I find that a little reductive. Doing PhD work in physics, for instance, will for some people mean a lot of experimental labwork. for others, it's basically a lot of applied math.
Anyway, if people really wanted money they wouldn't be doing math. Math professors are quite poor, compared to accountants, lawyers, bankers, and actors.
semic, my understanding from the far-too-many panels I've been to is something like: the market for math PhDs is fairly big but "sparse": most companies don't need any, and most that do only need one or two. But it seems that there's a lot of communication between the companies that need a few, and if you apply in one place but don't fit what they're looking for, they tend to know others that are hiring, and can give very strong recommendations.
Even getting the apparent quant jobs requires more stuff like math-related fields like stats/coding, and within math stuff more like numerical analysis. Algebraic geometry is not nearly as surefire
Consider the following elements with $K\geq 2$.
$\mathbf{X}$ is a $K\times 1$ real vector.
$\mathbf{P}$ is a $K^2\times K^2$ real symmetric matrix of the form
\begin{equation*}
\mathbf{P} =
\begin{pmatrix}
\mathbf{P}_{11} & \mathbf{P}_{12} & \cdots & \mathbf{P}_{1K} \\
\mathbf P_{12}^T & \m...
Either way, the classes you take if you're interested in math for its own sake seems to be different enough from those you take if you're interested in math-heavy industry jobs, enough that caution is important
Damin: the topic of study is rarely important. Everyone knows that an industry PhD is not doing the same sort of work they did for their thesis. You're valuable as someone who knows how to read the literature and commit to a problem for many years, not as someone who happens to know how to find stable solutions for nonlinear PDEs.
3
ah yeah, but for bachelors you might want to be a bit more careful
@user243301 What sort of approximation are you looking for? $g$ is given as a power series about $0$, that series converges absolutely and uniformly on the closed unit disk, so truncation of the series gives polynomials approximating $g$ pretty well. These polynomials are also the best approximations of $g$ in the $L^2$ sense.
@DanielFischer I am still struggling with mental problems. Still bad, still hanging in there, still sorting out my thoughts, still hoping for a miracle.
Eric: XD you get irritated by the immorality of the farming community and return back to math, later writing a book about your life in obtuse mathematics metaphors #reversegrothendeick
Students care more, I'm not held to some test that I need to teach for, and then when I get through that content spend all my time doing practice problems for that test, etc
@Daminark I taught in some average schools here, and I can tell you this: The classroom behaviour is atrocious. The worst behaviour you see on TV is actually the best in reality.
I mean I was shitposting in that particular example, I mean it doesn't need to have anything to do with, say, modern politics/governance. It's rather possible to stay mostly away from this sort of thing
Actually while I don't want to be pushy I'll just let this go to rest. Mentioned this before but I'm pretty sure the likelihood is higher that I die in a car accident than dealing with stuff like this so I'd rather not, I've had a rather bad aftertaste in all my previous interactions with this subject. I will fight to the death that $0\in\mathbb{N}$
I think other than naturals including zero or not, there is also the issue of whether rings have one or not, how to define the fourier transform, and how to write subset and proper subset.
The main reason I'm hesitant to say 0 should be in the natural numbers (beyond just not having learned it that way) is that it makes multiplicative stuff more annoying.
I am trying to show that any two open sets in $\Bbb{R}$ endowed with the finite complement topology intersect. Here is my attempt. Suppose we have two open sets $U$ and $V$ that do not intersect. Then $V$ must be contained in $U$'s complement, which is finite. But $V$ is infinite, which is absurd. So, $U$ and $V$ must intersect.
Does this sound right? If so, wouldn't it follow that $\Bbb{R}$ with this topology is not Hausdorff?
@Semi You know, when I see you in the chat, I know that I'm safe when doing physics:P You're sort of the Ted of Physics, I would say. Sorry for the cheesiness, but I still can't get over the fact how much you've helped me. (Physics is like a lost love to me, that I've finally found back, thanks to the physics chat and your help:P)
I just remembered while thinking about orientations right now that at one point while studying manifolds I was being utterly confused about orientation numbers. Feels a little weird now. :P
For some reason I couldn't comprehend the fact that positive and negative orientation of a certain basis of a vector space is just a random association of number, and could be done in any way whatsoever as there are exactly two equivalence classes of oriented basis.
Just curious, what's the difference between $1~\text{mol}^2$ (of dimension $\mathsf{N}^2$) and $(1~\text{mol}^2)\cdot N_{\text{A}}$ (of dimension $\mathsf{N}$)? Should I be asking this in chemistry.SE's chat instead?
I feel cheated for never having heard about this adjoining-i business before; it seems like a really obvious exercise I should have been forced to do in my algebra class :/
p can be written as sum of two squares iff p = 1 mod 4, and non-gaussian immediately means i can write them as a product of sum of two squares hence as a sum of two squares
Conjecture: If $Y$ is a subspace of $X$ endowed with the cofinite topology, then the topology Y inherits is cofinite. Proof: Since all of the closed sets in $X$ are finite, the intersection of a closed set in $X$ with $Y$, which will produce a closed in $Y$, is finite. Thus, all the closed sets in $Y$ are finite. Hence, if $U$ is open in $Y-U$ is closed and therefore finite.
By the way, this sounds completely false to me. I think I can even give examples with curves (as $N$) in $\Bbb R^2$ (as $M$) which bound nonconvex domains.
It's a very natural question, @Balarka. I need to think a moment. Of course, it's clearly false when $\mathcal N$ is itself $1$-dimensional and not a geodesic.
No, I want it parallel to the tangent plane of the torus on the top and instantaneously tangent to the top circle (perpendicular to the meridian). You always confuse meridians and parallels :P
The $c(t)$ should be interesting in that case. There are two arcs on the parallel curves (I hope that's the right diffgeo terminology?) and some extra stuff
Yeah I mean locally near the endpoint you contain the parallel curves (the actual geodesics going through the top circle). I don't think you contain the top circle either.
No, you're calling parallels the wrong things again. The meridians=profile curves are geodesics.
The actual geodesic starting tangent to the top circle wobbles periodically on the outer half of the torus, touching both top and bottom circles tangentially.
I don't think you ever did this exercise with Clairaut.
So I guess I'll do the calculus problem ... or maybe have Mathematica do it. I want to know when the chord joining the point on the line and a point on the torus is normal to the torus.
Oh, I misunderstood your line then. I thought it's the line which is tangent to the top circle and cuts the top circle diametrically. But it's the one which is orthogonal to the meridian. OK.
This seems a rather complicated scenario for a counterexample though.
Consider the following elements with $K\geq 2$.
$\mathbf{X}$ is a $K\times 1$ real vector.
$\mathbf{P}$ is a $K^2\times K^2$ real symmetric matrix of the form
\begin{equation*}
\mathbf{P} =
\begin{pmatrix}
\mathbf{P}_{11} & \mathbf{P}_{12} & \cdots & \mathbf{P}_{1K} \\
\mathbf P_{12}^T & \m...
