@TedShifrin Yeah, but it's easy though. $f : \Bbb R^n \to \Bbb R^n$ be cont. and define $x_k = f(x_{k-1})$, where $x_0$ is arbitrary. If $x_k$ converges to $a$, then $f(a) = a$. You just have to use that a sequence can't converge to two different things at once.
@Anush I originally thought that the matrix had elements from $[0,1]$. The braces were too flat. I noticed the braces on the vectors because you mentioned the probability of $2^{-n}$. After I wrote my answer, I looked back at the question and noticed that the elements of the matrix were also from $\{0,1\}$
@TedShifrin odd... SD is usually more overcast than LA, being against the ocean whereas LA is surrounded by mountains. Perhaps the mountains are holding the marine layer in.
@Anush I don't know if it has something to do with the newer MathJax, or with the fonts on my new laptop, but the braces are much harder to see as braces.
@Anush well, as I mentioned, I noticed the braces for the vectors so I knew there was a $2^{-n}$ probability arising from the vectors being equal, but thinking that the matrices had real entries, I thought that the probability that they had a non-zero null vector in $\{-1,0,1\}^n$ was $0$.
Well this is a weird question to ask here but since I have to give a talk on photo chemical smog I am asking anyways: Can we say stuff about smog by thinking on the lines of Brownian motion?
@MikeMiller Actually, I take back what I said. My space is homotopy equivalent to $K(G, n) \wedge K(G, m)$, the reason I wrote down above works just fine. But yeah, to prove that it's $(m+n-1)$-connected, the hint you gave is needed : $K(G, n) \wedge K(G, m)$ has no cells below $m+n$ dimensions. Thanks!
@robjohn I am sorry :/ I did google about FET and was looking at the wikipedia page, couldnt find the answers....so I asked you and I went to checking my other answers :/
first sentence: >The field-effect transistor (FET) is a transistor that uses an electric field to control the shape and hence the electrical conductivity of a channel of one type of charge carrier in a semiconductor material.
first sentence: >In an n-channel depletion-mode device, a negative gate-to-source voltage causes a depletion region to expand in width and encroach on the channel from the sides, narrowing the channel.
@TheArtist Later in life I finalized a university after dropping more, I know what you mean. I often found many of the studies a huge loss of time, and I was right. I was asked to write things on papers like 3-4 hours ceaselessly.
It was not for me this kind of stuff, and I felt often that the whole education system was such a big sh*t, nothing clever that kept me up.
Hours at uni should be like watching your favourite movies, maybe Star Trek? Well, yes, to make you return there, not just for the sake of finishing a school.
After hours one has to say "WOW, I'm looking forward the next hour!" As a professor, to come to hours and talk about your reseach, make kids curious about things you study, not just to talk about stuff known for hundreds years.
@TheArtist financial accounting - I felt attracted to it and I liked the way stuff was taught. Most of the time I was learning alone. I don't enjoy to stay in a room with many people, I don't feel comfortable.
I like very much to learn alone, without anyone around me.
@TheArtist Only If Euler and Ramanujan would admit that about me. I never think of the idea of being a mathematics, I simply do mathematics, and I do my best to do it like the giants previously mentioned. ;)
It's about the attitude towards mathematics, about the way to go, not about the final destination "I'm a mathematician now".
I like the way to go, and I enjoy so much to gather results, not for the sake of gathering results but often they show you how incredibly amazing mathematics is, an immense beauty that tends to $\infty$ (not that this makes much sense, but it is what I feel, huge amounts of pleasure that overwhelms you again and again)
I don't see anything so amazing like mathematics ... (that's true). It's bad many people will miss the huge pleasure given by mathematics for some reasons.
@Ramanewbie Rearrange to get $x^2 - 4 \leq 0$. Factorize to get $(x - 2)(x + 2) \leq 0$. Factor of two things negative $\implies$ one of the factors is positive and the other is negative. That gives you two cases : (i) $x \leq 2, x \geq - 2$ (ii) $x \leq -2, x \geq 2$. (ii) is of course impossible. Hence, $x \in [-2, 2]$.
@Chris'ssistheartist The reason why I think its 0 is because the $\lim_{n\to \infty} 1/{n^{\frac{1}{n}}}=1$ which makes the Integral limits from $1$ to $1$...
@BalarkaSen In fact I managed to get $(x - 2)(x + 2) \leq 0$ But then is there a formula to solve any $(x+n)(x+m)\le 0$ inequation, with n and m naturals ?
Let $s_n$ be a sequence defined recursively by $s_0 = \sqrt{x^2 - x}$ and $s_{n+1} = \sqrt{(x^2 - x) + s_n}$ for $n \in \mathbb{N}$, for some $x \ge 1$. Let $L_x = \lim_{n \to \infty} (2x)^n (x - s_n)$. Then, prove or disprove,
@TheArtist When $n$ is large enough you have $$\frac{1}{(x+y) (1+x y)}\approx \frac{1}{4}$$
So, all reduces to the calculation of $$\frac{1}{4}\lim_{n\to \infty} \left(\frac{n}{\log(n)}\right)^2 \int _{\large 1/n^{1/n}}^1\int _{\large 1/n^{1/n}}^1 \ dx \ dy$$
@TheArtist this is just a disguised well well known limit, so it remains to uncover it. I let you the pleasure.
It's in front of you, just focus on it a bit.
Note that $$\frac{n}{\log(n)}= \frac{1}{\log(n^{1/n})},$$ and I don't say anything more. :-)
@MikeMiller Is the $H$-space structure $K(G, n)$ obtains from realizing it as $\Omega K(G, n +1)$ the same as the $H$-space structure coming from the group multiplication on $G$?
@DanielFischer A Theorem I am studying states the following: Suppose $a \in \mathcal{A} := $ unital Banach Algebra and $\sigma(A) \subset \Omega$ open, then $\exists \delta > 0$ such that $\| x - a \| < \delta \implies \sigma(x) \subset \Omega$. Could you interpret this result as the spectrum can't suddenly expand but it can possibly suddenly contract?
@Moses You can't deduce that the spectrum can suddenly "contract" from that. That theorem only tells you the spectrum can't suddenly expand (or jump around).
