Call me lazy, but I'd like to make a new command to quickly display a norm, something like \newnorm{x+y} (to display \|x+y\|), whether I'm in the math environment or not.
You're right, perhaps that was not very clear. I want to be able to tell if an environment is in use in a certain area during compilation. Something that may return a boolean, to be used with some other package (I believe there is a if-then package or something like that... right?).
Hey all!
I'm trying to learn \LaTeX, and so far I'm doing fine. However, I would like to know if it is possible to be able to find out if a particular environment is in use at a certain point. Basically, I was wondering if something like:
\isInUse{math}
exists in LaTeX. Does anyone...
Hey all,
Consider the system X' = f(X), where X is a square matrix. I would like to solve this system with MATLAB, but all of its solvers require X to be a column vector.
Of course, it would be possible to rewrite X' and f(X) as column vectors, but I guess some efficiency would be lost. It...
Hey all,
Suppose I have a MATLAB function that returns two (or more) output arguments, but I only care about the second one, and I do not wish to assign the value of the first one to a variable.
What bugs me is when, for example, I solve an ODE:
[t, x] = ode45(blablabla)
and then...
Thank you both for this enlightment! I believe I understand much better now.
Ah, well, no I didn't :shy: I should have mentioned that I'm studying maths and not physics. Thanks again!
Fourier transform --> power spectrum
Hey all!
I've been learning about the discrete Fourier transform (and FFT too) recently. What I don't understand is why applying it to a signal gives its power spectrum. I am not really good in physics, so to me it just seems like a magical formulae, one...
Yeah, I've done all that stuff. Anyway Maple is here to help me :D
Are you sure b = angle ACE? It's angle CHJ on your diagram. Anyhow, I'm not sure I get that:
AC = HJ.tan(b) + HJ/tan(b) + x.tan(b) + x.tan(60°).
You're adding all the parts of the AC segment, right? In other words, CJ +...
This is perfectly right. I don't know if there's a "shortest proof", but your friend's certainly is short.
I'm somehow a newbie here, but I used to visit Tom's Hardware Guide's forums, and there's an interesting topic with... various "opinions" on the subject on it. Check it out...
Yup, you're certainly right, because AB is on the right bisector of the circumscribed triangle... Now I was thinking of finding some sort of relation between the distance AG and the radius of the circle, but since AG isn't really defined, I don't see what we could get out of it.
I also...