Why current is negative in this situation? Because EMF and current directions are the opposites?        &ndash; Schematic created using   Can someone give me a good explanation on relationship between current and EMF?
I am starting to see a lot of people claiming that convential current is 'wrong' because Franklin made an error when he first started investigating electrostatics, and that later scientists didn't bother correcting the mistake, but preferred to keep the 'convention' (here is a classic example: )  I always thought he didn't get it wrong. He said that current is positive in the direction that positive charge flows, and vice-versa. He of course had no way to know which side of two sticks behind rubbed actually gained or lost mass. So he wasn't wrong. What were you taught?  P.S. I can't help but feel we are lucky that he got it 'backwards', because clearly many people are confused about electrostastics (including the author of that text book!) and believe electricity has to involve electrons (an unfortunate name... why couldn't they have been named negatrons...)
Remove HTML tags from a String I have the following sequences $(a_n)$, $(b_n)=(a_{2n})$, and $(c_n) = (a_{2n-1})$.  I have to show that:   if $(b_n)$ and $(c_n)$ both convergence to $a$, then $(a_n)$ also converges to $a$; if $(a_n)$ converges then $(b_n)$ and $(c_n)$ also converge; there exist $(a_n)$ not converging but with $(b_n)$ and $(c_n)$ converging.   I am not sure how to solve this since I only know that $(b_n)$ > $(c_n)$ > $(a_n)$ on the terms how the sequences are defined. \selectcolormodel{gray} doesn't work for the duck hair, but it works for the other parts of the ducks, why?  \documentclass{ltugboat}  \usepackage{microtype}  \usepackage{graphicx}  \usepackage{ifthen} \newboolean{grayscale}    \setboolean{grayscale}{true}% for the black &amp; white version    %\setboolean{grayscale}{false}% for the colored version     \usepackage{tikzducks} \usetikzlibrary{positioning}  \newenvironment{mytikz}{%     \begin{tikzpicture}     \ifthenelse{\boolean{grayscale}}{\selectcolormodel{gray}}{}      }     {\end{tikzpicture}}  \usepackage{ifpdf}  \ifpdf  \usepackage[breaklinks,hidelinks]{hyperref}  \else \usepackage{url} \fi   \begin{document}      Why \textbackslash\texttt{selectcolormodel\string{gray\string}} does not work for the duck hair?      Black and white version:          \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz}      Colored version:     \setboolean{grayscale}{false}      \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz} \end{document} Integral equation problem What is "root" and how can I become it? When I ran a command, I got the following error:     You need to be root to perform this command.   What is root and how can I become it? For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$ "Much of it is" or "Much of it are"? I'm writing an college-entrance essay and for me it's imperative that I understand and implement the correct one of these two expressions:     "Much of it is" or "Much of it are"   If you need context:     I'm having trouble with my writing. The reason I'm having trouble is that much of it is about....   Should there be an "is" or an "are"? Error in pip: unsupported operand type(s) for -=: 'Retry' and 'int' I'm trying install the requests, but I can't do that. I'm using Ubuntu 16.04  I tried to install using  $ sudo pip install requests   and I get the following error   Collecting requests Exception: Traceback (most recent call last):   File "/usr/lib/python2.7/dist-packages/pip/basecommand.py", line 209, in main     status = self.run(options, args)   File "/usr/lib/python2.7/dist-packages/pip/commands/install.py", line 317, in run     requirement_set.prepare_files(finder)   File "/usr/lib/python2.7/dist-packages/pip/req/req_set.py", line 360, in prepare_files     ignore_dependencies=self.ignore_dependencies))   File "/usr/lib/python2.7/dist-packages/pip/req/req_set.py", line 512, in _prepare_file     finder, self.upgrade, require_hashes)   File "/usr/lib/python2.7/dist-packages/pip/req/req_install.py", line 273, in populate_link     self.link = finder.find_requirement(self, upgrade)   File "/usr/lib/python2.7/dist-packages/pip/index.py", line 442, in find_requirement     all_candidates = self.find_all_candidates(req.name)   File "/usr/lib/python2.7/dist-packages/pip/index.py", line 400, in find_all_candidates     for page in self._get_pages(url_locations, project_name):   File "/usr/lib/python2.7/dist-packages/pip/index.py", line 545, in _get_pages     page = self._get_page(location)   File "/usr/lib/python2.7/dist-packages/pip/index.py", line 648, in _get_page     return HTMLPage.get_page(link, session=self.session)   File "/usr/lib/python2.7/dist-packages/pip/index.py", line 757, in get_page     "Cache-Control": "max-age=600",   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/sessions.py", line 480, in get     return self.request('GET', url, **kwargs)   File "/usr/lib/python2.7/dist-packages/pip/download.py", line 378, in request     return super(PipSession, self).request(method, url, *args, **kwargs)   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/sessions.py", line 468, in request     resp = self.send(prep, **send_kwargs)   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/sessions.py", line 597, in send     history = [resp for resp in gen] if allow_redirects else []   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/sessions.py", line 195, in resolve_redirects     **adapter_kwargs   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/sessions.py", line 576, in send     r = adapter.send(request, **kwargs)   File "/usr/share/python-wheels/CacheControl-0.11.5-py2.py3-none-any.whl/cachecontrol/adapter.py", line 46, in send     resp = super(CacheControlAdapter, self).send(request, **kw)   File "/usr/share/python-wheels/requests-2.9.1-py2.py3-none-any.whl/requests/adapters.py", line 376, in send     timeout=timeout   File "/usr/share/python-wheels/urllib3-1.13.1-py2.py3-none-any.whl/urllib3/connectionpool.py", line 610, in urlopen     _stacktrace=sys.exc_info()[2])   File "/usr/share/python-wheels/urllib3-1.13.1-py2.py3-none-any.whl/urllib3/util/retry.py", line 228, in increment     total -= 1 TypeError: unsupported operand type(s) for -=: 'Retry' and 'int'   My python version is the 2.7 and the pip is 8.1.1 I'm trying to install drivers for my Wifi on my HP All-in-one desktop. I've been told to show the output of the command "sudo lshw -C network" so here it is:  *-network description: Ethernet interface product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller vendor: Realtek Semiconductor Co., Ltd physical id: 0 bus info: pci@0000:01:00.0 logical name: enp1s0 version: 15 serial: 48:ba:4e:5c:06:5f size: 100Mbit/s capacity: 1Gbit/s width: 64 bits clock: 33MHz capabilities: pm msi pciexpress msix bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=full firmware=rtl8168h-2_0.0.2 02/26/15 ip=192.168.1.21 latency=0 link=yes multicast=yes port=MII speed=100Mbit/s resources: ire:31 import:e000(size=256) memory: fea04000-fea04fff memory:fea00000-fea03fff  *-network UNCLAIMED description: Network controller product: Realtek Semiconductor Co., Ltd. vendor: Realtek Semiconductor Co., Ltd. physical id: 0 bus info: pci@0000:02:00.0 version: 00 width: 64 bits clock: 33MHz capabilities: pm msi pciexpress cap_list configuration: latency=0 resources: import:d000(size=256) memory:fe900000-fe90ffff   Output from "lspci -nnk | grep -A2 0280":  02:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8821CE 802.11ac PCIe Wireless Network Adapter [10ec:c821] Subsystem: Hewlett-Packard Company RTL8821CE 802.11ac PCIe Wireless Network Adapter [103c:831a] 03:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. RTS5229 PCI Express Card Reader [10ec:5229] (rev 01) PrintWriter and PrintStream never throw IOExceptions Some days ago I realized that  (as well as ) never throw an IOException when writing, flushing or closing.  Instead it sets an internal flag (trouble=true) when an error occurs. It's not possible to get the exact exception, but only if there was some exception ().  My question is: why would one want to have such behavior? Isn't that bad API design? The user who first created a tag wiki is treated as owner for suggested edits was declined by the user who created the tag wiki excerpt, who doesn't yet have the reputation to be a trusted user. (The site is still in beta, and the reputation necessary to be a trusted user is 4000.)        Should the user who first proposed a tag wiki, or a tag wiki excerpt be the one who can unilaterally decide about a proposed edit for a tag wiki? The user who first proposed it isn't necessarily the one who knows if the tag's purpose can be better described with different words, nor does that user necessarily know if the tag has been lately used differently from how it was first used. Since a tag wiki is different from other posts, there should be an exception to the rule "the author of a post can unilaterally decide about any edits."  The paradox is then that the creator of a tag wiki still needs two users to approve their suggested edit. That's what happened in .        The funny fact is that in the revisions for the tag wiki, or tag excerpt (as in the screenshot), the tag wiki/excerpt is said to be created from the Community user.        This bug is  definitely . Word for romantic partner you are living with but not married to Java is the key here. I need to be able to delete files but users expect to be able to "undelete" from the recycle bin.  As far as I can tell this isn't possible.  Anyone know otherwise? How to make a domain receiving email behind cloudflare? Indentation in tcolorbox package I'd like to have the same indentation within a tcolorbox as I have in the rest of the document.  How can I get it?  Thanks.  MWE:  \documentclass[a4paper]{book}  \usepackage{lipsum} \usepackage[most]{tcolorbox}  \begin{document}  \chapter{Example with \texttt{tcolorbox}} \lipsum[1-8]  \begin{tcolorbox}[breakable,enhanced] \lipsum[9-14] \end{tcolorbox}  \end{document}   The first block of text has the typical indentation, but that in the box doesn't. $A \oplus B = A \oplus C$ imply $B = C$? I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.  For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true.   I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:    I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.  On the other hand, I could be completely wrong about my reasoning.    Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well. Please require more reviewers on First Posts On the one hand, I see very few posts in the review queue when I visit.  On the other hand, I see a lot of pretty awful first posts sitting on the front page.  This leads me to suspect that robo-reviewers are blowing through the queue.  Given how empty the queue is, could you please increase the number of people who have to check in before a post leaves the queue?  Of course, some honey-pot solution to detect and disable the auto-clickers would be even better. How to perform a logistic regression for more than 2 response classes in R I want to predict the impact of oil price over a Colombian oil company's stock price. I plan to use a logistic regression for this with a categorical variable (Up or Down given the direction of the stock price). Here is part of my dataset:  Minute  ecopet  profit  sum_profit   direccion  cl1_chg   sum_cl1    direccion_cl1 571     2160     0       10           Up         -0.03     0.00      Down 572     2160     0        0           Neutral     0.07    -0.03      Down 573     2160     0       -5           Down       -0.08     0.04      Up 574     2160     0       -5           Down       -0.07    -0.04      Down 575     2160     5       -5           Down       -0.08    -0.11      Down 576     2165     0       -25          Down        0.00    -0.19      Down 577     2165     0       -25          Down       -0.05    -0.19      Down 578     2165     0       -15          Down       -0.17    -0.24      Down 579     2165     5       -15          Down       -0.06    -0.41      Down 580     2170     0       -20          Down        0.03    -0.47      Down 581     2170    -10       0           Neutral     0.04    -0.44      Down   My dependent variable is 'direccion'. But as you can see it has 3 response classes. I know that to implement a binary logistic regression in R the code is:  glm.fit=glm(direccion~direccion_cl1, data=datos, family=binomial)   I am working with intraday information and plan to predict what happens when the oil moves up/ down (in the previous 10 minutes) and how it impacts the stock price in the next 10 minutes.  Could anyone tell me how could I perform this? I don't really know how to perform the logistic regression with 3 response classes. Thanks! Identifying genes from a list of genes How to prevent applications from discovering my phone as being Rooted I have to install an application that has discovered my device to be rooted.  So this is a two part question.   How did it discover in the first place that my phone has been rooted How to prevent the application from discovering my phone as being rooted