@BalarkaSen I know that you will see this later but I think that after looking, Hatcher has way better examples that have helped me in the introductory chapters. So just a thought.
user105491
00:50
@BalarkaSen Although more adavanced than introductory algebraic topology, it suffices to prove that $S^n$ and $S^m$ are not homotopy equivalent. (Not that you don't know this, of course.)
julian, have you done a solid undergraduate algebra, like Artin, thoroughly? You should. And you too should not be skipping rigorous calculus (Spivak + my book). You all are in too much of a hurry and jump into the deep end when you shouldn't.
Well, sitting there telling yourself you can't do it is a terrible idea. I told @Anthony the same thing. Belief and love of the subject are important. I can only try to help you with approach and intuition.
Seriously, a friend is coming for dinner, but I'll be back later, unless I get banned again.
I don't mind taking it slow, it's just that nothing else slows down for my learning speed, which is the only frustrating part (e.g. classes moving into the next topic when I'm still behind 2 topics, etc.)
@TedShifrin Hi Prof Shifrin, do you by any chance have a list of homework assignments that went along with the MATH 3500-3510 course that is available on youtube?
@TedShifrin I was asking my topology prof today if I could do a review class for our topology to cover the stuff we learned since we have a midterm next week so it would be nice experience for me to make sure I have understood stuff 100 %
I think I will make it like 5 hr review class as I want to cover all of chapter 2 of munkres
quotient topology,Box topology,product topology, continuous functions, basis and subbasis of topology, etc
@Semiclassical, it's is important to give timely and constructive feedback. I've always tried to get stuff returned very fast. The grading is no less painful when you procrastinate.
Not linear alg, Karim. Multivariable + linear, integrated.
Can somebody please help me fill in a long-standing gap in my algebraic understanding? Consider $3t^2=27$. I was never taught that I need to divide by 3 before square-rooting both sides to get the right answer of $3$, rather than $sqrt(27)/3$. I'm dying to know what this rule is (more than just operator precedence).
Sometimes coefficients have weird properties and I used to get horribly, horribly confused over them (as if they were more than just a number). Some of those misinterpreted properties have propagated through my math. And I usually only apply these "properties" in the exact sort of equations where I first accidentally committed them to memory, which makes it woefully hard to hunt down. I do most of them right, but I've trained myself to do some, even basic, problems incorrectly. Huh.
At least I get to shoot down another misconception; thanks.
The idea is that both sides are equal, so they are 'the same thing', so whatever you do to one, you must change the other identically to keep them equal. So any operation must effect everythng on both sides, sorry if this is annoying
is there a notion of weak convergence in a normed space that doesn't have an inner product?
currently I know of weak convergence to mean $\lim_{n\to\infty} \langle x_n, x\rangle = \langle x,x\rangle$ for $x_n \rightharpoonup x$, so I wouldn't think so, but a classmate asked me this and I am not sure if there is some other notion
Math is vast and is continuously becoming more vast. How can a learner make sure that he is continuously growing his mathematical maturity while studying math formally?
@SanathDevalapurkar yeah, but obviously Julian would have to learn fundamental groups before homology and you cannot prove S^m and S^n are not homotopy equivalent without homology. you need to go step-by-step: jumping straight onto a certain fact would do no good for a beginner in algebraic topology.
so, I don't really see the point of stating that fact.
@JulianRachman monomorphism? epimorphism? how does that even relate to being or not being homeomorphic?
@Tien-ChengHuang Not sure how your beginning statement and the following question are related. For the latter you should be moving over the text you are reading with a critical mind. What are the implications of what you are reading etc. Most people don't have time for this admittedly
Hi can anyone answer this question?Solve initial value problem Q=a cos (1/24)t + b sin (1/24)t when Q (0 )=4 and Q '(0)=0. I've tried but the answer I get is incorrect
What could we change at the following definition so that $\text{EXP}(\mathbb{C})$ is a ring?
"We define EXP($\C$) to be the the set of expressions \begin{equation}\label{a} a=\alpha _0+\alpha _1e^{\mu_1z}+\dots +\alpha _Ne^{\mu_Nz} \end{equation} (beyond the `zero function', $0$, which we will consider to be also an element of EXP($\C$)), where $\alpha_0, \alpha _1,\dots, \alpha_N\in \C\setminus \{ 0\}$ and $\mu_i\in \C\setminus \{ 0\}$; in writing such an expression we will always assume that the $\mu_i$ are pairwise distinct."
For a compact Riemann surface $X$, let $\operatorname{Cl}^{1}(X)$ be the divisor classes of degree $1$. If the map $X \to \operatorname{Cl}^{1}(X)$ given by $P \mapsto [P]$ is injective, why then follows $X \not \cong \mathbb{P}^1$?
I tried to do this by contradiction, assuming that $X \cong \mathbb{P}^1$
Also one of the previous exercises gives that if $X \not \cong \mathbb{P}^1$ then there are no non-constant meromorphic functions $X \to \mathbb{C}$ with only 1 pole of order 1
(For those that have read my previous question): I have reduced it to finding is a non-constant meromorphic function $f: \mathbb{P}^1 \to \mathbb{C}$ with a root in $P$ and a zero in $Q$ for $P,Q \in \mathbb{P}^1$
at this point, though, my desire is less "find some interesting new result" so much as "put everything we've done up to know in a more precise language" a la OPs
so translation/transcription rather than creation, i guess
I might have mentioned that I went to a session at OPSFA chaired by Peter Miller on RH in relation to asymptotics for DEs, in particular nonlinear Schrödinger and related equations
I thought that stuff was fascinating, but I probably won't pursue it for a while
It is called abridged co-ordinates. It is generally used in conics/cubics to prove interesting theorems in line co-ordinates. The idea is to write a curve as $C = C_1 + \mu C_2=0$ where $C_1$ and $C_2$ are two other algebraic curves. The curve $C$ will pass through $C_1 \cap C_2$.
I am not able to find geometry books which explain the applications of this technique.
I am currently attending an undergraduate analysis course assigning Marsden's "Elementary Classical Analysis" as the textbook. If I want to complement my study by some chapters from baby Rudin, which chapters will you recommend?
@AntonioVargas no problem. i guess complement with chapter 1-4 of baby Rudin is enough. (in fact i read through this part years before and did a few exercises)
that study group on Reddit seems want to work through all exercises
Please help me solve this Fourier series and correct my solution if it is wrong. it's a non-periodic function which we need to write its Fourier series (even and odd) :
$
f(x)=\pi - x
$
;
$
0<x<\pi
$
I have reached cosine extension(even) as follows:
$
\phi(x)= \begin{cases} \text{$\pi-x$ ; $0<x
So no I haven't. It looks like you find a solution to a problem and then keep looking for a better one. This problem either has a solution or doesn't. I am just looking for the whole solution set (ideally).
Quick question. Right after the definition of topological $m$-manifold with boundary, I came across a statement: If a point $p \in M$ is mapped to the boundary in one chart, then it is true in every chart. Is this obvious? I tried to prove it and the only way I could figure it out was to take homology groups after removing the points in the images of the chart maps and showing they don't agree if one point was on the boundary and the other wasn't
@Agawa001, it is from assuming that the top row is 1,2,3,4,5 and finding that there are only 2 solutions. then there are 5! ways to permute the top row.
Actually, I think it is that simple. If one chart maps the point to the boundary, the image is an $m - 1$ manifold, if it doesn't, it's an $m$-manifold. By invariance of domain, they can't be homeomorphic, contradiction.
We consider the finite difference method for the approximation
$\left\{\begin{matrix}
-u''(x)+q(x)u(x)=f(x)\\
u'(a)=u'(b)=0
\end{matrix}\right.$
and let $K$ be the $(N+2) \times (N+2)$ matrix of the method. Let $v \in \mathbb{R}^{N+2}, v=\begin{pmatrix}
v_0\\
v_1\\
\dots\\
\dots\\
\dots\\ ...
I am revisiting a problem: Looking for a "ping-pong lemma" proof that $\langle x \mapsto x+1,x \mapsto x^3 \rangle$ (as automorphisms of $\mathbb{R}$) generates a free group on two generators
@PaulPlummer I have been thinking a bit about the cellular approximation theorem nowadays. It's a research problem in Kirk-Davis, chapter 1. I have a few ideas, but I can't get it to work.
$f : X \to Y$ be a map between CW-complexes. Then $f$ is homotopic to a map $g : X \to Y$ such that $g(X^n) \subseteq Y^n$, i.e., $g$ takes $n$-skeleton to $n$-skeleton for every $n$.
In fact, given any CW-complex $X$, this shows $\pi_k(X) \cong \pi_k(X^{k+1})$.
Where $x^{k+1}$ is the $k+1$-th skeleton of $X$.
It's because you can cellular approximate maps from $S^k$ to get it down to $k$-skeleton and you can cellular approximate homotopies, which are maps from $S^k \times [0,1]$, to get it down to $k+1$-skeleton.
There is a seminar today on: "Quasimorphisms on right-angled Artin groups", and I guess quasimorphisms are morhpisms $\phi : G \to \mathbb{R}$ where $\sup | \phi(a)+\phi(b)-\phi(ab) | < \infty$, which just looking at it seems like it is related to bounded cohomology... It is cool to see ideas after you learn of their existence (even if I still don't know anything about it)
I think it gives a lot of us discomfort. Sometimes it is nice though, because you get a feeling for why you would want to know the theorem in the first place without doing the hard work of proving it @BalarkaSen
@Agawa001, I see your saying they go hand in hand. Yes. Just wondering if you have a favorite problem that you use them both for that you have been working on.
@Agawa001, btw my computer went dead earlier that's why I suddenly dropped the conversation. sorry.
@BalarkaSen Haha. I have not been doing much geometric group theory... Sort of recently I proved that $C'(1/6)$ satisfy the linear isoperimetric inequality. That is if $w=1$ in a group then the min $n$ such that $w= u_1r_1^{\pm 1} u_1^{-1}...u_n r_n^{\pm 1} u_n^{-1}$ is called the area $A(w)$, where $r_i$ are relation in the group, satisfies the inequality $|w| \leq kA(w)$ for a fixed $k$. I got to go, maybe I can say more about it when I get back (the proof is sort of clever, but needs a bit of
Please help me solve this Fourier series and correct my solution if it is wrong. it's a non-periodic function which we need to write its Fourier series (even and odd) :
$
f(x)=\pi - x
$
;
$
0<x<\pi
$
I have reached cosine extension(even) as follows:
$
\phi(x)= \begin{cases} \text{$\pi-x$ ; $0<x
We consider the finite difference method for the approximation
$\left\{\begin{matrix}
-u''(x)+q(x)u(x)=f(x)\\
u'(a)=u'(b)=0
\end{matrix}\right.$
and let $K$ be the $(N+2) \times (N+2)$ matrix of the method. Let $v \in \mathbb{R}^{N+2}, v=\begin{pmatrix}
v_0\\
v_1\\
\dots\\
\dots\\
\dots\\ ...
