I have a question and I'm really struggling to grasp what I need to do. I have Rolle's theorem and Mean Value Theorem given to me, and the question follows;  Let $f(x)$ be a continuous function on $[0,1]$ such that $$\int_{0}^{1} f(y) dy = 1$$  Prove that then $f(x)=1$ for some $x\in[0,1]$.  I don't know if it's the way this is worded but I've never been so confused in my entire life! Thanks in advance.
We are asked the following:      let $f(x)$ be a continuous function on $[0, 1]$ such that   $$\int _ {0} ^ {1} \ f(y) dy = 1$$   prove that $f(x) = 1$ for some $x ∈ [0, 1]$   The question asks to take into account Rolle's Theorem and Mean Value Theorem.  I'm aware of the definitions, but I am really unsure how to answer this question and my tutors are not really responding till after easter.    Could someone please help? Why is there a "y" involved?  Thanks you
TikZ rectangular node with different rounded corners I am building a drupal site using Views, search API module and user-points module.  I created the "database server", and indexed many fields of nodes.  I created a search page using views and the search API, that present content (image, title etc.) In addition, I want to add to every content in this search view the name of the author that wrote it, his author picture.  I add the Indexed Content: Author relationship. Unfortunately, I can't find a way to add the author picture and the current total point to my view.  Can somebody advice how to do that? Researching English/Christian families in India during first half of 19th century? Modified files in a git branch are spilling over into another branch I am working on a git repository with a master branch and another topic branch. I have switched to topic branch and modified a file. Now, if I switched to master branch that same file is shown as modified.   For example:  git status in git-build branch:  # On branch git-build # Changes to be committed: #   (use "git reset HEAD &lt;file&gt;..." to unstage) # #       modified:   cvsup_current #   Switch to master branch  [root@redbull builder_scripts (git-build)]# git co master M       builder_scripts/cvsup_current Switched to branch "master"   git status in master branch  [root@redbull builder_scripts (master)]# git status # On branch master # Changes to be committed: #   (use "git reset HEAD &lt;file&gt;..." to unstage) # #       modified:   cvsup_current #   Why is that the file is shown as modified in master branch even though it was modified in git-build branch?   My understanding was that the branches are independent of each other and when I change from one branch to another the changes do not "spill over" from one branch to another. So I am obviously missing some thing here.   Anyone got a clue stick? Leaving Italy without paying a bus fine My wife and I got bus fines today in Bologna. We had only recently arrived and had purchased a ticket from where we were staying which was about 6 or 7 km from the city. The hotel receptionist explained the ticketing system to us and said that if we buy a ticket we could use it on as many journeys for 75 minutes.  We took the bus from our hotel to Bologna and then straight away got on another bus to get to central Bologna where all the sights are. When we got on this other bus we assumed that it was ok as we were still within the 75 minutes. However, when we tried to validate our tickets in the machine it flashed red like an error had occurred so we figured something wasn't right.   We tried to ask the driver about it or if we could buy a ticket but he just grunted and started to drive off. We therefore started to leave the bus at the next stop so we could try and buy another ticket but were ushered back on by a few men who turned out to be ticket inspectors!  They alleged that our tickets were not valid in the city and that we had to pay a 65 euro fine. They wanted to see our official documentation and so took details from our UK driving licences and issued us both with fines. They wanted us to pay there and then, but we explained we didn't have enough cash on us to do so and they then stated we could pay by card.   However, I still wasn't happy about doing that so he then said that we could pay later by going to an Italian post office; or if we don't pay within a certain time the fine will go up to 270 euros each and will be posted to us (he got our addresses off our UK driving licences)!  I am a bit miffed about this as we were noway trying to evade and as soon as we realised our mistake we tried to talk to the bus driver to buy a ticket or get off the bus, but he drove off and ignored us! I have a slight suspicion that it was a rouse as it is very strange that he did this and then out of nowhere bus inspectors got on and made a beeline for us!  If we go back to the UK without paying these fines and ignore the subsequent fines that they send us through the post (if they do send any), then what happens? Do they have any powers when we are back in the UK? a total fine of 500 euros for an honest mistake seems utterly outrageous.   Fine below ASP.NET MVC Relative Paths In my applications, I often have to use relative paths. For example, when I reference JQuery, I usually do so like this:  &lt;script type="text/javascript" src="../Scripts/jquery-1.2.6.js"&gt;&lt;/script&gt;   Now that I'm making the transition to MVC, I need to account for the different paths a page might have, relative to the root. This was of course an issue with URL rewriting in the past, but I managed to work around it by using consistent paths.  I'm aware that the standard solution is to use absolute paths such as:  &lt;script type="text/javascript" src="/Scripts/jquery-1.2.6.js"&gt;&lt;/script&gt;   but this will not work for me as during the development cycle, I have to deploy to a test machine on which the app will run in a virtual directory. Root relative paths don't work when the root changes. Also, for maintenance reasons, I cannot simply change out all the paths for the duration of deploying the test - that would be a nightmare in itself.  So what's the best solution?  Edit:  Since this question is still receiving views and answers, I thought it might be prudent to update it to note that as of Razor V2, support for root-relative urls is baked in, so you can use   &lt;img src="~/Content/MyImage.jpg"&gt;   without any server-side syntax, and the view engine automatically replaces ~/ with whatever the current site root is. How can I separate a separable function I want to block Facebook (the Facebook site) for concentrating purposes, but I don't want to break the share or like buttons on other pages (like a good article or a nice clip on YouTube I want to like or share). How can I do that? I use both Windows and Linux. Why is the sky never green? It can be blue or orange, and green is in between! I, like everybody I suppose, have read the  why the colour of the sky is blue:     ... the two most common types of matter present in the atmosphere are   gaseous nitrogen and oxygen. These particles are most effective in   scattering the higher frequency and shorter wavelength portions of the   visible light spectrum. This scattering process involves the   absorption of a light wave by an atom followed by reemission of a   light wave in a variety of directions. The amount of multidirectional   scattering that occurs is dependent upon the frequency of the light.   ... So as white light.. from the sun passes through our atmosphere,   the high frequencies become scattered by atmospheric particles while   the lower frequencies are most likely to pass through the atmosphere   without a significant alteration in their direction. This scattering   of the higher frequencies of light illuminates the skies with light on   the BIV end of the visible spectrum. Compared to blue light, violet   light is most easily scattered by atmospheric particles. However, our   eyes are more sensitive to light with blue frequencies. Thus, we view   the skies as being blue in color.   and why sunsets are red:     ... the light that is not scattered is able to pass through our   atmosphere and reach our eyes in a rather non-interrupted path. The   lower frequencies of sunlight (ROY) tend to reach our eyes as we sight   directly at the sun during midday. While sunlight consists of the   entire range of frequencies of visible light, not all frequencies are   equally intense. In fact, sunlight tends to be most rich with yellow   light frequencies. For these reasons, the sun appears yellow during   midday due to the direct passage of dominant amounts of yellow   frequencies through our atmosphere and to our eyes.      The appearance of the sun changes with the time of day. While it may   be yellow during midday, it is often found to gradually turn color as   it approaches sunset. This can be explained by light scattering. As   the sun approaches the horizon line, sunlight must traverse a greater   distance through our atmosphere; this is demonstrated in the diagram   below. As the path that sunlight takes through our atmosphere   increases in length, ROYGBIV encounters more and more atmospheric   particles. This results in the scattering of greater and greater   amounts of yellow light. During sunset hours, the light passing   through our atmosphere to our eyes tends to be most concentrated with   red and orange frequencies of light. For this reason, the sunsets have   a reddish-orange hue. The effect of a red sunset becomes more   pronounced if the atmosphere contains more and more particles.   Can you explain why the colour of the sky passes from blue to orange/red skipping altogether the whole range of green frequencies?  I have only heard of the legendary 'green, emerald line/ flash'     that appears in particular circumstances     Green flashes are enhanced by mirage, which increase refraction... is more likely to be seen in stable, clear air,... One might expect to see a blue flash, since blue light is   refracted most of all, and ... is   therefore the very last to disappear below the horizon, but the blue   is preferentially scattered out of the line of sight, and the   remaining light ends up appearing green   but I have never seen it, nor do I know anybody who ever did. Can't run a shell script So I have this script:  #!/bin/bash # Flips the screen (hopefully)  syntax_error=0 orientation=0  current_orientation="$(xrandr -q --verbose | grep 'connected' | egrep -o  '\) (normal|left|inverted|right) \(' | egrep -o '(normal|left|inverted|right)')" case $current_orientation in         normal)                 current_orientation=0         ;;         left)                 current_orientation=1         ;;         inverted)                 current_orientation=2         ;;         right)                 current_orientation=3         ;; esac  if [ $current_orientation -eq 0 ]; then         orientation=2 fi  if [ $current_orientation -eq 2 ]; then         orientation=0 fi method=evdev  # LENOVO S10-3t CHANGE ==&gt; Hard Coded my device number to 11!!!!!!!!  device=11  swap=0 invert_x=0 invert_y=0 real_topx=0 real_topy=0 real_bottomx=4020 real_bottomy=4020  case $orientation in         0)                 swap=0                 invert_x=0                 invert_y=0                 topx=$real_topx                 topy=$real_topy                 bottomx=$real_bottomx                 bottomy=$real_bottomy         ;;         1)                 swap=1                 invert_x=1                 invert_y=0                 topx=$real_topx                 topy=$real_topy                 bottomx=$real_bottomy                 bottomy=$real_bottomx         ;;         2 )                 swap=0                 invert_x=1                 invert_y=1                 topx=$real_topx                 topy=$real_topy                 bottomx=$real_bottomx                 bottomy=$real_bottomy         ;;         3 )                 swap=1                 invert_x=0                 invert_y=1                 topx=$real_topx                 topy=$real_topy                 bottomx=$real_bottomy                 bottomy=$real_bottomx         ;; esac  if [ $method = "evdev" ]; then         xinput set-prop "$device" "Evdev Axes Swap" $swap         xinput set-prop "$device" "Evdev Axes Swap" $swap         xinput set-prop "$device" "Evdev Axis Inversion" $invert_x $invert_y         xinput set-prop "$device" "Evdev Axis Calibration" $topx $bottomx $topy $bottomy         if [ $orientation = 2 ]; then                            xrandr -o inverted         fi         if [ $orientation = 0 ]; then                 xrandr -o normal         fi fi  #   It's for flipping screen in my Lenovo S10-3t. I copied it from the netbook's wiki page, and added the #!/bin/bash at the top. The filename is flipscreen.sh. How can I make it work? My windows directory is huge Messages on Macs don't work I've tried every suggestion I could find on Apple forums. Deleted .plist files, removed message caches, reset passwords, coordinated sign on/offs between my iPhone and my Macs, reset iPhone and reinstalled Mavericks on my Macbook Air. All pointless exercises.   I suspect it all started because I jailbroke my iPhone using Pangu. It may or may not have been the culprit, but my problems started a couple of days after I did it. All my phone conversations went green and stopped synchronizing with Messages apps on my Macs. Trying to send a new message from Macs would result in either "Your message could not be sent" error with a warning icon next to it, or:     XXX is not registered with iMessage.   So I restored my iPhone to factory defaults using iTunes. That didn't help immediately, but in a day or two, my conversations with other iPhone users went blue. So I am happy that works.  So now I sacrificed my Macbook Air to try and make messages work on it. I reinstalled Mavericks several times, each time trying different approaches: not immediately signing into iCloud, not associating an iCloud profile with login profile, etc. Nothing helps. I launch Messages app, it asks me to sign in (and I do, successfully), and then nothing goes through.   Sending messages to myself also works inconsistently. Yesterday, whatever I sent to myself (using phone number) would appear delivered to the same window, but not on my phone. Today I get "... not registered with iMessage" error. Simulate bad cell phone reception in the lab If $|f(z)|\leq |f(z^2)|$ on the unit disk, show that $f$ is a constant function. Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|&lt;1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.   My Try:  I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint? In thinking how to ask this question (somewhat) succinctly, I keep coming back to a Microwave Oven.  A Microwave Oven has a grid of holes over the window specifically designed to be smaller in diameter than the wavelength of the microwaves it produces, yet larger than the wavelengths of the visible light spectrum - this is so you can watch your food being heated without getting an eye full of microwaves.  The "realness" of electromagnetic waves seems indisputable - both from the microwave example above, and also because if I want to broadcast a radio wave with a certain wavelength, then I need to make sure I have an antenna of corresponding length to produce the wave I'm looking for.  Furthermore, we discuss and treat these waves as real, measurable "objects" that exist and can be manipulated.     Now, if I want to describe the behavior of my Microwave Oven in the framework of QM (let's pretend my oven is going to only produce 1 photon of energy corresponding to a wavelength of the microwave spectrum for simplicity) then I'll describe the behavior of that photon as a wavefunction that evolves over time and gives a probability distribution within my microwave that similarly does not allow the photon to pass through the safety grid and exit the oven cavity giving me a retinal burn.  The difference is, the wavefunction is never treated as something "real" in this description.  When the safety grid is described as working to protect you because it has holes smaller than the wavelength of the classical waves it's blocking, this is a useful description that seems to describe "real" objects/be-ables.  While it is possible to describe why an individual photon has low probability of passing through the same grid but extended physical properties such as wavelength (in space) are treated as non-real because we're dealing with a point particle and with behavior described by something we also treat as non-real (the wavefunction); it seems unclear to me why we insist this wave function which predicts behavior of physical measurements so well is somehow "non-real."    Put another way, if we have no problem treating EM waves as "Real," then why do we insist on treating the wavefunction that describes the same behavior as "unreal?"  I understand there is recent research (Eric Cavalcanti and his group for one) trying to argue this point, but as every respectable physics professor, I've ever encountered, has treated the wavefunction as an indisputably non-real mathematical tool, I needed to ask this community for an answer. AMT stock option scenario Try to do something vs. Try doing something Is there any differences between      Try to do   And     Try doing   Thank you in advence. How put color in all math mode? How to use x coordinate of a point and y coordinate of other point? Possible Duplicate:             How to use x coordinate of a point and y coordinate of other point? I need similar to  \coordinate (A) at (2,1); \coordinate (B) at (3,0); \coordinate (C) at (4,3); \coordinate (D) at (5,-1);  \draw (xA,yB) -- (xC,yD); Most textbooks introduce differentiable manifolds via . This has the advantage of being concrete, but the disadvantage that the local coordinates are usually completely irrelevant- the choice of atlas and chart is arbitrary, and rarely if ever seems to play any role in differential geometry/topology.  There is a much better definition of differentiable manifolds, which I don't know a good textbook reference for, via . This definition does not involve any strange arbitrary choices, and is coordinate free. Paragraph 3 in Wikipedia (which is the actual definition) states:   A differentiable manifold (of class $C_k$) consists of a pair $(M, \mathcal{O}_M)$ where $M$ is a topological space, and $\mathcal{O}_M$ is a sheaf of local $R$-algebras defined on $M$, such that the locally ringed space $(M,\mathcal{O}_M)$ is locally isomorphic to $(\mathbb{R}^n, \mathcal{O})$.   This confuses me, because I don't see why such a sheaf should be acyclic, or where conditions like "paracompact" or "complete metric space" or "second countable Hausdorff" are implicit. So either:   The wikipedia entry has a mistake (I would want to sanity-check this before editing the entry, because this is such a fundamental definition which thousands must have read). Somewhere in that definition, the condition that $M$ be paracompact is implicit.   Question: Should the definition above indeed require that $M$ be second-countable Hausdorff or paracompact or whatever? Or is it somehow implicit somewhere, and if so, where?    Also, is this definition given carefully in any textbook?  Update: I have editted the Wikipedia article to require that $M$ be second-countable Hausdorff. But I'm still wondering if there is a textbook covering this stuff, and whether requiring the sheaf to be acyclic might have worked instead as an alternative.