We only did problems where we showed that there was some direction where the limit wasn't the same, thus the function doesn't have a limit at that point..we never had to prove that the limit value was something
@Cristopher Recall that $(a+b)^2\geqslant 4ab$ (since $(a-b)^2\geqslant 0$), so that $$\frac{x^4y^4}{(x^2+y^3)^2}\leqslant \frac{x^4y^4}{4x^2y^3}=\frac 1 4x^2y$$
To show that $\mathcal{B}[a,b]$ is a vector space, it'd need to contain an additive identity I think, @Clarinetist. $f(x) = 0$ for all $x \in [a,b]$ is of course an additive identity of the set so we can say for sure that $\mathcal{B}[a,b]$ is a vector space.
that sounds like fun. can you be more specific about what you mean by "how it plays with certain classes of graphs" or are you not comfortable with that?
so the idea is something like that, even though the problem in general is absurdly hard, when you have extra structure you should be able to make a better algorithm?
@TedShifrin John Lee's textbooks all have exercises. And nearly all my graduate physics textbooks have exercises. And so do the economics books I have. Are mathematicians who write grad textbooks exercise-averse?
it's already been determined that the decision problem "$G$ has metric dimension $n$" is NP-hard for outerplanar graphs $G$, but we think we can get the strong dimension in polynomial time which would be pretty rad
the proof isn't so bad. you basically show the strong dimension problem is reducible to the minimum vertex cover problem
and this is doable by coming up with a graph that has the same vertices but different edges, and on which a cover will be a strong resolving set for the original
although somehow the way they proved metric dimension was np-hard on outerplanar graphs was to reduce it to the boolean satisfiability problem, which you can look up if you haven't heard of it and immediately notice that it has nothing to do with metric dimension
i think the way these things happen rather often is that people know how the proof 'should go', hit a snag, and make severe modifications so that the proof that should work really does work; but it might look way different now
they'll be a three line proof away from showing that some step of an algorithm can be done completely based on concrete observations and they'll relegate everything to case analysis anyway. this is so frustrating
i also get the feeling that referees tend not to like philosophical diversions in papers, like sections about why something is true or why one would think to prove it this way
That is annoying I agree. I myself would like to ask questions about ANN design, but am not sure where to ask that. AI would be a natural place for that, yes?
though unless I can make things work so I can configure the menus using the same syntax, it might be too much work to make it all look like I have it now
@TobiasKildetoft You can ask me if you need help with Mate. In Mate, you can have one or two panels, and you can add and remove things to them and move them around easily.
@JasperLoy Right, but I need specifically to be able to add menus to the top panel, which consist of submenus where each item points to a file on the computer
ohh, nevermind, it can be done. The stronger version os where the is some $x$ such that the constant sequence $a_n = x$ converges to everything, but for any $y\neq x$ the sequence $b_n = y$ converges to only $y$
@StanShunpike Sorry, Stan: I was thinking (but didn't type) more advanced graduate texts. First-year material generally has exercises. And it's not a universal thing. As I've opined before, I think exercises in large part make or break a book. I'm most proud of the exercises in my books.
When writing a longer document such as a thesis, is it "okay" to use the same index (such as i) in different contexts, as long as it is made clear to the reader which values the index can take / what its meaning is?
Yeah, but we're just making pretty colors now. No fun concentrated stuff yet, haha. Anyway, it's Friday, which means that I can do a bunch of things I've been planning to.
When there are consistent different index ranges, I will sometimes establish rules for a section by saying: Here $1\le i,j,k\le n$, $n+1\le \mu,\nu\le N$, etc.
I just started the proof.. the way I am going is this : Assume that some element is there such $y\in X$ $\{x_n\}$ does not converge to $y$ Therefore $\{x,x,x,......x\} $ does not converge to $y$ How should I go after this... Because since it is a constant function this would imply $x\neq y$ @Tobias
@Rememberme Well, then you probably want to look up the definition of convergence in general topological spaces (and try to see why this is the correct notion when compared to the metric space one)
I mean that Let (X,d) be a topological space and $\{x_n\}$ be some sequence in X which converges to x then for every open set $U$ containing x such that the set of point $\{x_n\}$ lie in the set $U$@Tobias
also, you should not require all the points of the sequence to be in the open set, as that is not what convergence is about (you should be able to ignore any initial segment)
not in general no. It is what is known as a topological embedding, meaning that it is a homeomorphism when you consider the codomain to be only the image
Hi. I asked a question last night, but when it was answered I was absent, so I wanted to thank you @PedroTamaroff for helping me out. Also thanks, @KhallilBenyattou for chiming in too :)
When it comes to two or more variables, we don't have L'hopital so the evaluation becomes more complicated sometimes, but also more challenging, in a good way
I was wondering, is there any interest in a "product" integral defined by $\displaystyle\int_a^b f=\lim_{n\infty}\frac{b-a}{n}\prod_{k=1}^nf\left(a+(b-a)\frac{k}n\right)$ ? (for strictly positive functions)
Lately I have been having back problems. To fix this I went to a physiotherapist who gave me some really good advice about how to fix my posture. She also said that I should avoid sitting. Therefore I have now rearranged my screen and my keyboard so that I can spend time by the computer standing instead of sitting.
Hi @anon Could I ask you something? Do you have an idea for the following? We have 1000 boxes in a row (at the positions 1, 2, 3, … , 1000). How can we show that we can paint them with two colors, for example red or blue, in such a way so that there is no arithmetical progression of length 20 from boxes that have all the same color?
There's a PhD in biology chat who insists P(U=a) for a uniform distribution on the reals[0,1] is zero, and that the integral of zero across [0,1] can be zero and the integral of the probability measure can be one across the same interval without causing a paradox. They are correct; probability measures are only sigma-additive and [0,1] is uncountable.
Why can't I define a uniform probability on integers and reconstitue the paradox that way?
That is, P(Z=a) is zero for integer a, but the integral across P(Z=a) for all integer a =1.
Because there are no paradoxes. What you're describing right there is precisely the reason there is no uniform probability on the integers - $\sigma$-additivity.
@MikeMiller I guess the integers are exactly the same size as other countably infinite sets. At the risk of asking to reformalize probability and measure theory, is there a simple reason why this is formalized this way? Strengthening the third axiom to uncountably infinite disjoint sets seems like a handy property.
If it's not easy to explain, I can just take decades of mathematicians word for it.
@Resonating Well, because we don't want it. The whole point of measure theory is to make rigorous the probability theory studied on $[0,1]$ (or on $\mathbb R$) in a more general setting; if we couldn't use the uniform distribution on $[0,1]$ as a measure there wouldn't be much point.
@Chris'ssistheartist I'm here, but I am having problems with my ISP. I am not able to access ajax.googleapis.com and without access to that site, much of SE is inoperable. I didn't notice that I had been logged off of chat. I am now connecting via my iPhone.
One reasonable argument would be to try to go in the other direction: well, why don't we reduce this to only having finite additivity? And I think the intuition here is just that you should be able to take the "limit" of the probabilities of the first $n$ events together. e.g. if I know what $\mu([1/n,1/(n+1)])$ is for all $n$, I ought to know what $\mu([0,1])$ is by just putting them all together - after all, I can count them all one by one.
@TobiasKildetoft An uncountably infinite subset. Countably infinite subsets with measure 1/2^n for instance can have finite measure in total, like mike's example. Are there things that can be summed across uncountably infinite sets to produce finite numbers? If we did have such an entity what would be the point of it? You couldn't do much math to it.
I'm beginning to understand how this is all laid out now. Thanks.
