815 reputation
113
bio website joshphysics.com
location Los Angeles
age 28
visits member for 1 year, 7 months
seen Aug 23 at 20:06

My new website: joshphysics.com

Currently a visiting lecturer at the UCLA Department of Physics and Astronomy.

Ph.D. theoretical high energy physics from UCLA.

BA/BS in physics/math respectively from UC Berkeley.


Mar
28
comment What extra assumption makes this transformation affine?
@Rahul Ah thanks! Does one need to restrict the field over which $V$ is defined?
Mar
28
comment What extra assumption makes this transformation affine?
@BISHD Great! One might think that such curiosity would warrant an upvote ;) ? Thanks for the edits btw.
Mar
28
awarded  Custodian
Mar
28
reviewed Approve suggested edit on What extra assumption makes this transformation affine?
Mar
28
comment What extra assumption makes this transformation affine?
@BISHD lol ok. I had hoped what I was asking was clear from the context, but I agree it's not entirely clear, so I changed the wording to "how weak one can make additional assumptions." How's that?
Mar
28
revised What extra assumption makes this transformation affine?
added 15 characters in body
Mar
28
comment What extra assumption makes this transformation affine?
@BISHD Yes I understand that. I am not assuming that $f$ is linear.
Mar
28
comment What extra assumption makes this transformation affine?
@BISHD Just a map.
Mar
28
asked What extra assumption makes this transformation affine?
Mar
8
comment Inverse function with Dirac Delta
@TZakrevskiy Could you clarify what "completely describes" means? What if you were to take $\{\delta_x\}_{x\in\mathbb R}$?
Mar
7
revised Inverse function with Dirac Delta
added 77 characters in body
Feb
24
comment $g''(t_0)$ where $g(t) = f(t, 1-t) , t_0 = 0$
@user129120 Your guess is as good as mine.
Feb
24
answered $g''(t_0)$ where $g(t) = f(t, 1-t) , t_0 = 0$
Feb
17
comment Inverse function with Dirac Delta
@Garrett See math.ucsd.edu/~wgarner/math4c/textbook/chapter2/…, for example. Notice also that saying "the" function $f(x) = x^2$ is misleading. One needs to both specify the domain of the function, and what it does to each element of the domain. If we define $f$ as the function whose domain is $[0,\infty)$ and for which $f(x) = x^2$ for all $x\in [0,\infty)$, then $f$ is invertible. But if we extend its domain to $(-\infty, \infty)$, then it is not invertible.
Feb
8
accepted Cochains: terminology
Feb
8
comment Cochains: terminology
@TedShifrin I most certainly will, especially since you're the second person who's suggested that to me, the first being a math grad student. I as a theoretical physics grad just learning this stuff, I really appreciate your help.
Feb
8
comment Cochains: terminology
@OlivierBégassat Thanks again.
Feb
8
comment Cochains: terminology
Thanks very much for this Olivier. Would you mind pointing me to a reference that discusses this stuff; I would very much appreciate it.
Feb
8
comment Cochains: terminology
@TedShifrin As a matter of terminology, would you happen to know if elements of $\mathrm{Hom}(M,\mathbb R)$ are ever referred to as cochains when $M$ is the $\mathbb Z$-module of chains? The $k$-form example you gave is precisely what I had in mind when I wrote this question; such objects arise naturally in physics (especially thermodynamics when modeling heat). By the way, I'm a fan of "multivariable mathematics," so I was, I must admit, somewhat star-struck when I saw your answer.
Feb
8
revised Cochains: terminology
edited body