I am confused on how LSTM learn from word embedding. I know that LSTM accepts 3D input (sample, timesteps, features). So, when we use embedding layer (word2vec) and we have 300-d vector representation, how does LSTM learn. I do understand that for a sentence, the words are timesteps. However, when we deal with vectors should the timesteps be same as vectors?  If it is true, then how can LSTM learn the sequence of words in that case to predict anything about sentence.
I recently read a lot about neural networks using GRU or LSTM units. There are many easy to use frameworks like tensorflow that do not even require high knowledge about programming. Unfortunately, I never really found good information on how the training of those networks work. Simple backpropagation might probably not work for gated recurrent networks or is just too inefficient for networks with such a high number of variables to learn.  So my question is:  What are the state-of-the-art algorithms used for the initialization and training of neural networks with GRU or LSTM units? I am not looking for frameworks to use, but for initialization and update equations for the internal parameters.
Is there some way to embed LaTeX source code in a PDF file? the numer of partial function between two sets in combinatoric way $A$ and $B$ two sets. $|A|=k, |B|=n.$ I need to prove in a combinatoric way that the number of partial function from $A$ to $B$ is $(n+1)^k.$ I thing the number is $Σ(n^i),$ for $i=0$ to $k.$ but I have no idea how to show equality, and how to prove it in a combinatoric way.  If someone can show me the way or guide me that will be great!  thanks for helpers. In terms of UX what do you guys think is the best approach when integrating a trial period for subscriptions?  My application offers a 15 day trial, at the minute the user can sign up without providing any payment information (Paypal Subscription). The account is then set to trial status and after 15 days the user will be redirected to an upgrade page.  Do you think this is the best approach? I am unsure whether or not to let Paypal manage the trial, this however means that the user will need to go through the Paypal payment system during signup. I think this may put some people off signing up.  On the other hand this approach may increase the number of conversions to a full account. Let $f : \mathbb R^n \rightarrow \mathbb R$ be a measurable function.  We know that $f$ is not necessarily locally integrable - but how hard can a measurable $f$ fail to be so? For example, is it true that $f$ is locally integrable outside of a negligible set $N \subset \mathbb R^n$?   Edit: In consideration of the comment by zhs, let me correct the phrasing of the question. I was originally wondering whether the 'singular' behaviour of a measurable function can be localized to a 'small' set.   A motivational example is $f(x) = 1/x$ over the real line. It is locally integrable in the complement of the origin. Hence, a conceivable phenomenon, which I had in mind originally when asking the question, could be that a measurable $f$ is integrable over open sets that are compactly contained in the complement of some negligible sets. I have unistalled and reinstalled the Ubuntu Software Center as per info I found in a similar thread and I got the same response about line 91 or something like that.  I just tried to upload a screen shot but since I'm new it won't allow me to.  I also can not figure out how to cut and paste anything so I have to hand type what the error screen says, both when I attempt to open the software center and nothing happens, when I try to enter commands into the terminal to uninstall, reinstall, whatever I get the same following:  COULD NOT INTITIALIZE THE PACKAGE INFORMATION An unresolvable problem occured while initializing the package information Please report t:his bug against the 'update-manager' package and include the following error message: 'E: Malformed line 91 in source list/etc/apt/sources.list (dist parse)  E: The list of sources could not be read.,  E: The package list of status file could not be parsed or opened.   How do I report bugs?  What can be done about this.  I have searched and everything everyone says to do leads me back to the same line error message.  So, I don't know how to get to line 91 in the source list; to tell you what it says. Sorry, I'm really new to this. That is what I need is to find out how to get there and fix what it says. I would really like to NOT have to re partition my hard drive and start from scratch, so I'm really looking forward to getting this problem solved. I need to be able to install new software. I have a third-party library which consists mainly of a large number of static (.a) library files.  I can compile this into a single .a library file, but I really need it to be a single .so shared library file.  Is there any way to convert a static .a file into a shared .so file?  Or more generally is there a good way to combine a huge number of static .a files with a few .o object files into a single .so file? Do Stack Overflow and other Stack Exchange sites have an app I can use? How to Convert Persian Digits in variable to English Digits Using Culture? Combinatorics, choosing 3 elements from 100 In how many ways can we pick three different numbers out of the group $1,2,3,\dots,100$ such that the largest number is larger than the product of the two smaller ones? (The order in which we pick the numbers does not matter.) Do not hide VLQ flag option from users who already flagged and may wish to retract System hides an option to flag Very Low Quality in few cases, namely when post is:        Because of that users who already flagged it when none of above conditions applied miss an option to .  Suggest1 to show this flag option to such users so that they could retract a flag.    1 Frankly I won't mind if this gets  because I doubt that it is important. This is posted mainly to document how system behaves and to provide reference for those who may wonder why they can't retract flag in some cases. Why am I getting black photos when I use the Manual feature on my DSLR? I was taking photos a few days ago, and I had it set to Manual. After about four photos, the rest started coming out black. I could see through my eyepiece clearly, and it took the photo normally, but when I reviewed it they were all black. Why is this true: The only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the identity Can anyone explain me please how to see this statement: the only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the Identity.  how can I prove formally that? or how can I see that only the Identity satisfy it? I added a user to a group, but group permissions on files still have no effect Anyone can optimized this code for better explanation and easy to understand? Is there an implementable (non-theoretical) way to prove if email address belong to real email account? How does the T-1000 see? In Terminator 2: Judgement Day, how does the T-1000 sense things? It's made of mimetic poly alloy; however, does it have a retinal scan camera like the T-X or T-800, or is just molecules that allow it to see what it does? In other words: let's say it is made out of nanobots. Would a group of these nanobots provide the capability of sight? Kind of like humans with cells that  make up the brain? I have always wondered whether James Cameron explained this on a special edition or if the topic was left up for insight from Terminator fans. From what I get from the whole mimetic poly alloy and some of the extra scenes, it seems to feel its way out to get information. Of course, it's very much not blind! Thoughts? Scale and Rotate Stopped Working [2.8] I've used Ubuntu before, but I'm still very new. I just installed 16.10 on a new computer yesterday, and the Wi-Fi has been considerably slower than my Windows laptop. I put the computer together mainly to play Dota 2, and I experience decent ping but constant 5%-10% packet loss while playing. Occasionally, after the computer has been left on for a while, the Wi-Fi connection will drop completely- I can still see networks in the drop down menu, including my home network, but my computer will not reconnect without a reboot. I've looked all over for a solution, but my inexperience has prevented me from getting anything to work. My Wi-Fi card is a Ralink RT2561/RT61, here's some info  The output of sudo lshw -C network   *-network                          description: Wireless interface         product: RT2561/RT61 802.11g PCI         vendor: Ralink corp.         physical id: 6         bus info: pci@0000:01:06.0         logical name: wlp1s6         version: 00         serial: 00:0e:2e:de:df:c0         width: 32 bits         clock: 33MHz         capabilities: pm bus_master cap_list ethernet physical wireless         configuration: broadcast=yes driver=rt61pci driverversion=4.8.0-22-generic firmware=0.8 ip=192.168.1.107  latency=64 link=yes multicast=yes wireless=IEEE 802.11         resources: irq:17 memory:dfef8000-dfefffff   Here is the output of sudo iwconfig:   *-network                          description: Wireless interface         product: RT2561/RT61 802.11g PCI         vendor: Ralink corp.         physical id: 6         bus info: pci@0000:01:06.0         logical name: wlp1s6         version: 00         serial: 00:0e:2e:de:df:c0         width: 32 bits         clock: 33MHz         capabilities: pm bus_master cap_list ethernet physical wireless         configuration: broadcast=yes driver=rt61pci driverversion=4.8.0-22-generic firmware=0.8 ip=192.168.1.107  latency=64 link=yes multicast=yes wireless=IEEE 802.11         resources: irq:17 memory:dfef8000-dfefffff   I'm at a loss, I'm very bad at this. If any additional info is needed I can provide it Why can't an enum value be fully qualified in a switch statement? I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$.  One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th edi, p442) is as following: $$ \langle x|[x,p]|y \rangle = \langle x|xp-px|y \rangle = (x-y) \langle x|p|y \rangle. $$ On the other hand,  $ \langle x|[x,p]|y \rangle = i \langle x|y \rangle = i \delta (x-y)$. Thus $$ (x-y) \langle x|p|y \rangle = i \delta(x-y). \tag{1} $$  We use $(x-y) \delta(x-y) = 0$. Take the derivative with respect to $x$; we have $\delta(x-y) + (x-y) \delta'(x-y) = 0$. Thus $$ (x-y) \delta'(x-y) = - \delta(x-y). \tag{2} $$  Comparing Eqs. (1) and (2), we identify $$ \langle x|p|y \rangle = -i \delta'(x-y). \tag{3} $$  In addition, we can add $\alpha \delta(x-y)$ on the right-hand side of Eq. (3), i.e. $$ \langle x|p|y \rangle = -i \delta'(x-y)  + \alpha \delta(x-y), $$ and $[x,p] = i$ is still satisfied. We can also add $$ \frac{\beta}{\sqrt{|x-y|}}\delta(x-y)$$ on the RHS of Eq. (3).  Here $\alpha$ and $\beta$ are two real numbers.  My question is, what is the most general expression of $\langle x|p|y \rangle$? Can we always absorb the additional term into a phase factor like Dirac's quantum mechanics book did?  Thank you very much in advance.