in fact it is a "natural" isomorphism for finite-dimensional spaces, but not even an isomorphism at all in general for infinite-dimensional ones. this wildly different behavior indicates dimensionality is a critical part
probably the easiest or most straightforward/available way of wrapping one's head around $(V^*)^*\cong V$ is with $V\cong K^{{\rm dim}_K(V)}$ and $(K^n)^*\cong K^n$ (in fact this shows $V\cong V^*$ when $\dim_KV<\infty$). unfortunately this does not touch naturality at all, and naturality is why the double dual is special and distinguished from the usual dual. It's not clear to me if naturality is indeed "obvious."
@PeterTamaroff Anyway, the point is, you can reconstruct the representation of v wrt to a basis of $V$ as a linear combination of these reports when $V$ is finitely generated.
@anon In the Wikipedia page it says "If $V$ is finite dimensional, it is isomorphic to $V^*$, but the isomorphism is not natural and it depends on the choice of basis"
another question : let $$ f : [a,b] \to \mathbb{R} $$ be a continuously differentiable function such that : $$ \int_a^b f(x) \ dx = 0 $$ prove that : $$ \left | \int_a^x f(t) \ dt \right | \leq \frac{(x-a)(b-x)}{2} M for all $$ x \in [a,b] $$ where $$ M = \max_{x \in [a,b] } | f'(x)|$$
Yes, but my question is - if I have a circle centered at the origin, the range of theta is 0 to 2pi... but if the circle is centered at say, (x,y)=(0,1), then the range of theta is 0 to pi. I just don't understand why it's still not 0 to 2pi
namely, any two points not on the same line through the origin will have different value of theta
@agent154 ithe range of theta should be 0 to 2pi all the same. unless you're hiding information from us. which is likely; users tend to filter what information they think is relevant for potential answeres, and occasionally do a poor job of filtering...
yes, sometimes I talk about math with my friends, and sometimes I use math words or talk about math words and ideas with my friends. generally I dumb it down to layperson-mode as much as possible. I don't talk about math with my family though.
I don't dare try to explain such things to laypeople... though I am rather fond of my knowledge that R is an uncountable set...
I have a couple of friends at school into math... one person I know is also doing a joint major in Computer Science and Pure Math... so I can talk to him about math humor.
@TobiasKildetoft I have seen some exercises given in the linear algebra course at Cambridge. (My colleague's son started studying there last year.) Some of them were very nice. There is a chance that they also have interesting ones for introductory calculus course.
Another way to find good and interesting exercises might be to go through highly voted questions at MSE in the tag of your interest.
@MartinSleziak yeah, it does happen (the linear algebra course here has one extra hard problem per week, and those are usually really nice and challenging)
except one which nobody including the lecturer was able to solve
The number of integers less then or equal to $x$ with coprime exponents in there prime factorization is, $$\sum_{n=2}^\infty\frac{\ln(x)^n}{\zeta(n)n!}+O(\sqrt{\ln(x)})$$
Can I invite a user to private chat somehow? There is a guy with too little rep to chat who keeps giving half or incorrect answers to a question, and the last one seems like he might be interested in learning
Presentations can be given by exact sequences. The statement that $V$ has presentation $R^m/AR^n$ is the same as the exactness of $$R^n\xrightarrow{A}R^m\to V\to 0$$
@user1 Yes, now I see that. I'm quite unfamiliar with the exact sequence. I don't know how to understand and manipulate it. I only know the definition, which would be messy.
@robjohn I should probably be studying for tests or something, or planning ahead for school. I usually leave everything till the last minute. Can I ask what secondary school you went to after high school?
@Chris'ssis If $m\equiv -1 \text{ mod 4}$, then for some rational number $q$ $$\pi^{m}q-\frac{\zeta(m)}{2}=\sum_{n=1}^\infty \frac{1}{n^m(e^{2\pi n}-1)}$$
Though $2013 \equiv 1 \text{ mod 4}$, still I think there might be some nice form for it
