@TobiasKildetoft However just to know that I'm at the very beginning of mathematical activity, self-educated, I have hundreds, thousands of new results in mathematics.
@AlpArslan No, taking sum of duals of each graded components preserves the dimension as being countable. If you just took the dual, you would get something of uncountable dimension
(think about it this way: if you have a circle and a circle of a smaller radius inside it in R^2, you can never perturb the smaller one without intersecting with the larger to pull it outside the larger one. But in R^3, you can move it up in that extra dimension to do that)
here's a hard question for you, to which I don't know a good answer
suppose someone gives you $y(t)$, $x(t)$ which are each power series in $t$; for convenience, take them both to have leading term $t$.
then $y(t)-x(t)$ is a power series in $t$ with leading term $~t^2$, which can in turn be removed by subtracting off an appropriate multiple of $x(t)^2$
@Semiclassical I'll think over it and if I find a nice way of putting that I'll let you know. This makes me think it might be connected to the Faà di Bruno's formula.
Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives, named after Francesco Faà di Bruno (1855, 1857), though he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast stated the formula in a calculus textbook, considered the first published reference on the subject.
Perhaps the most well-known form of Faà di Bruno's formula says that
d
n
...
I did some dimensional analysis in my head. It's essentially half of the circumference * radius. I imagine a semicircular length and the radius is the distance to the center from any point on the length.
for a sector of a circle there's no circumference tho
@SemiC how do they construct the area of a sector of a circle? Suppose I have a quarter of a circle. What is its area and can you intuitively tell me why it is so
@Mike so the book says because $T^2$ is $K(\mathbb{Z}^2,1)$, there is a bijective correspondence between homotopy classes of based maps $T^2 \to T^2$ and homomorphisms $\mathbb{Z}^2 \to \mathbb{Z}^2$. is there a theorem for this or should this be obvious
like I understand arc length is the infinite sum of lengths of line segments and it eventually breaks down to right triangles. What are the parts being summed here? I'm so bad with polar coordinates i dont get it @semiC
This needs proving; it's done in chapter 1 of Hatcher. The idea is "Realise every homomorphism by doing it cellwise and showinf you can extend cell-by-cell; same for showing maps inducing the same homomorphism are homotopic
@BalarkaSen Oh, but that's honestly obvious. Just make the homotopy transverse to the other submanifold and take preimage of the intersection. That's a 1-cobordism between the two intersections.
Well, one needs the transversality extension theorem, not just the transversality theorem
@Semiclassical i wouldn't quite say i feel down. i feel that not being able to think/having a dull day is more of a struggle than being depressed, because that's partially how i try not to be depressed.
@Semiclassical @BalarkaSen if you feel depressed with all the sources you have around, then what can I say about myself that I had to do math for a long period of time with very very low resurces?
Eventually, this fact brought me in this point. So never complain too much, seemingly bad situation can be full of rich positive effects over time.
"How would I know the right word for what I want? How would I know that actually I don't want what I want? Or that I actually don't want what I don't want? They are elusive things: the moment we name them, their meaning disappears, melts, dissolves like a jellyfish in the sun. My conscience wants vegetarianism to win over the world. And my subconscious is yearning for a piece of juicy meat."
I see, mr eyeglasses. Well, do a good job so that you can get a good rec and ultimately look for something you might prefer. I know a lot of math people go into business-type consulting, but I doubt you'd like that much, either.
A number of my former advisees/students have done it with just BS, mr eyeglasses. Computer skills and writing skills are, however, both somewhat important.
Then how to feel yourself depressed when you get the math results you appreciate daily? @Semiclassical My math doesn't allow me to feel depressed, you know, I don't have time for such a state.
I think that might be right, but they're still alive and kicking, @Semiclassic. I'm not saying I'd want to do it myself, but there are lots of different such entities.
I've got a repeat of a previous notation question that I've asked here before. I still don't 100% get the notation on $\mathbb Z/2\mathbb Z$. I understand what this set is (and that it's more than just a set), but by what mechanism does the module produce the result? I'm familiar with the modulo in terms of equivalence relations, but $2\mathbb Z$ isn't a relation, right?
@TedShifrin I believe I've received that explanation before, but I was always wondering why we set them to 0, it almost felt like an arbitrary decision. $A/B$ becomes set all of $B$ to the $0$ element.
But, isn't the accepted notion of $\mathbb Z/3 \mathbb Z$ would be to partition membership into $3 \mathbb Z$, $3 \mathbb Z + 1$, $3 \mathbb Z + 2$? By either original explanation, we should just be making all multiples of $3$ into $0$.
Of course not, @Axoren. When I used to teach this, I would use colored chalk and color code the different equivalence classes. And then draw a picture of $\Bbb Z/3\Bbb Z$ with three elements, one dot for each color.
@TedShifrin If $X$ bounds $M$, make $M$ transverse to $Z$ by a homotopy with boundary also transverse to $Z$. Intersecting that perturbed copy of $M$ with $Z$ gives a manifold with the same # of bd points as $X \cap Z$ mod 2.
@Semiclassical one thing because I noticed that: some prefer to say that they don't listen to me, but actually read all I say, and it's simple: how to answer my questions? Ted? No way.
@Semiclassical but I don't care what and who does here. I simply don't care. If someone is curious in some math I talk about it's OK. Otherwise, it's OK again.
When I taught applied math (back in 1986) I had fun teaching some of the asymptotics stuff (stationary phase, in particular). I thought Bender/Orszag had some great material.
@Semiclassical Do you you how to calculate, say, $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots -\frac{1}{n^2}\right)^2$$