Need help with visa form My wife and I will be travelling to the UK together. My wife depends on me. In my application I completed the income part.  In my wife application:  In the part "planning to spend money" in the UK I will put 40% of my total cost of the trip. For example, I put GBP 3,000 in my trip cost. Now I will put for my wife GBP 1,200. Is this ok?  Secondly, in my wife's application I don't know what to put in "Spend money in each month" because she is dependent and I spend in my home monthly. I just give her pocket money or for purchasing some food items. Now what should I put in this field?
How to apply for a UK standard visitor visa as a family? I am an Indian citizen residing in the United States. My wife and I are planning to travel to the UK for a week. We plan to apply for the UK standard visitor visa from the US. I was filling out the application and I have a few questions.   Travel History - The application asks for travel history to the United States, Canada, Australia and New Zealand. Considering that we're resident aliens in the United States, should we consider all the times we've traveled back to the US from India? Travel History - Should we also list out the times we've traveled to India from the US? Cost of the trip - Hypothetically if we plan to spend a total of ~4,000 USD on this trip, should I split up the cost and write down 2,000 USD on my application and 2,000 USD on my wife's application? Income - From what I've read on the forum, my wife must say that "someone else" i.e. me - is paying for her costs and mention my GWF number in the remarks on her application. Would that be sufficient considering that I'll be sending all income documents with my application? Savings - We have joint savings accounts. Should I mention the same amount on both our applications? Posting the documents - Is the up-to-date advice to send all documents in one packet? Or would sending them in separate packets at the same time be better? My thought was to annotate my wife's packet with my GWF number if we were to send them separately. Bank statements - We get e-statements. Would printing those out be enough?
the following code will only make the qed not appear in the second proof. However, it it possible to make the qed symbol never appear at the end of the proofs without having to specify it locally every time?  \documentclass{article} \usepackage{amsthm} \begin{document} \begin{proof} This has the QED symbol. \end{proof} \begin{proof}\renewcommand{\qedsymbol}{} This hasn't. \end{proof} \begin{proof} And this has it again. \end{proof} \end{document} Should academic papers necessarily carry a sober tone? I was reading  titled "Optimal Symmetric Rendezvous Search on Three Locations." While talking about the history of search problems, the author mentions the following anecdote in passing.     In 2007 a letter writer to the Guardian newspaper queried, “I lost my wife in the crowd at Glastonbury (a music festival). What is the best strategy for finding her?” A reader replied, “Start talking to an attractive woman. Your wife   will reappear almost immediately.”   While I found it quite amusing to read this, I do not often come across papers with such witticisms. Is there an unwritten rule about the tone of sobriety that is considered appropriate in academic papers? Are jokes or anecdotes fine as long as they do not appear forced? Quickly create a large file on a Linux system How can I quickly create a large file on a Linux () system?   will do the job, but reading from /dev/zero and writing to the drive can take a long time when you need a file several hundreds of GBs in size for testing... If you need to do that repeatedly, the time really adds up.  I don't care about the contents of the file, I just want it to be created quickly. How can this be done?  Using a sparse file won't work for this. I need the file to be allocated disk space. Is “of ” necessary in “all of ”? Listen to all your fans   Name all the states   vs      Listen to all of  your fans   Name all of  the states   What part of language is of  in these examples? Is it necessary or optional, correct or incorrect? Numbering of appendices in the backmatter of a book I'm using the standard book class. If I write  \documentclass{book}  \begin{document}  \frontmatter \tableofcontents  \mainmatter \chapter{foo} \section{Donald}  \backmatter \chapter{bar} \section{Mickey}  \end{document}   I expect to have chapter bar numbered as appendix A and section Mickey as A.1. Instead chapter bar is not numbered and section Mickey is 1.2, following section Donald. If I insert the appendix command to obtain  \documentclass{book}  \begin{document}  \frontmatter \tableofcontents  \mainmatter \chapter{foo} \section{Donald}  \backmatter \appendix \chapter{bar} \section{Mickey}  \end{document}   then chapter bar is unnumbered and section Mickey is numbered as .1.  What am I doing wrong? Why does the loss/accuracy fluctuate during the training? (Keras, LSTM) I've been using gksudo nautilus and sudo nautilus through Alt+F2.   What's the difference? They look very similar! Birthday problem: why is this solution wrong? This question is about the birthday problem: the probability that in a group of n people, at least two of them have the same birthday ().  An easy way to calculate the probability is to calculate first the probability that no two people have the same birthday.  Let's say that I want to calculate the probability that in a group of 20 people, NO two people have the same birthday.  So, for 20 people with different birthdays, I can choose the first birthday in 365 ways, the second in 364 ways and so on, while for 20 people who can have the same birthday, I can choose the first birthday in 365 ways, the second also in 365 ways, and so on. At the end:  $p={365 \cdot 364 \cdot 363 \cdot ... \cdot 346 \over 365 \cdot 365 \cdot 365 \cdot ... \cdot 365}\approx 0.59$  This is the right method and I understand it. I don't understand why the following method is wrong:  The probability that in a group of 20 people NO two people have the same birthday is the ratio between the combination without repetition of 20 birthdays and the combination with repetition of 20 birthdays:  $$p={C_{365,20} \over C'_{365,20}}={\binom{365}{20} \over \left(\binom{365}{20}\right)}={\binom{365}{20} \over \binom{365+20-1}{20}}=\frac{{{365!}\over {20!\cdot(365-20)!}}}{{(365+20-1)!}\over{20!\cdot{(365-1)!}}}\approx 0.35$$  I understand the mistake is in the denominator (as the numerator is the same of the other method, after simplifying that 20!), but why? Isn't it right to calculate the k-combination with repetition of 20 birthdays?  Thanks for helping! When doing calculations on very large numbers where integral data types such as double or int64 falls short, a separate class to handle such large numbers may be needed.    Does anyone care to offer an efficient algorithm on how best to do this? How do you retrieve items from a dictionary in the order that they're inserted? Is it possible to retrieve items from a Python dictionary in the order that they were inserted? Is there any way to block websites without Root access I am searching for a way to block websites without having root access. Is there any way to do it? Any help will be appreciated. Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof:  $$ |d(x,a) - d(y,a)| \leq d(x,y) $$ for any $a\in A$ - but we cannot just replace it by $|d(x,A) - d(y,A)|\leq d(x,y)$ since the minimum (or infimum in general) can be attained in different points $a\in A$ for $x$ and $y$, so we only have that $$ |d(x,A)-d(y,A)|\leq d(x,y)+\sup\limits_{a,b\in A}d(a,b) $$ which does not mean continuity. Cannot move std::unique_ptr with NULL deleter to std::shared_ptr? My adoptable storage is full. How can I move the data to a bigger MicroSD card? I've got a Moto G 2014 LTE running CM13. When I found out about the whole 'adopt SD card as internal storage' thing, I was quite excited, as the actual internal storage on this phone is tiny. I set up a 16gb micro SD, the biggest I had at the time, and have been using it with minimal problems.  Fast-forward to this day, and I now have a 64gb micro SD that I want to use instead of the 16gb one. However, I'm at a loss as to the best way to copy the data over onto the new one, or if it's even possible. The ideas I've got so far include:   Titanium Backup all [system] apps, factory reset, format new SD and then restore (should work, but a massive pain) NANDroid the SD card onto a USB OTG stick, then restore to the new one (pretty sure this is a bad idea and won't work, but...) Copy all the stuff onto my computer via my phone through a file manager of some sort (I have a linux/linux/windows triple boot, so file format isn't a problem), format the new SD into phone, and copy back (can't be that simple, can it? Hidden/inaccessible files?) Copy the files onto USB OTG with a root file explorer app, reformat the SD as portable, format the new one as adopted, and then copy them back (could work, except for the point below)  with the command adb shell sm set-force-adoptable true, use a root browser to copy all the data over from the old SD, format the old SD as external and remove it, and take the new SD and put it in the SD card slot (seems overly complicated, and also suffers the below pitfall)   The other thing that worries me is mount points: will the apps with externally-stored data know to look at the new SD card for it? Will I have to remount it in a specific location with the shell each boot (or use an init.d script perhaps)? I suppose that it can always be accessed through /sdcard or /storage/emulated/0, but I'm not 100% sure what goes on behind the scenes.  I've got recent NANDroid backups, so if anyone can eliminate the options that definitely won't work (or say which one[s] probably will), I can try them and report back in case someone stumbles across this question in the future.  EDIT: So, my phone further bit the dust and got to the point where it got into a pretty major bootloop. Ended up having to flash a whole new ROM. As such, I probably won't be able to test the full image/decryption/etc. method that I was originally planning. However, I will write up what I've discovered so far in case anyone wants to take up this particular method in the future:   Image the old SD with dd as follows: sudo dd if=/dev/mmcblk0 of=~/SDCardBackup.img, where /dev/mmcblk0 is the location of the sd card as obtainable through lsblk Restore the image you just made to the NEW sd card : sudo dd bs=4M if=~/SDCardBackup.img of=/dev/mmcblk0 (you may have to unmount it first) From here on in, I was following  (thanks @beeshyams!) tutorial, which explains the next steps way better than I ever could have done. The plan was to decrypt the storage and resize the main partition with gparted. I had no idea whether it would work or not, but I felt it was worth a try. However, when I came to run the actual decryption command using the key I'd extracted, I got an error (which I believe was my shoddy linux skills more than anything).   As I say, I have no idea if the above method would work or not even if everything did go to plan. Also, feel free to correct any of my commands above if I mis-typed them (I'm on Windows currently, so this is all from memory.) If this info helps anyone, then good. But until then,  should work fine.  Thanks, Jackdafish Intuitive reason why the Fisher Information of Binomial is inversely proportional to $p(1-p)$ It confuses/blows my mind that the Binomial has variance proportional to $p(1-p)$. Equivalently, the Fisher information is proportional to $\frac{1}{p(1-p)}$. What is the reason for this? Why is the Fisher Information minimized at $p=0.5$? That is, why is inference most difficult at $p=0.5$?  Context:  I'm working on a sample size calculator, and the formula for $N$, the sample size needed, is an increasing factor of $p(1-p)$, the result of a variance estimation in the derivation. Conjugate class in the dihedral group Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map?   I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles.   Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then   1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same  2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$   3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$  Or is there a different/better way to think about this problem? How do I find the caller of a method using stacktrace or reflection? Strange error defining a macro with arguments for math mode Many econometrics textbooks (e.g. Wooldridge, "Econometric analysis...") simply write something similar to: "If the population model is $y = xB + u$ and (1) $\text{Cov}(X,U) = 0$; (2) $X'X$ is full rank, then OLS consistently estimate parameters $B$". I've worked out the math behind consistency but I'm a bit lost in interpreting the meaning.  We're talking about consistent estimation, but estimation of what? Please tell me if the following points are correct:   If we have random sample $X,Y$ and $X'X$ is invertible, then we can always define Best Linear Predictor of $y$ given $x$. And then OLS always consistently estimates coefficients of Best Linear Predictor (because in BLP we have $\text{Cov}(u,x)=0$ from the definition). Bottom line: we can always interpret OLS estimates as coefficients of BLP The only question is whether BLP corresponds to conditional expectation $\text{E}(y|x)$. If it does (for which we need $\text{E}(u|x) = 0$), then we can interpret OLS estimates as partial effects.   What is wrong with this or what am I missing? I don't get the point of stating the assumption of $\text{Cov}(u,x) = 0$, if this assumption is by definition fulfilled in case of Best Linear Predictor. And if we want to give structural interpretation to BLP coefficients (partial effects) we need stronger assumption of $\text{E}(u|x) = 0$ anyway. Is there any other interpretation we may have with zero covariance?