@DanielSank Download the Windows 10 installer onto a USB stick, wipe the disk and do a fresh clean from scratch install and it will gibve you no more problems.
Apple built Darwin on mach to make it easier to switch architectures (something they've done three times in the history of the company and twice for "macintosh").
"They reported this to the Windows developers, who disassembled SimCity, stepped through it in a debugger, found the bug, and added special code that checked if SimCity was running, and if it did, ran the memory allocator in a special mode in which you could still use memory after freeing it."
"In fact if you poke around in the AppCompatibility section of your registry you’ll see a whole list of applications that Windows treats specially, emulating various old bugs and quirky behaviors so they’ll continue to work."
"For example, a lot of developers used to try to make their Macintosh applications run faster by copying pointers out of the jump table and calling them directly instead of using the interrupt feature of the processor like they were supposed to.
Even though somewhere in Inside Macintosh, Apple’s official Bible of Macintosh programming, there was a tech note saying “you can’t do this,” they did it, and it worked, and their programs ran faster… until the next version of the OS came out and they didn’t run at all. "
This article is always a fun read
"a veritable panoply of external dependencies each one of which is going to be a huge headache when you ship your application to a paying customer and it doesn’t work right. The technical name for this is DLL Hell."
Inside Machintosh was full of warnings not to take the current implementation as a license to do things. It sometime took years for an update to kick cheaters in the butt, but it happened many times.
The no. 1 problem with Windows is that MS have to let OEMs do their own installs and piss about with the install as they do it. Even the big names like Dell overload the Windows install with crap, and the box shifters do installs that should earn them a long spell in jail.
This doesn't affect Apple because they have complete control over the hardware and OS, and it doesn't affect Linux because everyone does their own install;s.
There has been an experiment where infinite quantum linked rings were created to make a quantum knot.The structure is topologically stable and the knot can't be untied without breaking the rings.
If these rings have energy and allow the structure to be stable, wouldn't there be infinite energy ...
Microsoft grew up during the 1980s and 1990s, when the growth in personal computers was so dramatic that every year there were more new computers sold than the entire installed base. That meant that if you made a product that only worked on new computers, within a year or two it could take over the world even if nobody switched to your product.
"When they tried to “End Of Life” Windows 98, it turned out there were still so many people using it they had to promise to support that old creaking grandma for a few more years."
@EmilioPisanty I hope I'm not muscling into your conversation but I did notice a number of new users trying to vandalize the site by posting garbage questions, answers or comments.
@ZeroTheHero no worries. Even I can't see deleted users (presumably mods can? who knows), and I could only find the old question because I voted to close it.
"Have you ever heard of SEMA? It’s a fairly esoteric system for measuring how good a software team is. No, wait! Don’t follow that link! It will take you about six years just to understand that stuff."
One of the article on that blog is from 2000 and it has a link to a software company, except it doesn't exist anymore and now the link is to a tree removal company
"The story goes that one programmer, who had to write the code to calculate the height of a line of text, simply wrote “return 12;” and waited for the bug report to come in about how his function is not always correct. "
@ACuriousMind For a Dirac spinor $\psi$, what are $u$ and $v$, if they are defined by $(\gamma^\mu p_\mu - m)u=0$ and $(\gamma^\mu p_\mu+m)v=0$? Are they the Weyl spinors which constitute the Dirac spinor? Or are they the sums and differences of the Weyl spinors like $\psi = (u,v) = (\psi_L + \psi_R, \psi_L - \psi_R)$?
The latter are normally written as $\psi = (\phi, \chi)$, and the former are usually called $\psi_L$ and $\psi_R$, so I believe they are neither.
@Bass They are the positive and negative frequency solutions. For any constant Weyl spinor $\xi$, $u = (\sqrt{p\cdot \sigma} \xi,\sqrt{p\cdot \bar{\sigma}}\xi)$ and $v = (\sqrt{p\cdot \sigma}\xi,-\sqrt{p\cdot \bar\sigma}\xi)$ are such solutions.
@Bass The terminology is a bit confusing here - one should distinguish between Weyl and Dirac spinors, which are just elements of their specific group representations, and there are the spinor-valued fields. $u$ and $v$ are specific spinor-valued fields, but not special as mere spinors.
@Bass You need to be more explicit whether you're writing down spinors or spinor-valued fields here. First, the $u,v$ depend on the choice of a Weyl spinor $\xi$, so they're Dirac-valued fields $u_\xi(p)$ with a Weyl-spinor parameter $\xi$.
@ACuriousMind OK, I'm trying: $\phi$ and $\chi$ are just spinors, they are defined by their transformation under the Lorentz group. $u$ and $v$ are spinor fields, i.e. solutions to the Dirac equation.
Now, if you look at my formulae up there, for the choices $\psi_{L,\xi}(p) = \sqrt{p\cdot \sigma}\xi$ and $\psi_{R,\xi}(p) = \sqrt{p\cdot \bar\sigma}\xi$, then you have indeed that $u_\xi(p) = \psi_{L,\xi}(p) + \psi_{R,\xi}(p)$ and $v_\xi(p) = \psi_{L,\xi}(p) - \psi_{R,\xi}(p)$.
I asked a similar question in the math section but no one answered , hopefully I will find the answer here.I am an engineering student and I often face mechanics problems which involves usings concepts like infinitesimal and limits . I have a clear idea of those concepts ,however applying them on...
There has been an experiment where infinite quantum linked rings were created to make a quantum knot.The structure is topologically stable and the knot can't be untied without breaking the rings.
If these rings have energy and allow the structure to be stable, wouldn't there be infinite energy ...
