Let $x \in \Bbb R$ be arbitrary and let $m = m(n) \in \Bbb Z$ be such that $y := x - \frac m n \in \left(-\frac 1 n, \frac 1 n \right)$. Observe,
$$
f_n(x) =
f_n \left(x - \frac 1 n \right) =
f_n \left(x - \frac 2 n \right) =
\dots =
f_n \left(x - \frac m n \right) =
f_n \left(y \right)
$$
Since ...
I propose the following: Choose x<y arbitrarily. Since $f_n'$ are uniformly continuous on [x,y], given $\epsilon>0$ there is a $\delta>0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. Choose $N$ such that 1/N <$\delta$. Let P_N:={m: x+m/N <y}. P is bounded from above and therefore has a maximum element M.
are you using uniform continuity of f on all of R here? you start by saying " . . on [x,y]" but the statement about u, v doesn't seem limited to u, v in [x,y] (which is fine it's just an obstacle to a clean read)
M>= (y-x)N-1 so that $y-x-1/M<=1/N<\delta$. Now, $|f(x)-f(y)|=\lim_n |f_n(y)-f_n(x)|=\lim_n|f_n(y)-f_n(x+M/N)+f_n(x+M/N)-f_n(x+(M-1)/N)+...+f_n(x+1/N)-f_n(x)|=\lim_n|f_n(y)-f_n(x+M/N)|\le \epsilon$. Since this is true for every $\epsilon>0$, it follows that $f(x)=f(y)$.
@leslietownes yes.
I want it to be non empty. I think I should put some more restrictions on N for that to happen.
@leslietownes no, I am using the fact that since f_n's are continuous on compact set [x,y], f_n's are uniformly continuous on [x,y].
So I make P_N non empty by choosing 1/N < min (delta, y-x).
instead of choosing 1/N < delta.
with nonemptyness, Well ordering works.
In the exam, this question came with a hint: You may use the following fact: If $f_n\to f$ uniformly on R and $x_n\to x$, then $f_n(x_n)\to f(x)$.
Oh, I mistyped. I should have said: $|f_n(u)-f_n(v)|<\epsilon$ whenever $|u-v|<\delta, u,v\in [x,y]$.
While solving 2nd-degree differential equation using Auxillary equation (m^2 - 4m + 3) I got general solution as c1 e^x + c2 e^3x = y.. But if I still use order reduction then I got another solution as e^ (-2x)/2. Should It be included in general solution ?
Can someone maybe give me a short example where we use tensor products in practice? We have only seen the mathematical aspect without discussing the usefulness in "real life"
Suppose that $\{f_n\}$ is a sequence of continuous real valued functions defined on $\mathbb R$. Suppose that the sequence converges uniformly to a function $f$ and the following holds for all $x\in \mathbb R$ and for all $n\in \mathbb N$
$f_n(x+\frac 1n)= f_n(x)$.
Then, it is to be shown that $f...
The probability measures here are Borel probability measures on $\mathbb{R}^n$.
I am a bit confused between measures and their associated densities (i.e. w.r.t Lebesgue measure) mostly with regards to the continuity equation. In particular I seem to get two different solutions when I solve the eq...
I "just" want to know that this integral is projection onto this eigenspace. I kind of know nothing about functional calculus and so on but I know what unbounded operators are...is there a clean path to this result? preferrably in someone's notes or books?
Sorry, the direct product $A\times B$ is the set of ordered pairs; the free product $A* B$ is a set of finite words of elements in $A$ and $B$ up to some relations
That means: a function $G\to H\times K$ is the same data as a pair of functions from $G\to H$ and $G\to K$; a function $G*H\to K$ is the same data as a pair of functions $G\to K$ and $H\to K$
(assuming those are all group homomorphisms)
(if they're just regular set functions, the correct notions would be set product and disjoint union)
In any case, for $\Bbb Z_4*\Bbb Z_6$, first we need a disjoint way of writing the elements of each
Let's say $\Bbb Z_4=\{0,1_4,2_4,3_4\}$ and $\Bbb Z_6=\{0,1_6,\dots,5_6\}$
(it's OK for the identity to be the same in each)
Then an element of $\Bbb Z_4*\Bbb Z_6$ could be something like,
$3_42_61_45_6$
or $3_41_61_63_4=3_42_63_4$
If two adjacent parts of the word are from the same group, you can combine them
@AkivaWeinberger The last part is very helpful! I am in paricularly looking at quotient $\BBb Z_4 * \BBb Z_6 / N$, where $N$ is normal subgroup of the free product defined.
ah yes, sorry. The normal subgroup is defined as $<i_4(a) i_6(a)^{-1}| a\in \Bbb Z_2>$, where the homomorphisms are $i_4 : \Bbb Z_2 \to Z_4$, and $i_6: \Bbb Z_2 \to Z_6$.
@AkivaWeinberger Proof: Act by $\mathrm{SL}_2(\Bbb Z)$ on $\Bbb H^2$, and look at the orbit of $i$ and $e^{2\pi i/3}$. Connect $i$ and $e^{2\pi i/3}$ by an edge, and throw in the orbit of this as well. Bicolor it, with red on the orbits of $i$ and blue on the orbits of $e^{2\pi i/3}$.
Then you get the Cayley graph (with standard generators) of $\Bbb Z_2 * \Bbb Z_3$, and the action of $\mathrm{SL}_2(\Bbb Z)$ has stabilizer $\Bbb Z_4$ for the red vertices, $\Bbb Z_6$ for the blue vertices and $\Bbb Z_2$ on the edges connecting blue and red vertices.
The quotient graph (the graph of groups @AlessandroCodenotti mentioned) is a single edge.
I will use the one-step subgroup test.
Since $A,B\le G$, they are themselves groups and their identity is the identity in $G$. Thus $e\in A\cap B$. Hence $A\cap B\neq\varnothing$.
Since $A$ and $B$ are abelian, all elements $x$ of $A\cap B$ commute with all elements of $ A$ and of $B$. But $G=AB$...
Also yeah I checked - if $x^y=y^x$, both rational, and $x<y$ (so they're not equal), then there's a natural number $n$ such that $x=(1+\frac1n)^n$ and $y=(1+\frac1n)^{n+1}$
That's why, if you graph $x^y=y^x$, the trivial branch and the nontrivial branch seem to intersect at $(e,e)$
Can anyone see where I made a mistake? in this question math.stackexchange.com/questions/4474646/… it seems like if I solve the continuity equation for a measure and then transform to a density I get a different answer if I solve the equation for a density straight away.
I have to prepare a presentation of 90 min about modular representations and the cde triangle until monday and I dont have a clue of almost anything the book is talking about. Any handy survival tips?
During the study of the convergence of $\iint_{\{x \ge 0,y \ge 0\}} \frac{\arctan x^3}{x^4+y^4}dxdy$, my lecturer used polar coordinates and splitted the integral in the sum of two integrals: one in $B_1((0,0))\cap\{x \ge 0,y \ge 0\}$ and one in $\{x^2+y^2 \ge 1\} \cap \{x \ge 0,y \ge 0\}$.
He then said that the function $f(\theta)=\cos^4 \theta+ \sin^4 \theta$ has a positive minimum $k>0$ attained at some $\phi \in [0,\pi/2]$ for the extreme value theorem, hence we can conclude that the two integrals are convergent.
He claim the convergence because the integral in $\{x^2+y^2\ge1\}$ is less than $\int_1^{\infty}\pi/(2r^3)dr \cdot \int_0^{\pi/2}\frac{1}{\cos^4 \theta+\sin^4 \theta}d\theta$ and the integral in $B_1((0,0))\cap\{x \ge 0,y \ge 0\}$ is less than $\int_0^1 dr \cdot \int_0^{\pi/2}\frac{\cos^3 \theta}{\cos^4 \theta+\sin^4 \theta}$.
@robjohn Oh, I like that. Integration by parts is one of my favorite tools. It is (in my opinion) the best way to obtain Taylor's Theorem, and my graduate PDEs courses could easily have been retitled "IBP: The Musical".
My question is: do we need to bound $\frac{1}{\cos^4 \theta+\sin^4 \theta}$ from below with $\frac{1}{k}$ because we must be sure that the integrals in $d\theta$ are finite by excluding the possibility? If yes, isn't it enough to notice that the function $\cos^4 \theta+\sin^4 \theta$ is never $0$ because cosine and sin never vanished simultaneously?
I suspect this latter reasoning isn't enough because, even if they both don't vanish, I can't exclude the possibility that the denominator becomes small and so that the function is unbounded. Is this latter reason the reason why my lecturer int…
T_01: more seriously it's always helpful to focus on key definitions and examples than general results. if the book is mostly general results and you have the time it might help to find resources for a different audience. google found me a senior thesis on this topic that might be more readable than a book. some lecture notes or slides somewhere might be even better.
Yeah I read that senior thesis by arun debray (you mean that?) It made the base definitions more clear to me, however, I am not able to understand the examples. Also, the examples are on character level and I dont know how to "pull that back" to the module level of things
@robjohn I'm too lazy to read through all of the answers---do any of them mention the identity theorem (i.e. a complex analytic function defined on a set with a limit point has a unique analytic extension)? As far as I can tell, that is, perhaps, a relevant thing to say.
the beauty of mathematics is that you can't fake it
it's not like poetry, where you can gaslight your audience into intellectual submission. in math, theyll chew you out in no time. especially if your prof is the expert on the subject matter
T_01: any hope of walking in and "now we're discussing modular representations of the ABC triangle," drawing an equilateral triangle on the board, and saying that the rotation gives rise to a "representation" of the integers "modulo" 3?
@JoeShmo There was a Numberphile video a few years ago which essentially claimed that $\sum_{n=1}^{\infty} n = -\frac{1}{12}$. This "result" was demonstrated by manipulating divergent series in a way that doesn't really make sense. But the video got spread widely around, so the number $-1/12$ comes up a lot in contexts where folk don't really grok what is going on.
I just watched a lecture where a professor proved that buying the Euro with the Dollar and buying the Dollar with the Euro can both have a positive expected value given the same probabilistic model...
Is this somehow identical to the necktie paradox, or is something else going on?
@TedShifrin Hi, a question if I may: I have done Exercise 8.4-18 in the book Multivariable Mathematics but in the back of the book the solutions to (a) and (c) are missing (solutions to cases (b) and (d) are present and I get the same answer). For (a) and (c) are the following solutions correct? 8.4-18(a): $4\pi$ 8.4-18(c): $4\pi h\left(\frac{1}{\sqrt{a^2+h^2}}+\frac{1}{5}((a+h)^{5/2}-h^5\right)$. Thanks
@TedShifrin this is my reasoning for 8.4-18(a), and I think it makes sense: 8.4-18(a): We parametrize by $\mathbf{g}:(0,\pi)\times (0,2\pi)\to\mathbb{R}^3,\ \mathbf{g}\begin{pmatrix}\phi\\ \theta\end{pmatrix}=\begin{bmatrix}a\sin\phi\cos\theta\\ a\sin\phi\sin\theta\\ a\cos\phi\end{bmatrix}$ $\int_{S}\eta=\int_{S}\left(\frac{1}{(x^2+y^2+z^2)^{3/2}}\left(x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy\right)\right)=\int_{(0,\phi)\times (0,2\pi)}\left(\frac{1}{a^3}\left(a^3\sin^3\phi\cos^2\theta+a^3\sin^3\phi\sin^2\theta+a^3\cos^2\phi\sin\phi\right)\right)d\phi\wedge d\theta=\int_{(0,\phi)\times …
@TedShifrin ah, there was a mistake in a change of variables I did while computing the integral for the two "lids" of the cylinder, right at the end of the calculation. Now I get $4\pi$ also for this one
Basic question, I know how to generator symplectic isotopies: take a function $f : M \to \Bbb R$ on the symplectic manifold $(M, \omega)$, then look at the symplectic gradient $H_f$. This gives tons and tons of symplectic vector fields, and I can flow along that to do whatever I want
What about contact isotopies? I expect a similar answer
Say $(M, \alpha)$ is a contact manifold. If $f : M \to \Bbb R$ is a function, I can restrict $df$ to the contact distribution $\xi = \ker \alpha$, and then take $(d\alpha)^{-1}(df|\xi)$, because $d\alpha$ is a symplectic form on the fibers of $\xi$, and thus defines a pointwise isomorphism which I call $d\alpha : \xi \to \xi^*$ by abuse of notation
This is a vector field $X \in \xi$. So flowing along this will definitely preserve the contact distribution.
I guess I can take a linear combo with a Reeb vector field, and that gives me all the directions of freedom needed?
before posting my question on main, I reckoned I might ask here first. so I have a question about the proof that for any group $G$ there exists a space whose fundamental group is $G$
Let $G$ be a group. Write $G$ as the quotient of a free group $F$ with generators $\{g_i\}_{i\in I}$ by a normal subgroup $N$ with generators $\{r_j\}_{j\in J}$. Consider the bouquet of circles $$ Y'=\bigvee_{i\in I} S^1. $$ It is known that the fundamental group $\pi_1(Y')=\star_{i\in I}\mathbb Z\cong F$, where $\star$ denotes the free product. For every relation $r_j$, there is therfore a corresponding element in $\pi_1(Y')$, so we can choose basepoint preservings maps $$ \widehat{r_j}\colon S^1\to Y'.
I am studying elementary Algebraic Topology recently. I have seen that a topological space is identified with a group. We are telling the group as Fundamental Group. So every topological space $X$ and for any $x \epsilon X$ there is a fundamental group $\pi_1 (X,x)$. I have a question about the c...
Suppose you have some bouquet $B$ of circles, and you attach a single cell by some map $f : \partial D^2 \to B$ from the boundary of the cell $D^2$
The neighborhood of the 1-skeleton and the interior of the $2$-cell intersects in a little annular nbhd of $\partial D^2$ in $D^2$.
This intersection piece has fundamental group $\Bbb Z$, because it's an annulus. So I have $\Bbb Z \to \pi_1(\mathrm{int} D^2)$ and $\Bbb Z \to \pi_1(B)$
Two maps from the intersection piece to the two open sets I covered the thing by
The first map is clearly 0, $\pi_1(\mathrm{int} D^2) = 0$
@BalarkaSen Once this is clear, just verify that the second map $\Bbb Z \to \pi_1(B)$ is equivalent to the map $f_* : \pi_1(\partial D^2) \to \pi_1(B)$, $f$ being the attaching map
