@KannappanSampath I don't know why the OP didn't upvote the answers that they accepted. It it is the answer they are looking for, one would think that it is useful. I have upvoted two of them (because they deserve it), but the OP should also (or they seem to be saying that their question is worthless).
@robjohn About your "shameless plug" don't you need to know in which direction the approximation is? :-).
I'm often quite confused with these approximations if one wants to compute limits, if it is a lower bound and that goes to zero, we know nothing, right?
@JonasTeuwen The shameless plug was referring to the reference in that answer to the other answer (about $\frac{\sin(x)}{x}$). However, to answer your question, since $\sqrt{1-x}\le1$, we have that $\frac{x}{1+\sqrt{1-x}}\ge\frac{x}{2}$.
No. Why do you get that each time a zero appears that the subsequent terms are a multiple of the sequence starting at zero? In particular, why is it $3$ if you do it mod $5$? (+1 btw)
I had thought I would have to wait until midnight because the number in the rep audit only counts entire days with \ge 200, but the badge arrived about 10 minutes after the qualifying upvote.
@HenningMakholm It takes 50 times at 200 points, and I capped earlier this morning taking me to 16 days at 200+ points (out of the 50 needed for Epic).
So out of the 100 combinations of one digit and the next, 60 belong to one large cycle. Then there are 20 in the cycle staring at (0,2), and 3 in the cycle starting at (0,5), and 1 in the cycle starting at (0,0). What are the last 16 pairs?
@robjohn I'm still pondering your Fibonacci answer. How do I "see" that the period mod $2 \cdot 5$ is $LCM(20, 3)$? It reminds me of cyclic groups but I somehow can't make the step in this case for some reason... (Maybe just because I'm very tired.)
@ymar I've been meaning to tell you this: How about you write something nice about yourself on your profile? How you assess your own skill doesn't matter on this site, so I would not write it.
@ymar Well maybe off-putting is the wrong word. But it bothers me if people write bad things about themselves. Why don't you write something nice about yourself instead? Or put a funny quotation or some ascii art or anything else you can think of.
@MattN Well, I don't think nice things about myself really. But I'm not attached to what I've written, so I'll delete it so it doesn't bother other users.
@MattN Perhaps, but I don't really mind who I am. I just know I'm a bad student because I know how much I know and how long it takes me understand stuff. It is bad that a fourth-year student doesn't know how to integrate.
@MattN I'm not sure. I think it's good to evaluate your position in whatever situation. Iff it doesn't make you bitter that is. I don't think I'm bitter.
I realize this might be a bit of a loaded question, but what does it mean to say that "a topology on a set $X$ is the same as the topology on a set $Y$"?
@ymar Unfortunately, in my annoyingly slow paced intro topology class, we haven't yet discussed homeomorhisms yet. I realize the question is vague, but how does one determine that topologies are "the same" ?
@ymar So, with the limited tools I have available (mostly set theoretical tools). I need to show a 2-way containment? An open set in one topological space is contained in the other, and vice-versa?
@DavidK We have two topological spaces $(\mathbb R^2,T_1)$ and $(\mathbb R^2,T_2)$. You have two show that $T_1=T_2,$ that is, indeed, that if a set $A\subseteq\mathbb R^2$ is in $T_1$ then it is in $T_2$. And the other way around. If a set $A\subseteq\mathbb R^2$ is in $T_2$ then it is in $T_1$.
What about a set $U\in (X,T_1)$ where $U=\{(0,0),(1,0),(0,1)\}$ where $(0,0),(1,0),(0,1)$ are just points of $\mathbb{R}^{2}$. I can't seem to 'construct' this set from two sets in $PP\mathbb{R}_0\times PP\mathbb{R}_0$.
Well really I mean from two sets in $PP\mathbb{R}_0$
@ymar But can't they also be cartesian products of finite sets that all contain the real number 0? Like your example above?
@ymar This was my original thinking on the problem. How could one obtain an open ball in $\mathbb{R}^2$ from the product of two sets in $\mathbb{R}_0$? Didn't seem possible, but...
@DavidK The containment $T_2\subseteq T_1$ is clear. If a set is a sum of "rectangles" with one vertex in $(0,0)$, it must contain $(0,0)$. The other containment is interesting. We can take any set that contains $(0,0)$ now. We need to find one that isn't a sum of such rectangles. You said an open ball. Did you mean with the center in the origin? This one is also in $T_2$ (any one that has $(0,0)$ in it is for that matter). Can you see why?
@ymar Yes I meant any open ball that contains the origin, not necessarily centered at the origin. And yes, I can see that such a ball is open in $T_2$. These rectangles don't necessarily need to have a vertex at $(0,0)$ right? Just a sum of rectangles that "together" contains the point $(0,0)$?
@DavidK No, they don't have to have $(0,0)$ as a vertex, but these are enough.
@DavidK Note that nothing bad happens when the rectangles intersect non-trivially
@DavidK Oh, but no, it's not enough that the sum of rectangles contains the origin.
The basis of $T_2$ is the set of rectangles which are Cartesian products of sets containing $0$. As such, each of those rectangles must contain the origin.
@ymar Right, so any 'rectangle' union the origin is also open. Or the Cartesian product of any sets which both contain $0$. So the rectangles could be 'dis'-connected from the origin but still contain it. Loosely speaking.
@robjohn I've looked at the Chinese remainder theorem now. I finally understand. I'm quite glad that it uses a theorem and is not just obvious (since it was not obvious to me). Glad I read your answer, thanks for pinging me with it.
@robjohn One can also see that the period of the Fibonacci sequence modulo n is always the longest possible period of any cycle that satisfies the recurrence mod n. (And indeed every period divides the period of the Fibonacci sequence). Namely, starting from any (a,b) what we get is the sum of a times the (1,0)-sequence plus b times the (0,1) sequence.
@ymar Yes I am. I'm trying to construct that set by unions and intersections of cartesian products. But not having much success, and I think I see why...
Well that set is clearly open in $PP\mathbb{R}_{(0,0)}^{2}$. But I can't seem to construct that set from unions and intersections of open sets in $PP\mathbb{R}_{0}^{2}$.
Your answers are on harder questions. Those gather fewer votes.
What can I say? I'm just shameless about picking all of the low-hanging-fruit that comes around when I'm out to get a badge. But I'll slow down a bit from now, I think.
A few random facts before I declare the bragfest over: (a) on the second day I ever posted on MSE, 280 rep were eaten by cap gnomes. (b) running total of rep eaten so far: 3810.
Well, 1e-15 is effectively zero compared to 1e3 (in the exact sense that when you add the two numbers in double precision you get the large one back unchanged).
@JonasTeuwen I was computing $\operatorname{li}(x)$ for this answer. There was no request for more information, so I never posted the long integrations.
Yeah, I just figured that part out. Now I have to finish writing my notes. This gives me the proof of $A$ is a W-group if and only if $\mathrm{Ext}^1(A,\mathbb Z)=0$.
