00:00 - 20:0020:00 - 23:00

12:00 AM
It's fine I went for a walk and I don't feel the urge to delete it anymore
I understand it is annoying when the answer turns out to have been deleted when you try to submit it
Speaking about politeness, that is why I don't ask questions on the site, I just forgot about it.
I'll stick to trying to help people asking

2 hours later…
2:01 AM
0

Let $f$ be a bounded analytic/holomorphic function on $\Bbb{D}$ such that $|f(z)| = 1$ for almost every $z \in \partial \Bbb{D} = S^1$, where the unique circle is given the standard Lebesgue measure (this is the definition of an inner function). Is it possible to show that $|f'| \le 1$ on $\Bbb{D... 2:25 AM Hello 2 hours later… 4:06 AM Find the image of$S_2=\{z \in \mathbb C||z-1|\leq 1\}$under$w=1/(z-i)$. Is my solution logically correct? I have doubt in$|w-(1+i)|<|w|$onwards will squaring this term alter the image? 4:21 AM If S is a connected complete regular surface and define$diam(S) = \sup\{d(p,q)|p,q\in S\}$where$d$is a intrinsic distance. If$diam(S)$is finite then$S$is compact. I need to show$S$is closed. So if$\{x_n\}$be a sequence in$S$that converges to$x$. Then I want to show this sequence is d-cauchy so that$x\in S$. So suppose not. Then, there is$\epsilon>0$such that$d(x_i,x_j)>\epsilon$for any$i,j$. I want to show$d(x_1,x)$is infinite. How can I? 4:37 AM 4:56 AM Anyone can solve it? Euler is here @love_sodam maybe sorry, I can't In case there's any machine learning enthusiasts: 3 There was some recent media reporting about a purported Google breakthrough on applying machine learning techniques to tackle the protein folding problem, as told for example in this news article, DeepMind AI handles protein folding, which humbled previous software. Unfortunately there is not muc... @love_sodam Where does$x$live? 5:12 AM @TedShifrin I think I can handle this. You don't need to answer. Then why spam the room with it several times? does anyone know something about the sum $$\sum_{k=1}^{n}k!$$ So beats Carlsen, what are the consequences? I converted that sum to $$\int_{0}^{\infty}\frac{e^{-x}(x-x^{n+1})}{1-x}$$ Don't know what to do with this integral Well, what are you trying to do with this sum/integral? There's a lot that can be said about just about any mathematical expression you could put out there, a lot of it being potentially vacuous in light of some particular goal. You have to be more specific. 5:26 AM closed form Ah, gortcha you probably mean gotcha Yeah, that one I don't know. I would assume that this is something someone has studied before, though, so it's likely in some database somewhere. (Also, gortcha is an intentionally misspelling of gotcha, yes. It is meant to imitate how I often choose to pronounce the non-word in real life.) The closed form is Hello, did somebody has some text with some examples about taylor series expansion in maps from R^m to R^n? I need to solve one of this for a problem, but I haven't found any text that could help (every text I've found online only works from R^n to R) 5:34 AM $$(-1)^n\Gamma(n+2)(!(-n-2))-!(-1)-1$$ Don't know how to prove this @big_GolfUniformIndia You just do it for each component function separately. It works fine. the ! sign before a number is subfactorial But then I'll have a Taylor polynomial at each component? Yes, if you remember that$\mathbf f$is a vector function, all of its derivatives will be too, so the usual formula works for the vector function. But you can understand it by looking at each component separately. Lemme ask it on main Really want the proof 5:41 AM @TedShifrin I just realize how to solve it I've seen the subfactorial before, but I'm unsure about how it works. Haven't actually worked with it. I'm not sure about that form, though, because plugging$n=4into it, I'm getting a complex result. I don't think mathematica can give a wrong answer Oh, it very much can, make no mistake about that. It has a tendency to lift things to the complex plane, and that can potentially lead to errors (I don't know the specifics of how mathematica works, but I've encountered this before) 0 I was trying to get a closed form of the sum $$\sum_{k=1}^{n}k!$$ Mathematica gives the answer $$(-1)^n\Gamma(n+2)(!(-n-2))-!(-1)-1$$ Here the ! sign before a number is the subfactorial. Here is my try: \begin{align} \sum_{k=1}^{n}k!&=\sum_{k=1}^{n}\Gamma(k+1)\\ &=\sum_{k=1}^{n}\int_{0}^{\infty}x... 5:46 AM blackpenredpen yaaay He's awesome Another question Does someone know about evaluating non periodic continued fractions? Many years ago, when Mathematica was a baby, it got some basic Calc II integrals just wrong. You would have realized that I research on a combo of combinatorics, NT and analysis non periodic continued fractions have no general formula, I think *general Let me ask a special case @TedShifrin I'm glad they've improved it. I don't think I would have ever come to use it if I had that kind of experience 5:51 AM I recall being amused. Here is a continued fraction representation of\pi$: This can easily be proved by expanding pi There is no pattern in the numbers I don't think so, but is there a fully rigorous proof of this? Then I'm not sure what it means to say "here is a continued fraction representation." What is the doubt? It's just like saying$\pi = 22/7$. It's accurate only as far as you go. I don't care about accuracy I just know it's right :) 5:54 AM I think it's time to ignore you. There is no rigorous proof Oh no So,$\pi$is the limit of these continued fractions as you let them get more and more accurate To put some light onto this, I have another YT video for you okay thanks So the list of people who ignored me is getting bigger gradually 5:59 AM How can I show: let$S$be a connected regular surface. If every closed and bounded subset of$S$w.r.t d (intrinsic distance) is compact then$S$is complete Don't take it personally. This place gets a lot of traffic from lots of people with lots of ideas. People who are regulars here tend to be fairly busy and have to make decisions quickly about what they dedicate their time towards. okay No worries but one disadvantage I would get less answers in chat 6:14 AM @LeonhardEuler look, if everyone else is also getting ignored, then the people aren't ignoring "you" 2 6:53 AM If$k$is a finite field and$\omega:k^* \to \mathbb{C}^*$is a homomorphism s.t.$\omega^2\neq 1$, why does$\sum_{d\in k^*} \omega(d)^2 = 0$? 2 hours later… 8:29 AM @copper.hat Ok 8:45 AM is$\{f \in C^\infty_c(\mathbb{R}^n) | \int_{\mathbb{R}^n} f(x)\ dx = 0\}$dense in$\{f \in L^1(\mathbb{R}^n) | \int_{\mathbb{R}^n} f(x)\ dx = 0\}$? @JoeShmo I think this is likely it sounds like it might be right, but it might not be.. how do I approximate an$L^1$function with$0$integral with a$C^\infty_c$function with$0$integral? How are you generating these fractal images? 8:50 AM xaoS ah, I should have kept it a secret @JoeShmo try a smooth approximation, then modify somehow so that the smooth function has integral$0$. Don't know if that will work, but could be an idea. @LeonhardEuler Danke schön Bitte schön it ain't working.. 8:56 AM @JoeShmo if you show exactly, where it goes wrong, and what other approaches you have tried, this could make for a good question on MSE the formula$z^2+c+z$generates this fractal: this is better than Mbrot any other weird formulas? And$\exp(z)+cgives this: I call this pacman Not much to zoom in Are there any analytic NT books to read after Apostol's? It covers most of ANT But still, I want to learn more 9:38 AM Can someone verify my answer: 0 We have \begin{align} \lim_{n\to\infty}\left(\frac{1}{a}+\frac{2}{a^2}+...+\frac{n}{a^n}\right)&=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{a^k}\\ &=\sum_{k=1}^{\infty}\frac{k}{a^k} \end{align} Let1/a=x,so we have \begin{align}\sum_{k=1}^{\infty}\frac{k}{a^k}&=\sum_{k=1}^{\infty}kx^k\\ &=x\sum_... 10:12 AM It is bad to link all my questions and answers here, but this one is really important for me: 0 A fractal is usually defined to be a self similar shape (this is the informal definition). But, the formal definition is: A fractal is a set for which the Hausdorff dimension strictly exceeds the topological dimension (formal definition). Before proceeding to the question, I will tell what this... Pardon me for linking so many posts here 10:50 AM Does an abelian category always have all limits? or is this a stupid question 11:09 AM @EdwardEvans Since pre-abelian categories have all finite products and coproducts (the biproducts) and all binary equalisers and coequalisers (as just described), then by a general theorem of category theory, they have all finite limits and colimits. okay thanks :P First time I properly answered something here Maybe second time Ah this property is called finite completeness you learn something new every day 11:24 AM thanks for directing me to the relevant wikipedia page Fine if I have a DVR A and its field of fractions Frac A, how do I show that there is no non-trivial extension that is a DVR B such thatA\subset B\subset Frac A$? 11:39 AM @LeonhardEuler: so that formula is based on the fact that $$!(-n)=\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=0}^{n-2}k!$$ Ok @robjohn thanks for your answer which makes it not quite so interesting, since to compute$!(-n)$, you need to compute the sum you were looking for. Are you sick @robjohn ? @Astyx why do you ask? Why did you turn green? 11:41 AM Maybe he hates christmas, he's the grinch 15 hours ago, by amWhy @robjohn Thinking about it, the mean square, ala The masked grinch who stole Christmas who be a good match for you. Consider green instead of orange! haha ok @Astyx I also had a green avatar for St Patrick's Day. @robjohn is a shape-shifting creature Just a joke don't take personally @Astyx think about it in terms of the valuation; if$a \in A$then$|a| \leq 1$. If$B \neq A$then there is a$b \in B$with$|b| > 1$. Now pick an arbitrary$x \in K$and play around until you can conclude that$B = \operatorname{Frac} A$11:45 AM What is |.| ? The valuation ? that's your valuation right I usually write it v(.) if you wanna think about it in terms of$v$then$v(a) \geq 0$and$v(b) < 0$What are your favorite mathematical papers? v determines |.| and |.| determines v so it doesn't really matter 11:46 AM And why but I like |.| better for some reason I would be interested in them Mine is Riemann's paper on number of primes less than a given quantity How are they related ? Is || the exponential of v ? I don't think I've seen this notation before v = -log |.| OK! Thanks 11:47 AM npnp Riemann's paper was fantastic Just to check, every element not in the maximal ideal is invertible, because the ideal generated by that element has to be A, so contains 1? Yeah, DVRs are local I love reading articles Papers with funny names have been mentioned here "Scott is not always sober" Now I want to know about y'all's favorite paper @Astyx if your element is not in the maximal ideal and not invertible then it's contained in some other maximal ideal 11:51 AM Just for fun right Btw, I have got something These cups look very good: @Leaky HSÜEEMÜHIR is the German analogue of STFGMPID stoppppppp 11:54 AM How beautiful cups @EdwardEvans ok Was it so bad? it's just lame Ok Lame=no excellent Excellent=no lame Lame+no lame=excellent +no excellent Lame(no+1)= excellent(no+1) Lame=excellent :P Okay I am going to stop 12:13 PM Hello everyone, I have a question math.stackexchange.com/questions/3930308/… 3 Let$H$be the Hawaiian earring, i.e. it is the union of circles$S_n$with centre$(1/n, 0)$and radius$1/n$. By a universal cover I mean a covering$p\colon X\to H$(with$X$connected) such that for any other covering$p'\colon X'\to H$, there is a map$f\colon X\to X'$such that$p'\circ f =...

Just wondering
Is there a closed form of $$\sum_{k=1}^{\infty}e^{-x^2}$$

You don't have any $k$ inside the sum

12:30 PM
ah sorry it is n
oh sorry again @Rithaniel it is x
Didn't notice

12:50 PM
Well, I'd start looking towards geometric series, maybe

$\sum \exp(-k^2)$ from k=1 to k=infinity has a a closed form in terms of the Jacobi third theta function

ok
@Rithaniel that won't work, I think

1:11 PM
I need help understanding the asymptotic expansion of $$\int_2^x \exp\bigg(\frac{1}{\log(t)}\bigg)~dt$$
0

Consider two matrices acting on points in $(0,1)^2$ in the real plane, $$h_s=\begin{pmatrix} e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix}.$$ and $$g_s=\begin{pmatrix} 1-e^{-e^{s}} & 0 \\ 0 & e^{-e^{-s}} \end{pmatrix}.$$ How do you compose these transformations? I tried to do $h_s... Anyone know how to compose two matrix transformations? You just multiply the matrices, right? I don't know honestly, but I'll try that :) Just a quick doubt. Does$\sum_{p^m}$mean the same thing as$\sum_{p} \sum_{m=1}^{\infty}$? 2:15 PM -1 Let$f$be a bounded analytic/holomorphic function on$\Bbb{D}$such that$|f(z)| = 1$for almost every$z \in \partial \Bbb{D} = S^1$, where the unique circle is given the standard Lebesgue measure (this is the definition of an inner function). Is it possible to show that$|f'| \le 1$on$\Bbb{D...

1 hour later…
3:21 PM
Hello
anyone good at diff. geo.

3:33 PM
0

I want to prove that there does not exists a complete surface of revolution with constant Gaussian curvature $K=-1$. To do this, I classified the surface of revolution with $K=-1$: If $\gamma(t)=(x(t),0,z(t))$, then the surface revolution from $\gamma$ has $K=-1$ if and only if $x(t)= c_1e^t+c_2e... 4:28 PM A topological generator g of a group G is one such that$\overline{\langle g \rangle} = G$? woops yes cool ty lmao ut 1 in$\widehat{\Bbb Z}$I'm just doing my obligatory monthly skim of classical Iwasawa theory before I get distracted by math that I actually have to do 4:41 PM @Rithaniel yes, you were right about multiplying the matrices together to get the composition of the transformations. Ah, excellent. Glad you came to your answer 5:25 PM How does the locally ringed space approach to differential geometry differ from that other approach? 6:02 PM I don't understand the main obstacle(s) in encoding the metric in terms of the rings of analytic functions 6:57 PM Hello i have a question, but it's not being answered: about weak convergence. 0 So the function is $$u_n(x)=\begin{cases} a&\mbox{ if } x\in (\frac{i}{n},\frac{i+1/2}{n}], i=0,\dots,n-1, \\ b&\mbox{ if } x\in(\frac{i+1/2}{n},\frac{i+1}{n}],i=0,\dots,n-1.\end{cases}$$ in$\textit{L}^1((0,1))$and i want to show$u_n \rightharpoonup (a+b)/2$. I think i have successfully proven... 7:14 PM Hi All. I want to understand insight of this theorem of group theory.$\bf{Theorem}$Let G be any group . Let$a \in G$have order$n$. Then, for any integer$k$,$a^k = e$if and only if$k = qn$, where$q$is an integer.$\bf{Proof:}$Suppose that$n$is the order of$a$and, for some integer$k$,$a^k = e$. By the division algorithm, there are integers$q$and$r$such that: The important thing to understand is what the definition of order of an element is.$k = nq + r , 0 <= r n$Yes i know the order, how to find it. But i don't understand the theorem meanings here. Tell me what the order is. Pls explain in. if$n$is the order of group why we take$k$here. Because$a^n = ee = a^k = (a)^{nq + r} = (a^n)^q\cdot a^r = e\cdot a^r = a^r$What is this step here$a^r = e$what is meant by that??? Why we use Euclid algorithm here??? You never answered my question. It is the most important thing. 7:29 PM Order of group is number of elements in the group O(G) = n No. That's the wrong order. What is the order of an element$a\in G$? In cyclic group order of element = order of group, then it is the generator of group. power of an$a \in G$If$a^n = e$then$n$is the order of group where$a \in G$. Under given operation No. Pls explain. You have NOT given me a correct definition. If I tell you that$a^6 = e$, you do not know that the order of$a$is$6$. 7:32 PM yes 6 is the order of group. No. You MUST learn PRECISE, correct definitions. Ooooo... Why i am calculating order this way. I have found right answer. Also my text book said this. Pls share and explain me. The order of$a$is the SMALLEST positive integer$n$for which we know$a^n=e$. least positive integers Yes. But every word is super important. You cannot do math if you are sloppy. 7:35 PM Yes what is meant by that. Pls explain me by example why least positive integer is here. Suppose$a^3=e$. Then it is also true that$a^6=e$. So what is the order of$a$? 3 is the order of group in both cases. NOT group. But how can we get a^6 No, you actually do not even know the order is$3$. It could be smaller. 7:36 PM Is$\bigcup_{n \in \mathbb{Z}} (2^n, 2^{n+1}] = (0, \infty)$? Or is it$[0,\infty)$? How would you ever get$0$, @user330477? second one Huh? Heya, @Thor. What? @TedShifrin We would also not get$\infty$, right?$\infty$is not a number; it's just a symbol. 123, you better pay attention to your definition of order of an element and understand it completely. 7:38 PM I think things mixed up. @TedShifrin pls tag comment by name. THanks @Ted but$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \cdot 8 = \infty$Pls explain me how can we get order of element bigger than$n$, where$n$is the order of group. @TedShifrin Thank you. You're so helpful, @Lukas. 7:40 PM I never said you could do that. You still are not paying attention to details, @123. Suppose I tell you$a^8=e$and$a^6=e$. What do you know about the order of$a$? @TedShifrin I tried to understand but did not understand. Work on my last question. I think 2 is the order. Can you prove it? because$a^6\cdot a^2 = e\cdot a^2 = a^2 = e$7:43 PM OK, good. But does that say that the order is$2$? It's cdot Sorry typo error Yes, sure, no problem. my question is that. how can we get order of element bigger than order of group. @123 we define the order to be the smallest positive integer$n$such$a^n=e$, because there are many$n$such that$a^n=e$. If you have$a^n=e$for some$n$, then you can only conclude that$n$is the order of$a$if you also know that no smaller positive integer will work You cannot, but you still keep NOT understanding the definition. You cannot say that the order is any integer$m$for which$a^m=e$, and the order of the GROUP is something entirely different. Unless this is a cyclic group generated by$a$. 7:45 PM @TedShifrin If we don't get order of element bigger than order of group. We don't need least positive integer idea. This is confusion. Yes, smallest ... Ah, it's been 8 years since I last taught abstract algebra. Retirement is bliss :D @123 ... If$a^8=e$, does that mean the order is$8$? @TedShifrin Not necessary we need to look order of group. than we decide. Hey @Ted NO. NOTHING to do with the order of the group. You need to stop saying this same wrong stuff. Are you talking ONLY about the cyclic subgroup generated by$a$? @TedShifrin Ookay. Because i learnt that we need to check if order of element is same as order of group than it is the generator. 7:48 PM But most of the time you do not have a cyclic group, but you still talk about orders of elements. So if the order of$a$is$6$, can it happen that$a^{10} = e$? Yes order of element different depend how$a^n = e$. English is obviously not your native language, but you still MUST use words precisely. Sorry for my bad english. Yes this is not my native languange. No it can not happen. if order of$a$is$6$. It means$a^6 = e$. Than$6$is the order of that element. NO. I've corrected you ten times. It does mean$a^6=e$, but it means MORE than that. Ookay. pls correct me. How we decide the order. 7:54 PM I have told you and Lukas has told you. @TedShifrin Pls patient i am naïve in this. I think Ted is being very patient here But you MUST read and understand what we say. I'm not going to repeat the same thing 10 times. I've already repeated 5 times. :D yes. You even said your text book said least (I said smallest). You CANNOT ignore this over and over and over. 7:55 PM @TedShifrin least positive integer you said. Pls explain the criteria. How to know the least positive integer. You explain what it means. If I say that the order of$a$is$6$, why is that different from just saying$a^6=e$? @TedShifrin Because i don't understand how can i calculate least positive integer for this situation. If$a^6=e$, tell me what the order of$a$MIGHT be. I think 2 is least positive integer. No,$1$is, actually. Could$a$have order$1$? 7:58 PM @TedShifrin This is my confusion. How to calculate least positive integer. Could$a$have order$2$? Could$a$have order$3$? Could$a$have order$4$? Could$a$have order$5$? Could$a$have order$6$? Could$a$have order$8$? Yes order$1$has identity element. Yes$a$can have any order. NO, wrong. I told you that$a^6=e\$.

00:00 - 20:0020:00 - 23:00