3:16 AM
@Thorgott So, which one should I pick? Rudin is terse, and I thought classic books may be having clear explanation.

4:10 AM
Hi everyone

4:40 AM
@Knight: Rudin is way too hard for you. Are you taking an actual course that has a textbook? You were reading Spivak a week ago. That's an excellent source. Other options are Mattuck's Introduction to Analysis, which I like a lot, and Abbott's Understanding Analysis.
Hi @Yuvraj. What's up?

@TedShifrin struggling with a small limit question!

What is that?

In probability, what is the meaning of $(X,Y)\sim N_2(0,0,1,1,\rho)$? I believe this is a bivariate normal, but I don't see why we have 5 parameters.

limit actually I need to find limit x tends to $0^+$ [1+[x]]^{2/x}$@TedShifrin @Yuvraj: If$x$is a small positive number, what is$[x]$? Oh, wait, you're using way too many brackets. What are you doing here? I thought$[x]$meant greatest integer$\le x$, but you have way too many brackets. Please make this clear. 4:57 AM It is gif @TedShifrin Each bracket is gif Oh, OK, it is what I thought. But$[x] = 0$for all small$x>0$. So what is$[1+[x]]$? Is 1 Yes. So the answer is ... It's more interesting if you do $$\lim_{x\to 0^*} (1+x)^{2/x}.$$ Do you know that? Yes sir I wrote that$e^2$but answer was 1 LOL, the one I just gave you is$e^2$, but of course the answer is$1$to your question. 5:02 AM How? You just told me that for$x>0$small, we have$[1+[x]] = 1$. Always. So what is$1$to any power? What's the limit of the constant function$1$? Equal to constant So we're done. 5:04 AM Sir actually there is one more part Where x was tended to$\infty$For that also [x] =x So you're asking$\lim\limits_{x\to\infty} (1+x)^{1/x}$? Yes Is the$1$relevant there? You can do it without the$1$first and then see if it changes anything. What's the answer? 1 is not relevant right ? You'll check that in a minute. Without it, what's the limit? 5:07 AM 1 OK, how'd you do it? [x] will be equal to x as limit x tends to infinity Adding one make no change No, that doesn't make sense. But you can use squeeze to deal with properly. Ah! I do not know how to apply it?$x-1<[x]\le x$. 5:13 AM Yes$\lim \limits_{x \rightarrow + \infty} (1+x)^{1/x} = \lim \limits_{x \rightarrow + \infty} e^{1/x \ln (1+x)}$So, if you can show that$x^{2/x}$and$(1+x)^{2/x}$have the same limit, you're done by squeeze.$1$is not relevant here. @abhas_RewCie ok @TedShifrin that was a easy to analyse how do we do it for a function like 1/n^2+2n function @Yuvraj If you understand$\epsilon - \delta$definition, then you can make more sense of what's happening here... 5:17 AM @abhas_RewCie no I am not aware @Yuvraj I'd rather say prefer books like IA Maron's - Problems in Calculus of One Variable They have some easy examples and good olympiad style examples. @abhas_RewCie got it @TedShifrin are you there? Yes, back. I don't understand what your question is. 5:27 AM @TedShifrin limx tends to$1^-$Huh? You wrote stuff with$n$s. Are you missing parentheses? I was asking about how do we apply squeeze theorem For complicated function To the original question? I wrote that. Let$f$be a function such that$-3 \le f(x) \le 5.$Find$\lim_{x\to2}((x-2)^2f(x)+1)$Like this one There's no universal answer. You have to think. 5:32 AM @Yuvraj if$f$is continuous, then the answer is$1$Well, what can you say about$(x-2)^2 f(x)$? @abhas_RewCie f is continuous You don't need continuity at all. @TedShifrin ok? That's important. 5:33 AM It would be leess than 5(x-2)^2 Use the inequalities they gave you. And greater than -3(x-2)^2 Or$\le$, and .... Yes, and the justification is that .... ? No wait it is 1 Yeah, I'm ignoring the$1$. What would you do if they had$x-2$instead? 5:38 AM In the question? Yes. What made your inequalities work? It would not change anything I feel? Well, you didn't answer my question just above. What made your inequalities work? This one Yes 5:43 AM Ok both the function just above and below have the se limit As our function If$-3\le y\le 5$is it always true that$-3u\le yu\le 5u$? Yes I feel so nu, only$u \geq 0$Very important. So if$u$isn't known to be positive, what's the best estimate to give?$u=0$5:49 AM @abhas_RewCie yes yes! No, @abhas. @abhas_RewCie it is absurd for u=0 Hint: What is the biggest$|yu|$can be? 5u @TedShifrin$+\infty$That's biggest, I've heard...^ 5:52 AM @abhas_RewCie there is no information about u okay If u can be anything if u=(-) then Honestly, I don't understand what's the whole conversation about... will have to go... :) bye :) @TedShifrin yes ? I want the answer in terms of$|u|$. 5:55 AM 5u I said (Here$u$mght be$x-2$in our problem.) What's the biggest$|y|$can be? Oh,$5|u|$is right. good. I missed it above, but you needed abs value. Now you can do squeeze with the abs value. What is the abs value sir ? Sorry what does it mean? You know absolute value. I just got tired of typing. Enough for me tonight. It's late here. Night! Good night sir thanks for helping me out @TedShifrin 6:15 AM @Lelouch You gave some example where the 4-point condition holds wrt a fixed basepoint but the guy isn't hyperbolic. Can you reiterate what it was? I am not sure if that works 3 hours later… 8:58 AM @BalarkaSen Sorry, in retrospect the "counterexample" is BS because if that holds for a particular point with$\delta$, then the space is$3 \delta$hyperbolic. Maybe that's not$\delta$hyperbolic, but that would be somewhat pathological counterexample ig. I mean, if the basepoint condition holds with$\delta$, then the four point condition holds with$3 \delta$Anyway, if you don't mind summarizing, can you please mention which theorems from BH he proved today ? I had algebra 2 endsem today and I am feeling bit sick so I didn't attend todays class. There was a MSc First year guy I would have asked, but he's not attending the class today either. 9:28 AM @Lelouch Right, I just noticed that. If that condition holds at a single basepoint then it's$2\delta$-hyperbolic, there is no counterexample. He proved exponential divergence. 1 hour later… 10:40 AM and what did he do in the second half @BalarkaSen ? hyperbolic => linear isoperimetric inequality @TedShifrin The word "lim" stands in english for limit, but in german we say Limes. And it comes from the romanian empire: "von den Römern angelegter Grenzwall zur Befestigung der Reichsgrenzen." They for protecting the empire is called Limes. ok cool, thanks 11:15 AM @Suisse du meinst "The Roman Empire" 11:44 AM @TedShifrin Regarding this reply: Yes you said it quite correctly, it’s hard for me not terse. I shouldn’t blame a book which is so widely accepted just because I cannot get much from it. No, I’m not taking any Real Analysis course from any nationally/internationally recognised institutions, I’m doing it on my own and take help from people over here. Finally, thanks for that Mr. Mattuck’s book, I really like him I watched his lectures on Differential Equations (available on MIT.OCW) and in reality his way of teaching was very digestible. (I’m still using Spivak, but it’s a first course Calculus book and I have quite completed it much) Rudin definitely is terse I think it's simply not suited for self-study anyhow 12:07 PM Prof. Strang once said “We may not be Newton or Leibniz or Gauss or Einstein, but we can share some part of what they created” Sometimes, it seems true! 2 hours later… 2:33 PM So, the notion of zero-set (in$\mathbb{CP}^n$) of a polynomial in$(n+1)$makes sense variables makes sense only when the polynomial is homogeneous. Apparently, the converse is also true. That is, if$F$is a polynomial in$(n+1)$variables and$F(x_0, \dots, x_n) = 0 \iff F(\lambda x_0, \dots, \lambda x_n) = 0$, then$F$is homogeneous. Should this be easy? Any hints please. 2:57 PM I can't tell if I should write out my thoughts or not. What I did so far is write$F$as a sum of homgeneous pieces and take derivatives in$\lambda$and evaluate at$\lambda = 0$. Oh whatever. Write$F(x) = \sum F_i(x)$where$F_i$is homogeneous of degree$i$. Then$F(\lambda x) = \sum \lambda^i F_i(x)$. Supposing$F(\lambda x) = 0$for all$\lambda$, plugging in 0 shows that$F_0(x) = 0$; taking derivatives in$\lambda$and plugging in 0 you find that$F_1(x) = 0$; second derivatives show that$F_2(x) = 0$, etc. So$F(\lambda x) = 0$for all$\lambda$iff$F_i(x) = 0$for all$i$. So your question is: "Suppose$Z(\sum F_i) = \bigcap Z(F_i)$. Is all but one$F_i = 0$?" Is it possible to show by elementary means that the sum of two matrix commutators is a matrix commutator? That is, say$M = AB - BA$and$M' = CD - DC$, can we somehow explicitly show that$M + M'$is a commutator?$A, B, C, D$are all$n \times n$square matrices over$\mathbb R$. 15 If the ground field$\mathbb{F}$is$\mathbb{R}$or$\mathbb{C}$, the following gives an elementary proof. Clearly every commutator has zero trace, so it suffices to show that every real or complex matrix with zero trace is a commutator. First, every traceless matrix is$\mathbb{F}$-similar to a ... @MikeMiller That seems a bit complicated. Would it possible to express$M + M'$as a commutator just in terms of$A, B, C$and$D$? 2 Something like (A + B)(C + D) - (C + D)(A + B) (but that obviously doesn't work) No. 2 I'm not going to prove it though. You can try. @MikeMiller What's the reason? 2 3:13 PM I believe I said that I'm not going to prove it for you, but that you could try. @MikeMiller Oh, I thought that maybe you had some insight (short of a full proof) that allowed you to make the claim. :) In any case, I don't know how to prove these sort of things so I'll probably have to wait till I learn some more algebra. 3:38 PM I'm writing the actual keywords of the question: Let$f$be a polynomial with integer coefficients. Define$a_1 = f(0);~~a_2= f(a_1) = f(f(0))$and$a_n = f(a_{n-1})$for$n \geq 3$. If there exists a natural number$k \geq 3$such that$a_k = 0$, then prove that either$a_1 = 0$or$a_2= 0$The$a_i$s that are mentioned in the problem, are those the coefficients of the polynomial or just some random sequence? I don't understand your question. They are defined in the very sentence you just wrote. Okay! I thought they were the coefficients of the polynomial$f(x)$. 3 hours later… 7:16 PM Let A be a square diagonal matrix (n by n). Let the only elements on its diagonal to be +1 or -1. Alternating sign is allowed. For example, if n = 2. We consider A = [1,0; 0, -1] to be an ideal one. But I noticed A = [-1,0; 0,1] is as same as A = [1,0 ; 0, -1]. Question: If A (n by n) does not equal to the Identity matrix. How many distinct such matrix A exist? Please answer this in terms of n.' This is question I saw in some website , I didn't understand what is meaning of this statement over here "I noticed A = [-1,0; 0,1] is as same as A = [1,0 ; 0, -1]". that doesn't make sense 7:34 PM does anyone know a good resource for learning about currents? linear functionals on compactly supported differential m-forms I'd assume Krantz & Parks is a good resource 7:49 PM Federer :3 we all thought this, but you didn't have to say it Fremlin @Alessandro uncannily rhymes with Gremlin 8:10 PM what tf is wrong with the stars ? someones starring crap etc the fault in our stars lol lmfao 8:41 PM @robjohn Hi Sir . 9:11 PM @Binod hello @robjohn Sir you told$x/2 +\sin(\theta)$is not a one-one function. I know it's correct , but how do I show it mathematically , without the help of a graphing calculator . yeah cause you can't inject$\Bbb R^2$into$\Bbb R$continuously @Binod any injective function must be strictly increasing or strictly decreasing so its derivative, if it exists, must be always >= 0 or always <= 0 @LeakyNun I am not sure what you mean by that.. I was making a remark based on the fact that your function has two variables @LeakyNun Okay ... but I have attempted questions where if f(x) is a polynomial function$f(x)+ sin(x)$is one-one . Can you elaborate a bit more ? 9:23 PM you probably mean$f(x) + \sin(x)$@LeakyNun Yes ,edited . so in that case see the actual message that I pinged you for the actual answer 5 mins ago, by Leaky Nun @Binod any injective function must be strictly increasing or strictly decreasing 5 mins ago, by Leaky Nun so its derivative, if it exists, must be always >= 0 or always <= 0 Okay , so it's not a one-one function as derivative changes it's sign multiple times. exactly But suppose if i have$Q(x) +sin(x)$, after differentiating I will end up with a$Linear(x) + cos(x)$, how can i solve it ? Is it by default one-one ? 9:28 PM well a linear polynomial must go to -infty in some direction and +infty in the other direction @Binod$f(0)=0$,$f\left(\frac{2\pi}3\right)=\frac\pi3+\frac{\sqrt3}2$,$f\left(\frac{4\pi}3\right)=\frac{2\pi}3-\frac{\sqrt3}2$. Note that$f\left(\frac{2\pi}3\right)$is greater than either of$f(0)$and$f\left(\frac{4\pi}3\right)$. I got a simple question. I am trying to show$cl(E) \subset E \cup E'$. To show$x \in cl(E)$is in$E$at least. I wrote that for any open set$N$, the intersection$N \cap E$is non empty, but$N \cap E \subset E$and I can take the open set to be as smalls possible. I just want to confirm if this step is right @Binod A continuous, injective function cannot have an interior maximum. @robjohn Can I make an assumption if the value of f(x) does not decrease when x goes from 0 to$2\pi$, it will be one-one in the domain R ? (In general period of the trigonometric function ) @Binod If it has period$2\pi$, then that sounds good. You also have to make sure that it does not stay constant on any interval so non-decreasing is not enough 9:37 PM @robjohn Thanks . Hi @robjohn Hi everyone. I was doing a question on limits . $$\lim_{x\rightarrow0} [(x+1)/5]$$ []: greatest integer function, can i take 5 out of the limit ? Not in general, because the [] function is not a linear function. Why would you want to, anyhow? @TedShifrin no particular reason . Try doing this problem correctly and then doing what you want with the 5. Do you get the same answer? 9:50 PM @TedShifrin hey Ted! how r u? hi @Stan I have another question, this one is a bit vague. I've been reading some theorems in measure theory and it looks like half the time the point of "almost xxxx" is there just because it can be inserted. Is it just to let everyone know "this can be made weaker"? Or stronger, you mean? Depends if it's a hypothesis or a conclusion. hypthoesis usually Then the result is stronger. 9:51 PM @TedShifrin Sir is this the same as above :$[\lim_{x\rightarrow0} (x+1)/5]$? @Bhavay: Not necessarily. Why don't you just work the problem as it's written? Take$x$with$|x|$very close to$0$and evaluate$[(x+1)/5]$. I got a simple question. I am trying to show$cl(E) \subset E \cup E'$. To show$x \in cl(E)$is in$E$at least. I wrote that for any open set$N$, the intersection$N \cap E$is non empty, but$N \cap E \subset E$and I can take the open set to be as smalls possible. I just want to confirm if this step is right. My def of closure is if x is in$\cl(E)$, then for any open set N(x), it also contains a point in E. What you said is just wrong. @TedShifrin but then i don't understand what's the difference btw the two . "To show$x\in \text{cl}(E)$is in$E$at least." Is totally wrong. @Bhavay: Because you're treating [] as a continuous function, which it is not. 9:54 PM @TedShifrin thank you for your comment about the bilinearity of the inertia tensor. really saved me a lot of mistakes worked perfectly when i tried a few simple examples Try what you said if you remove the$1$and the$5$. Is it true that$\lim_{x\to 0} [x] = [\lim_{x\to 0} x]$? @Stan: It should work perfectly :P Knowledge is powerful, eh? @TedShifrin Sir is this the method for question 1 or question 2 ? is the latter stuff i wrote correct? @TedShifrin truly truly amazing. i had no idea how powerful vectors were til i did this project. i mean, i knew they were, but literally every single formula i learned in my calc class at caltech i've used on this project @Bhavay: I stated the question in that line, didn't I? 9:57 PM @TedShifrin Ah , i see they are not .$\lim_{x\to 0} [x] = D.N.E $, while for$[\lim_{x\to 0} x]=0$. @Hawk: I can't read the rest if you're trying to prove something wrong. Are you saying those words prove the wrong thing? @Bhavay: Precisely. It's all about continuity. Now,$[]$is continuous at$1/5$, so it will work in the original problem, but you're making dangerous mistakes if you don't write a justification in every case. i wasn't going for a full proof. You don't understand what it means to prove$x\in A\cup B$. You can't prove it's in$A$. @TedShifrin I see , thank you for the help. I have one more doubt though.. @Stan: Of course vectors are powerful, both in physics and in math. Did you do some of those geometric proofs in my linear algebra book in the first chapter? I forget. 10:01 PM @Binod I hope that helps. okay so what i left out was the other part. I meant if$x \in cl(E) \implies N(x) \cap E \neq \empty$. If$x \not\in E$, then x \in$E'$. If$x \in E$, then we are done @TedShifrin hey there... @Hawk: If you want to prove that$A\cup B\subset C$, then you can do two cases. Show$A\subset C$and show$B\subset C$, but this direction you just cannot separate. Aha, @Hawk, now you're writing something that makes sense. @TedShifrin I did. Your book is terrific :) no i m thinking about the other inclusion$C \subset A \cup B$. i was trying to argue how it might also be in "A" if it is not in "B" that's why i left out the accumulation set in the beginning 10:02 PM What you wrote did not make sense. But now your thinking is clearer. is the new thing i wrote complete for$C \subset A \cup B$? You start with a point of the closure. If it belongs to the set already, we're fine. If not, you use your definition to conclude that it's a limit point. Therefore the point belongs to the union ... that's what i wrote No, go reread what you wrote. Part of becoming a serious math student is being self-critical and rereading. "$x \in cl(E) \implies N(x) \cap E \neq \empty$. If$x \not\in E$, then x \in$E'$. If$x \in E$, then we are done" bein in the limit point means still that N(x) \cap E\{x} is nonempty 10:08 PM That's what you wrote after I complained loudly.$\lim_{x\rightarrow0}\sqrt{(x^2+5x+7)} -\sqrt{x^3+9x^2+8} $, My book write since the effective degree of the polynomial is not same$\rightarrow \infty - \rightarrow \infty = D.N.E$. But isn't$\rightarrow \infty - \rightarrow \infty $an indeterminant form and we should tried manipulating it first before reaching to the conculusion that the limit d.n.e ? I wrote a summary in words, rather than in symbols. yeah so we are saying the same thing Yes, @Hawk. I just want you to be careful not to write the nonsense you wrote the first time. OK, we're done with this. yeah the first one i wrote i skipped some details 10:09 PM @Bhavay: That looks like garbage to me. That's in an actual textbook? No, @Hawk, the first one was WRONG. Just admit it. cuz i wasn't sure if i need a longer argument for the set E. but thank you. I explained twice why it was wrong. no i know its wrong @TedShifrin Tbh it was in a youtube video .. OK, so don't say "I skipped some details." @Bhavay: You get what you deserve if you trust YouTube videos. (Perhaps that includes mine.) 10:10 PM youtube vids are gonna be what u have to depend on this semester.... Sir what's your take on the above question ? @Bhavay: Be patient. You wrote$x\to 0$. Do you mean$x\to\infty$? and probably the whole year.... Even if you have remote learning, isn't your professor doing lectures or presentations of some sort, @Hawk? opps , sorry i meant$x\to\infty\$10:12 PM honestly the process is so confusing for me i don't even know what is going on OK, now is it legitimate to say the limit is$\infty$or$-\infty$? Does your course allow that to be an answer? For example, do you say that$\lim_{x\to\infty} x^2$DNE or$=\infty$? anyways ciao @Bhavay gl with ur progress. might return later @TedShifrin DNE . Because with the problem you posted, I would say the limit is$-\infty$. DNE is very vague .... it allows for infinite limit, negative infinite limit, or no limit at all. So what happens if you do your usual conjugate trick with the square roots? @TedShifrin Are you saying limit is$- \infty\$ , as rate of decreasing of the second polynomial is faster than the first one ?

10:15 PM
Yes, if one refers to such things. You can see that with the conjugate stuff for sure.

@TedShifrin Thank you sir.

10:30 PM
@TedShifrin: I have posted the egg with some more graphics.

Cool, @robjohn. I'm in the middle of something right now but shall look.

@Thorgott springer.com/gp/book/9780817646769 is this the book you speak of?

yeah

Thanks, have you read it? is it any good?
@Thorgott

10:48 PM
I've read parts of it (though not the chapters dealing with currents) and another book from them and I usually find their exposition to be very clear and digestible. Your mileage may vary, of course.

@Thorgott Krantz is a good teacher and writer. I had him for my undergrad analysis, complex analysis, and Fourier series courses.

very cool. thanks guys

Hi there :)
How do you read this formula with the argmax?

1

2015 was a milestone year for AI--"deep learning" was validated in a very public way with AlphaGo. However, at the time, the question was raised: "What else is deep learning good for?" 5 years later, I want to gauge: How is deep learning applied to real world problems in 2020? What real world ...

11:03 PM
I know that this O with a belt is a "weak leaerner model"
the gay T is the parameter
soo how to read that? hmm and what does the star over the j on the O with belt means?

ah, that's cool

@nbro wanted to upvote, but no reputation :( -its a "good" question. I think the next step is somehow to connect all those small AI models together.. they still don't get context.

@Suisse There is or was a lot of hype around deep learning, but is it really being used to solve important real-world problems? Well, I know of at least one important application, so I will provide an answer, but I would like to see more interesting answers and use cases.

@nbro I only know Tesla self driving thing. And also recognizing cancer on CT images. I am currently doing "predictive maintenance" with random forest. Then you can count all speech recognition which is used.. but I don't see more realworld stuff there to be solved by AI.. there are a lot for sure

11:50 PM
I just provided an answer ;)
@Suisse Btw, feel free to create an account on AI SE! :)

@nbro there is an AI SE?? :D cool

Yep! I am a moderator there :)

cool :) .. why am I a noob everywhere I go? hehe