How can I display keystrokes while recording my screen?
Windows utility to render which key I am pressing on-screen
From official :     Due to a head injury, Cyclops is unable to shut off his optic blasts   at will and must therefore wear ruby quartz lenses to block the beams.   From the same page:     Cyclops possesses the mutant ability to project a beam of heatless   ruby-colored concussive force from his eyes, which act as   inter-dimensional apertures between this universe and another.   When his eyes are always blocked by ruby-colored beam of concussive force, how does he see? Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ What is the solution of $\cos(x)=x$? There is an unique solution with $x$ being approximately $0.739085$. But is there also a closed form solution? For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.   My attempt: Let  $$\begin{align*} f_n(x) &amp;= \frac{\ln\left(1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(\displaystyle\frac{1-2 \left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\ &amp;=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}. \end{align*}$$ Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But  $$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider $$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$ So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$  It looks too easy, is there any crucial lack?  $(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof. Transformer VA Ratings, Power factor, and Pulsed current What exactly goes into a transformer's VA rating? At a fundamental level it seems that there has to be a limit to what the VA rating can represent.  For example if I draw 2 Amps-RMS from the secondary of a 30VA rated transformer at 15 Vrms this makes sense as the maximum Complex Power I can draw. Still, at a power factor of 1 my maximum current in the transformer would be sqrt(2) * 2 Amps.   But what if my power factor is abysmally low (ie in a DC rectifier) due to harmonic distortion and while I still draw 2A-RMS I have brief current spikes of 15 Amps? Surely the transformer's magnetics couldn't handle these spikes as I'd expect the core would saturate.  My question is what is the practical limit for VA ratings of transformers? Would brief current spikes that reduce to only 2A-RMS in this case be fine? Obviously the secondary/primary copper losses play a part in the rating but how do the magnetics factor in?  I ask this because I am designing a simple bench power supply using a 125VA transformer with a 24Vrms secondary and am trying to figure out how much power I can draw before the current spikes into the inductor are too much for the transformer. I also don't want to have to resort to active PFC as I don't want to risk messing up any HVAC PCB design.  Thank you dev/sda1: clean, 552599/6111232 files, 7119295/24414464 blocks   After I turn on my laptop this message appears. The system never boots, the message just stays there. If a black body is a perfect absorber, why does it emit anything? I'm trying to start understanding quantum mechanics, and the first thing I've come across that needs to be understood are black bodies. But I've hit a roadblock at the very first paragraphs. :( According to :     A black body (also, blackbody) is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.   OK, that's nice. It's an object that absorbs (takes in itself and stores/annihilates forever) any electromagnetic radiation that happens to hit it. An object that always looks totally black, no matter under what light you view it. Good. But then it follows with:     A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation.   Say what? Which part of "absorbs" does this go with? How can it absorb anything if it just spits it right back out, even if modified? That's not a black body, that's a pretty white body if you ask me. Or a colored one, depending on how it transforms the incoming waves.  What am I missing here? I want to learn Linux Kernel and I've downloaded the latest kernel version. Now I want to configure Linux with make menuconfig and I'm receiving an error:     Arch/x86/Makefile:184  Compiler lacks asm-goto support...   Can somebody help me? String Comparison in Java Recently, I am trying to create 3D models for my own game project. However, when I do some researches on game models, I come up with the term - game-ready models and I could not find any clear definition online. Could anyone tell me about what exactly is game-ready models and how to know if my 3D models (created by Blender mainly) is suitable for games.  From what I know by far, quad topology is preferred in Blender. However, for most of the game engine, triangle topology is preferred (Please correct me if I am wrong). Does it mean I have to manually change the models in Blender from quad topology to triangle topology before importing to the game engine?  Thanks for any helps. Show $R \setminus S$ is a union of prime ideals I'm stuck on the following question:     Let $R$ be a commutative ring with $1$, and $S \subseteq R$ a saturated multiplicative set (that is, $1 \in S$ and $x, y \in S$ if and only if $xy \in S$).  Show that $R \setminus S$ is a union of prime ideals.     Here's my attempt so far.  If $a \in R \setminus S$, then the ideal $Ra$ does not intersect $S$, because if $ra$ were in $S$ for some $r \in R$, then $a$ would have to be in $S$.  Thus, it follows by Zorn's lemma that there exist ideals $I$ of $R$ which are maximal with respect to the property that $I \cap S = \emptyset$.  Let $I$ be any such ideal.  If I can show that $I$ is prime, the conclusion will follow.  Let $xy \in I$.  I want to show that $x$ or $y$ is in $I$.  Suppose $x \not\in I$.  Then $I + Rx$ intersects $S$, so there exists $s \in S, a \in I, r \in R$ such that $s = a + rx$.  Then $sy = ay + rxy \in I$, so $sy$, and hence $y$, cannot be in $S$.  Similarly if $y$ is not in $I$, then $x \not\in S$.  So we are reduced to proving the following is impossible: $xy \in I$, $x$ and $y$ are not in $S$, but they aren't in $I$ either. How to print the full NumPy array, without truncation? Do extension fields always belong to a bigger field? Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?  Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials? Related to Meta Super User efforts:  and Meta Server Fault efforts:  We already auto-remove negatively voted unanswered old questions automatically after 30 days, network wide, with no human intervention required. I was thinking we might extend that to remove old, unanswered zero score questions... based on the following criteria:  fewer than (question age in days * 1.5) views 0 score or lower, or 1 score or lower if the original author is deleted no answers 1 comment or fewer asked more than 1 year ago (thus based on creation date, not last activity date, so , or a user editing it, does not give a doomed question 365 more days of zombie &quot;life&quot;)  See the results of this query on:  (10509) |  (787) |  (650) This query has been refined based on the comments, and this is the final version that will roll out tonight. Every site in the network (except metas) will automatically delete old questions that meet this criteria from this point forward. How can I initialize normal array member variable in initializer list? Can I use 'in' instead of 'at'? -- "I have more books in my house." I am learning English using the Duolingo English course, one of correct solution was      I have more books at my house.   Can I use      I have more books in my house.   If it is OK, what is the difference between both? Is the electromagnetic spectrum discrete? Stack Exchange's heavy reliance on external scripts The program ed, a minimal text editor, cannot be exited by sending it an interrupt through using Ctrl-C, instead printing the error message "?" to the console.  Why doesn't ed just exit when it receives the interrupt?  Surely there's no reason why a cryptic error message is more useful here than just exiting.  This behavior leads many new users into the following sort of interaction:   $ ed hello ? help ? exit ? quit ? ^C ? ^C ? ? ? ^D $ su # rm -f /bin/ed    Such a tragic waste—easily avoidable if ed simply agreed to be interrupted.  Another stubborn program exhibiting similar behavior is less which also doesn't appear to have much reason to ignore C-c.  Why don't these programs just take a hint? Prove that if for every $x \in \mathbb{R}^N$ $Ax=Bx$ then $A=B$