1:20 AM
for $r(t) = 2t\vec{i}+3t^2\vec{j}, t=1$, if asked to find the principal unit normal vector, it's $N(t) = \frac{T'(t)}{||T'(t)||} = \frac{r''(t)}{||r''(t)||}$ or in the 2D case, you can swap the x and y components of $T(t)$ and make one negative, but I got two different answers depending on the method
Yes, you need to see which way $T$ is turning. Also, your final formula is not always right.
wait yeah.. should be $T'(t) = \frac{r''(t)}{||r'(t)||}$ since $T(t) = \frac{r'(t)}{||r'(t)||}$ so $$N(t) = \frac{\frac{r''(t)}{||r'(t)||}}{||\frac{r''(t)}{||r'(t)||}||}$$
I'm so rusty how would one solve $\int \sqrt{4\cos^2 t+1}dt$ ?
1:43 AM
@Obliv Still wrong. For arbitrary parametrizations, it’s more complicated..
The integral is a mess. You’re not rusty. You never knew how to do this.
well the context is finding the length of the plane curve $r(t) = 2\sin t\mathbf{i}+\mathbf{j}$ over the interval $[\frac{\pi}{2},\pi]$
@mick You state that $(1 + ax)(1 + bx)$ is of the form $1 + ab x + ..$, but it should be $1+(a+b)x+\dots$
I assume it's $$\int_{\frac{\pi}{2}}^{\pi}||r'(t)||dt$$ which gives that expression above I think
Oh yes I treated $||r'(t)||$ as a scalar, but it's not necessarily that @TedShifrin
oh jeez I forgot $\mathbf{j}$ turns to $0$
nvm
@robjohn Nice egg :-)
1:58 AM
Let $(x_i)_i$ be a possibly countably infinite collection of real numbers. Does there exist a twice differentiable function $f:R->R$ such that $f(x_i)=f(x_j)$ for all $i,j$ and $0\leq f'\leq 1$?
this is a question our prof told us to think about, not a hw problem btw
im stuck
how does $0 \leq f' \leq 1$ even make sense
is that the bounds of its range?
I must add $f$ is not zero function
@Obliv this just means $0\leq f'(x)\leq 1$ for all $x$
$f'$ is derivative
ah okay
idk what $(x_i)_i$ means either
@copper.hat Thanks! It’s the same one from last year. I don’t know if I can come up with a new spring-themed avatar each year. This year, I’ve had a bit more to deal with, so definitely reuse is in order.
@Obliv just a subset of $\mathbb{R}$
2:10 AM
interesting, so the first derivative is bound between $0,1$ does that influence the 2nd derivatives bounds?
yes, by MVT
cool, so this is a slowly growing function then
meaning the original function converges to some point I guess
2:52 AM
@robjohn Hope all is well. Reuse is good :-).
So far, so good
3:13 AM
$H_0(X,A)= C_0(X,A)/Im(.)= Z${path components of X not intersecting A}
So if $H_0(X,A)=0$,then every path component of X intersects A.
And converse is also true.
@Thorgott
I have used the fact that $C_0(X, A) =C_0(X)/C_0(A)=Z${$\sigma : \Delta^0\to X, Im \sigma$ not a subset of A}
And then idea similar to showing that H_0(X) = Z{path components of X}.
My question has 3 reopen votes.

3 hours later…
5:51 AM
@robjohn As I said, Rudin proved it in theorem 10.13.
He first proved the theorem when $p\in\triangle$. In the next case when $p$ is a vertex he argued that we can make the sides as small as possible. So, by ML inequality, the integration is zero. Next when $p$ is an arbitrary point from $\triangle$ then consider the three triangles with one vertex $p$ and other vertices from the original triangle $\triangle$.
6:08 AM
That's more generally true. If $f$ is bounded on a deleted nbd of a point that is not assumed to be analytic priori, then $f$ is actually also analytic on that point.
6:23 AM
Write $X = \bigcup_i X_i$ as a union of path components then show $H_0(X,A)\simeq\bigoplus_i H_0(X_i,X_i\cap A)$. Now use that $Y\subset Z$ is nonempty if and only if $H_0(Y)\to H_0(Z)$ is surjective
6:57 AM
Thanks @onepotatotwopotato. I am trying to understand the later half of it. $Y\subset Z$ is non empty iff H_0(Y)-->H_0(Z) is surjective. I'm not familiar with this result.
Oh I should assume $Z$ is path connected.
Recall $H_0$ is just counting path components. Any $x\in Z$, $H_0(x)\simeq H_0(Z)$.

4 hours later…
10:45 AM
@Koro That is correct
11:04 AM
@robjohn thanks you are right
12:01 PM
Question: Let $p$ be a prime and $\frac{1}{p}=0.\overline{00331}$ is octal ( base $8$) representation. Then what is the order of $2$ in the group of units of the ring(in fact field) $(Z_p, +, •)$ ?
$\frac{1}{p}=0.0033100331...$
12:22 PM
$15$
Did you really need math mode for that, @robjohn?
@anak Why not?
$\text{I was just wondering.}$
If one is reading the question with MathJax, why not answer in the same.
If I have to write some words inside of , \text{} is the usual choice? e.g. $\{A:A\ \text{is a finite set}\}$. I need to use \ to add some space between symbols and words.
12:37 PM
I usually just start with a space in the \text{} function. For example, $x^2\text{ this word is a space away}$
Oh that's a trick I can use later
@SouravGhosh multiply $\frac1p$ by $2^{15}$ and you get $1+\frac1p$, so the order is a factor of $15$. Multiplying by $2^3$ and $2^5$ are easily seen not to give the same fractional part, so it has to be $15$
and you can just look at the powers of $2$ mod $151$ ;-)
I am out for a bit over an hour. BBL
1:38 PM
What does a basis look like for polynomials q_1 = 2 , q_2=3+2x is it like (a,b) where a is the constants and b is the coefficients of x?
and this extends to higher dimensions with coefficients of higher degree vars?
so B = [(2,0),(3,2)]?

1 hour later…
2:47 PM
The open set $\Bbb C - 0$ is homeomorphic to the annulus but is not biholomorphic to any annulus whose inner radius is $1$ and whose outer radius is $r>1.$ Why is this?
8

Suppose $A=\{z\in \mathbb{C}: 0<|z|<1\}$ and $B=\{z\in \mathbb{C}: 2<|z|<3\}$. Show that there is no one -to-one analytic function from A to B. Any hints? Thanks!

Ah thanks
Take $X=(0,1)^3.$ Fix points $p,q$ s.t. $\text{dist}_3(p,q)=\sqrt{3}.$ Construct a smooth regular foliation of $X$ with $(3-1)-$dim. leaves which are topologically $(0,\sqrt{3})\times S^{3-2}$ accumulating to $p,q.$

How is this question is equivalent to the existence of a smooth foliation of $\Bbb R^3$ whose leaves are all diffeomorphic to the open annulus $(0,1)\times S^1?$ I can visualize the former but not the latter unfortunately.
I know that the open annulus is diffeomorphic to the punctured real plane, but I don't see the process of getting the open annulus from the leaves of $X$
Oh wait you just deform the cylinder topologically into the annulus?
Then if $S=\Bbb R^2 - 0$ is a class of surfaces foliating $\Bbb R^3$ I think you could take each leaf in the class and take the punctures of each leaf to be located at the origin of $\Bbb R^3.$ And the leaves curve downward for $z<0$ and curve upward for $z>0.$

1 hour later…
4:24 PM
what's latex for a big curly brace, but under a statement?
\underbrace
$\overbrace{\text{here is an expression}}^{\text{some kind of label}}$
$\underbrace{\text{here is an expression}}_{\text{some kind of label}}$
thanks
@Xander How did the radar question go?
Still waiting for one student to take the test.
Ah.
5:09 PM
Would anyone sufficient in the use of the CAS program "Maple" be able to tell me how to write a "binomial coefficients" in the form it is written in most textbooks?
https://postimg.cc/S2z3VxZf

Here is a link to view the format of the binomial coefficients in which I strive to write myself in Maple.
5:29 PM
Is it just me or does early transcendental functions by larson feel more like a brochure than a textbook?
the examples are fine but some of the theorems seem unmotivated and unexplored
I am enjoying my Diff eq textbook, but this one not so much. Maybe i'm just impartial to the content

2 hours later…
7:09 PM
Is there a name for rings where $2$ has an inverse?
@mr_e_man In an abstract ring, what is "2"?
I came up with a new way to view transcendentality
woah
7:24 PM
Basically algebraicness is based off of solutions of polynomials with rational coefficients. But it doesn't say this is how you have to define polynomial. Usually it's just assumed that polynomials are defined in the usual way. But you can define polynomials as relative to the space you're in. And then transcendentality is then viewed as embedding polynomials from one space into the other
the roots to the polynomial automatically are transcendental with respect to the "foreign" space you've embedded it into
@XanderHenderson - Obviously $1+1$... I assume any ring has a multiplicative identity.
@mr_e_man I believe that i have previously made my objections to the phrase "obviously" known. If someone felt the need to ask, I am not sure why you need to be condescending and claim that it is obvious...
I am a bit confused by what is meant by centralizer here. The set S is not a subset of the group $SU(2)^{\otimesn}$, so what does centralizer mean here?
To be clear, to my understanding the fact that $S$ is not a subset of $G$ means that the usual definition of centralizer of a group (cf. Dummit and Foote) certainly cannot be applied here
7:41 PM
I am confused about Wikipedia's description of an injective function and a bijective function. They write:
injective function: "every element of the function's codomain is the image of at most one element of its domain"
bijective function: "each element in the codomain is an image of exactly one element in the domain"
I know there's a difference between the two types of functions, but are these descriptions different? They read very similiar.
bijective means surjective and injective
so the description of bijective should certainly encapsulate the description of an injective function because a bijective function is by definition injective
@schn - "At most one" allows the possibility of none. "Exactly one" does not allow that possibility.
Thanks for the replies!

2 hours later…
9:54 PM
can someone help me here
0

Let me consider the circle homeomorphism $T:\Bbb{R}/\Bbb{Z}\rightarrow \Bbb{R}/\Bbb{Z}$s.t. $T(x)=x-\frac{1}{2}$ if $x\in [0, 1/2]$ and $T(x)=\frac{x}{2} -\frac{1}{4}$ if $x\in [1/2,1]$. Now I want to find a lift $F:\Bbb{R}\rightarrow \Bbb{R}$. This means $F$ needs to be continuous and $\pi(F(x... 10:40 PM Does anyone know if the user here is referring to a standard definition of$S_3$? math.stackexchange.com/a/4076525/755010 I'm trying to figure out what$a_{\sigma(i)}$would map to nvm, figured it out 1 hour later… 11:49 PM Let$K$be the Klein 4-group$\{e, a_1, a_2, a_3 \}$. Let$\{1,2,3\} = \{i,j,k\}$. Then$K$has the property that$a_ia_j = a_k$. This notation is fantastic for proofs i would have never thought to use the ambiguity of$\{ 1,2,3 \} = \{i,j,k\}$to give a general definition of the Klein 4 group well, implicitly, we suppose$\{i,j,k\}\$ is ordered, so maybe we should specify it is unordered to tip off the reader we're using this idea
why stop there? maybe you can spend a full 2000 words explaining the multiplication in a four-element group
then gratuitously insult the reader in the last paragraph
turn it into a logic puzzle somehow. "of the last __ statements, exactly five are lies. statement 6 is not a lie"