but I would prefer it if those people were also to invent their own terminology at this point, instead of changing the already established concepts
well, not necessarily invent their terminology as much as just make it aware what people usually mean by those things and what those terms meant historically
I feel like courses are kind of a scam really, not sure what u guys think; Theyre fast, gloss over results, and it is, atleast for me, impossible to stay ahead of the class in my reading of the book which results in me taking in the material for the first time in the class, which I hate
I feel like i would never get the familiarity with a theorem / result by looking at the proof immediately after the theorem, which is what usually happens in a class setting
I know somebody who basically insists on a similar anti-course position. They take a lot of time learning a few things in depth, reading, exploring, etc. All well and good, but they're at odds with the demands of studying and that person will probably not go on to a research career in mathematics, if they can bring themselves to graduate eventually
@SineoftheTime this sounds more like a local problem than one about courses
as in, per se
sure, people to better and worse jobs at teaching, no doubt about it
@BenSteffan you make good points, but I cant help but wonder, would you not, to some extent, lose some "dexterity" in certain topics? It could really be just a "me" problem, but I have to play with definitions, theorems corollories etc to make myself familiar with the thing
not to defend pedagogical incompetence (of which there is a limitless supply) but there's definitely value in not having learning be self directed and having to say "well whatever this is, i just have to deal with it" as a backstop to all of the choices that might otherwise be involved in learning
the point of studying mathematics at a university is not just to "learn mathematics," it's to learn techniques and skills for how to navigate mathematics, proof skills etc.
@VladimirLysikov the function $d(t)$ represents a varying distance between two objects. The whole point of this constraint is to avoid collision. I believe the function is smooth in this sense.
a lecturer tells you "these are the features of the field, here's some main techniques, here's some exercises, here are the important results" and after that you can somewhat manage on your own
@SineoftheTime oh, I think I misunderstood your initial message
sure
norms have changed a lot about that
@Sahaj hello :)
I get the frustration about courses, and being behind, etc. We all do, because we've all experienced it.
@SineoftheTime thank you for the link. Yes. I came across this lemma while searching for differential inequalities. What I see in texts is $\dot{u}(t) \leq a u(t)$ but mine $\dot{u}(t) \geq -a u(t)$ but I'm not sure if it falls in the same group.
As I was saying before, talking with almost all professors of the "Analysis" sector, I've told them that despite being able to do the exercises, etc I don't feel having a deep understanding. They agreed saying that now professors pretend less from the students
@nickbros123 Assimilating knowledge takes a long time. Not only that but mathematical knowledge also rapidly diverges into different mathematical fields. Perhaps if more theoretical math could be crammed into high school etc. it wouldn't be as much of a problem.
I feel like a lot of time gets wasted on doing nothing in high school et cetera that could be somehow made useful
@CroCo denote by $f(t)$ the solution in the case you have equality, i.e. $\dot f=-af(t)$, then $f(t)=e^{-at}$. Now $\frac d{dt}\left(\frac{d(t)}{f(t)}\right)=\frac d{dt}(d(t)e^{at})=e^{at}(\dot d+ad(t))\ge 0 $
Maybe no one has tried to systematize it because they thought, why would we put more work into this, where the average joe will keep on being average...?
it's a running joke here already for a professor to say "you all probably know this from high school already" when introducing, like, the dual space in first term linear algebra here
@Jakobian i suggest asking yourself, (1) how much do you think you would need to be paid per hour to teach math to literally anybody who comes in the door and (2) how much does your country, or any country, pay teachers
@leslietownes I am just saying that going from "mindless calculation" to more "teach how to think" is going in direction that's getting at least a little bit more social sciency than math currently is. Where you can't just get away with doing the same things over and over again, but need to actually think how to teach people
the reason math education systems prioritize symbolic calculation is that it's relatively easy, which is not to say that it's easy, to train people to do that
personally, i'd need more than what my state would pay a starting teacher, to "teach people how to think"
but i might accept that hourly rate to teach them the quadratic formula as a black box
and so it doesn't mystify me when my state's free public education gives me one thing and not the other
it definitely sucks, but it doesn't astound me
which is an interesting side point, there's value in memorizing, and mechanically executing rules, and if people were better at that, the world would probably be a better place
i think i got 2x2 matrices at one point in my high school education
not long enough to think of matrices as first class objects in a world that i was a part of, but long enough
for example, if you click on my profile, you will have 11.9 k rep points that's not on math.stackexchange, but on all communities in total
on math.stackexchange I have 10.9 k rep points
which I would have more if it counted the communities I joined and wasn't active in, so I suppose the 100 or so starting points aren't counted into that
@SineoftheTime, I've tried to follow your suggestion and this is what I got.
Problem: If $d'(t) \geq -a d(t)$ then $d(t) \geq d(0) e^{-at}$ where $a\in\mathbb{R}^+$. Proof: let $y(t)=e^{-at}$ and taking the time derivative of $y(t)$, we get $y'(t)=-a e^{-a t}=-ay(t)$. It is evident that $y(0)=e^{0}=1$ and $y(t)>0$. Now we compute this $$ \begin{align} \left( \frac{d(t)}{y(t)} \right)' &= \frac{d'y-dy'}{y^2}, \\ &\geq \frac{(-ad)y-d(-ay)}{y^2}, \\ &\geq \frac{-ady+ady}{y^2}, \\ &= 0. \end{align}
@CroCo denote by $f(t)$ the solution in the case you have equality, i.e. $\dot f=-af(t)$, then $f(t)=e^{-at}$. Now $\frac d{dt}\left(\frac{d(t)}{f(t)}\right)=\frac d{dt}(d(t)e^{at})=e^{at}(\dot d+ad(t))\ge 0 $
@CroCo looks good but since I'm lazy I'd have used the product rule, since $d/y=de^{at}$
@SineoftheTime I calculated the Hessian at the points and I find that at the points $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2})$ they are relative minima, instead at $(0,0)$ the Hessian is zero so (as the professor said) I just have to say that I can't say anything
@leslietownes ok, maybe my writing reveals my confusion :) maybe it is better to say $X\geq 0$ a.s. means simply $P(X<0)=0$, but which in turn means that $P(X=0)=x$ and $P(X>0)=y$ such that $x+y=1,0\leq x,y\leq 1$.
so for example if the set on which X = 0 had measure 1/2 and the set on which X > 0 had measure 1/2 it would be true that X >= 0 holds a.s. despite neither X = 0 nor X > 0 holding a.s.
psie it might help to rephrase that a little. P(X < 0) + P(X = 0) + P(X > 0) = 1 because the events partition the space and P( ) is a probability measure. so that'll always happen independent of any hypothesis on X, and then if you were to know that X >= 0 a.s. that would tell you that P(X < 0) = 0 so the other two terms add to 1
and you choosing to call them x and y for some reason is up to you :)
I have silly question. Earlier today I asked about the implication "$f>0$ a.e. $\implies\int f>0$", which is a consequence of "$f=0$ a.e. $\iff \int f=0$". I'm in a situation where I want to prove $$\text{"}X\geq0 \text{ a.s.}\implies \mathrm {E}[X\mid\mathcal G]\geq 0\text{ a.e."}.$$ The proof proceeds by contradiction, setting $G = \{{\rm E}[X \mid \mathcal{G}] < 0 \}$ and assuming $P(G)>0$.
Now, at one point in the proof I have to use the fact that a negative function has a negative integral, but the above equivalence from Folland's book applies to nonnegative functions only, and he has developed the whole theory using nonnegative functions. I'm not sure how to go about proving what I want for nonpositive functions.
What I want is: "$f<0$ a.e. $\implies \int f<0$".
I feel like I have to go back two section, rewrite the definition of the integral of a simple function, rewrite the definition of the integral of a nonpositive function, etc.
any of this kind of thing could be derived from first principles and there's an issue of, why would you do it that way if you had some other way
if you had the other thing as a black box just apply it to g := -f which is positive a.e. and the desired thing about the real number int g implies the desired thing about int f
@leslietownes that's clever :) so if $g$ is negative and we know "$f>0$ a.e. $\implies\int f>0$" holds for nonnegative (measurable) functions $f$, then we apply the implication to $h=-g$ and we get $-\int g>0\iff \int g<0$. Thank you!
When I use the command urldate in LaTex, it requires for me to fill in the year, the month and the day, so YYYY-MM-DD. This format makes definitely sense to me; first comes the year, then the month of course and then finally the day. Totally logical. However, when I compile, it displays it as MM/DD/YYYY, which I cannot fathom. Does it make sense to you?
... ... ... that is obviously not the best date format. my brain is fried
@psie I don't know anything about the origins of the biber citation format, but I could imagine that, as so many other things, was invented by some person in their basement many many years ago, and that they decided on one date format for parsing over any other for reasons only they knew
the output format on the other hand is... not standard american, but certainly standard something
