I am stuck on the following question which I came across in a recent exam paper:  Let $f \colon [\pi,2 \pi] \to \Bbb R^2$ be the function $f(t)=(\cos t,\sin t)$.Then which of the following are necessarily correct? The options are:  There exists $t_0 \in [\pi,2\pi]$ such that $f'(t_0)=\frac{f(2 \pi)-f(\pi)}{\pi}$  There exists no $t_0 \in [\pi,2\pi]$ such  that $f'(t_0)=\frac{f(2 \pi)-f(\pi)}{\pi}$  There exists $t_0 \in [\pi,2\pi]$ such that $||f(2 \pi)-f(\pi)|| \le \pi ||f'(t_0)||$  $f'(t)=(-\sin t,\cos t) \,\,\forall t \in [\pi,2\pi]$.    MY ATTEMPT: Clearly option 4 is correct but I am not sure how to prove/disprove for other options. Looking at options 1,2,3 I think mean value theorem has a role to play.  Can someone explain me with some details. My edit:  Option 3 appears to be true since $||f(2 \pi)-f(\pi)||=||(1,0)-(-1,0)||=||(2,0)||=2$ and $\pi ||f'(t_0)||=\pi \,\,\text{since}\, ||f'(t)||=||(- \sin t,\cos t)||=1$ and so $||f(2 \pi)-f(\pi)|| \le \pi ||f'(t_0)||$ is true. Option 1 makes no sense after putting the values of $\pi, 2\pi$ in $f$. So option 2 appears to be correct. Am I going in the right direction? Is there any other elegant way of tackling this problem?
$f:I=[\pi,2\pi]\to \mathbb{R}^2$ be given by $f(t)=(\cos t,\sin t)$  which of the following are necessarily correct?  $1$. $\exists t_0\in I$ such that $f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))$  $2$. $\nexists t_0\in I$ such that $f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))$  $3$.$\exists t_0\in I$ such that $\pi||f'(t_0)||\ge ||(f(2\pi)-f(\pi)||$  $4$. $f'(t)=(-\sin t,\cos t)\forall t$  $4$ is clearly true, $1$ is true by Mean Value Theorem as $f$  satisfies all the conditions of MVT? have no idea of $2$ and $3$, Thank you for help
I'm new to Linux and am trying to set up Python / Django on my machine!  I installed Python 3.3 from the source file and it was compiled into /usr/local/bin. Then I created a symbolic link between /usr/bin/python and /usr/local/bin/python3, so that whenever I invoke python from the command line it uses the latest version.  Now I am trying to install MySQL Python and I got the following output:  apt-get install python-mysqldb  Reading package lists... Done Building dependency tree Reading state information... Done python-mysqldb is already the newest version. The following extra packages will be installed:   apt-listchanges python-apt Suggested packages:   python-glade2 python-gtk2 python-apt-dbg python-vte python-apt-doc The following packages will be upgraded:   apt-listchanges python-apt 2 upgraded, 0 newly installed, 0 to remove and 142 not upgraded. 3 not fully installed or removed. Need to get 0 B/394 kB of archives. After this operation, 250 kB of additional disk space will be used. Do you want to continue [Y/n]? Y Traceback (most recent call last):   File "/usr/bin/apt-listchanges", line 28, in &lt;module&gt;     import apt_pkg ImportError: No module named 'apt_pkg'   Any ideas on how to fix this?  Or any tips on how to clean up this install (if this one is too broken)? How to display a video in a web page using html,php When searching for the desired SE site, the down arrow key (down) should work in order to choose the site one is looking for and hit enter to visit it, instead of using the mouse point to choose it. Would be pretty convenient. Install Headers on Kali apt-get install linux-headers-`uname -r` Reading package lists... Done Building dependency tree Reading state information... Done E: Unable to locate package linux-headers-4.9.0-kali3-amd64 E: Couldn't find any package by glob 'linux-headers-4.9.0-kali3-amd64' E: Couldn't find any package by regex 'linux-headers-4.9.0-kali3-amd64' When is mapping $g$ to $g^{-1}$ a group homomorphism? It just means that the map maps the identity to the identity and inverses to inverses. So does that mean it's only a homomorphism if the inverse is itself? Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.   I see  that,  $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} &lt; \sum_{k=1}^{n} \frac{1}{n} = 1 $.   Thank you. Use of 'prototype' vs. 'this' in JavaScript? What's the proof that the Euler totient function is multiplicative? That is,     why is $\varphi(AB)=\varphi(A)\varphi(B)$, if $A$ and $B$ are two coprime positive integers?   It's not just a technical trouble—I can't see why this should be, intuitively: I bellyfeel that its multiplicativity should be an approximation at most. And why the minimum value for $\varphi (n)$ should be $ \frac{4n}{15} $ completely passes me by. Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$ FaceTime not working on iPhone 6 Plus purchased from Dubai I bought my iPhone 6 Plus in Dubai. I am using the phone in Sri Lanka and I cannot use the FaceTime. How can I use FaceTime while in Sri Lanka? I have looked for similar questions here on stack exchange. The closest example to this that I found is . That question assumed less than perfect absorption of material lining the inside of the sphere, and the effect of thermally re-radiating it inside the sphere.  A lot of the answers focused on that much of the thermal radiation would not be reflected back to the star, and is likely to be reabsorbed by the sphere at another location. On the other hand, this question asks what would happen if the Dyson sphere, using mirrors, deliberately refocus a significant portion of the rays directly back at the star. Why are high white balance temperatures redder when warmer objects are bluer? I know it's dumb to ask this type of questions. But, I still don't get this. I've read the "Black body" story and know that when it's heated, it gives off colors; 'warm' colors (red→orange→yellow..) that have low color temperatures, at low temperatures, (I mean, when it is cooler than afterwards) right? And, the black body gives off 'cold' colors (white→blue..) that have high color temperatures at high temperatures (when it is hotter than before), right? Correct? So there're two, yes, very dumb, questions and the latter one is dumber..     Why are that 'warm' colors said to have low temperatures? And 'cold' colors vice versa? So the redder colors have lower c/t and bluer colors have higher c/t, then   why does the camera's color temperature - adjusting slider has the bluer colors on the left, and the redders on the right?       Please, forgive me for my ignorance... Expectation inequality for a set and its subset. Given that a function $g(x)$ is a monotone increasing function. Its domain is an interval [c,d]. There are two sets $a_j \in A, j= 1,...,J$ and $b_k \in B,k=1,...,K$  on this interval, which satisfy $c_z \in C = B$\A and $c_z \geq a_j$ for all $a_j \in A$. For example, the interval $[c,d]=[0,8]$,$A=\{0,1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5,6,7\}$. In this case, does the following inequality hold? If yes, could you give me some hints to prove that?  $$\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)$$ Why Cooper or Dr. Brand can't send message? In 2018, Kyle Reese (a normal civilian who wasn't part of Resistance) was on the top in the hit list of Skynet. It means that Skynet of that time had information from future about Kyle Reese being father of future leader of human resistance. Right?  Why didn't Skynet kill Kyle Reese instantly after his detection kicking John Connor out of existence? Why Session object destruction failed Why do employers require you to spread your 401(k) contributions throughout the year to get the maximum match? I think the title says it all. I can't see any reason employers have to incentivize employees to contribute to their 401(k) evenly throughout the year. Instead couldn't they use a relatively simple formula that is based on how much of their income the employee has contributed to their 401(k) thus far for the year? If that or their current contribution is over the threshold, they get the match this pay period, even if they didn't contribute anything.  EDIT: To clarify, my employer matches 100% of the first 3% of my contribution and 50% of the next 2%. The match is added every pay period along with my contribution. I am pretty sure there is no "true-up" feature, so if I were to hit the $17.5K limit early, and not be able to contribute at least 5% of my salary, I would not get the full match. So I have to be careful to make my contributions evenly throughout the year to get the maximum match. Just trying to figure out my employer's motivation for requiring this. I stand up and I look at two parallel railroad tracks. I find that converge away from me. Why? Can someone explain me why parallel lines seem to converge, please? How can I stop sending my hostname via DHCP? Ubuntu sends my computer's configured hostname to any DHCP server it tries to use. How do I stop it from sending a hostname? Does RSA work for any message M? I decided to read the original RSA paper A Method for Obtaining Digital Signatures and Public-Key Cryptosystem because of a question I had about RSA (which is not the question I'm about to ask, but may be a question on here at some time) and came across something I had never seen before in any textbook description of RSA (Note: this comes from the beginning of ).  for any integer (message) $M$ which is relatively prime to $n$, $M^{\phi(n)} \equiv 1\pmod{n}$  The part I don't recall ever hearing in class, or coming across in a text book, or seeing in any description of RSA I've ever read is that $M$ and $n$ must be relatively prime. So, I decided to play around with a toy example of RSA to see what would happen. I use $p=13$, $q=31$, $e=7$ and $M=2p=26$ (for completeness: $d=103$, $\phi(n)=360$, $n=403$). I observed the following using my toy instance of RSA: $M^e\equiv M\pmod{n}$, $M^e\equiv 0\pmod{p}$ and $M^e\equiv M\pmod{q}$. Do these equivalences hold for all $M,p,q$ when $M$ and $n$ are not relatively prime? How would I prove this? What is the real-world effect of this?