i think "convex" is doing most of the work here, but i will defer to copper on that.
thats a really interesting observation though, because quantitative properties of the normed space like the modulus of convexity can be pretty bad. basically, arbitrarily bad, according to this theorem.
i think there's a theorem kinda like this about which functions can be a modulus of convexity. its the absolute limit of what you would expect.
> Suppose Rick's mother wants to give 5 whole fruits to Rick from a basket of 7 red apples, 5 white apples, and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Rick's mother can offer 5 fruits to Rick is
i agree that this is a creepy problem, why is it coming at me with this stuff about rick and his mom like i need to know about that relationship. i don't walk up to people i don't know on the street and give them that kind of energy.
last night i gave munchkin a peanut butter and strawberry jam sandwich, four wheat thins, maybe 8 almonds, a piece of chocolate, a handful of raspberries, and four strawberries.
she's like a roman emperor, when she's done with one thing she just shouts out some other thing that she wants.
one time she responded to my wife asking her about her day with one word: ALMONDS!
an explicit argument should proceed along the following lines: show that for $t<r$, we have $tx$ in the interior of $A$ (this will require a bit of work), then use continuity to get an estimate on the limit from one side. repeat same thing with exterior to get an estimate from the other side and put them together.
something like this holds in greater generality: if $C$ is convex, compact and $x$ is interior, the map from the standard unit sphere to $(0,\infty)$ sending a $y$ to the unique $t$ such that $x+ty\in\partial C$ is continuous
(you can drop compactness and get a continuous function into $(0,\infty]$, even)
@Koro If we consider a 3 by 3 matrix B, whose ith row is ($l_i,m_i, n_i$). Then note that $BB^T =I$. So $B^TB=I$ (because $B^T$ is the inverse of B). If we look at the entries of $B^TB$ and compare them with the entires of I. This totally gives me the result. This is what you meant, right? Thank you soooooooo much!!!!!! **I do get it now!, And I was able to understand it completely, thanks to you** I am facing a deficit of words to thank you... 😊 I remain indebted to you.
The definition in Armstrong is as follows (this is my understanding of the definition) : $Z_p$ acts on S^3 (viewed in C^2), the quotient space $S^3/Z_p$ is called a lens space and is denoted by L(p, q).
No mention of q is what confuses me. There is an alternative definition of L(p, q) in the exercises. The exercise has explicit mention of q.
But the exercise is to show both these spaces homeomorphic. But I don't understand this because the first definition is not clear to me.
According to the first definition: L(p, 1)= L(p, n) for all $n\in \mathbb N$.
I know two definitions of the Lens Space $L(p,q)$:
Take $S^3\subset\mathbb{C}^2$, and consider the $\mathbb{Z}_p$ action $T(z_1,z_2)\mapsto (\omega z_1, \omega^q z_2)$, where $\omega = e^{\frac{2\pi i}{p}}$ and $T$ is the generator of $\mathbb{Z}_p$. Then $S^3/\mathbb{Z}_p$ is $L(p,q)$.
Take tw...
Can anyone please help me with this: Prove that if a normal subgroup of $A_n$ contains even a single $3-cycle$ it must be all of $A_n.$ I am not getting how to solve this. I know there is a particular thread in this site math.stackexchange.com/questions/4141444/… concerning this topic, but I dont have any idea about the lemma used there. So I dont get the solution there at all. Can anyone suggest me a solution for this....
i don't think it engages with the material in a way that lets you formulate generalizations of the material. it is probably fine for the basics of point-set topology, but so would any other topology book. same way chapter 2 of rudin is a kind of decent topology book.
A p-set made me prove the following: \begin{align} \sum_{k=0}^\infty\frac1{2^k}\binom{n+k}k&=2^{n+1}\\ \sum_{k=0}^n\frac1{2^k}\binom{n+k}k&=2^n \end{align}I know how to do the first one (in fact $\sum_{k=0}^\infty\binom{n+k}kx^k=\frac1{(1-x)^{n+1}},$ so we may substitute $x=\frac12$), but I have no idea how to do the second one - at least, not in an elegant way. I managed to do an ugly induction proof.
at some point you have to say 'who cares lol' and focus on stuff that you actually care about for some other reason than someone said it's in a textbook you should read or a body of material that someone says you should cover. for me personally, armstrong did not pass that threshold.
Munkres does have that and I learnt from Munkres but look at me getting stuck at pretty much every exercise on quotient topology in Armstrong ( I feel some progress in me but I want to reduce no. of times of getting stuck).
Armstrong discusses universal property of quotient topology a bit differently (I understand that that is same As Munkres). But I think the way it was said in Armstrong will never make me forget it. I absorbed it /learnt it :).
Consider the sequence $0<a_1<1$ and $a_{n+1}=\cos a_n$ for $n\ge 1$. I showed by induction that $0<a_n<1$ for each $n\in\mathbb{N}$. Now I want to show that it is Cauchy, can someone check my reasoning about Cauchyness, please? Let $\epsilon>0$. Since $\lim_{k\to+\infty} 2(\sin 1)^k=0$, there exists $k_\epsilon\in\mathbb{N}$ such that for each $k\in\mathbb{N}$, $k \ge k_\epsilon \implies 2(\sin 1)^k<\epsilon$. So, in particular, it is $2(\sin 1)^{k_\epsilon}<\epsilon$.
Let $n,m\in\mathbb{N}$ such that $n>k_\epsilon$ and $m>k_\epsilon$. It is $|a_n-a_m|=|\cos a_{n-1} \cos a_{m-1}|=2\left|\sin\frac{a_{n-1}+a_{m-1}}{2}\right|\cdot\left|\sin\frac{a_{n-1}-a_{m-1}}{2}\right|$.
Since $0<a_n<1$ for any $n\in\mathbb{N}$, it is $0<\frac{a_{n-1}+a_{m-1}}{2}<1$ and so $0<\left|\sin\frac{a_{n-1}+a_{m-1}}{2}\right|<\sin 1$ and $|\sin x|\le|x|$ for each $x\in\mathbb{R}$, it is $2\left|\sin\frac{a_{n-1}+a_{m-1}}{2}\right|\cdot \left|\sin\frac{a_{n-1}-a_{m-1}}{2}\right|\le \sin 1\cdot|a_{n-1}-a_{m-1}|$.
@SineoftheTime I see, unfortunately I don't know anyone currently enrolled there. The masters program is supposed to be very good from what I've heard though. Where are you studying at the moment if you don't mind sharing?
I take a rectangle $[0,1]\times [0,1]$. I define ~ as follows: (0,y)~ (1,y) for every y in [0,1], (x,1)~ (x,0) for every x in [0,1], (x,y) ~ (x,y) for all x,y in (0,1). Then $[0,1]\times [0,1]$/~ is Mobius strip.
@Thorgott is it the following: a particle starts moving at the yellow boundary and follows the edge through red, purple and then meets back at the yellow boundary point where it started its journey. So the locus of that particle is the boundary circle ?
@AlessandroCodenotti I'm studying at the university of Ferrara, near Bologna. I tried to see the programm of the master at Pisa, but it's not clear. I'd like to have additional informations
I'm very far from actual analysis, so I'm not sure which places in Italy are good, apart from Trieste which is supposed to be very good for analysis and mathematical physics
But I don't know if I'd like to go to Pisa, expecially I'm intrested in the programm. However, I've heard that on average 0 students are admitted at the master
I define the following action of Z on $R\times [0,1]$: ($\forall n\in \mathbb Z, \forall x\in \mathbb R, \forall t\in [0,1]$) n. (x, 0)= (x+n, 1) , n.(x,1)=(x-n, 0), n(x,t)=(x,t) for all x in R, t in (0,1).
but this looks wrong as 0.(x,0)= (x,1), which should have been (x,0) so this . is not an action.
If x ∈ Q and x = sup{q ∈ Q | q > 0, q^2 < 2} then x > 0 and x^2 = 2.
Proof: Let E equal the set on the right hand side, and suppose x ∈ Q such that x = sup E. Then, since 1 ∈ E and x is an upper bound for E, 1 ≤ x =⇒ x > 0. We now prove that x^2 ≥ 2. Suppose that x^2 < 2.
Define h=min{1/2,(2-x^2)/(2(2x+1))}.
I am curious why h is defined is it is.
Epically why is there 1/2? and why compare it with (2-x^2)/(2(2x+1))?
@Koro think geometrically, you want each $[0,1]\times[0,1]$ to become a Möbius square and all of those to be identified, that means you should twist the second-coordinate every time you traverse an entire interval in the first coordinate
as for the topology, in the first case, the open sets are of the form $U\times\{0\}\cup V\times\{1\}$ with $U\subseteq X$ and $V\subseteq Y$ open respectively
in particular, $X\times\{0\}$ and $Y\times\{1\}$ are disjoint open subsets of the disjoint union and they are homeomorphic to $X$ and $Y$ respectively
I proved it like this: Take the quotient map p: R-->R/Q. Let $p^{-1}(U)$ be an open set in R for some U subset of R/Q. Suppose that U is non empty. There exist rationals a and b, a<b, $(a,b)\subset p^{-1}(U)\implies p(a,b)\subset U, p(a,b)=\{[c]: c\in (a,b)\}$. Take any r in R. There exists an s in Q, r-s is in (a,b). It follows that there is a $c_r$ in (a,b), $[c_r]=[r]$. It follows that $R/Q=\cup_{c\in (a,b)} [c]\subset U$, whence $U=R/Q$.
Hence, the only open sets in R/Q are $\emptyset$ and R/Q.
@AkivaWeinberger This actually comes up in differential geometry in a somewhat unimportant place. But you'll find that and proof here.
@CuriousMind Of course, supposing the sup of $E$ is in $\Bbb Q$ is a mistake! You know it cannot be. In general, it's better to figure out your own proof than to read someone else's. You need $h>0$ so that $(x+h)^2<2$, showing that $x$ cannot be an upper bound of $E$ at all. Now just figure out your own $h$ that will work. You need $2xh+h^2<2-x^2$.
@Gwyn This seems way too messy to me. Can't you show that $|a_{n+1}-a_n|\le c|a_n-a_{n-1}|$ with $0<c<1$. Then a geometric series easily shows that the sequence is Cauchy.
@Curious It's like most limit proofs. You need a preliminary assumption to get bounds when you have something more complicated than a linear function. You want to bound $h(2x+h)$. I would just do something simpler. We know $x<2$. If we restrict $h<1$ (which of course is ridiculously big), then $2x+h<5$. So you just need $0<h<(x^2-2)/5$.
Thus, I need $h<\min(1,(x^2-2)/5)$. Like in "all" limit proofs.
They are doing the same thing, but leaving $x$ in there and using $h<1/2$.
To be honest, I could not begin to spend an hour reading it.
I think bogging down in trig identities is a mistake.
We need just to know that we're in a closed interval where $|\sin|\le c<1$ and then invoke the Mean Value Theorem. That's always the tool you should think of.
@TedShifrin by the way, I came across your lecture video on YouTube about matrices where you explained row operations in detail. I gotta you explained it really well, helped me review matrices
It was only when I started teaching these things that I understood that certain elementary matrices commute and can be combined nicely. This makes figuring out $LU$ much easier :)
ted is spreading fear, uncertainty, and doubt about the soundness of lesliecoin. probably because he is himself in arrears to his publisher for the lavish advance they gave him on multivariable mathematics, which they have yet to recoup, because his book sucks.
@Gwyn you might consider $\cos(\cos(x))-\cos(x)=2\sin\left(\frac{x+\cos(x)}2\right)\sin\left(\frac{x-\cos(x)}2\right)$ and think about contraction mappings.
Even at MIT there were numerous students who had to take 18.100 (Rudin real analysis) numerous times to get a good grade when I was an undergrad in the early 70s. Soon thereafter, they created a more accessible version of the course — presumably intended for students not planning to go to do a PhD in math.
My professor for the course was a probabilist (now deceased) who never drew a picture (just like Rudin) and who spent the first weeks of the course "motivating" Dedekind cuts by trying to define arithmetic carefully with infinite decimals. Suffice to say, this was not my favorite math course during my career.
I actually never taught that course in all my years. I had a lot more fun teaching analysis in the context of Spivak and then my multivariable math book. That was something like 27 years of teaching analysis. :)
Suppose that I take a Mobius strip and identify its boundary circle to a point, then am I right in saying that the resulting identification is two disks connected at a point?
I used to sit near the first row or in the first row but now I prefer to sit close to the last row as I find attending classes at my college a complete waste of time. So sitting there helps me self study and the attendance issue is also resolved. If there is no attendance in a class here, I avoid attending that class.
@ペガサスSeiya Of course. I had to produce individually printed exams for one student (with 36-point type, or something like that) because his vision was so bad. This guy now has a PhD in CS and is himself a professor! :)
@Koro What I described above is the usual description of $\Bbb RP^2$.
One way of checking that this is correct is that one of the common descriptions of the projective plane is that it's a Möbius strip with a disk attached to the boundary circle.
I see why it will be true: I take a rectangle $[0,1]\times [0,1]$ and identify the top side to a point, and the bottom to another point- we get a disk.
@robjohn How about, just for Feb 14th, you change from frown to smile. Just that day?? Maybe, and of course, you are free to ignore such suggestions :-)
@Thorgott I think in general one can say that: If X is compact Hausdorff, A is closed in X, then X/A is homeomorphic to one pt. compactification of (X-A).
@robjohn What I wanted to ask was: is this police a private police (like some firm provides the security staff) or this police is the govt. police (being paid by the US govt.) and can make arrests?
I know two definitions of the Lens Space $L(p,q)$:
Take $S^3\subset\mathbb{C}^2$, and consider the $\mathbb{Z}_p$ action $T(z_1,z_2)\mapsto (\omega z_1, \omega^q z_2)$, where $\omega = e^{\frac{2\pi i}{p}}$ and $T$ is the generator of $\mathbb{Z}_p$. Then $S^3/\mathbb{Z}_p$ is $L(p,q)$.
Take tw...
