with one I'm on, I'll pretty much have to taper if I'm to function while getting off of it. (if I miss a dose in the morning I notice it by the afternoon, and I'll sleep awful in the evening)
He's a postdoc who is piggybacking off some other instructor who also does not know what he is doing. I felt bad for him today though. He was explaining mod arithmetic and somebody critisized how he taught it, and laughed at him..
These days in low level classes, most homeworks are on computer and one gets instantaneous feedback. Great for computation. Bad for writing mathematics.
Nate: Maybe you could get together with other students and politely request a different policy. Ask the undergraduate coordinator/undergraduate department head, if necessary.
@TedShifrin That's fair, I remember seeing the course outline or one of the Harvard math courses that had a take-home final which made me think if it was done regularly
Yeah I feel bad for the postdoc. I think he knows a lot of mathematics and probably would be considered like many of you here who has a real love for math. But he has to do things based on what the other instructor is doing and seems to be trying his best. And that's really all that matters.
I'm pretty sure the main difference between the 8th and 7th edition of Stewarts text is that the colors are different and the exercises have been rearranged (but not changed)
Yes, for sure. But it seems that mathematics teachers here really want to save students money, my next semester math teacher (tyana something..) assigned readings from a textbook freely available online from the UWO library!.
I know I sound like an author when I say this, but in truth book prices are ridiculous, but for the most part they've inflated like everything else (college tuition, automobiles, housing, etc.) ... at least in the US.
He was talking about how $f(xy)=f(x)+f(y)$ (or something similar, suspend your disbelief) and someone argued that the equation was pointless and my teacher said that he thought the equation was nice and had to add it to his box of tools.
@TedShifrin I've finally accepted that there's no hope of an exact solution to the integral curve problem in the general case. I've tried way too many methods of playing with the equation, but it always comes down to an unsimplifiable diff eq. Can you think of which numeric approximation methods would be best for approximating $\frac{dp}{dt} = F(p(t))$?
I feel like I'm doing things backwards, though. Originally, I wanted an exact form of $p(t)$ so that I could perform optimization on it. Now, I'm numerically approximating it, so that I can optimize on the approximate, and then hope that the optimum I find is approximate of the true optimum.
If you want to take a few minutes to TeX up a cogent statement, I'll happily send it to one of my smart ex-students who's writing his Ph.D. on numerical PDE.
I'll see if I can write something up. The benefit of having a one-parameter closed form is that it's efficient to calculate, which makes solving minimization problems computationally easy. Without that, any additional overhead we incur trying to approximate the one-parameter closed form must remain less computationally expensive than gradient descent OR it must be more precise/accurate given the same amount of computation spent (if we ran gradient descent for as long, which is better?).
@TedShifrin This will take me some time to formalize, but that's the gist of the problem.
But if we identify $X_{\alpha} = X^*$ as the author says, then the definition is circular since $X^*$ would have to inherit the subspace topology from the disjoint union
And if we don't, then $U \cap X_{\alpha}$ is always disjoint
So every subset of the disjoint union would be open in that case
I'm trying to integrate (x^2-3x+6)/(x^3+8x) via partial fraction decompositualization, But, I end up with A(x^2+8)+Bx^2+Cx = (x^2-3x+6)/(x^3+8x) and I'm not sure its possible to solve...
I remember having a nice time figuring out why the Hawaiian earring philosophically and mathematically should not be the wedge of countable many circles.
@TedShifrin Disjoint union of circles with the origins identified is exactly the same as the countable wedge of circles, isn't it?
The reason if you remove the origin from earring you don't get the disjoint union of intervals is because the Hawaiian earring is the one-point compactification of the topologists' comb.
