Triple checked it and went over it with my professor aaaaaand it's one of those things that's really simple once you've solved it but an ass to actually think of, so I'm good, but thank you for the advice. :)
Really quick question guys, so I'll just jump right in, is there a more $i$ looking version of $i$ - like you might use for inclusion map or identity map? It's like a $j$ but curves to the right.
@Chris'ssis: I have stayed out of this, but I truly think you need to be more humble and show respect for mathematicians with degrees, publications, and academic jobs, who inhabit this chat. You owe @Tobias a serious apology. In particular, you are not the arbiter of what is serious mathematics.
next semester you know I will be learning algebraic topology the professor it will be directed reading class, so I will learn the material on my own the professor told me each week he will make me present what I learned for infront of him and his grad students.
@TedShifrin I always learned on my own for example when I was 14 I learned programming in C and C++ on my own from books so studying from books became a habbit for me like I learn better from books than from professor
I'm going to meet up this Saturday with a guy who was a student in my 300-person multivariable calculus class at MIT in 1981. He is now the president of the National Society of Black Engineers and I haven't seen him in ... 34 years.
And Balarka is doing stuff as a 15-yr old that I didn't learn until graduate school. You don't need to feel like you should compete with them. You need to be you.
Okay, I've got a silly question. A derivative (for $f:\mathbb{R}\rightarrow\mathbb{R}$) is defined as: (given an $x\in\mathbb{R}$) $\forall\epsilon>0\exists\delta>0[0<|x-a|<\delta\implies|\frac{f(x+a)-f(x)}{a}|<\epsilon]$ I can easily use this to show $\forall\epsilon>0\exists\delta>0[0<|x-a|<\delta\implies|f(x+a)-f(x)|<\epsilon]$ but that's not quite the definition of continuity. Unless I can say "continuous at a limit" or something. What am I missing?
Wait, I know it's differentiable @TedShifrin I can't say anything about $x=a$ itself other than "as a limit it exists", $f(x)/f(x)$ has this but is discontinuous at 0...... grr, I could do this 2 years ago.
You don't realize what geometry entails yet. Complex analysis is beautiful, but if you do all the proofs at the graduate level, there's a lot of serious analysis.
At the undergraduate level, they skip most of that.
Thanks @TedShifrin I had made some really weird mistake. it's actually trivial because it's =0 as you mentioned and the point is in the domain. (I stand by the $0<|h|<\delta$ bit, because the quotient isn't defined for $t=0$)
@Ted Also, I literally just noticed that I will be multi next year because I will be taking the AP BC exam this year so should I learn ahead or maybe learn some complex analysis?
@Julian: I still want you to learn single variable rigorously and then multivariable rigorously, as opposed to the engineering course, but still learning computations.
Okay last one, and I hope this isn't daft. What do we actually define limits on? When I encountered them the "space" on $\mathbb{R}$ was never mentioned. It can't be just a metric space because limits in discrete metric would make no sense.
@TedShifrin I'm not "convinced" at the first countable bit. I would totally believe that for any NORMED space you could define limits and differentiability. A metric lacks...the scalar part (discrete metric for example) (how can I find out without pestering you?)
@TedShifrin I wouldn't say that limits where you have the discrete metric say make any sense. But in my mind if you have a (complete, as you said) normed space, I can totally believe you get limits (derivatives and limits of functions, you know the classic real analysis)
@TedShifrin if you consider the directional derivative for a moment, it matters not if you use $\Vert\cdot\Vert_\infty$ or $\Vert\cdot\Vert$ in the denominator, because these are equivalent norms, unless every first countable topological space is norm-able(?) (if it's metricisable, this isn't that big of a jump)
....if I have a "question-quota" please don't answer this as I'll find a way to survive. Did you change the topic after the ")" there? Because I cannot make any sense of anything after the ")"
What's an example of a finite dimensional function space other than the space of polynomials of degree $\le n$? I'm looking to motivate change of basis (so I'm avoiding $\Bbb R^n$ for this) and I'm getting tired of talking about polynomial spaces.
I'd like to know, Is there an linear(additive), continuous, homogeneous, functional that is not uniformly continuous? What are the advantages if there isn't one.
Today I did something my math professor hadn't seen before. I used L'Hospital's rule in "reverse" (by finding an antiderivative instead of a derivative).
This was the homework problem: Let $f$ be differentiable on an interval $A$ containing 0, and $(x_n)$ be a sequence in $A - \{0\}$, with $(x_n)\to 0$. If $f(x_n) = 0 \forall n\in \mathbb N$ and $f$ is twice differentiable at 0, show that $f''(0) = 0$. (Well, that's part b, part a was showing that $f(0) = f'(0) = 0$)
Here was my proof: $f$ is continuous, so $$f(0) = \lim_{x \to 0} f(x) = \lim_{n \to \infty} f(x_n)\\ f'(0) = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{n \to \infty} \frac{f(x_n)}{x_n} = \lim_{n \to \infty} \frac{0}{x_n} = 0$$. $$f''(0) = \lim_{x \to 0} \frac{f'(x) - f'(0)}{x - 0} = \lim_{x \to 0} \frac{f'(x)}{x} = L$$ For some $L$, because $f$ was twice differentiable at 0. By L'Hospital's rule, $$L = \lim_{x\to 0} \frac{f(x)}{\frac12x^2} = \lim_{n\to\infty} \frac{f(x_n)}{\frac12x^2} = 0$$
Of course, I had to note that I actually met the conditions for L'Hospital's rule.
Why would not being locally contractible imply its double isn't locally contractible?
The cantor set points have very different looking nbhds when you take the double (this is not to be confused with the standard doubling of manifolds with boundary).
If you have locally noncontractible at the bad point, then any nbhd of the bad point would be noncontractible. Double it - any nbhd of the bad point in the double also contains some nbhd of one ball. That's not contractible.
i faintly recall that there is some theorem saying that the first non-vanishing homology group is also the first non-vanishing homotopy group or something. is there something like that?
so just construct the homotopy $F : S^\infty \times I \to S^\infty$ between id and the null map by nullhomotoping each $S^{n-1}$ in $S^n$ at each stage
@skullpetrol I saw your comment. It is uncalled for. Kit is just doing her job as a moderator. I don't think Rob and Cerb do much to moderate the room as room owners. They should not have been appointed in the first place.
Hello fellow Math.SE users! I am fairly new to the Maths part of SE and I am looking for guidance of search terms because I am certain my question is already answered somewhere on here. What I am looking for is a way to shorten the distance between two points by an absolute amount and get the coordinates of the new point.
Clarification: I have two points with x and y coordinates, I need to plot a line that stops short of one of the points. How would I get the coordinate of the new point?
@Alendri: Use vectors. Say you are given two distinct points $P = (x_P, y_P) , Q = (x_Q, y_Q)$ in $\mathbb{R}^2$, any point on the unique line containing both of them can be written as $$\begin{pmatrix} x_P\\ y_P \end{pmatrix} + t \cdot \overrightarrow{PQ},$$ for some $t \in \mathbb{R}$. Note that if you take $t = 0$, you get back $P$ and if you take $t = 1$, you get $Q$ (or more precisely their position vectors). Hence every point between the two is given by a $t \in (0,1)$.
Note that you can easily compute the distance from such a point to $P$, it is given by $$|t \cdot \overrightarrow{PQ}|,$$ so if you want to find a point with a given distance $d$ from $P$, you can just solve an equation for $t$. The same works for $Q$ too, of course.
@Huy Alright, I will try to figure that out, thanks a lot! Do you happen to know of a question/answer to the same end? As more people tend to read those than this chat, so if I need to refer to a solution I can point them to the correct place.
@Alendri: I'm not sure if I understand what you mean but if you insist on asking a question on the main site, you could ask about how to parametrize points on a line given by two points, I guess.
Well, I was thinking I cannot be the first to ask this question. There may already be a question about it answered which I have been unable to find, rather than asking it again and having it closed as a duplicate. You are clearly much more familiar with the terminology than me so I thought you might know of such a question.
For a number $\alpha \in \mathbb{R}$, define $f(x) = x^{\alpha}$ for $0 < x \leq 1$ and $f(0) = 0$.
I am tasked with computing $\int_{[0,1]} f$ in both the bounded and unbounded cases.
I assume that they mean when $f$ is bounded? I'm a little unsure about that...
In any case, I know that in th...
@JessyCat If $\alpha \geqslant 0$, then $h$ is bounded (by $1$). Then you can directly integrate. If $\alpha < 0$, then $$h_M(x) = \begin{cases} M &, x \leqslant M^{1/\alpha} \\ x^\alpha &, x \geqslant M^{1/\alpha}. \end{cases}$$ Compute $$\int_{[0,1]} h_M = \int_0^{M^{1/\alpha}} M\,dx + \int_{M^{1/\alpha}}^1 x^{\alpha}\,dx$$ and let $M\to +\infty$. For the last part, note that $h \leqslant M$ and $h \leqslant f$ is equivalent to $h\leqslant \min \{M,f\}$.
@AntonioVargas: I'm just looking for some software to have all papers, lecture notes, books etc organized and to read them whenever I want (Windows, MacOS, iOS). Mendeley works but it often lags and sometimes even crashes, and for some reason hyperrefs within PDFs don't seem to work for me, which is really frustrating.
@TedShifrin It seems to me that you also talked about my math some time ago, isn't it? I think you're not qualified to talk about my math, not you, not @TobiasKildetoft, and not others like you that talked about my math here either. But you know what? I might be wrong. Then let's do like that: I prepare pen and paper, I post some thousand of questions from personal research and I take notices from you according to the solutions you provide to me.
@TedShifrin I also have papers published in respectable journals, and many paper waiting to be published, and I don't even count tons of proposed problems.
@Chris'ssistheartist: If you think people like Ted/Tobias are not qualified to talk about your math, what makes you think you are qualified to talk about their math?
I'd like to know, Is there an linear(additive), continuous, homogeneous, functional that is not uniformly continuous? What are the advantages if there isn't one.
For a number $\alpha \in \mathbb{R}$, define $f(x) = x^{\alpha}$ for $0 < x \leq 1$ and $f(0) = 0$.
I am tasked with computing $\int_{[0,1]} f$ in both the bounded and unbounded cases.
I assume that they mean when $f$ is bounded? I'm a little unsure about that...
In any case, I know that in th...
