continuity of $f ' $ if $f$ is differentiable on $(a,b)$ can we say that $f'$ is continuous on $(a,b)$?  I tried some functions and it seems that we can say so but I'm not sure.
Discontinuous derivative. Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.
Yesterday, I have received two : Revival, and Self-Learner. However I was only notified of the Revival badge, because apparently the Self-Learner badge is not considered "".  What badges have their notification suppressed if the user has a 200-rep account somewhere on the Stack Exchange network, i.e. they won't trigger a new item in our achievements dialog? Changing Locale within the app itself Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value $x$) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.  If you choose to swap, your expected value should be the same, as you still have a $50\%$ chance of getting the higher card and $50\%$ of getting the lower card.  However, the other card has a $50\%$ chance of being $0.5x$ and a $50\%$ chance of being $2x$. If we keep the card, our expected value is $x$, while if we swap it, then our expected value is: $$0.5(0.5x)+0.5(2x)=1.25x$$ so it seems like it is better to swap. Can anyone explain this apparent contradiction? Changing capitalization of filenames in Git From everything I've read about quantum mechanics and quantum entanglement phenomena, it's not obvious to me why quantum entanglement is considered to be an active link. That is, it's stated every time that measurement of one particle affects the other.  In my head, there is a less magic explanation: the entangling measurement affects both particles in a way which makes their states identical, though unknown. In this case measuring one particle will reveal information about state of the other, but without a magical instant modification of remote entangled particle.  Obviously, I'm not the only one who had this idea. What are the problems associated with this view, and why is the magic view preferred? I am confused whether drift velocity depends upon area or not . I checked answers for this question but all of them were cancelling each other out . from the equation $$I = neAv$$ It is clear that drift velocity should depend upon area but at the same time don't you think that by increasing the area of cross section of a conductor should have resulted in increase in drift velocity please help me out with it in detailed answer of how should i go with all the possible cases and explanation ,thanks :) How do you change your hostname in Windows 7? Not the computer name, but the hostname. Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$ Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!! TypeError: 'str' does not support the buffer interface When abstract logic circuits get translated to the electrical equivilant, how are conversions from 0 to 1 handled? I've searched around and there's lots of good information out there, but I'm a bit of a n00b when it comes to the precise meanings of words like current, voltage, amplify, resistor, diode etc. So what at first presents itself as a simple explanation quickly turns into a wikipedia tabfest of ever increasing complexity. With that in mind please forgive my probably inaccurate use of these terms in this post.  The basic problem for me is that gates on a logic circuit appear to be creating power from nothing, which is impossible. For example imagine two NOT gates connected in sequence: To my uninitiated mind what seems to be happening is that the first NOT gate cuts the power to the second NOT gate, which responds by magically conjuring new power out of thin air and sending it on it's way. The only explanation I can come up with feels a bit redundant and absurd (That every single gate on the diagram is implied to be directly connected to a power source)  The closest I've come to an answer is something to do with changing the "signal" from "high" to "low" but again, I'm not familiar enough with the language. It still suffers from the same problem, if you've got a low signal how is it that a NOT gate can turn it into a high one without directly drawing on a power source?  To wrap up and present the question in all it's rookie glory: How do logic gates physically convert 0 to 1 without violating the law of conservation of energy? I am stuck in a question for Algebraic Structures:  I need to prove: If given any ring $R$, for any $a$ in $R$ $a\cdot a=0$ then $a+a=0$.  (of course $+$, $*$ and $0$ regarding $R$)  I didn't find any question regarding this question. IANAL et al. is usually the start of answers to [legal] questions on any SE site (e.g. ). You usually end up with personal opinions which sound sensible and the accepted answer is often the one closest to the OP's hopes. But it's probably often not correct. I know the situation can be rather complicated and even differ by location, but it would be really helpful if the SE team were providing official lawyers (indicated maybe by using a § in the username, forbidden for everyone else, just like the mod ♦) to provide some orientation that one could actually rely on beyond much doubt... C++ performance of accessing member variables versus local variables Is it more efficient for a class to access member variables or local variables? For example, suppose you have a (callback) method whose sole responsibility is to receive data, perform calculations on it, then pass it off to other classes. Performance-wise, would it make more sense to have a list of member variables that the method populates as it receives data? Or just declare local variables each time the callback method is called?  Assume this method would be called hundreds of times a second...  In case I'm not being clear, here's some quick examples:  // use local variables class thisClass {     public:         void callback( msg&amp; msg )         {             int varA;             double varB;             std::string varC;             varA = msg.getInt();             varB = msg.getDouble();             varC = msg.getString();              // do a bunch of calculations          }  };  // use member variables class thisClass {     public:         void callback( msg&amp; msg )         {              m_varA = msg.getInt();              m_varB = msg.getDouble();              m_varC = msg.getString();               // do a bunch of calculations         }      private:         int m_varA;         double m_varB;         std::string m_varC;  }; Is it time for Stack Exchange, Inc and the community to legally separate amicably? A sci-fi book or TV show about a tribesman discovering he or she actually lives in a huge spaceship Calculating $\int_{- \infty}^{\infty} \frac{\sin x \,dx}{x+i} $ I'm having trouble calculating the integral  $$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$  using residue calculus. I've previously encountered expressions of the form   $$\int_{- \infty}^\infty f(x) \sin x \,dx $$  where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$  How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work Where are the proprietary NVIDIA drivers in 16.04? The proprietary NVIDIA drivers "disappeared" with my latest update:    The option of "Proprietary drivers for devices" is correctly set under the "Ubuntu Software" tab. I have the graphics-driver PPA (ppa:graphics-drivers/ppa) correctly added to my sources.  Any ideas on how to bring the proprietary drivers back?  PS1: I need them because some of the games I play under wine don't work without them, for whatever reason.  PS2: Also, this feels like it should have a straightforward answer, but I've been looking online for a while without luck. Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$ Good and easy way to share files on local machine I would like to have a directory that has following properties:    Many users can copy files into it  These files can be deleted/changed by these users (user A can delete/modify file that was copied into this directory)    it cant be done using normal file permissions (because permissions are retained on copy).  Here is what I found on the net:        Some use cases:    Sharing music on local machine  Simple git repository sharing (just make a bare repository writeable to many people) --- i know that there are solutions like gitosis Allow many developers to modify test instance of php app without giving them root (i guess they would copy files) --- I'm leading a team of nonprofit junior developers and I need to keep that one simple!    EDIT  AFAIK setting SGID bit is not enugh, it only affects newly created files --- and basic workflow for these use cases ivnolves copying and other operations (which cleave file's gid unchanged) Meta duplicates don't require an answer. Don't suggest there is one when that's not true This request was sparked . The vote to close as a duplicate on meta, in contrast to the main site, does not require for the duplicate to have an answer. Yet, the closure message still says:     This question has been asked before and already has an answer.    ...even when this is not the case. Perhaps we should either only show this particular sentence if there is indeed an answer, or remove this altogether for meta.   Note: I fear this might be a duplicate request, but I was unable to find one. (Not even one without an answer :) )