I visit it - and mostly StackOverflow - a lot lately; it often offers quick solutions to the problems I tend to run into at my IT job. But I think their atmosphere of self-moderation and downvotes is sometimes a bit ... erm, is "harsh" too harsh a word? As others have said, it's quite useful as...
Cool, it looks like we're an example to other sites, and not the least ones at that: http://meta.math.stackexchange.com/questions/16734/are-they-really-learning-something?cb=1
Oh, I just realized that probably "D" stands for "Distance", "C" is the speed of light (we normally use a lower-case c for that) and "T" for "Time".
In that case: yes, this is how you would convert distance to time. For example, you will find that 1 second corresponds to 300,000 km (taking c =...
Yes, the speed of light is a speed - it has units of m/s. For any speed, if you multiply a time by it you get a distance and if you divide a distance by it you get a time. Of course the big deal with the speed of light is that, unlike your car on the freeway, it is constant throughout the...
Which has level sets at z = Ax + B for some suitable constants A and B. I.e. the function is linear and does not have a minimum.
I would generally tackle these problems using Lagrange multipliers by the way, if you haven't seen them check them out, they are cool ;)
Can you show us what your Lagrangian looks like, that you use to derive these equations of motion?
You denoted the constraint by ##L##, but I don't think that expression is your Lagrangian because I don't see a kinetic term.
Since you have already calculated what M does to your axes (basis vectors), try figuring out what happens to a little square with corners on (0, 0) and (1, 1).
What happens to its area, for example? (At this point you may want to calculate the determinant of the matrix).
If you go from the...
I know lavinia, but they're not part of the real numbers, and the usage of statements like "inf - inf = 0" strongly suggests to me that the OP is not ready to be introduced to the mathematical details without a conceptual understanding of why infinity can not be treated that way without...
I suggest the OP stop treating infinity as a number, like 2, that you can multiply by another number, like 7, and get a sensible answer.
In "standard" mathematics (i.e. up to somewhere in an Analysis 2 course at University level) infinity is not a number, but a concept. It is the result of a...
The problem is they are really a quantum phenomenon so if you don't know about discrete energy levels and fermions it will be hard to explain properly.
It's been a while so I hope someone won't come along and tell me this is nonsense. But if I recall correctly, the reason is they live in different spaces. If you have a boson state ##|b\rangle## and a fermion state ##|f\rangle## then a combined state would be ##|b, f\rangle = |b\rangle \otimes...
You can make it more precise by defining "If ##a \approx 1## then X" to mean "There is an ##\epsilon > 0## such that X for all ##a## satisfying ##|a - 1| < \epsilon##."
Then ##a < 1/2## contradicts ##a \approx 1## because for any candidate ##\epsilon##, you can find a value ##a## such that...