i only ever opened one, by krantz. it would have been better experience if i knew more g--m--r-. it is probably good if you know some. thankfully i never needed to use it.
ted: berkeley's fundraising goons just phoned. 7:40 pm.
Given that f is continuous on $(0,\infty)$ and that $f(x)\le f(nx)$ for every x and for every positive integer n then $\lim_{x\to \infty} f(x)$ is to be shown to either exist finitely or infinitely. Any hints for this ?
@maths But if this is where you’re doing graduate work, you need people to learn from and advise you. Also some complex algebraic geometers would be good.
There are alot of people working on something called mirror symmetry, which I think uses alot of complex geo... my advisor is a differential geometer though
Ted: Monotonic hint is really nice. Thank you! With that $x_n:=f(2^nx)$ is monotonic so either tends to a limit $l$ or diverges to $\infty$. In the former case, it remains to show that for every $y_n\to \infty$, it follows that $f(y_n)\to l$. That is,it remains to be shown now that $f$ does not oscillate as x goes to infty.
I had forgotten the more modern recommendations in the 4th edition. I did supply them all. Including me, Munkres, Guillemin & Pollack, Artin, Dummit/Foote, Stein/Stakarchi, etc.
Integrated course in multivariable and linear algebra, quite in the style of Spivak’s Calculus, but with more applications.
Say $M$ and $N$ are smooth manifolds. Show that $C^\infty(M)$ (the set of smooth functions on $M$) is ring-isomorphic to $C^\infty(N)$ iff $M$ and $N$ are homeomorphic
Find a rational $\alpha\in\Bbb Q$ such that $\sqrt\alpha\notin\Bbb Q$, and such that the degree of $\sqrt[{\Large4}]\alpha$ over the rationals is $2$ (rather than the expected $4$)
ted, today my daughter had to be separated from one of her friends because they were playing in day care when they were supposed to be working on an activity.
the idea of sneaking a toy to school in her lunchbag is actually very clever. and she apparently hid it from the teachers, or they pretended not to notice, because we didn't get an email home about it.
i was pretty normal as a teenager but i still broke a lot of rules. i had a good sense for what i could get away with. i can already see this in my daughter.
i'm generally into scifi and monster/horror movies and have a few 'creative' friends who are tangentially involved in the industry. probably they followed him and then i began to do so.
She met a lawyer who had more than 40 years experience and worked for top firms, and he said after speaking to my sister that she had the potential to overcome any challenge
i have a friend who speaks a huge number of languages well enough to get by. i thought he was fluent and he may have given that impression. then i heard him speak spanish one time and realized that his proficiency was maybe no better than mine. people are very polite and work with you if you make an effort.
@AkivaWeinberger ok, thanks. At some point of time at college, I used to know hiragana or katakana (I don't remember which one of them is easier to write, I think hiragana is easier to write) letters so thought I'd finish what I started once. :)
I'm supposed to evaluate the expression $\nabla_{\theta}((\partial_{\phi})^{\theta))$ and i'm not sure what the expression in the connection is other than $g^{\mu \theta}\partial_{\mu}$
Show that there is no injective conformal map from the punctured disk $\{z:0<|z|<1\}$ onto the annulus $\{z:1<|z|<2\}$. To show this, suppose there is a conformal map $f$ between them. Then $f$ is bounded near $0$, we can extend it to holomorphic map $\tilde{f}$ on open disk. Now, $\tilde{f}$ is a nonzero holomorphic map on a simply connected domain so it has square root $g^2 = \tilde{f}$. Does this imply any holomorphic function between annulus has square root? (so that it makes a contradiction
I am reading about localization in fourier series.Author uses something like this.
$g\in R[a,b]$ then $\frac {g(x_0-t)+g(x_0+t)}{2sin\frac{t}{2}}$ is also integrable. Can you please explain from what theorem this follows?
@AkivaWeinberger It's not hard to be fluent in two different languages, just be born outside of English speaking country and learn English ;) But that's not because they're exceptional in languages or anything, I think you knowing Hebrew or Spanish (at a decent level I assume) is better than those people speaking English and their native language fluently. Anyway, people are just different so no need to beat yourself up about it.
@LeakyNun Yes and that's where a contradiction arise. So I wonder if the existence of square root $g$ implies holomorphic function between annulus has a square root.
Clairaut's Theorem: $f_{xy} = f_{yx}$ whenever the mixed second derivative exist and are continuous.
Why need continuity?
Can't both be discontinuous but have same value for given input?
I haven't prove this theorem but I am curious about the continuity criteria. I think I can prove it with limit definition and my guess is when you perform partial derivative first time and if there is discontinuity in second time the the first derivative is discontinuous...
Hmm it seems like I have example.
Sorry I had vague thought. No needed to reply to this comment I found the solution.
it's a good thought. a lot of textbooks have counterexamples. the web is probably littered with them too. other hypotheses on f sometimes supply the continuity even if it is not expressly assumed.
I am reading about localization in fourier series.Author uses something like this.
$g\in R[a,b]$ then $\frac {g(x_0-t)+g(x_0+t)}{2sin\frac{t}{2}}$ is also integrable. Can you please explain from what theorem this follows?
