I posted a question on StackOverflow:    Then I got spam from a 3rd party trying to get me to go to their site to get an answer, instead of SO. I know SO can't stop the email spam from a 3rd party, but is there someway to find if they have any SO accounts and nuke this f'ing spammer?
Well today as I was going home I got a strange email, in my private inbox (email associated with S.O.) asking me if I had solved a question on S.O. that I didn't ask. To be more precise I had given an answer.   After reading:       and several other email related issues on S.O. I decided to let you all know about this because probably I am not the only one.   So the mail subject was:     Re: Deleted facebook account - Cannot access my apps?   which refers to   And the body (masked last name of sender):      Hi Jimmy,      Just wondering if you ever got this question answered.  If not, maybe   a brief chat with a html wizpert could help: link   .      Good luck getting this resolved!      Cheers,      Michael      ------------------------- Michael W%^%^GF Founder, CEO @ Wizpert   Now I know that my email address is not visible and I read about the MD5 regarding gravatar but I don't think this is the case. I thought about some methods and sure it's possible via metadata, github, etc for someone to get my mail, but I do not dare to think that this has reached an automated level, so that some service could easily get the email of S.O. members.   I didn't get annoyed by the message, since the CEO took the time to contact me :-p but someone else could get annoyed by these kinds of messages, since they come in the important inbox of gmail.   What is more strange is that Wizpert seems to be a nice site and also is active in the Careers 2.0 if I am not mistaken.   Another thing that crossed my mind is that since Wizpert needs unanswered questions to promote their services, possibly this is an action of Wizpert trying to attract people into their services.   Well that's it, I hope I did good to let you know (maybe I did't include the correct tags).
sudo: unable to resolve host   I think I need to edit the /etc/hosts file to make an entry for 127.0.1.1 PC name.  But how can I do that if I can't access sudo? It seems to have gone since I updated to 13.10.  I have tried nano, which still doesn't work. How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$? I can show that the following limit exists but I am having difficulties to find it. It is $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}$$ Can someone please help me? I am currently using 19.04 and I got "No new release found for 19.10" AES Mix Column Transformation Section 3.4.3 "The MixColumns Step"(page 39) of , it states     The columns of the state are considered as polynomials over   $\operatorname{GF}(2^8)$ and multiplied modulo $x^4 + 1$ with a fixed polynomial $c(x)$.      The polynomial $c(x)$ is given by   $c(x) = \mathtt{03} \cdot x^3 + \mathtt{01} \cdot x^2 + \mathtt{01} \cdot x + \mathtt{02}$      Let $b(x) \equiv c(x) \cdot a(x) \pmod{x^4 + 1}$. Then         How this polynomial multiplication modulo $x^4+1$ is converted to Matrix Multiplication? I am going to Santa Monica for 4 weeks to polish my English in July and would like to volunteer at the Planetary Society located in Pasadena.  I am a Swiss citizen and have a biometric passport. However, I don't have any special visa, as there is no visa needed for the school I am attending and using the Visa Waiver Program, I am allowed to stay up to 90 days in the US.  Am I allowed to do volunteer work during my stay? How do I show a function on 2-adic units is continuous (using the 2-adic metric)?   I'd be happy to learn the general rule or definition. But in particular I need to show that $f(x)=\dfrac{3x+1}{2^{v_2(3x+1)}}$ is continuous at all odd numbers.  Since the function is isometric in $\lvert\cdot\rvert_2$, i.e. since $\lvert f(x)\rvert_2=\lvert x\rvert_2$ every orbit of the function is duplicated, multiplied by any power of $2$.  This can therefore be formulated in various ways; as a function though the odd numbers, through the dyadic rationals, but the following seems good to work with.  If we let $n$ be a positive integer and $k$ be the power of $2$ we can define:  $f(2^k(2n+1))=(3(2n+1)+1)\cdot2^k\cdot\lvert3(2n+1)+1\rvert_2$  So $f(2^k(2n+1)=(3n+2)\times2^k\times\lvert3n+2\rvert_2$  Here is what I have so far:  I think to prove continuity in the 2-adic metric I need $\lvert x_n-x\rvert_2=0\implies\lvert f(x_n)-f(x)\rvert_2=0$  I think $k$ and $n$ are independent so I think I can examine them independently.  Taking $k$ to infinity brings both $x$ and $f(x)$ to zero so that seems to satisfies the continuity requirement.  Moving on to $n$; this seems to be an exercise in proving convergence within odd integers which are in a sense a subset of the 2-adic units.  I know all Cauchy sequences in these converge to 2-adic units but not much more than that.  However I do have a little insight into this particular function.  For example if we examine the inputs $x$ which map to any given output of $f(x)$, it can fairly easily be shown that these $x$'s take the form of a set $\left\{4^mp+\dfrac{4^m-1}{3}:m,n\in\mathbb{N}\right\}$ so, at least for any given output $f(x)$ the inputs always converge to $x=\frac{-1}{3}$ as the intervals between them become large powers of $2$.  In fact the orbit of the function $a(2^kx)=2^k(4x+1)$ on variation of $x$ and holding $k$ fixed is in a sense orthogonal to $f(x)$; which is a restatement of the above except not in closed form. How can I make a plane's edges taper inwards? How can I deform one edge or multiple edges of a plane or cube so they are tapered in and rounded equally? TikZ arrow emanating from node at an angle for a distance How do I draw a line of length l and at an angle of a from a circular node N in TikZ? What is a field, really? Remove "Set as desktop background" from Windows context menu When I right click an image file in Windows, one of the context menu options is "Set as desktop background". I know some entries can be , so I'd like to know if it's possible to remove the "Set as background" option.  Note that while  can be found under HKEY_CLASSES_ROOT\, the "set as background" option seems to be located somewhere else. Conclusively stop wake timers from waking Windows 10 desktop How can I get the Cartesian product (every possible combination of values) from a group of lists?  Input:   somelists = [    [1, 2, 3],    ['a', 'b'],    [4, 5] ]   Desired output:  [(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4), (2, 'a', 5) ...] File /etc/ld.so.nohwcap is missing in Debian 7 What is the file /etc/ld.so.nohwcap used for?. In my machine (Debian 7), this file is not found. Therefore, Which package or library this file belongs to? Fixed point of a Function (monotonically increasing) I recall from complex analysis a while ago that the series $$\sum_{n=1}^\infty \frac{z^{n!}}{n}$$ has radius of convergence 1, diverges on roots of unity, and converges elsewhere on the circle. It's simple enough to see that it has to diverge on roots of unity: consider $z$ such that $z^k = 1$ for some $k \in \mathbb{N}.$ Plug this into the series and you see that past the $k$'th term, $z^{n!} = (z^k)^m = 1$ for some $m \in \mathbb{N}$; hence, the series becomes harmonic and must diverge.  But I don't recall how to show that the series converges for other values on the circle. If this is simple I'd like just a hint, if it is involved I would prefer a sparse outline of the proof. Thank you! Move EFI System Partition to another drive If I install the Dev Preview 3 on my Nexus 5 right now, will I get an OTA to the stable build or will I have to flash it as well? A geometric sequence has its first term equal to $12$ and its fourth term equal to $-96$.   How do I find the common ratio? And find the sum of the first $14$ terms I know standing in line to get a visa is never fun, but was wondering if any Americans had experience getting a standard tourist "L" visa in Hong Kong.   I'm leaving on a RTW trip in less than two weeks and I plan on being in China around June. So I either need to get a 6 or 12 month multi-entry visa before I leave the USA - which would entail booking all my flights and accommodation and securing an invitation letter in the next few days - or just fly to Hong Kong before going to China in June.   Hong Kong seems like the simplest answer, but China usually doesn't allow Americans to get visas outside of their home country (HK seems to be an exception). I just wanted to know if any Americans had done it successfully.  Update: The Chinese Visa Office in HK currently has this notice on their website: "If you don't reside or work in Hong Kong permanently, you are required to apply Chinese visa from the Embassy or Consulate-General of Peoples' Republic of China in your resident country." It seems practices may have changed. Any Americans with experience care to contradict this?  Link: How to ignore deprecation warnings in Python