Java - Error Message: 'Cannot Convert from void to String
What does "Incompatible types: void cannot be converted to ..." mean?
How can I require an optional Perl module if installed? I have Perl code which relies on Term::ReadKey to get the terminal width. My installation is missing this module, so I want to provide a default if the module isn't present rather than throw an exception. How can I conditionally use an optional module, without knowing ahead of time whether it is available. # but only if the module is installed and exists use Term::ReadKey; ...  How can I accomplish this? What makes a Feynman diagram real or virtual? Simple question: as the title says, what makes a real Feynman diagram real, and what makes a virtual diagram virtual? Or in other words, how do I tell whether any given diagram is real or virtual? I've never gotten a really satisfying explanation of this. I would imagine it has something to do with virtual particles, but all internal propagators are virtual particles and I know for a fact that having internal lines doesn't make a diagram virtual. Why does an conditionally convergent infinite series, when deranged, have a different limit? $\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit? First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g. \begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*} one gets the sum of first n terms as different functions, but the series's limit remains. One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason. Are there other explanations? Is the use here of 'countably many' is proper? I installed Ubuntu alongside Windows but I did not disable secure boot as someone told me it could cause problems with boot loader. So when installation finished and I rebooted, the system went straight into Windows. I went to Ubuntu IRC Support and they told me to turn off the secure boot and reinstall Ubuntu. I did that but my system still boot into windows.  This is a picture of partitions on my system:    This is my first Ubuntu install on my new laptop. I followed everdaylinux guide to install it and I used something else option. Number of possible combinations of x numbers that sum to y I want to find out the number of possible combinations of $x$ numbers that sum to $y$. For example, I want to calculate all combination of 5 numbers, which their sum equals to 10.  An asymptotic approixmation is also useful. This question seems to be very close to number partitioning, with the difference that a number can be 0. See:    All possible partitions for sum 10 and 3 positions that can be zero, are 63 possiblities: (numbers shown as 3 digits)  019 028 037 046 055 064 073 082 091 109 118 127 136 145 154 163 172 181 190 208 217 226 235 244 253 262 271 280 307 316 325 334 343 352 361 370 406 415 424 433 442 451 460 505 514 523 532 541 550 604 613 622 631 640 703 712 721 730 802 811 820 901 910 EM algorithm gaussian mixtures- derivation I'm trying to make sense of a derivation I'm following from the lecture notes of Stanford's ML course.  Specifically the notes are here:   I'm particular I'm interested in the step where they are optimizing the lower bound of the log-likelihood by finding the derivative of that function w.r.t the mean (top of page 7). I'm a bit stuck on going from line 1 to line 2 in that sequence.  To copy here (using abbreviated notation for space):  $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\log\frac{\alpha_j \exp\psi_{j}^{(i)}\phi_j}{w_j^{(i)}}  \qquad(1) $$ $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\psi_{j}^{(i)}\qquad(2) $$ where $\psi_{j}^{(i)} =-\frac{1}{2}(x^{(i)}-\mu_j)^T\Sigma^{-1}_{j}(x^{(i)}-\mu_j)$ and $\alpha_j=\frac{1}{(2\pi)^{n/2}|\Sigma_j|^{1/2}}$.  The quantity $\mu_s$ is the mean (of a multivariate gaussian distribution) for class $s$.    In short, I'm not sure how they get from (1) to (2).  From just changing the log of (1) I see:  $$ \nabla \mu_s \sum_{i=1}^{m}\sum_{j=1}^{k} w_j^{(i)}\log \left(\alpha_j \exp\psi_{j}^{(i)}\phi_j\right) - w_j^{(i)}\log w_j^{(i)} \qquad(1) $$  To get to their form (2) my impression is that they take pre-emptively discard terms that would contain derivatives of $w_j^{(i)}$ (w.r.t. $\mu_s$).  At least if you assume that much, you get to (2).  However, as they show earlier in that PDF,  $$ w_j^{(i)} = p(z^{(i)}=j | x^{(i)}) =\frac{p(x^{(i)}|z^{(i)}=j)p(z^{(i)}=j) }{\sum_j p(x^{(i)},z^{(i)}=j) } $$ i.e. this quantity is the posterior distribution of the class labels.  Intuitively, it seems like $\partial w_j^{(i)}/ \partial \mu_s$ should be non-zero.  After all, changing the location of the mean $\mu_s$ would change your posterior probability for a particular class given the same data.  (Right?).  Just from a mechanical point of view, $\partial w_j^{(i)}/\partial \mu_j$ would seemingly have some non-zero terms since $p(x^{(i)}|z^{(i)}=j) \sim \mathcal{N}(\mu_j, \Sigma_j)$.  If I work through the algebra, I get  $$ \frac{\partial w_j^{(i)}}{\partial \mu_j} = w_j^{(i)}(1-w_j^{(i)})\frac{\partial \psi^{(i)}_{j}}{\partial \mu_j} $$  What am I missing here?  If I take the log-likelihood (without even thinking about EM algorithm),    $$ \ell = \sum_{i=1}^{m}\log \sum_{j=1}^{k}p(x^{(i)}|z^{(i)}=j)p(z^{(i)}=j) $$ and differentiate w.r.t $\mu_s$, I correctly get  $$ \mu_s^* = \frac{\sum_{i=1}^{m}w_s^{(i)}x^{(i)}}{\sum_{i=1}^{m}w_s^{(i)}} $$  so I'm comfortable with the eventual result, just not the intermediate steps. tense and time and composite sentence If the eating meal starts at 7:00 AM and finishes at 8:00 AM. the drinking liquor starts at 7:00 AM and finishes at 7:50 AM.  can I say that :   I drank the liquor when I ate the meal. I ate the meal when I drank the liquor.   Is I need to use progressive tense ? I am currently working my way through "A Concise Introduction to Pure Mathematics" by Martin Liebeck, and have been stuck on an excersise for a couple of days now. The question reads:  "Show that for an integer n $\geq2$, the period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$."  I've been trying to solve this without making any substantial progress. And yes, I have seen explanations that when performing a long division the decimal expression either terminates to $0$ or eventually repeats. But I haven't seen any proof of this that I've been satisfied with, even less do I know how to formalise this type of proof. I've tried to write out long division with arbitrary digits, but that obviously turns out to be too much of a hassle and basically impossible to draw any conclusions from. On my work computer I have two USB hard drives that I use rarely. They have a power save mode that sends them into sleep after a couple of minutes of them being idle.   Whenever I open a context menu on a file, the drives are woken up (most likely caused by the "send to" handler).  So I eject the drive, but I can't find a way to get it back, other than unplugging and replugging it in.  Is there a way to unmount the drives, and then remount them only when I actually need them? (On Windows 7 Ultimate.) Quote environment with reference at the end right I was confused initially about what the pins mean next to a topic in the new help center.  "those articles are "featured" articles; they are meant to draw attention to the most important site policies or most frequently encountered issues."   If the pin is used to denote an important topic, it should at least be noted as such. Also a pin generally means to me that this is something that I personally selected.  Could their meaning be clarified somehow? How do I speed up my design work? I've been a designer for 6 months, having previously had a 1st class honours degree in Graphic Communication. My issue is that I am a Very good designer, but I work too slow. It's beginning to drive a rift between myself and the creative director who desperately wants me to work quicker. I've always been a workaholic even in university, I put in 110% all the time, really work my ass off, but I'm still not fast enough to complete projects quickly, my work is good, but it's not efficient.  Do you guys have any tips to help me speed up? Why can "trans" be replaced with an x? I can't think of an example, so I may be wrong about this, but I think I've seen people replace the prefix "trans" as in transport with an x. "Cross" makes sense, as in "railroad crossing", and I guess "chris" is sort of a punny extension, the extension being like "chris cross" and the example being "xmas", but why trans? $A^A$ in category of graphs (reference is Lawvere/Schanuel, Session 31, Ex. 1)  I'm trying to calculate the exponential object $A^A$ and its evalution map $e \colon A \times A^A \to A$ in the category of graphs, where $A$ is the "arrow graph" (ie. one arrow and two dots).  In the following, $D$ is the graph with one dot and no arrows, $1$ is the terminal object in this category (graph with one dot and one arrow, the loop).  So far I have:   The points of $\mathbf{1}\to A^A$ correspond to the maps $A\to A$ (via two standard isomorphisms), and since $\mathbf{1}$ is the loop, and there is one map of graphs $A \to A$, there is one loop in $A^A$. The dots $D\to A^A$ correspond to the maps $A \times D \to A$, of which there are four, hence four dots in $A^A$. The arrows $A \to A^A$ correspond to the maps $A \times A \to A$, of which there are four, hence four arrows in $A^A$.   But I'm stuck on how to put these together to constitute $A^A$ and its evaluation map. Can you explain the "Axiom of choice" in simple terms? The locus of the intersection point of two perpendicular tangents to a given ellipse For a given ellipse, find the locus of all points P for which the two tangents are perpendicular.   I have a trigonometric proof that the locus is a circle, but I'd like a pure (synthetic) geometry proof. Environment variables when run with 'sudo' How many actuations are "too many actuations"? When looking at used camera bodies, a question that comes up a lot, is "how many shutter actuations, or clicks, does it have?"   What I am wondering is, how many is too many. In other words, is there a reference for knowing when a sensor is more likely to go? How should I judge a camera body with 20000 vs 5000? Is that a significant difference, or is that not even relevant.  Obviously this might something that is dependent on model, so if there is some information based on model that would be even more interesting.  One of the guys at Canon Tech Support said that this is completely irrelevant for the Rebel series cameras, but he is obviously slightly biased. :) Do I need to change my ESTA address when visiting the US? Apparently, you can speed up your computer by adding more RAM. Is the amount of ram to be added is directly or indirectly related to the amount of hard-disk  space, processor speed, Motherboard etc... Can the memory of a RAM be compared with Virtual Memory and Cache Memory?  Note: About my PC Configuration: Intel core i5 processor, Motherboard DP55WB 4GB RAM, 700 GB Hard-disk space.