Validation of Email in JavaScript
How to validate an email address in JavaScript
Conky Widgets are Opening in their Own Windows I just upgraded to 16.04 and I'm getting all my customization set up again, but it seems that most Conky widgets are opening in their own window, instead of just appearing on the Desktop; although some are working correctly.  For example:  LSD (working properly):    Seamod (not working, opening in own window):    Since it's not affecting every single widget, I assume it could be an issue with the conkyrc with the widgets that aren't working, but this also never happened in 14.04 so I wasn't sure if maybe there was some compatibility issue in 16.04. My Google-fu has failed me in finding a solution.  If anyone can point me in the right direction, it would be much appreciated! :) This is relating to the movie. Frodo finally relinquishes his desire for the ring, and it melts in the fires of Mount Doom. The next scene shows the eruption of Mount Doom and the destruction of Barad Dur. Gandalf looks up with tears in his eyes.  I have read the books, and can't seem to remember this being explained or described. I'm just wondering why they depicted Gandalf as being so emotional about the event, when throughout the story it depicts Gandalf as having unparalleled wisdom, faith and knowledge. Out of all the Ainur he listened to Eru's song the longest and contemplated it's meaning.   It strikes me that Tolkien portrayed Gandalf as all knowing, and has ultimate faith that Frodo would carry out the task at hand. So why was he so emotionally surprised that it happened? Especially when he had seen far more surprising things happening throughout the War of the Ring and in the movies he was depicted as emotionally immovable, an almost cold hearted awareness of the evil that lurks within the world.   I understand this is probably going to be down-voted as it has alot of opinion based statements. The TL:DR is: Was Gandalf described in the book as being so emotional at Sauron's fall? Or was this simply for effect in the screenplay? Fermi in his lecture asserts:     The second law of thermodynamics rules out the possibility of constructing a perpetuum mobile of the second kind.   So, this means there can be no machine which just transforms all the heat energy gained by cooling surrounding bodies into mechanical work.  But how does actually the Second Law prohibit this?   The hot reservoir provides heat energy to the system. Does it cause a decrease in entropy of the universe(system + hot reservoir)? How? In order to receive heat wouldn't the system have to be cooler than the reservoir? If so, then entropy increases as the heat energy gets expelled from the reservoir at a higher temperature than the temperature the system receives the heat energy. Is it so?  I'm not getting how the Second Law nullifies the existence of this machine. Could anyone please explain how the Law prohibits the perpetual machine of the second kind? Are PHP short tags acceptable to use? Unfortunately, despite having tried to learn regex at least one time a year for as many years as I can remember, I always forget as I use them so infrequently. This year my new year's resolution is to not try and learn regex again - So this year to save me from tears I'll give it to Stack Overflow. (Last Christmas remix).  I want to pass in a string in this format {getThis}, and be returned the string getThis. Could anyone be of assistance in helping to stick to my new year's resolution?    Related questions on Stack Overflow: In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. How does item level and rarity affect the possible attributes? Inequality involving Euler-Mascheroni constant I would like to solve the following problem from 'Real Analysis and Foundations' by S. Krantz.  Let $a_n = 1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log{n}$ and $\gamma$ be the Euler-Mascheroni constant. Show that $$\vert a_n - \gamma \vert \le \frac{C}{n}$$ for some universal constant $C&gt;0$.  I have no clue how to solve this problem. From searching various sites it seems that $C=\frac{1}{2}$ but I am not sure with this, too. Thanks in advance. If $f:U\to V$ is holomorphic and $f'(z)\neq 0$ for all $z\in U$, then$f$ is locally bijective. I am trying to solve the following problem:     Let $f:U\to V$ be a holomorphic function such that $f'(z)\neq 0$ for all $z\in U$. Show that for all $z_0\in U$, there exists a disc $D_\varepsilon(z_0)\subseteq U$ such that $f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))$ is bijective.   Atempt:  I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let $z_0\in U$. Since $f'(z_0)\neq 0$, there is a disc $D_r(z_0)\subseteq U$ such that $$f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)$$ where $h$ is holomorphic on $D_r(z_0)$ and $h(z)\neq 0$ for all $z\in D_r(z_0)$.   Let $0&lt;\varepsilon&lt;r$ to be defined later and fix $w\in D_\varepsilon(z_0)$. To show that $f$ is injective in $D_\varepsilon(z_0)$ is to show that the function $f(z)-f(w)$ has exactly one zero in $D_\varepsilon(z_0)$. But $$F(z):=(z-z_0)h(z)$$ has exactly one zero in $D_\varepsilon(z_0)$, so we might want to apply Rouche's Theorem with $F$ and $$G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)$$ to conclude that $F$ and $F+G$ have the same number of zeros and hence $f$ is injective. But that means that we need to find $0&lt;\varepsilon&lt;r$ such that $|G(z)|&lt;|F(z)|$ on $\partial D_\varepsilon(z_0)$. That is, $$|w-z_0||h(w)|&lt;|z-z_0||h(z)|$$ for all $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$.  Can we find such $\varepsilon&gt;0$?  It seems intuitive since if we expand $h$ in a power series at $z_0$, $h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$, then $$|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n &lt; |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.$$ for $w\in D_\varepsilon(z_0)$ and $z\in\partial D_\varepsilon(z_0)$. But this is of course not a proof. I'm considering buying a solid state drive for my Unibody White MacBook 2010. The trouble is, I have no idea what to look for. What kind of connector am I looking for? (SATA?) What size? I'm told to get the kind that Apple uses so that I don't void my AppleCare warranty. What model/brand(s) would that be? A book suggests that we should not interchange "so" and "so that". "So" means "therefore", and "so that" means "in order that". However, it seems to me that in many cases they don't actually have a difference. For example:     Alice got up at 5:00 A.M., so that she could drive her son to school.   If I substitute "so that" with "so", this sentence presents the exact same causal relation to me:     Alice got up at 5:00 A.M., so she could drive her son to school.   I must have missed something here, since English is my second language. Could somebody explain the difference? How compatible are my old Nikon lenses? I'm looking at buying a new SLR body. I have an old film Nikon entry-level SLR (about ten years old) with a kit lens. If I buy a Nikon body, how likely is it that my old lens will work on it? Would I be better off buying a new lens? Calculate work done by a hovering helicopter over time This is likely to be very simple, but...  How does one calculate work done by a hovering unmoving aircraft over time?  As in work in Joules.  In this scenario, to remain hovering the aircraft has to exert a force that counters gravity (which would be its mass times G, pointing in opposite direction of gravitational pull).  However, work is defined as newton per meter, and the aircraft does not move.  Meanwhile the battery charge or fuel is going to be expended, so the energy is spent.  The other question here:   is dealing with conservation energy and not calculating drain/work performed by hovering. This is the proof, which I mostly understand except for one bit:  You have $h_1 \in H_1$ and $h_2 \in H_2$.   We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore  $$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $$  and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$  $$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$$ $$ h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$$ $$ h_2^{-1}h_1h_2 = h_1 \{1_G\}$$   Multiply both sides on the left by $h_2$ giving us  $$ h_1h_2 = h_2 h_1 $$  The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$." Peggy would like to prove to Victor that she knows the discrete logarithm of $y$ based $g$; that is, she knows $x$ such that $y = g^x \bmod p$. One round of the interactive proof protocol consists of the following steps.   Peggy picks random $k \in \mathbb Z/(p−1)\mathbb Z$, computes $t = g^k \bmod p$, and sends $t$ to Victor. Victor picks random $h \in \mathbb Z/(p−1)\mathbb Z$ and sends $h$ to Peggy. Peggy computes $r = (k − hx) \bmod (p − 1)$ and sends $r$ to Victor. Victor verifies that $t = g^r y^h \bmod p$.   The interactive protocol can be converted into a noninteractive zero-knowledge proof by choosing and making public a collision-resistant hash function $H$, and changing the second step of the interactive protocol to the following: Peggy computes $h = H(y, t)$. Then the noninteractive proof consists of $(t, h, r)$, which can be verified as follows: $$h = H(y, t), \qquad t \stackrel?= g^r y^h \bmod p.$$   What is the problem if in the non-interactive proof the hash $h$ depends only on $y$? That is, $h = H(y)$, and the proof consists of $(t, h, r)$, which can be verified as follows: $$h = H(y), \qquad t \stackrel?= g^r y^h \bmod p.$$ What is the problem if in the non-interactive proof the hash $h$ depends only on $t$? That is, $h = H(t)$, and the proof consists of $(t, h, r)$, which can be verified as follows: $$h = H(t), \qquad t \stackrel?= g^r y^h \bmod p.$$ I'm having some trouble proving that the Gaussian Integer's ring ($\mathbb{Z}[ i ]$) is an Euclidean domain. Here is what i've got so far.  To be a Euclidean domain means that there is a defined application (often called norm) that verifies this two conditions:   $\forall a, b \in \mathbb{Z}[i] \backslash {0} \hspace{2 mm} a \mid b \hspace{2 mm} \rightarrow N(a) \leq N (b)$ $\forall a, b \in \mathbb{Z}[i] \hspace{2 mm} b \neq 0 \rightarrow \exists c,r \in \mathbb{Z}[i] \hspace {2 mm}$ so that $\hspace{2 mm} a = bc + r \hspace{2 mm} \text{and} \hspace{2 mm} (r = 0 \hspace{2 mm} \text{or} \hspace{2 mm} r \neq 0 \hspace{2 mm} N(r) \lt N (b) )$   I have that the application meant to be the "norm" goes: $N(a +b i) = a^2 + b^2$, and I've managed to prove the first condition, given that N is a multiplicative function, but I can not find a way to prove the second condition.   I've search for a similar question but I have not found any so far, please redirect me if there's already a question about this and forgive me for my poor use of latex. Convert a Cannondale SM1000 (about 1991 model) to disk brakes I've got an older, about 1991, model Cannondale SM1000 mountain bike purchased new.  Would it be worth it to convert to disk brakes and if so what would be the process? Looking for story title - younger gen waters down age-slowing drink of older gen, crammed into small apartment I remember a short story in which multiple generations of people being crammed into a small apartment in a city. The magical drink that either reduced or completely stopped aging was something like a soda that came in a six pack. The story focused on visual aspects of the drink at some point. The younger family members complained that their elders were living too long and apparently decided to water down the soda, I think that conclusion was ultimately left up to the reader. Many thanks for any help tracking down this story, might be one of Kurt Vonnegut's perhaps? How do you add CSS rules (eg strong { color: red }) by use of Javascript? In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation.