds that $m^e \equiv m \pmod n$ As to the question, how many such messages exist, this question has already been asked and answered here:  However, I do only understand the answer up to a certain point Suppose that $n$ is only 1 prime. Then it holds that the number of unconcealable messages is $\gcd(e-1, p-1)+1$ This is because in $m^e \equiv m \pmod  p$ , either $m = 0$ or $m \not= 0$. In the latter case $m$ can be expressed as $g^t$, where $g$ is a Generator in $\mathbb{Z_p}$ $g^{t*e} \equiv g^t \pmod p$ $g^{t(e-1) \bmod p -1 } \equiv g^{0 \bmod p -1} \pmod p$ $t(e-1) \equiv 0 \pmod {p-1} $ has exactly $\gcd(e-1, p-1)$  solutions, so the total number of unconcealable messages is, together with $m=0$ , $\gcd(e-1, p-1)+1$ When we now look at real RSA, which has two prime factors in its modulus, it holds that the number of unconcealable messages is $(\gcd(e-1, p-1)+1)\cdot(\gcd(e-1, q-1)+1)$ Why is that so? I suppose it has to do with CRT, but I just cant understand why. Double calculation producing odd result Frea is trying to kill me I've been trying to play the Dragonborn DLC for Skyrim on the PS3. In order to advance the story, I need to talk to Frea and her dad. But every time I go near the village they try and murder me for slaughtering the village back when I was like level 15 (I'm level 43 now).  I've tried yielding but they won't accept it and I don't know how to make them stop, all I want to do is finish the story. Does anybody know how to fix this issue? Stop asking me if Community auto-flags were helpful Select longest continuous sequence I am trying to construct a query in PostgreSQL 9.0 that gets the longest sequence of continuous rows for a specific column.  Consider the following table:  lap_id (serial), lap_no (int), car_type (enum), race_id (int FK)   Where lap_no is unique for each (race_id, car_type).  I would like the query to produce the longest sequence for a given race_id and car_type, so it would return an int (or long) that is highest.  With the following data:  1, 1, red, 1 2, 2, red, 1 3, 3, red, 1 4, 4, red, 1 5, 1, blue, 1 6, 5, red, 1 7, 2, blue, 1 8, 1, green, 1   For car_type = red and race_id = 1 the query would return 5 as the longest sequence of the lap_no field.  I found a similar question  however my situation is a bit more straightforward.  (I would also like to know the longest sequence for a given car_type for all races, but was planning to work that out myself.) Difference between Loading a class using ClassLoader and Class.forName Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K A similar question to mine was answered here on stackexchange:    However, I am unable to follow the formulas so perhaps someone could show the calculation and the way they arrived at it to answer this question.  My card game is very similar, except the counting of cards goes all the way from 1 (Ace) through to King and then restarts.  If you can make it through an entire deck of cards without hitting the same ranked card you win.  So, for example, if you call out Ace - 2 -3 - 4 - 5 - 6, etc. and hit "7" when you've just called out "7", you lose.  What is the chance of winning this card game?