what is the font used to designate Z (ring of integers), Q (field of rationals), R (field of reals) and C (field of complex)?
I know what my symbol or character looks like, but I don't know what the command is or which math alphabet it came from. How do I go about finding this out?
I am using a Collection (a HashMap used indirectly by the JPA, it so happens), but apparently randomly the code throws a ConcurrentModificationException. What is causing it and how do I fix this problem? By using some synchronization, perhaps?  Here is the full stack-trace:  Exception in thread "pool-1-thread-1" java.util.ConcurrentModificationException         at java.util.HashMap$HashIterator.nextEntry(Unknown Source)         at java.util.HashMap$ValueIterator.next(Unknown Source)         at org.hibernate.collection.AbstractPersistentCollection$IteratorProxy.next(AbstractPersistentCollection.java:555)         at org.hibernate.engine.Cascade.cascadeCollectionElements(Cascade.java:296)         at org.hibernate.engine.Cascade.cascadeCollection(Cascade.java:242)         at org.hibernate.engine.Cascade.cascadeAssociation(Cascade.java:219)         at org.hibernate.engine.Cascade.cascadeProperty(Cascade.java:169)         at org.hibernate.engine.Cascade.cascade(Cascade.java:130) Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI. Which year does Spider-Man: Homecoming take place in? My supervisor asked me to find out which distribution represents a particular situation.  I have a VoIP generator that generates calls "uniformly" distributed between callers. This means that the volume per caller distribution is "almost" uniformly distributed between a minimum and maximum. So by running a test with 10000 users and a min value equal to 30 calls per week and a max value equal to 90 calls/week i obtain that not all the users respect this limits: we have some users that generate &lt;30 calls and some other users that generate >90 calls. It is clear that the obtained distribution is not uniform.  The situation is this:    He said that i have to perform a sort of numerical process in order to find some formulas that could define this distribution. Initially, as wrote before,we wanted to obtain a uniform (min,max) distribution (the green area in the figure) but this is not the case as proved with chi-square test.  Moreover the curve in figure is not symmetric, the probability of call generation below 30 call/week and greater that 90 call/week are not identical (it is high for 90calls/week). The variability of the number of generated call increases with the increasing call generation rates.   "Actually implementation of this distribution is nothing but assigning different call rates in a range for users in domain which indicates implementation of several delta functions. As the call rates increases the variability of the generated calls also increases with the average call rate and this leads to the asymmetric behavior of the curve." [cit. from the Voipgenerator documentation]  Someone can help me?I think that now i cannot use Q-Q plot because i don't know which theoretical distribution i have to use in order to compare it with my empirical data.  Sorry if I have stressed with a similar problem a few weeks ago, but initially we thought we could change the implementation, but now we cannot. Hence i have to discover the type of the distribution i obtained and i don't know how can i do this. If a Windows shop moves "everything" to the cloud, does it still need Active Directory? Taking a spin off of this question:  in a new direction for 2014.  Taking into account a basic Windows infrastructure:   domain controllers Exchange 2007/2010/2013 Sharepoint SQL File Servers / Print Servers AD Integrated DNS AD authenticated 3rd party devices (let's say 802.1X for networking and maybe some content-filtering, etc.) AD/LDAP authenticated "administrative" functions on IT apps/hardware/etc. perhaps some KMS stuff throw in a CA if you'd like home grown apps 3rd party in-house apps   Now, let's rip it all out and decide we are going to the cloud.  We've contracted to move Exchange/Sharepoint/File Services to Office 365.  SQL will now be hosted as well on  something like Azure.  We've gotten away from the need for AD-DNS and simply run everything via a simple Windows DNS server.  We still need 802.1X and would like SSO if possible to our various cloud apps.  Home grown and 3rd party in-house apps would likely stay, but have the ability to use internal user databases instead of AD authentication  The question is...do we really need Active Directory at all?  Or more to the point, AD on-premise or even hosted via Azure or similar (ADFS) or running ADDS on a hosted VM through Azure or similar.  Could/Should we look to something else like a 3rd party SSO option such as  or similar that can provide SSO functionality even if it is as simple as LastPass or similar for each user?  What kind of legitimate needs does AD fulfill if everything else in the cloud?  Could a MS-centric infrastructure get away with not having AD at all if they move everything that previously relied on AD to SaaS offerings that didn't rely on AD authentication? Do fontsize environments exist? Without any doubt there are fontsize switches in the form of control sequences (described  for instance.) But, from time to time I ran into code examples where environment versions of those were used, e.g. \begin{small}...\end{small}. After testing those successfully I began to incorporate them into my TeX vocabulary.  A few days ago @GonzaloMedina commented on one of my answers that those environments do not exists and constructions relying on them work by pure coincidence. Then I tried to find out more about those ominous environments. Ominous because they don't seem to be somehow documentated. (The only thing I found that tried to explain them in a somehow systematical manner was some Wikibooks resource.)  Are these environments real or not? "System" storage on macOS Sierra is 470GB! Prior to the spell cast on Mjolnir, did Thor control lightning on his own? We have given isosceles triangle $ABC$ with baseline $AB$. Point $M$ is midpoint of AB. Draw perpendicular line to side $AC$ through $M$ which intersects $AC$ in point $H$. Let $P$ be midpoint of $MH$. Show that lines $BH$ and $CP$ are perpendicular to each other.  I also have "hint": Construct height $BD$ to side $AC$ and look at the triangles $ABD$ and $MCH$.  Analytically there is no problem but I get stuck in geometrical approach to solution. As I understand, the different voltage between two object is caused by the total different charged between two object. Thus, the higher extra electron between two opposite charged object, the higher voltage between them, isn't it? So if i have two battery or capacitor connect in parallel, the negative connect to negative, the positive connect to the positive. Caused the charge from each plate to be double. So that's as my expected, it must be double voltage, but ironically why it's out of my expected and the reality is vice-versa? A hard problem on exponential integration Suppose $a : [0 , 1] \to \Bbb R$ is an infinitely smooth function. For $\lambda\ge1$, define $$F(\lambda) := \lambda \int_0^1 e^{\lambda t} a(t) \, dt.$$ If $\sup_{\lambda\ge1}|F(\lambda)|\lt\infty$, then $a$ is the identically zero function. Below are the some results I have derived:   Derivatives of any order of $a$ vanishes at $t = 1.$ For any $\delta\lt1$, $a(t)$ has a zero in the open interval $(\delta , 1).$  The same is true for all the derivatives of $a(t)$. This tells us that the $n^\text{th}$ derivative of $a$ has infinitely many distinct zeros for all natural $n$. If $a$ is analytic, then I can show that $a(t)\equiv 0.$ If $a(t)\ge0$ on $[0 , 1]$, then it is obvious that $a\equiv 0.$   I would appreciate any hint.  Actually $(1)$ and $(3)$ follows from $(2)$. For $(2)$, suppose that $a(1)\gt0$ (the case that $a(1)\lt0$ can be argued in the same way as the following), then there exists $\delta\lt1$ such that $a(t)$ is strictly positive on $I = [1-\delta , 1]$. Then write $F(\lambda) = \lambda\int_0^{1-\delta}e^{\lambda t}a(t)dt + \lambda\int_{1-\delta}^1e^{\lambda t}a(t)dt$. Now there exists $c\gt0$ such that $a(t)&gt;c$ on $I$ since $I$ is compact, so that the second term is bounded below by $c (e^{\lambda}-e^{\lambda(1-\delta)})$. But the first term is only $O(e^{\lambda(1-\delta)})$, so this contradicts the fact that $F(\lambda)$ is bounded. Morever, using the same idea and integration by parts, one can see that $n^{th}$ derivative of $a$ must vanish at $1$. Since the latest versions (ICS+?) no longer have mass storage, is there any viable alternative to MTP?  The protocol seems inherently broken and the decision to regress to this format seems like a very 'iphone-esque' move (dumb things down, take away power from the users).  There seem to be numerous problems  with files not showing and it seems to me that transferring a file should be simpler than requiring a system reboot - I am unreasonably demanding that way.  I'm aware of tools like Wi-Fi file transfer but often I find myself in areas without wifi, with just my laptop, phone and a usb cable. The other day, for example, I was one a plane, which is an environment decidedly non-conducive to using a wireless system. Why did Harry's father emerge before his mother in HP and the Goblet of Fire? In Harry Potter book 4 (Goblet of Fire) Harry's wand 'links' with Voldemort's during their duel, and Voldemort's wand begins 'spitting out' each spell it had cast, in reverse order.  This produces false 'ghosts' of people he had killed with .  In the US printing, James Potter appears, followed by Lily.  If the spell order was supposed to be reversed, why did Harry's father's "ghost" emerge before his mother's, since it's repeatedly stated (before and after) that he died first? "a ten years old boy" or "a ten year old boy" 1) a ten years old boy is sitting on the couch.      2) a ten year old boy is sitting on the couch.      3) a ten-years old boy is sitting on the couch.      4) a ten-year old boy is sitting on the couch.   Which is correct? Why do some games allow offline Steam achievements? I know of  where it is asked if you can get Steam achievements offline. The accepted answer to that was no.   But I do earn achievements for some Steam games while playing in offline mode!  For example, right now I am earning achievements for Cave Story+ as I play in offline mode. In other games I also earn achievements while playing in offline mode. Not just when I am disconnected but still in "online" mode as might be suggested in the linked question.  But what is really bugging me is that this does not happen with all games that have Steam achievements.  So, is allowing offline achievements something that Steam allows, the game devs allow, or is it something decided by a magical elf in the steam servers rolling some d20 dice around?  Edit:  As per request, I was able to attain visual evidence:    You should be able to see the achievement-get in action while in offline mode.  What is worth noting, (and something that I forgot to mention before) is that I only started earning achievements after I had went into online mode with the game installed. This is why I was able to earn this achievement with the ones after unlocked. I had started the game for the first time in offline mode, and did not notice that I was not earning achievements. When I had connected once, I did not have to start the game while online. I was able to earn achievements offline.  I also must state publicly: I do not play on easy. This was for testing purposes only people. It was for Science! Is there a difference between copy initialization and direct initialization? no wifi in Ubuntu 15.04 Is there a way to change the process group of a running process? Is there a way to change PID, PPID, SID of a running process? It would make sense for the answer to be no, but I'd like to make sure. Is "can not" unambiguous? I always try to remember to use "cannot" when applicable as in "I cannot take an umbrella" (because I do not have one).  I also thought that the problem with "I can not take an umbrella" was that it was ambiguous: it could also mean that I could decide not to take an umbrella. Trying to write unambiguously most of the time, I would therefore never use it.  Now I see it claimed that it is in fact not ambiguous, and only carries the second meaning. Usually I would accept such advice at face value and start using "can not" when the second meaning is unambiguously meant, but this comes from a style guide I have other reasons not to respect very much.  Does "can not" have a single unambiguous meaning?  (I have read  but my question is more about current usage. The length of the answer to that question does confirm my prejudice a bit, though) My question is simple, why do we believe  to be absurd? Is the Ultra-violet catastrophe incorrect or is it only because we can not create or know of a perfect emitter? I am a bit confused as to why Quanta was necessary.