@Mike, I do think it's terse. I think the exercises are absolutely its strong point. ... Oddly enough, someone on Amazon complained that my Multivariable Math book is too "talky" or "wordy," when almost everyone else complains that I'm too terse.
I asked my TA if he could come up with an explanation, @Studentmath. I'll let you know if he tells me one.
Also, @Studentmath, I realized I was an idiot ... Ross doesn't emphasize indicator functions nearly enough. For the exercise of $X=$number of $1$'s in $n$ rolls, $Y=$number of $2$'s in $n$ rolls, he asked for the covariance. I told my students to compute $E[XY]$ by conditioning. But the problem is very easy with indicator functions. And I didn't think of it before I wrote the problem set.
At UCLA this is also the first proof-based class. I'm mostly excited because I hear the upper division students are much more motivated than my calculus class...
@TedShifrin I think you would have to work really, really hard to write a book on fundamental groups that's 2/3 as long as Hatcher... or a book on singular homology and cohomology...
Or a book on the amount of homotopy theory he does.
@Ted I was lucky to have a great explanationary book attached, made by my own university, which I have lost. It had tons of exercises solved in it via indicator functions, and many exercises given generally had to be solved that way.
@Ted They are my first-go-to method when it comes to a bit complex expectation/variance problems, I think they will get how useful they are just with a couple of questions directing them to use that indeed
I wonder why Ross doesn't emphasizes them. He does leave a remark here and there, but not more
torus is understandable. just take a 0-dim cell complex and then form a 1-dim cell complex by making up the fundamental domain and then go toa 2-dim cell complex by thinking of the previous fundamental domain as the boundary of some open set and identify
@Ted actually, if they know how to solve things the harder way, they might just appreciate them more now, if they do keep doing probability they will obviously get used to them one way or another
@BalarkaSen You attach the boundary of the n-cell by the map $S^{n-1} \to \{*\}$. What this does is identify every point on the boundary to a single point, i.e., collapsing it to a point.
OK, now go write down the actual cell structure we just talked about. How many cells are there in various dimensions, and what are the attaching maps to the skeleta of lower dimension? (i.e., what are the attaching maps of the 2-cells to the circle?)
When I was learning this I only got to talk to a professor about it twice a week. This gave me plenty of time to work out examples, like why the sphere is a cell complex, before asking... Maybe I should only talk to you twice a week :)
I solved the case for the non-homogenous constant coefficients case and I wondered if there is a way to find a general solution to a non-constant coefficient case. I don't know how to approach this at all, the substitution $y(x)=\frac{\log (v(x))}{r(x)}$ gets problematic immediately.
Hey @amWhy!!! I want to show that the algebraic set $V$ of $K^n$ is irreducible iff $I(V)$ is a prime ideal. That's what I have tried:
We suppose that $I(V)$ is a prime ideal and $V$ is reducible. That means that there are two non-empty subsets $V_1,V_2$ of $V$ such that $V=V_1 \cup V_2$. Then, $I(V)=I(V_1 \cup V_2)=I(V_1) \cap I(V_2)$. We have that $V_1 \subset V \Rightarrow I(V) \subset I(V_1)$ and $V_2 \subset V \Rightarrow I(V) \subset I(V_2)$ How can we find a contradiction?
Hello guys. A moderator on a math forum pointed out errors in some stuff. But I can't quite see what he means, and he's not going to answer me anymore, probably. First, he said that both @Balarka's and my proof of this lemma :
If $\displaystyle \lim_{x \to \infty} \prod_{n\le x} f(n)$ and $g(m) \sim h(m)$, then $$\displaystyle \lim_{m\to \infty} \frac{\prod_{n\le g(m)} f(n)}{\prod_{n<h(m)} f(n)}=1 $$
are wrong, and retain the same error. More precisely, he said the lemma itself is incorrect, namely I need more conditions to reache the desired conclusion.
(ok, apparently that linky thing doesn't work here)
@anon Perfect, thanks. So you say fixing that last product, and explaining it better is enough for a valid proof? Any idea as to what that moderator referred to? In the other paper, probably there is an error when I say "implies" (damn professor of mine, wondering why I still tell him anything), he said there is an easy counterexample to it, but I haven't really tried to figure it out as I had the other one.
@Committingtoachallenge It is the second lemma of the paper.
@anon Well, let's say it isn't. In fact, the actual thing I care about later on in the proof, is like $ a(n) \prod f(n) $, where $\prod f(n)$ alone is equal to $0$, but multiplied by that infinite function it becomes non-zero, so I think there's no problem
@PedroTamaroff @KajHansen Yes, because $x \in A \rightarrow x \in K \Rightarrow x \in A \cup B \rightarrow x \in K$, so $A \cup B \subset K$, right? :)
Suppose you want to count the number of permutations with cycle type $(ab)(cd)$ in $S_5$.
First step: pick a $4$ subset of $\{1,2,3,4,5\}$. This can be done in $\binom 54=5$ ways.
Next, pick a $2$ subset of $4$, this can be done in $\binom 4 2=6$ ways, and this determines the pair $(ab)$, which are the two elements you chose, and also the other pair $(cd)$, which is the pair you didn't choose.
Finally, you need to count repetitions: inside the parentheses, $(ab)=(ba)$, and $(ab)(cd)=(cd)(ab)$ since the cycles are disjoint.
