@Astyx This is typical in US calculus courses as well (one makes a copy of the real axis and marks intervals where the derivative is $+$ and $-$, and then, say, on the underside, where the second derivative is $+$ and $-$). We don't have a fancy word.
Hi. I'm having trouble understanding a problem statement. It says to consider, for $n\geq 3$ a space $X$ given as the wedge sum of (n+1)-circles, where there is one central circle, and then n circles tangent to this circle around it, which are wedged together pairwise to form a nice chain of length n, and each of these are each wedged at the nth roots of unity. It says that multiplication by roots of unity restricts to an action on Z/n on this space, such that the quotient map X-->(Z/n)\X is...
a covering space
(Hopefully that description is clear, since I have a picture instead of a description)
What I don't understand is, I assumed the free action by Z/n just takes you from the circle which is attached at an nth root of unity, to the circle attached at the other nth root of unity
And it seems that (Z/n)\X is just a pair of circles (the main one, and whichever tangent one you want)
but then you have n points in the fibre over any tangent circle, but only one point in the fibre over any point on the main circle, that isn't a root of unity (but X seems to be connected, so this doesn't make sense, since there should be a well defined number of sheets over any open)
Basically my question above is, how would you interpret the induced Z/nZ action? If I take it to mean that it transports you between the circles at the roots of unity (by the identity), but does nothing to the main circle, then that doesn't actually appear to make the quotient map a covering map, which it apparently is, so presumably that isn't the correct free (Z/nZ)-action induced by the n-th roots of unity. (Also if it wasn't entirely implicit, the main circle is the unit circle in C)
Hmm, maybe I'm being dumb, and the action just comes from the n-fold symmetry, so that the action does apply to the main circle (by taking those arcs between roots of unity to other arcs between roots of unity), that fixes everything probably
I want to prove that if $P \subset Q$ then $$ U(f,P) \geq U(f,Q)$$
My attempt: Let’s assume that $Q$ contains just one more point than $P$ that is $$ P = (t_0, t_1, t_2, ... t_{k-1} , t_k, .... t_n) \\ Q = ( t_0, t_1, .... t_{k-1}, u , t_k , ... t_n)$$
Upper sum on $[a,c]$ is $$ U_1 = M (c-a) $$ and upper sum on $[a,b] \cup [b,c]$ is $$ U_2 = M_1 (b-a) + M_2 (c-b)$$ Where $M_i$ are the supremums in the i th interval
But how do we prove that $M \geq M_1 \\ M \geq M_2$$
But how do we prove that $$M \geq M_1 \\ M \geq M_2$$
@Knight that means that both $\sup\limits_{x\in[a,b]}(f(x))\le\sup\limits_{x\in[a,c]}(f(x))$ and $\sup\limits_{x\in[b,c]}(f(x))\le\sup\limits_{x\in[a,c]}(f(x))$
@TedShifrin Just day dreaming. Thinking something dumb things like proving some series to be {whole, Rational, Natural ......} trying to make theorem. Very dumb idea.
@Ted is right. Formulate the problem first before trying to prove it. It is true that 1 + 1/2 + ... + 1/n is never a natural number for any n >= 2, neither is 1/n + 1/(n+1) + ... + 1/(n+m) for any n >=1, m >= 1.
In general for sum of reciprocals of natural numbers to be an integer there are growth restrictions on the denominator. Sum of reciprocals of natural numbers which sum to 1 are well-studied; look up Egyptian fractions and the greedy algorithm.
@TedShifrin I dunno as time progresses I feel more attracted to classical math. I absolutely enjoyed thinking about Hilbert's theorem a few weeks back.
@BalarkaSen when you said classical math I was thinking of Euclidean geometry and other high school maths that are more concrete than the s*** I'm doing
@LeakyNun It's a good exercise to compute homology of the homotopy fiber of $\Bbb{CP}^n \to \Bbb{CP}^\infty$ using Serre spectral sequence, by the way. I worked it out last night without cheating.
I have to read Chapter 9, pseudospherical surfaces and Backlund transform. Speaking of, I never finished the Backlund computation. I will write it down sometime today
" I translated it like this" Doesn't give lots of information. It is your assumption.
And your assumption tells that you got a vector field which is 0. So indeed you will get 0 divergence and 0 curl Since There is no rotation and divergence going on.
If the special solution of a an inhomoginous differential equation is made of linear combination of two functions, is then each function also a solution for the equation?
If we write the general solution for a differential inhomoginous equation as a superposition of a the inhomogionous solution and the homoginous. Does then each one of the terms that we added in itself a solution for this equation?
I have theoreticla physics and we didnt really do integration and such deeply but i need to know some basics to work
@TedShifrin A Conversational Introduction to Algebraic Number Theory: Arithmetic Beyond Z is the book you were talking about?
@TedShifrin looks like the book u mentioned about hardy is better since it looks like it covers almost all the stuff covered in book mentioned by BalarkaSen
and it covers fermat's last theorem ? looks interesting since my high school teacher told me his arse was kicked by reading his proof and theorem.
But I think I should learn abstract algebra and group theory before that monster.
Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the rotation around the origin with angle $\alpha$ and let $\sigma_{\alpha}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the reflection about a line through the origin that has angle $\frac{\alpha}{2}$ with the $x$-axis.
Let $v\in V$ and $\alph...
Another one is when you take the sequence where the $n$-th functions graph is a triangle of width $1/n$ and height $n$ (say, left basepoint at $0$ always)
The moral of those two counterexamples is that you don't want your sequence to "escape to infinity" in some way
And this is what is being precluded by assuming there is some integrable majorant
@Thorgott You increased it to measure theory lol. Yep for uniform convergent it is trivial just need to use some property of integral. For other cases I need to learn Measure Theory.
But I used more complicated proof to prove that crap. Will take time to convince you. 😂😂😂
I don't know what video games should I play now. Every video games I played just feels extremely boring. I have feeling that every progress I make will be temporary and nothing will be eternal. Just having the feeling of one day everything will diverge and become dark. Everything is just illusion of time.
What kind of stable singularities should maps $f : M^3 \to N^3$ between $3$-manifolds have? There should in general be a surface worth of fold points, lines worth of cusp points along those fold surfaces, and those lines meeting at swallowtail catastrophes
I would be surprised if there are other possibilities
You should imagine it as, consider two "oppositely oriented" cusps so you can run them togather and cancel by a homotopy $f_t : \Bbb R^2 \to \Bbb R^2$ so that $f_0$ has no cusps and $f_1$ has two oppositely oriented cusps
The movie of this has a cusp catastrophe at the point it cancels
Let's say $f(0) = 0$ for convenience. Stuff from calculus says: if $f'(0) \neq 0$ (or in higher dimensions $\nabla f \neq 0$) then $f$ looks like a straight line. To rephrase, there is a diffeomorphism/change of variables $\varphi$ of the domain so that $f(\varphi(x)) = L(x)$ is a linear function.
You also know from calculus that if $f'(0) = 0$ but $f''(0) > 0$ then $f$ "looks" like the parabola. More precisely (in 1D, I will not bother phrasing it in higher dimensions) we have that there is a diffeomorphism $\varphi$ of the domain so that $f(\varphi(x)) = x^2$.
What you're seeing is that you can describe totally the local behavior of a function given some constraints on their derivatives.
A singularity is the local behavior of a function at a critical point --- we consider two singularities equivalent if there is a diffeomorphism of the domain taking one to the other
So we see that the simplest singularities are local mins / local maxes but they get much more complicated
Balarka is talking about a specifically nasty kind of singularity which I think comes from a map R^3 -> R^3
A non-mathematical detour: Have you ever looked at ruffles in your bedsheet? Mostly they will look like paraboloid (a "fold"), or picture like the one I linked above (called a "cusp"). Why don't other kind of ruffles occur? This is because these are the only kind of ruffles/singularities which are stable under small deformations of the bedsheet - this is what Whitney proved a long time ago
It's impossible and extremely hard to see other kinds of singularities in nature
Mike gave a precise formulation of what singularities of maps $\Bbb R \to \Bbb R$ looks like ($x^2$). Those will be the most common ruffles in bedsheets of 1D creatures
Let $\delta_a:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the rotation around the origin with angle $\alpha$ and let $\sigma_{\alpha}:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the reflection about a line through the origin that has angle $\frac{\alpha}{2}$ with the $x$-axis.
Let $v\in V$ and $\alph...
