I have this sum: $1+3+5+...+(2n+1)$, where $n$ is a natural number. I have to calculate it and interpret geometrically. Well, it's easy to find out that it equals $(n+1)^2$. But how to interpret it geometrically? I don't think it's about a graph(parabola). could you please give me at least an idea?
Prove that for all integers $n$, $n \geq 1$,   $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$   How would I go about proving this?
HP PSC 1410 all scanner's images are black I have a HP PSC 1410 printer with scanner inside.  I've been scanning documents and images with it for the last couple of years.  Recently my scanner has stopped performing as excepted and i can't scan or copy images\documents.  My problem is that every images\document i'm trying to scan - the result from the printer is a black page.  I've been googling about this issue and couldn't find any helpful stuff beside replacing the whole printer - which is something i'm trying to avoid.  Any ideas ? I've read something about the shipping lock of the scanner but couldn't find any information about how to check it.. What does "more buyers than sellers" in technical analysis of stocks actually mean? I often hear/read technical analysts looking at volume and price action on a chart and claiming that there are "more buyers than sellers" or vice versa.  Since every trade requires a buyer and a seller, how can volume/price signify more of one than the other?  What does that actually mean?  When I hear that there are "more buyers," I get the impression they actually mean more "professional" or "institutional" buyers, but I could just be imagining that meaning.  But wouldn't this "more buyers" situation actually represent a smaller number of large buyers placing multiple orders? (and those orders would be matched by a larger number of smaller investors on the other side of the trade?)  Maybe I have my assumptions backwards and "more buyers" is what they say when they actually mean "more retail buyers and less institutional buyers"?  Or maybe the expression has nothing to do with number of buyers/sellers at all and is just some arbitrary technical analysis situation?  What situation is it that they're trying to describe?  If the "smart money" traders had placed larger orders, then there would have to be equally large sellers on the other side of that large trade, right?  I know trades often get broken up, but if the trades had been broken up it would be equivalent to "large traders placing multiple orders" scenario described above.  If the trades were not broken up, then there would not be an imbalance in the "number" of "smart" vs. retail buyers/sellers, right?  UPDATE/EDIT: I usually see this stated in reference to technical indicators such as RSI.  I don't typically hear people saying "there are more buyers than sellers" in the situations you guys are describing with more unfilled orders on one side of the (potential) trade (though that would make more sense if that was what they were talking about).  See for example: After converting to nodes in order to render using Cycles I've no idea where to enable the z-transparency. I guess the image textures alpha channel needs to be pluged somewhere but I've no idea which node would be suitable.   The question is how could I correctly setup the nodes to have the same effect as when using z-transparency in BI?      The materials original settings for BI are:    Which renders like:    I had no luck with DarenWs suggestion: Some irreducible character separates elements in different conjugacy classes Let $x$ and $y$ be elements that are not conjugate in $G$.  Then there is some irreducible character $\chi$ such that $\chi(x) \not = \chi(y)$.   Clearly the "irreducible" part isn't important, since any character can be written as the sum of irreducible characters, but I'm having trouble going beyond that.  I'd appreciate a good hint over a full answer, and I'd be most interested in a way to construct a group representation $\varphi:G \to GL(V)$ of $G$ such that the character of the representation takes different values on $x$ and $y$. Using Illustrator CS6 to create shape out of multiple paths I am very new to Illustrator, in fact I am only experimenting with it in an attempt to help my friend develop a logo for a new business venture. I managed to draw out the shell of the logo using the line tool and modifying the lines with the anchor tool to get the shape I wanted.  However, I would like to somehow take all of these paths and create one object, or symbol which I can then apply a drop shadow, fill and various other effects too. Currently when I try to do this I get unwanted effects...  I'd appreciate any help on the subject. Given the differential equation $$\dot x = -2x + u$$ determine the optimal control law $u = - kx$ that minimizes the performance index $$J = \int_0^{\infty} x^2 \, \mathrm d t$$  My approach was to find the state feedback $k$. But since the value of $R$ (positive semidefinite Hermitian) is not given, that means $R=0$. How do I determine the optimal control for this system where $R=0$? Why does the sum of the reciprocals of factorials converge to $e$? I've been asked by some schoolmates why we have $$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$ I couldn't say much besides that the $\Gamma$ function, analytic continuation of the factorial, is defined with an integral involving $e$. Then I also know that actually $$ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$ Is there a reason for these facts?  P.S. I added the tag "intuition", please remove it if you think it is not pertinent. pyvenv vs venv vs python-virtualenv vs virtualenv and python 3 malikarumi@Tetouan2:~$ pip install virtualenv     Collecting virtualenv     Downloading virtualenv-12.0.7-py2.py3-none-any.whl (1.8MB)     100% |################################| 1.8MB 330kB/s   malikarumi@Tetouan2:~$ pip freeze     (a lot of stuff, but not virtualenv)  malikarumi@Tetouan2:~$ virtualenv testvenv1     The program 'virtualenv' is currently not installed. You can install it by typing:     sudo apt-get install python-virtualenv   What is going on here? is python-virtualenv == pyvenv? Isn't that still broken? Does original virtualenv still work with python? if venv (too many name variations!!!) is part of the standard library, , why am I being told to install it?  And when I did try to install it, I got:  malikarumi@Tetouan2:~$ sudo apt-get install python-virtualenv     Reading package lists... Done     Building dependency tree            Reading state information... Done     The following extra packages will be installed:     python-colorama python-distlib python-html5lib python-pip python-setuptools   At which point I aborted, because installing python3-pip just before that had given me   Setting up python3-setuptools (3.3-1ubuntu1) ... Setting up python3-pip (1.5.4-1) ...   And I wasn't sure if the extra packages would overwrite them or go onto python 2.7.  I read that python 3 is supposed to be the default going forward. Since both 2.7 and 3.4 are there, and have separate commands, how do I not only make sure Python 3 is my default, but that anything I install goes there and is used by it instead of 2.7?   BTW, I tried sudo apt-get python3-virtualenv and got: E: Unable to locate package python3-virtualenv To evaluate limit of sequence $\left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$ How do I evaluate the limit of the following sequence $$a_n  = \left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$$   I tried to take log and then using cauchy $1^{st}$ theorem of limits but I coudn't do it.  Could someone kindly help? GZip of drive image created by using `dd` command I created yesterday a drive image using GZip.  I opened the 51.3 GiB file named ssd.img.gz to see one icon thats a package of 3.9 GiB, yes, thats THree point Nine Gigabytes in size and has the name of the image as ssd.img  That does not seem right to me, has the image creation worked or not?  Yes the drives have data and an operating system on it.  Should I have used cat instead? only I did and it kept on saying that it had nothing to do.  Would using the | zip instead be a better option?    ** EDIT **  The command used was:  mark@zotac:~$ sudo -l Enter password:&lt;entered password to get in to root&gt; root@zotac:~# dd if=/dev/sda | gzip -c &gt; "/media/mark/Seagate External Drive/ssd.img.gz   the terminal spent the best part of 10 Hours at building the image, something in itself that I also found suspect, why so log when copying the files to a folder that I wanted, some 40GiB took 45 minutes...  The source drive /dev/sda had no mounted partitions.     ** EDIT **  Just to clarify, I am trying to make an image of the drive to back up later if things go wrong with a repair, the drive itself needs to be a file on a drive as the target drive partition is using almost all of the drive, the partitions can't be resized on it without formatting it and I am not about to lose 1.4TB of data.   So the dd if=/dev/sda of=/&lt;target&gt; isn't what I need, what I do need is an image of the drive that I can restore the image from to the drive should the repair go wrong.  So what command do I use, dd or cat? Why does | zip keep telling me it has nothing to do when used with cat but no error or warning with dd $$ A $$  Light is travelling away from the absolute center of the Schwarzchild black hole (i.e. none of that spinning/stretching field) at $\ C $, and so shouldn't $\ \Delta C$ need to be greater than $\ C $ in order for the light to collapse into the black hole, therefore violating the relativistic limit of $\ C $?    It seems like light would just 'hover' right in the center of a black hole then. Because if it goes away from it, the event horizon pulls it back, and if it goes into it, the velocity change must be greater than $\ C $ which violates general relativity; so technically there's no way for the light to move away, and so it would just be still.  The thing is that the light must move at the speed of light or general relativity doesn't work. It's massless... But in the center of a black hole, if the above is true (it can't move or escape), then it is not travelling at the speed of light. So it cannot be still.  It cannot be spinning in a constant orbit either, because that would require velocity in an outward angle. And it cannot move in an outwardly fashion, only in an inwardly one.  Is it spinning in an infinitely decaying spiral towards the center?  If so what happens to it if and when it reaches the direct and complete center, i.e. The case of it being 'shot' from the center.    $$ B $$  If you shot light at an angle towards a black hole, but not directly to the center of it  then that light from the outside will eventually become the same as the light that I said was shot from the center to the outside; because in both cases, they end up in an infinitely decaying spiral.    If so it would end up exactly like the light in the above paragraph, so then we need to figure out what happens to that light to figure this one out, or vice versa.    $$ C $$  If you shot light directly towards the center of a black hole it would have no spiral to decay into, so therefore it must stop at some point in the center of a black hole. If it stops though, it loses it's momentum and violates general relativity again because it's energy "disappears".    If so then it violates conservation laws. So I'm assuming this simply is not true.    Is the light travelling into a black hole somehow still travelling at the speed of light, in every direction possible where that direction keeps it facing towards the center?  If so, why is there not a point at which the only thing left for the light to do is stay still, since all other paths would increase the distance between it and the black hole's center, and therefore become Case A. And if there isn't, what would happen if there was?  Also, if it's a time and observation answer where it supposedly takes and infinite amount of time for C to hit the center, wouldn't it be the case that the black hole is 'eating' all of the available light in the universe and keeping it in stasis?  Also, what happens in a Kerr black hole then? Does it just rip the field of time/gravity at an angle such that the light would never get closer than the center? When rolling percentages, do 1d100 and two d10s (percentiles) share the same probabilities? Cannot select vertices from different object while in edit mode Whenever I enter edit mode, with 2 separate meshes (cubes), I can only select vertices/edges/faces of 1 mesh.  Example (the screen shot did not include the mouse cursor):    Also, when I enter edit mode with both meshes selected, one of the meshes has a red border and cannot be selected (it also has this border in Object mode): Other than portability, are there any other advantages (or disadvantages) to storing apps on external storage? I'm a citizen of an EU member state traveling to a small city in Vermont. Looking at flight data, it seems that it would be quite a bit quicker to fly into Montreal (Canada) and then drive a rental car to my final destination in Vermont (and back), rather than getting a connecting flight into a small local US airport. I am slightly worried about the trouble I might get myself into by crossing a land border twice, though.  Can anyone give me an idea of what I'd be getting into in terms of:   Crossing time Discomfort Additional paperwork required   For both the outbound (Canada/US) and the return leg (US/Canada) of my journey, when compared to crossing the border at an international airport?  Also, do I need to take any special actions when I'm arranging a rental car if I intend to drive it over to the US and back? In a photoelectric effect we remove electrons from a metal using high energy photons, the work function is the minimal energy required for this effect.  My question is why doesnt the work fucntion depended on time? The more electrons we take out from the metal, the more ionized it becomes (if we remove a lot of negative charge, it will turn positive), and as a result we would expect to need more energy for this effect (the possitive charge of the metal would make it harder to remove the remaining electrons).  Why does the work function remains constant no matter how much electrons we take out? Recently I found a claim saying that  $$  \int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2  $$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.   Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$ \int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x $$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the , but there are some subtle differences. Find the sum of angles without trigonometry? How to enable exFAT in Ubuntu 14.04 Figuring out algorithm/formula for item scaling in WoW