@JM Just going through transcript, this is incorrect: the message was auto-removed within a couple minutes, I did not personally remove or retract it. I just presumed with some of Jordan's statements around the time, my sentiments were fair game.
btw that group theory question I had turned out to be trivial @DylanMoreland :)
If $|X^U|=|X^V|$ for any $G$-sets $X$ with $U,V\le G$, we can choose $X=G/U$. This has a fixed point under $U$, so there must be a fixed point under $V$, ie $VgU=gU$, or equivalently $g^{-1}Vg\subseteq U$. Reversing roles tells us they're the same size, so they must be conjugate.
Thanks. I think Asaf did some re-evaluation and decided not to stay partly because he viewed the chatroom as a time waster and had other things he wanted to move on to.
I think Jyrki might have fully answered my question using contradiction, but I have to check a couple of things: (1) Are rank $n-1$ matrices all in the same orbit under GL action? (2) Can lower-rank matrices all be decomposed as products of diagonal $n-1$ rank matrices (and GL again I guess)?
hey @Eugene
@robjohn Indeed, that's why I took a screenshot of the whole thing and posted it.
@experimentX Sure, it's obviously related. The Jacobian doesn't tell you anything about global positioning though, it only has information about infinitessimal rates of change, and the local geometry around a single point
Let $ (\hat i, \hat j, \hat k) $ be unit vectors in Cartesian coordinate and $ (\hat e_\rho, \hat e_\theta, \hat e_z)$ be on spherical coordinate.
Using the relation, $$ \hat e_\rho = \frac{\frac{\partial \vec r}{\partial \rho}}{ \left | \frac{\partial \vec r}{\partial \rho} \right |}, \hat e_\t...
@JasperLoy, In this case rate is a percent, for example .05, so if base is 105% of something, I know what the original something is with base/1.05, but I want to know what was added to that something, but not sure if it is possible to do it shorter than I have already done.
the point is, if you know how to transform A-coordinates to B-coordinates, and you know how to transform B-coordinates to C-coordinates, you can get A-coordinates in terms of C-coordinates. the algebra can get messy, though.
if you actually try to explicitly calculate the inverse matrix, you're going to get a det(A) factor that comes out. then when you find the principal minor determinants, and multiply them (like in a dot product) with the cylindrical basis vectors, you'e going to get the same expressions.
I am trying to solve the differential equation $y' = x + y$ but the book tells me to make the change of variable $u = x + y$ what do I do? I know I can't just write $y' = u$ do I change it to $u' = u$ or what?
@StartupCrazy Half of the numbers seem to be larger than their neighbors; the other half are smaller than their neighbors. Perhaps each of those groups has a nice internal progression?
hmm...let me see if can explain this. the differential equation y' = y can be turned into the homogeneous one y' - y = 0, which is easy to solve. but in y' = x+y we have that pesky "x" term, so we want to get rid of it.
but in $y' = x+ y$ we only get rid of one term to replace it with another, so there are two functions? Tw dependent variables and an idependent? And this is okay because I "solve" them seperatively?
well, the whole goal is to find out "which" function of x y is. it's like a murder mystery, and the differential equation is our only clue.
we establish that y' = u' - 1, so that's our "new" left side. and our "new" right side is: u, so we have: u' - 1 = u. look, no pesky "x"'s all by themselves.
Anotehr thing I am confused about if I have $\frac{dy}{dx} = y + 1$ What can I do with it? I can only subtract y do I get $dy-y$ and I can't integrate that can I?
expanding the LHS of the desired form, we get $p y'+p'y = \mu y' -\mu y$. Thus $p=\mu$, and since $p'=-\mu$ we can also say $\mu$ satisfies the differential equation $\mu'=-\mu$. This can be solved as $\mu=e^{-x}$.
Oh, okay, then disregard my explanation of the integrating factor method.
What you're supposed to do then is: find one particular solution (it will be of the form y=ax+b), and then also find the general solution to the homogeneous equation y'-y=0. Then add the particular solution to this general homog solution and you will wind up with the general form of the solution to the inhomogeneous equation $y'=y+x$.
Actually, if we are to have sophistication-graded tags for set-theory, then it's a bit crazy that transfinite induction belongs in elementary set theory (explicitly according to the tag excerpt). How is one supposed to tag actually elementary set stuff, I wonder?
@Eugene i don't see anything wrong with it as a question. i can't say i fully understand it, but if there's an error made in your question, it seems like it would be instructive for an answer to point this out, and i cannot see how down-voting is called for.
@DavidWheeler oh seriously the downvoting isn't an issue. recently i've been frequently downvoted without explanation. i'm sure there's a troll somewhere.
@Eugene But I don't think there is much relevant overlap between ordinary induction in $\mathbb N$ and the transfinite stuff. Formally one is a special case of the other, of course, but the questions one could ask about either are very different.
@DavidWheeler In your answer, could you edit it to include the matrix with entries from $\frac{\partial f }{\partial \Bbb{x}} h(x)$ and $\frac{\partial f }{\partial \Bbb{y}} h(x)$
