Slow response on GitHub Pages with NameCheap
How do I decrease first byte time with GoDaddy and GitHub pages?
I recently setup a new computer for my parents with windows 10 home. I made two local accounts for them - no microsoft accounts. I also would like to keep them without a password. Problem is now that windows automatically signs in with the last used account. No lock/login screen.  I tried a number of answers from similar questions here but none worked. Since it's a home version there are no group policies and the method using task scheduler does not work because accounts with no password are not allowed to create such tasks.  Are there other options to force windows to show a user list on startup without setting passwords for the accounts? How does the Higgs mechanism work? How can I recover a deleted file on Android? I have to design a circuit in which I have to transfer data through lasers (serial transmission of data). Now what I was thinking was to check if the laser light is on then '1' is passed and when the laser is off '0' would be passed. Now I need to check if a 1 or a 0 was passed.   How would I check for this and another problem I'm facing is if it's on it may keep on passing 1's. How would I check for this? I though of using LDR's (Light Dependent Resistors)? Would this be okay or it there a better method? How does commands like ls know what its stdout is?  It seems ls is operating different depending on what the target stdout is. For example if I do:  ls /home/matt/tmp    the result is:  a.txt b.txt c.txt   However if I do  ls /home/matt/tmp | cat   the result is (i.e. new line per result):  a.txt b.txt c.txt   The process is passed a file descriptor 1 for stdout right? How does it determine how to format the result? Does the file descriptor reveal information? meaning of the word "in" in context It is from . It is at around 8 minute and 39 second. Here it goes:     Culinary cultural fusion is all the rage; more novels translated from languages then ever before, although few are read, and in the surest sign of cultural globalization, football, the world's game, finally has reached America.   I guess the sentence would make sense if the preposition in were dropped out. Could you please tell me what did the host use it for? Does the PS-Vita support PSP games Is the PS-Vita backwards compatible with PSP games?  Some games I would like to play include Final Fantasy 7: Crisis core, Jak and Daxter: The Lost Frontier, Coded Arms, Tekken: Dark Resurrection, and Secret Agent Clank. The laptop is dual booting with windows 7 and ubuntu 11.10.  For 11.10 to boot, I had to add 'nolapic' in GRUB boot options.  I downloaded 12.04LTS LiveCD yesterday and made a CD and a bootable USB stick.  With both USB and LiveCD, occasionally I can boot into 12.04.  But many times, the boot process stuck at some point.  For USB boot, sometime it works.  But other times, it seems to stuck right after seeing the message:  NMI watchdog enabled. takes one hw-pmu counted [0.408590] #2   The number in [] changes.  Some times it stuck at #2, and sometimes at #3.  For LiveCD boot, sometime it works.  But when it stuck, it stuck at different places.  At times, the boot process go directly to a blank screen with a blink '_' at up left corner.  Sometime it went a bit further, after showing options to try/install ubuntu then stuck with a blank screen with blink '_'.  Anything I can do to get around of this? We all know some curves can be described by $y=f(x)$ and some surfaces can be described by $z=f(x,y)$ However, there exists curves and surfaces which cannot be described by those, such as a circle and a sphere. Therefore, we introduce parameterized vector equations, which can describe them. For example, circle: $\vec r(t)=r\cos(t)\hat i+r\sin(t)\hat j$ sphere: $\vec r(u,v)=\rho\cos(u)\sin(v)\hat i+\rho\sin(u)\sin(v)\hat j+\rho\cos(v)\hat k$ However, all curves described by $y=f(x)$ and  $z=f(x,y)$ can be parameterized. For curves, $$\vec r(t)=t\hat i+f(t)\hat j$$ For surfaces, $$\vec r(u,v)=u\hat i+v\hat j+f(u,v)\hat k$$ Therefore, I think this suggests the set of all parameterized surfaces (or curves) is the super-set of the set of all surfaces (or curves) described by $z=f(x,y)$ (or $y=f(x)$). Is that correct? Now, here comes the the real challenge. A curve can also be described by an implicit function $f(x,y)=0$ and a surface can also be described by an implicit function $f(x,y,z)=0$ I have 3 questions regarding this.   Can all surfaces (curves) described by an implicit function be parameterized? (If yes, then what is the general way?) Can all surfaces (curves) described by parametric vector equations be represented using implicit function? (If yes, then what is the general way?) Compare the set of all parameterized surfaces (curves) and the set of all surfaces (curves) represented by implicit function. (which is which super-set?)   Sorry for the use of nontechnical terms. I use them because I don't know the technical ones. I have only started learning vector calculus last year in university.  EDIT: I think my question is not too clear, so I will give an example of writing the surface $f(x,y,z)=0$ into $\vec r(u,v)$ We want to parameterize a sphere. $$x^2+y^2+z^2-\rho^2=0$$ Let $x=\rho\cos(u)\sin(v)$, $y=\rho\sin(u)\sin(v)$, $z=\rho\cos(v)$, $$\rho^2\cos^2(u)\sin^2(v)+\rho^2\sin^2(u)\sin^2(v)+\rho^2\cos^2(v)-\rho^2$$ $$=\rho^2\sin^2(v)(\cos^2(u)+\sin^2(u))+\rho^2\cos^2(v)-\rho^2=\rho^2-\rho^2=0$$ I want to know if there is a general way of finding $x=x(u,v)$, $y=y(u,v)$ and $z=z(u,v)$ for any given $f(x,y,z)=0$ MySQL ERROR 1045 (28000): Access denied for user 'bill'@'localhost' (using password: YES) How to respond to "How much funding will you bring?" as an interview question for a faculty job? How to extract and install .deb without the command dpkg? I am running Debian jessie stable in order to understand how the .deb package interacts with the system. What is the correct way to extract and install a packge_name.deb without the command dpkg? What is the use of a persistence layer in any application? I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$  I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:  $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$  So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:  $$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$ But observe that I need to center it at $z_0=1$ so I rewrote it to:  $$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$  I can factor out $\frac {1}{2z}$ and use the basic identity to get : $$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$  The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.  Can anyone help me figure this one out? What am I doing wrong? Which is correct: "has died" or "died"? Resonances in high energy physics I still do not understand what a  precisely is. Is it exactly the same as a particle? Or only an excited state?  And why does it make a peak in some diagrams? And which diagrams? I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $&lt;_\mathbb R$?   Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set. Problem about angle in isosceles triangle bounding a $C^1$ function that upholds $f'(t)\leq 1-f(t)^2$ I have this problem: I'm used to compile and run .c files with CodeBlocks. If I do that from my Desktop (ext4 partition) it works without problems, but doing this from my secondary hard drive (NTFS partition) makes CodeBlocks to say permission denied while running a .c file.  I have read it's because of the partition type is it correct? But I don't know how to solve it. I tried to edit fstab (sudo gedit /etc/fstab) adding this line:   /dev/sda /media/Dati ntfs user,exec 0 1   but after rebooting Ubuntu I had a warning on boot about a problem mounting a drive, that's because I set KDE to auto mount my Dati drive on boot.   So I should auto mount my NTFS drive with execute permission? How can I do that?