@hhh: That doesn't sound like a problem I'd like to help with (too much thinking, too little interest), but you might as well try and nab someone here.
@anon No, I was agreeing that the scope of $y$ was in the integral. The $y$ outside the integral is not a dummy variable, unlike the one inside the integral.
I figured out something potentially of interest about MSE. If you click the star to favorite a question, then follow a link, then backspace, the star will be open for clicking again. If you do click it, it will tell you it favorited again. However if you refresh it turns out you unfavorited it.
@robjohn : The motivation for that question (jump discontinuities for $f:\mathbb{R}^2\to\mathbb{R}$) is that I want to generalize this class of functions to functions of the form $f:\mathbb{R}^2\to\mathbb{R}$....I want to move in that direction!
@robjohn: The bug doesn't appear on FF; when you backspace it reloaded the page with the star yellow'd. I think it has something to do with cacheing settings (in addition to the SE system not knowing the difference between starring and unstarring from the browser alone.)
Same with IE. It's all about the browser settings.
Bwahaha.
Bill did actually say "Aw Snap!," even if he was only quoting the error.
@anon wlog suppose that $t_0\gt 0$. Suppose that $x^2-t^2\lt 0$. Then $x$ is decreasing when $x$ is negative, but why is the case that $x(t)\lt 0$ for $t\gt t_0$?
@leo: Suppose we make a diagram with an x- and t-axis. At each point we will put an arrow pointing either upwards or downwards depending on whether $x(x^2-t^2)$ is positive or negative. This is an intuitive picture of why $|x|<|t|$ has to hold true given $|x_0|<|t_0|$. Now we have to translate it into a proof...
@anon thanks. I'll try. The next exercise is like this one but with $x'=txf(t,x)$ $f$ continuous and non-negative and some other things... Thanks by your help
for a function $f:\mathbb{R}^2 \to \mathbb{R}$ having a jump discontinuity like the one we have discussed earlier, at a point $P$, the size of the jump at $P$ is given as $fJ_P : [0,pi) \to \mathbb{R}$, which means the size of a jump at a point is a function.
...Now my question is "Are there any restrictions on the behaviour of the function $fJ_P$, meaning should it always be smooth or differentiable or continuous are can it be nowhere continuous and stuff like that because we need $f$ to be smooth everywhere except at $P$. @robjohn
btw, we can construct functions $f,g$ defined on a bounded interval $I$ so that $f\leq g$ but $f'(t)\gt g'(t)$ for some $t\in I$. Is there an example with a behaviour like this but over all $\mathbb R$?
@anon Number theory is really nice. Simple statements, too hard proofs. I don't have read much about it, just simple stuff, but I hope take a course on it the next semester or the next year
Yeah, it's not a homework problem, rather a statement I am using a slightly harder problem. I was just reviewing my solution, and it struck me as strange, just like when you repeat the word "golf" to yourself too many times.
*I am using in a proof for a slightly harder problem
@Ilya Oh, don't feel any need to look. I think I need to work some more on the n-dimensional case. I was doing the last part while on my walk this evening. Joriki brings up something that I was worrying about, but forgot about when I was out.
@MattN looks like it, eh? :) Seriously: it was upvoted during the night and I had a look and found the "point you to ... pointers to ..."-sentence distinctly awkward.
reading a book where such notation is used. They define for a class $\mathcal P$ of sets $\mathcal P_\delta$ to be the class of sets obtained from $\mathcal P$ by taking countable intersections
@tb: you know, I feel better with minimal-sort-of-definition of $\sigma(\mathcal P)$ than with $\mathcal P_\sigma$ having doubts if the latter is formal :)
The latter is formal enough, I think: they are very useful and the minimal number of $\sigma$'s and $\delta$'s you need to add gives a good indication of "how complicated" a set is.
@anon there's probably more truth to that than one might imagine... However, the reflex: "I want to figure this out on my own" (for whatever reason) is probably a healthy one.
Well, since $T^{k-1} \neq 0$ there is $v$ such that $T^{k-1}v \neq 0$. Then $v, Tv,\ldots,T^{k-1}v$ are linearly independent but only one of them is an eigenvector, namely $T^{k-1}v$.
(and the subspace spanned by $v,\dots,T^{k-2}v$ does not contain an eigenvector).
@tb Third time you showed me this. But, I would like to rephrase my question: How did you learn linear algebra then? (I think, I am not at ease with this as I am with when thinking about groups or metric spaces or some such thing.)
@KannappanSampath I had a basic course which covered enough of the fundamentals for me to figure out whatever else I needed. I only really understood what's going on when learning functional analysis.
@ZhenLin I decidedly disagree. How do you solve ODE's then? This alone is important enough to justify one month of suffering in every basic linear algebra course.
Here's one weak form of JNF I saw a couple weeks ago: let $T$ be a non-invertible endomorphism of a finite-dimensional vector space $V$. Then there are subspaces $U$ and $W$ such that $V = U \oplus W$, $T$ acts nilpotently on $U$, and $T$ acts as an automorphism of $W$.
I imagine what you should do is put all of the needed definitions in the article as a prerequisite section, create red links associated to them so that hardcore wikipedians will fill up more articles and strengthen it
I say "anyway" to segue into a topic slightly different from what I just said previously. It's poor form to use the same form of segue, though - like saying "anyway" over and over again - but I do it when I don't feel like expending the mental energy to look for a different word from last time.
@ZhenLin Funny that you said that... Jordan decompositions, being rather messy to compute numerically, aren't terribly useful in the applied setting either.
By defective, I remember you mean that there are less linearly independent eigen vectors (geometric multiplicity) than the algebraic multiplicity, am I right? @JM
The diagonal elements could be scalars, or Jordan blocks. If the block diagonal matrix is actually diagonal, you have the diagonalizable case. If you have actual Jordan blocks present, then you have a defective/nondiagonalizable matrix...
@tb: whenever you're here, just wanted to clarify. Any $G_\delta$ subset of a Hilbert cube is top. complete. So, that means that although $(0,1)$ is not a complete space under the usual metric, we just need to treat it as a metrizable space rather than a metric space - so we can choose a consistent metric (different from the usual one) which makes $(0,1)$ complete, right?
I used $(0,1)$ for the sake of simplicity, but I guess that to be formal I have to take a cylinder $(0,1)\times \prod_{k=1}^\infty[ 0,1]$
@Kan: I've just written a question, so if you have any ideas - please tell me.
