If watts exceeds rating?
Choosing power supply, how to get the voltage and current ratings?
Is it possible to split micro USB to USB cable so two cables (data + and data -) go to the PC with USB and two other power cables go to power brick? How to avoid interference in wireless communication? I am working on wireless communication system. We are using around 10 pairs of transmitter and receivers. We are using atmega16 microcontroller for encoding and decoding by USART ports.  Now we are able to transmit the data and receive the same in the receiver end, but there is a major problem, when we are finding the 2 transmitter data coming at the same time. The receiver is unable to get it due to interference.  Suppose one transmitter sends "SENDA" at the meanwhile another transmitter sends "GETTS", at that time the receiver can not receive proper data. As all the transmitters and receivers are working in same frequency, so this interference is occurring. How can I resolve this issue? In mathematics, we like talking about idempotent matrices. In a masters program that I am currently enrolled in, I have heard one professor pronounce it one way, and another professor pronounce it another way. The  explains it better than I could:  Is there a correct pronunciation for idempotent? Is it like omnipotent (om-nip-o-tent), so id-EMP-o-tent, or is it more like the two seperate words idem + potent I'd say EYE-dm-POT-nt, but then I speak Brit. Charles Matthews 07:57, 30 Jun 2004 (UTC) Ditto, but then I speak Australian ;) Dysprosia 08:01, 30 Jun 2004 (UTC) Actually the British is more like EYE-dm-PO-tnt, I guess. Charles Matthews I'd just like to pedantically point out that it would be pronounced i-DEM-po-tent because because syllable onsets are maximized. daesotho 20:33, 21 Oct 2004 (UTC) The American pronunciation puts primary stress on the second syllable ('dem') and secondary stress on the fourth syllable ('tence'). Could somebody more familiar with IPA than I am please add this to the article as an accepted pronunciation? My American dictionary says |ˈīdemˌpōtənt| for the adjective. I speak American (non-natively) and I wouldn't stress the second syllable for itempotence. My only question is, right now it says /ˌaɪdɪmˈpoʊtəns/ -- but can't it be stressed on the first syllable too? Should we add /'aɪdɪmˌpoʊtəns/ as an alternative? -- 87.160.141.177 (talk) 14:50, 3 June 2011 (UTC) And by the way I think it's silly giving a pronounciation guide as this is just imposing a particular accent. (For what it's worth I pronounce idempotent with a short 'i', as in the word 'id'). Alex Selby I agree with Alex Selby - the pronunciation guide should be extended to show the variation, or removed. Strictly, there's no reason for the first i to be long (neuter idem has a short first syllable in Latin), and I can't find any authoritative source which advocates the pronunciation given. Ms7821 (talk) 22:59, 15 July 2015 (UTC)  Is there a &quot;correct&quot; pronunciation for this word? Add 1Password extensions support to the SE iOS app Please implement the URL scheme to call 1Password for password lookups on the iOS app.   a URL scheme for switching to 1Password to search item titles for a custom term, usually a service name like “twitter” or “evernote”. This speeds up the process for the user so they can copy something like their password and get back to your app  That will ease sign up speed using long, complicated or unmemorable passwords for many users of both apps. In the two years since the above blog was published there are now updated instructions for implementation and code on github: How to prove $\sum_{k=1}^{n}F_k = F_{n+2}-1$ by induction when $F_n$ is the Fibonacci sequence Let $F_n$ be the Fibonacci sequence where  $F_0$ = 0  , $F_1$ = 1 and  $F_n$ = $F_{n-1}$ + $F_{n-2}$.                   I want to prove the following by induction. $$\sum_{k=1}^{n}F_k = F_{n+2}-1$$ Here is what I have so far.  Can anybody tell me if I am right?  Base case = n=1  $F_k$ = 1 = LHS  $F_{1+2}$ - 1 = 2 - 1 = 1 = RHS  Statement is true for n=1.  Assume statement is true for n=i.  $$\sum_{k=1}^{i}F_k = F_{i+2}-1$$  Prove statement is true for n=i+1.  $$\sum_{k=1}^{i+1}F_{k} = F_{((i+1)+2)}-1$$  $F_1$ + $F_2$ +...+ $F_i$ + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{i+2}$ - 1 + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{((i+1)+2)}-1$ = $F_{((i+1)+2)}-1$  Statement holds for n=i+1 Jutifying: irrational + rational = irrational I have the below list of files  aro_tty-mIF-45875564pmo_opt aro_tty-mIF-45875664pmo_opt aro_tty-mIF-45875964pmo_opt aro_tty-mIF-45875514pmo_opt aro_tty-mIF-45875524pmo_opt   that I need to rename to  aro_tty-mImpFRA-45875564pmo_opt aro_tty-mImpFRA-45875664pmo_opt aro_tty-mImpFRA-45875964pmo_opt aro_tty-mImpFRA-45875514pmo_opt aro_tty-mImpFRA-45875524pmo_opt In a simple battery + resistor circuit, what form of energy is lost from the electrons upon exiting the resistor? I will give this question a little context. Firstly, as I understand it, as soon as I "close the switch" on a circuit, electric current pretty quickly establishes a steady state where, at any given cross section along a wire, the average kinetic energy of that slice is both constant and equal anywhere along the wire. (Please correct me if I am misunderstanding).   If the average kinetic energy is constant anywhere throughout the wire in the circuit (which confuses me, as I would think that the electrons should all accelerate after exiting the resistor since they are no longer being impeded as much), that means that the average velocity is constant. So, in the example of a simple circuit with a battery of v Volts and a resistor of o Ohms, my question is the following:  Because a 'voltage drop' is known to occur across the resistor, what type energy is being "traded" to generate the heat that radiates from the resistor? I would think that the reflexive answer is "electrical energy"...hence the VOLTAGE drop (the sacrificed energy is clearly not kinetic, as the velocity is constant everywhere). However, I find this confusing. Does this mean that if I had a positive test charge, it would be easier for me to bring it to the beginning of the resistor as compared to bringing it to the end of the resister?     Further, if the correct answer IS "electrical energy", why exactly DOES electrical energy get "dissipated" as the electrons pass through the resistor? I always here the comment Oh! It's because all the electrons are running into densely packed crap which impedes their flow but that to me sounds like a reason for their KINETIC energy to be reduced. However, clearly that's not the case. So, ultimately, I guess the real question is:  **What goes on in the resistor that is literally removing electrical energy **. I am sure that this is a quantum mechanics question, but my quantum mechanics knowledge is not terribly strong. If I could get an answer that is devoid of crazy wave function equations, I would greatly appreciate it. Thanks! Money making in Runescape I wish to show that $\mathbb{Q}(\sqrt2, \sqrt3)$ is not isomorphic to $\mathbb{Q}(\sqrt[4]2)$.    What I did so far was to show that $\mathbb{Q}(\sqrt2,\sqrt3) =\text{span}\{1,\sqrt2,\sqrt3,\sqrt6\}$.  Assuming they are isomorphic, denoting $f$ the isomorphism, then $f(1) = 1$, so $x^2-3$ has a root in $\mathbb{Q}(\sqrt2,\sqrt3)$ so $\sqrt 3\in\mathbb{Q}(\sqrt[4]2)$. ($\sqrt2 \in \mathbb{Q}(\sqrt[4]2)$, $\sqrt[4]2\cdot\sqrt[4]2=\sqrt2)$.   Thus under the assumption, $\mathbb{Q}(\sqrt2, \sqrt3)\subseteq \mathbb{Q}(\sqrt[4]2)$. As a vector spaces of the same dimension over $\mathbb{Q}$, it means that $\mathbb{Q}(\sqrt2, \sqrt3)= \mathbb{Q}(\sqrt[4]2)$.  Now all is left to do it to show a contradiction: $\sqrt[4]2\in\mathbb{Q}(\sqrt2,\sqrt3)$, then $\sqrt[4]2 = a+b\sqrt2+c\sqrt3+d\sqrt6$ so $2=(a+b\sqrt2+c\sqrt3+d\sqrt6)^4$, I believe that opening the expression, I would find out that it can't be an element of $\mathbb{Z}$. However this is a really inelegant way (if it really works), and I wish for a better way to do so.  An other approach: $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt3)$ by the irreducibility of $x^4-10x^2+1$ over $\mathbb{Q}$ (an some dimension theorem usage). Also, $x^4-10x^2+1$ splits in $\mathbb{Q(\sqrt2, \sqrt3)}$.   Is there a way to show that this polynomial doesn't split over $\mathbb{Q}(\sqrt[2]4)$? with different method then assuming that $\sqrt3\in\mathbb{Q}(\sqrt[4]2)$ and getting a contradiction? Maybe the fact that $\sqrt[4]2$ has complex roots might help? How to avoid perceived circularity when defining a formal language? Suppose we want to define a first-order language to do set theory (so we can formalize mathematics). One such construction can be found . What makes me uneasy about this definition is that words such as "set", "countable", "function", and "number" are used in somewhat non-trivial manners. For instance, behind the word "countable" rests an immense amount of mathematical knowledge: one needs the notion of a bijection, which requires functions and sets. One also needs the set of natural numbers (or something with equal cardinality), in order to say that countable sets have a bijection with the set of natural numbers.  Also, in set theory one uses the relation of belonging "$\in$". But relation seems to require the notion an ordered pair, which requires sets, whose properties are described using belonging...  I found the following in Kevin Klement's,  (pages 2-3).  "You have to use logic to study logic. There’s no getting away from it. However, I’m not going to bother stating all the logical rules that are valid in the metalanguage, since I’d need to do that in the metametalanguage, and that would just get me started on an infinite regress. The rule of thumb is: if it’s OK in the object language, it’s OK in the metalanguage too."  So it seems that, if one proves a fact about the object language, then one can also use it in the metalanguage. In the case of set theory, one may not start out knowing what sets really are, but after one proves some fact about them (e.g., that there are uncountable sets) then one implicitly "adds" this fact also to the metalanguage.  This seems like cheating: one is using the object language to conduct proofs regarding the metalanguage, when it should strictly be the other way round.  To give an example of avoiding circularity, consider the definition of the integers. We can define a binary relation $R\subseteq(\mathbf{N}\times\mathbf{N})\times(\mathbf{N}\times\mathbf{N})$, where for any $a,b,c,d\in\mathbf{N}$, $((a,b),(c,d))\in R$ iff $a+d=b+c$, and then defining $\mathbf{Z}:= \{[(a,b)]:a,b\in\mathbf{N}\}$, where $[a,b]=\{x\in \mathbf{N}\times\mathbf{N}: xR(a,b)\}$, as in  question or  on Wikipedia. In this definition if set theory and natural numbers are assumed, then there is no circularity because one did not depend on the notion of "subtraction" in defining the integers.  So my question is:     Question Is the definition of first-order logic circular?   If not, please explain why.   If the definitions are circular, is there an alternative definition which avoids the circularity?   Some thoughts:   Perhaps there is the distinction between what sets are (anything that obeys the axioms) and how sets are expressed (using a formal language). In other words, the notion of a set may not be circular, but to talk of sets using a formal language requires the notion of a set in a metalanguage. In foundational mathematics there also seems to be the idea of first defining something, and then coming back with better machinery to analyse that thing. For instance, one can define the natural numbers using the Peano axioms, then later come back to say that all structures satisfying the axioms are isomorphic. (I don't know any algebra, but that seems right.) Maybe sets, functions, etc., are too basic? Is it possible to avoid these terms when defining a formal language? Ubuntu gets stuck in a login loop My Ubuntu is stuck in a login loop when trying to enter my desktop. When I login, the screen gets black and soon after that the login screen comes back.   I've read that the problem might be caused by an error depending on the graphics, here's my graphics card: ATI Radeon 7670M Make Blender usable on high resolution screen Moment generating function of the natural sufficient statistics of Gamma distribution I have $X_1,X_2 ... X_n$ Gamma-distributed r.v with density:    First of all I showed, that Gamma distribution belongs to exponential family and can be represented in form     I found, that for Gamma distribution $B(\theta) = log(\theta)$ where $\theta$ is $\alpha/\beta$  Then we know, that moment generating function of the random variable $T(X)$ is given by $$M(t)=\exp\{B(\theta+t)-B(\theta)\}$$  Natural sufficient statistics for Gamma distribution is:    But I didn't understand what should I do next? Because Natural sufficient statistics is two dimensional it confuses me:  $M(T) = exp[log(\theta + T) - log(\theta)]$  and $T = [s,t]^\intercal$ The category Set has the property that any morphism f can be decomposed into an epimorphism $e$ and a monomorphism $m$ with $f$ = $m\circ e$. (The intermediary set object is the image of $f$.)  Is there a name for this construction?  Is there a name for this property of categories?  In Set, the intermediary object is unique up to isomorphism. Is this the case for all categories with the property? What's the reasoning behind phrase "dissertation submitted in partial fulfillment of the requirements"? Phrase "dissertation submitted in partial fulfillment of the requirements" seems strange. Can one submit dissertation in full fulfillment of the requirements? I am looking to create realistic car movement using vectors How do I control how Emacs makes backup files? Emacs puts backup files named foo~ everywhere and I don't like having to remember to delete them.  Also, if I edit a file that has a hard link somewhere else in the file system, the hard link points to the backup when I'm done editing, and that's confusing and awful.  How can I either eliminate these backup files, or have them go somewhere other than the same directory? Why does the weighing balance restore when tilted and released "I think him to be about 50" or "I think he is about 50"?