Mathematics - 2018-10-24 (page 1 of 2)

user280247

00:05

I'm not sure i understand you but something I come up with Is that we should not waste time, it implies doing what you like -although it is a belief...; now even what you like can become a heavy thing if you Must do it. And it may be even worst: you will grow up and be unconvinced about the path you're following. For the first one, we should not be too ambitious, society is not that bad. For te second one, just pay attention to the present time...@Adam

user280247

I think it is somehow related to your comment...

user280247

Now to something important (?): is there any beauuutiful book about strange numeric systems?

Jeff

00:29

If $\sin(t)=\cos(t-\pi/2)$ and $\cos(t)=\sin(t-\pi/2)$ then isn't $\tan(u+\pi/2)=\sin(u+\pi/2)/\cos(u+\pi/2)$. Then if $t=u+\pi/2$ it becomes $\cos(u)/\sin(u)=\cot(u)$. But wolfram alpha says it's $-\cot(u)$. Where's the negative sign come from?

Semiclassical

@Jeff it's $\cos(t)=\sin(\pi/2-t)=-\sin(t-\pi/2)$

Jeff

Oh.... duh

Is $\tan(t)=\cot(\pi/2-t)$ correct? I think so. If so , then $\tan(u+\pi/2)=\cot(u)$ and not $-\cot(u)$.

@Semiclassical ^ ^

Semiclassical

00:44

first equation is right. but in that case $\pi/2-t=\pi/2-(u+\pi/2)=-u$

so you get $\tan(u+\pi/2)=\cos(-u)=-\cot u$

Jeff

Oh brother. so careless :(

Semiclassical

gotta be careful with those minus signs, ya

Jeff

@Semiclassical Yup.... rest of the prob works now. TY.

Semiclassical

np

Seninha

01:40

Hi, anyone knows a site talking about inductive definition of matrix type?

Adam

@santimirandarp It's really more of a fleeting childish emotion that I get less and less nowadays, whereby I will fairly bitter and self centred, and it's much easier to blame society than to look at how things have gone and say "ok well this is all entirely due to your own character flaws and mediocrity" as for your second question about a beautiful book, I really can't think of many as far as titles, I am trying to build a good hardcover collection, but as far as in depth focus

i am trying to commit to the graduate text i chose for analytic number theory, as in suppose i was actually at university still, a lecturer isn't selecting 50 titles for you to work from, you learn a lot more by going from the one text somehow i think at least for graduate level

user280247

03:03

@Adam I see..; so there is no solution, I mean, society give value to some things. For many of them there are even 'natural' reasons. Those skills are taken as 'good' thing; and indeed, we feel bad and call it 'fail' when we can't reach those high standards. Even if we forget about self centredness, that won't change. Possibly being not self centered'd help to avoid suffering, that's important I think.

user280247

And finally, you can't complain with nature because of the skills it gave you (or anyone); we can just try to enjoy what we do...To accept our own limits is a sign of cleverness, by the way...

Ilya Gazman

03:17

0

Q: How long will it take to rotate a moving body in space?

While decorating your spaceship for Halloween, you have been hit by a meteor. Now you at the outer space moving with the speed of $v_i$. Luckily your spacesuit is equipt with a rocket engine, all you need to do is to set the force power and direction and for how long you wish it to be applied. Fo...

Save Halloween, help me answering this question

user525966

anyone know first order logic terminology?

Adam

03:39

@santimirandarp sure really I'm very grateful to at least have the capacity to attempt to understand things I'm sure I don't need to tell you how rewarding the moments are when you see something in mathematics clearly and realise just how beautiful it is, I know for me personally I will never be truly satisfied with myself, but I do get depressed in the circumstance people ask me where all of my time goes, and I say I study prime numbers, then they ask what they apply to in real life,

and I normally say something like encryption, and then they are often
gestering for me to teach them how to hack facebook, and its that draining feeling of the thought of going to the length of explaining that a profession in computer science to the standard for which the number theory i study applies would most definately require an additional lifetime i simply dont have, and for that matter i really am not interested in "hacking" anything anyway

a lack of "real life" evidence of the efforts of study sometimes makes me lose my temper in short

user280247

04:15

@Adam uhmm, but where can we apply music, for example mozart? I'd try just showing some interesting things of prime numbers which 'hooked me up' when I started. I do not know very much of maths, but usually people ask me 'Hey you are the chemist' and I say; well I don't have idea why your soap tastes bad...But then I think, How can I tell them what I love about chemistry? And I just tell it...and usually people end saying oh why I didn't studied chemistry...XD

user280247

@Adam hope it doesn't bothers you...

Secret

For me, unless I am talking to a politician, I will not emphasise much on what the application of something is. I will just do whatever I like when possible

For example, my passion on infinity is simply because they are fun and weird. Whether they can ever be applied to something in real life will naturally come when it comes

Daminark

04:37

YES

Okay so

For my rep theory pset one problem was to take the action of $S_4$ on the edges of a cube, take the associated permutation representation, and decompose it to a direct sum of irreps

Note that this is 12-D, aka painful

My way of doing this was to just compute the character and then use the character table. Still took a while because I was bad at geometry and had to enumerate stuff, but alright finally found the characters... and it didn't work, when I ran it through I didn't get integers

Was just raging for a while because I did all that work for nothing

Turns out I wrote the character of the identity as 8 instead of 12, since I had just done the same problem but for the vertices

And once I changed to 12 it worked

:DDDDDDDDDDD

Rant over

Stan Shunpike

04:53

anyone heard of lagrangian duality?

Adam

05:09

@santimirandarp no not at all actually Chemistry was my first love almost two decades ago, although due to various circumstantial factors I decided that the best course of action was to steer away from it, mostly because I am the most respectable of members of West Australian society and there have been several blatant attempts of defamation of character regarding such things,

and mathematics is completely unrelated to chemistry and completely inapplicable, demonstrating my good Christian nature

Isana Yashiro

05:57

@Pig hi, how are you

I suddenly found that category theory is so interesting, because of a talk on youtube

But I don't even know in details what category theory does, but it sounds like about composing thing and find some identity

And I still can't find the intuition behind $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$, given that $f:A\to B; B_1,B_2\subseteq B$, could anyone help me?

(lhs) because when first do the intersection and consider its inverse image, all possible domain that hit them are consider

(rhs) now consider $B_1,B_2$ separately, the part $B_1\cap B_2$ are mapped twice to the domain, but this doesn't matter.

is this correct?

No, it still need some more details

now consider the part $B_1-B_2$ and $B_2-B_1$, will then hit to the same place? no, since $f$ is a function.

@Secret: hi, how are you

do you know something about category theory?

If anyone could provide some idea about this problem I will be very happy: math.stackexchange.com/q/2944374/390226

ÍgjøgnumMeg

09:16

@Fargle without wanting to sound completely(ha) clueless, don't you just need a complete metric space to do calculus?

Fargle

@ÍgjøgnumMeg Frankly, I don't know. Sometimes I just say things to sound smart

ÍgjøgnumMeg

@Fargle are you me?

Sharath Zotis

$\text { Using an } \epsilon - \delta \text { argument, prove that } \lim _ { x \rightarrow 0 } \cos \left( \frac { 1 } { x } \right) \neq 0$

Im having trouble with this could anyone help me?

I am trying to show this by contradiction

so basically I know if $ 0 < |x| < \delta$ then $|cos(\frac{1}{x})-L| < \epsilon$

I know that when $x$ is small $\cos \left( \frac { 1 } { x } \right)$ oscillates between -1 and 1

Fargle

@ÍgjøgnumMeg Maybe? I thought I was a little bit more together than that...

Sharath Zotis

and this is how I achieve a contradiction but I am not really sure how to get this or what $\epsilon$ to pick

ÍgjøgnumMeg

09:26

hahaha

Sharath Zotis

anyone have any ideas?

Fargle

@SharathZotis If I told you to find a $\delta$ that works for $\epsilon = \frac{1}{2}$ in this problem, could you do it? Why or why not?

Sharath Zotis

hmm I think I just thought of it

let x be some sufficiently large solution to cos(x)=1 or -1, then one of |1-L| and |-1-L| must be greater than 1

is this sufficent to show?

*$cos(\frac{1}{x}) = 1 or -1

*$cos(\frac{1}{x})$ = 1 or -1

@Fargle is this sufficient?

Fargle

@SharathZotis A small nitpick: |1-L| or |-1 - L| would both cap out at exactly 1. But yes, using the fact that cos is periodic, and then picking the points $x = \frac{1}{n\pi}$ for sufficiently large $n$, should take care of it.

Oops, I can tell I'm still not awake yet

And strictly speaking you wouldn't have to use or even mention periodicity, just the fact that you're gonna hit $1$ and $-1$ arbitrarily close to $0$ for cos(1/x)

Secret

10:13

Basically the central idea is, the part of the domain that maps by f to the intersection must contain parts of the domain that maps to $B_1$ and $B_2$ respectively

BAYMAX

Any help in these two linear algebra question?

Secret

and I don't think I knew enough category theory to solve actual category theoric problems

BAYMAX

4

Q: Dimension of $W_{2}$?

Let $A = \begin{bmatrix} 1 & -1 & -5 & 1 & 4\\ -1 & 2 & 8 & -3 & -4\\ 3 & -1 & -9 & 0 & 4 \\ 2 & 2 & 2 & -5 & -10\\ 0&-3&-9&5&13\end{bmatrix}$ Now we define the subspace $W_{1},W_{2}$ of $A$ as follows - $W_{1} = \{X \in M_{5 \times 5}| AX = 0\}$ $W_{2} = \{Y \in M_{5 \times 5} | YA =0\}$ I ...

linear-algebra matrices matrix-rank

2

Q: Injectivity and subspaces of $g_{1},g_{2},g_{3}$?

This is a follow up question to - Dimension of $W_{2}$? Let us define $B = \begin{bmatrix} 2 & -1\\ -3 & 1\\ 1 & 0\\ 0 & -2\\ 0 & 1\\ \end{bmatrix}$ $C = \begin{bmatrix} -1 & -2 & -1 & 1 & 0\\ -2 & 1&1&0&1\\ \end{bmatrix}$. Let us define $g_{1}: M_{2 \times 5}(\Bbb{R}) \rightarrow M_{5 \times ...

linear-algebra matrices matrix-rank

abenthy

10:45

Does somebody know what the automorphism group of $PGL(2,\mathbb{R})$ is?

Clearly, every automorphism $\varphi \colon GL(2,\mathbb{R}) \to GL(2,\mathbb{R})$ descends to a map $PGL(2,\mathbb{R}) \to PGL(2,\mathbb{R})$ because $PGL(2,\mathbb{R})$ is the quotient of $GL(2,\mathbb{R})$ by its center.

Unknown MathMan

11:16

How $\mathbb{R}/\mathbb{Z}$ quotient group will look like?

Secret

It's a circle

Unknown MathMan

Yeah that also given in the excercise. It's isomorphic to a circle. But, how you can realize it? I think it needs some work to show it.

Secret

You are basically trying to find the cosets of the form $x\Bbb{Z}$ such that $\Bbb{R}$ is partitioned by them

Since under the quotient all integers belong to the same equivalence class, it follows that you can only have $x \in [0,1)$

and $[0,1)$ can be mapped to a circle by $e^{2\pi i x}$

Unknown MathMan

@Secret How ? :-(

Not getting

Secret

$\Bbb{Z}$ is a ladder rung in $\Bbb{R}$ and is periodic with period 1. You can get different $x\Bbb{Z}$ as long $x$ is not integer multiple of the period

Anjani Gupta

11:26

Stack the intervals [n,n+1) over [0,1) to see the identification.

In this 0 and 1 are identifies, so this is S1

Unknown MathMan

@Secret Thanks for the picture, what the vertical lines signifies here?

Secret

...,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,...

Anjani Gupta

I have a problem and I have posted it here but no answer yet. Can you guys look at it?

-1

Q: Orbit map may not necessarily be a homeomorphism.

Let $R$ (real line) be a topological group acting continuously on topological space X and let x be a point in X such that its stabilizer is trivial. Then the orbit map from $R/Stab(x)$=$R$ to $Orb(x)$ is not a homeomorphism if $Orb(x)$ is dense in X. I know that this is a continuous bijection. S...

group-theory group-actions

Fargle

@UnknownMathMan So, for the quotient group, you're stipulating that if two elements "differ by an integer", they're treated as the same in the quotient. This means that, for example, all of the elements ..., -1.5, -0.5, 0.5, 1.5, 2.5, ..., are the same in the quotient.

What this means is that [0,1) can be taken as a "representative set", and if you "go to the right" past 1, you'll "wrap back around" to zero.

But that's exactly the structure of a circle! If you go forward enough on a circle, you eventually wrap back around to where you started.

Unknown MathMan

@Fargle That makes sense!

@Fargle but how we can mathematically write this? can you show by mathematical argument please?

Fargle

11:39

As Secret mentioned, the isomorphism is given by $\overline{r} \mapsto e^{2\pi i r}$---and this is well-defined precisely because of how you've taken the quotient, and the fact that $e^z$ is periodic with period $2 \pi i$.

(Where the function I'm talking about is mapping from the quotient group to the unit circle in the complex plane)

@UnknownMathMan To be a little more clear: the way you'd mathematically write this is by constructing an isomorphism between the quotient group and the circle group, which I'm thinking of as being the unit circle in the complex plane. You have to show it's well-defined, and then show that it's a homomorphism, and then show that it's bijective. Those steps shouldn't be too difficult, I leave them to you.

Unknown MathMan

@Fargle Yeah, I am getting the mapping part, but what is making problem is treating $[n,n+1)$ same as [0,1).

Anjani Gupta

@unkn

ÍgjøgnumMeg

Alternatively hit $\Bbb R \to S^1$ with an isomorphism theorem to get as little intuition as possible

Anjani Gupta

You see that 0,1,2,3,..,n,.. are in the same coset.

Unknown MathMan

Yeah

Fargle

11:45

@UnknownMathMan Well, the reason is precisely because of how the quotient group is defined, right? It's the group where the elements are $r + \mathbb{Z}$---i.e., the cosets of $\mathbb{Z}$. What I'm doing is choosing to represent this coset by whichever element lies in $[0,1)$.

Anjani Gupta

Similarly pick any number x between 0 and 1 and it will lie in the same coset as n+x for any natural number n

So just cut the real line from [1,2) and place it on top of [0,1)

Fargle

And then, because of the periodicity of $e^z$, we can kind of see that if $r' \in r + \mathbb{Z}$, then $e^{2\pi i r} = e^{2\pi i r'}$, so we can just work with this map on representatives, rather than on the entire real line.

Anjani Gupta

Continues doing so..

Unknown MathMan

@Fargle Got it :-) Thanks

Fargle

To put it briefly, every real number is "integer part + fractional part"---e.g. $2.5$ is $2 + 0.5$, $-3.6$ is $-4 + 0.4$. In the quotient, you identify all the integers with zero, so you're only considering the fractional part.

Unknown MathMan

11:49

Thanks to all

Fargle

No problem.

Anjani Gupta

Do all uncountable sets have a bijection between them?

Unknown MathMan

@AnjaniGupta Can you be clear what you mean by cut? Do you mean their structure are same under this quotient?

Or in other meaning belong to same equivalance class

Anjani Gupta

Yes, since they are just representatives of the cosets in the quotient.

Unknown MathMan

Ok

Fargle

11:52

@AnjaniGupta No---both $\mathbb{R}$ and $\mathcal{P}(\mathbb{R})$ are uncountable, but there can be no bijection between them by Cantor's theorem.

Power sets always have strictly larger cardinality.

Anjani Gupta

P(R) projectivized R**2?

Fargle

No, power set, sorry

i.e. set of all subsets

Anjani Gupta

Oh okay, Ill read the proof. Did you see the question I tagged a few moments ago?

Fargle

I did, but unfortunately I don't know how to answer it. Hopefully someone else does.

Anjani Gupta

oh no issues. In cantor's theorem cardinality refers to inclusion order relation, right?

Fargle

11:58

@AnjaniGupta Cardinality is an extension of the normal notion of "size" for finite sets. Two sets have the same cardinality iff there exists a bijection between them. For finite sets you say that if a set has $n$ elements, it has cardinality $n$; if it is countable, it has cardinality $\aleph_0$; if uncountable, it's a little more complicated.

No inclusion has to take place---in our case, a set is never a subset of its power set (though there is a simple identification with a subset of the power set, i.e., the subset of singletons)

Anjani Gupta

Yeah, the inclusion order relation says the same thing in a way? Just that a copy of set sits inside the range, and not the same set itself.

Fargle

Yeah, that's one way to look at it.

You can essentially define the order relation as that $|X| \leq |Y|$ iff there is an injective function $X \to Y$.

Anjani Gupta

precisely :)

Fargle

I believe the most standard proof of Cantor's theorem is basically a much more set-theoretic version of the diagonal argument for $|\mathbb{R}| > |\mathbb{N}|$.

Secret

[Random]
Now how on earth to reason about:
Real = irrational part + rational part

Anjani Gupta

12:05

I saw the injection proof which seemed very easy. Send x to singleton x.

Secret

It is true that the following mesh will not bump into itself whenever it is shifted by an irrational:

But clearly, the coset cannot be the irrationals else it will be measurable, in direct contradiction to vitali's theorem

11

Q: Visualizing quotient groups: $\mathbb{R/Q}$

I was wondering about this. I know it is possible to visualize the quotient group $\mathbb{R}/\mathbb{Z}$ as a circle, and if you consider these as "topological groups", then this group (not topological) quotient is topologically equivalent to a circle. But then, what does $\mathbb{R}/\mathbb{Q}...

group-theory topological-groups visualization

Fargle

@AnjaniGupta That will show that $|X| \leq |\mathcal{P}(X)|$, but it won't show $|X| < |\mathcal{P}(X)|$. For that you'd have to show that no injection is a bijection, not just the example you've given.

Secret

This means, there exists at least two irrationals $r,s$ such that $\Bbb{Q}+r \cap \Bbb{Q}+s\neq \varnothing$ unless $r,s \in V_{t}$

Fargle

To illustrate this, the function $f : 2\mathbb{N} \to \mathbb{N}$ by $f(k) = k$ is an injection that isn't a bijection, but these two sets still have the same cardinality.

Secret

but with partial overlaps like these, do the cosets commutes anymore...

ÍgjøgnumMeg

12:18

the density of $\Bbb Q$ in $\Bbb R$ is what makes it hard for me to imagine what it would look like

@Secret it's almost like getting chewing gum stuck in your hair, where your hair is $\Bbb R$ and $\Bbb Q$ is gum

Secret

$\Bbb{Q}$ actually resembles some kind of fractal. It is $\Bbb{I}$ and its subsets that is where most of the headache is

Anjani Gupta

@Fargle but we can see that none on the element goes to a 2-tuple. So its surely not a surjection.

Silent

Let $p$ be prime, and let $Z=\{z\in \Bbb C: z^{p^n}=1 \text{ for some } n\in\Bbb Z^+\}$.
I see that this is indeed group, but, I wonder if we can replace '$p$ is prime' with '$p$ any integer' ?

ÍgjøgnumMeg

@Secret yeah? In what way?

Fargle

@AnjaniGupta Yes, but the fact that one injection fails to be a bijection doesn't mean that the sets have to have different cardinality. For example, $g : 2\mathbb{N} \to \mathbb{N}$ by $g(k) = \frac{k}{2}$ is a bijection---the fact that $f$ was an injection that failed didn't rule out the existence of $g$.

Anjani Gupta

12:25

Every domain element goes to the singleton of that element. Which implies you can pick WLOG any not 1 tuple, say (1,2) and it will have no pe image

Oh sorry, yeah you are right.

But it's disturbing me logically yet.

Fargle

So the theorem, despite the fact that it's in some sense "classic" now, is rather non-trivial---it ends up showing that there is no bijection from a set to its power set.

The details of it are very tricky, because you end up working with a set defined such that its elements $x$ satisfy $x \notin f(x)$---and that's just gross to look at or think about, frankly.

ÍgjøgnumMeg

@Silent how did you show it was a group?

Silent

@ÍgjøgnumMeg I showed that it is subgroup of group of roots of unity. $Z$ indeed contains identity, if $x,y$ are in $Z$, then $(xy)^{lcm (p^k,p^l)}=1$ and $x^{-1}$ lies indeed in $Z$.

I think that in $(xy)^{lcm (p^k,p^l)}=1$, to get $lcm (p^k,p^l)=p^k$ (wlog) we need p prime, right?

hmm

am i correct?

Secret

12:42

@ÍgjøgnumMeg So basically, it looks like the fractions of the form $\{\frac{1}{n},1-\frac{1}{n}\}$ repeatedly squished between the gaps formed between them in each level, and it gets progressively more distorted the further down they go

As far numerical checking goes, level 1 are all irreducible fractions with denominators $[n+2,\infty)$ and in level 2 are all irreducible fractions with denominators $[2n+1,\infty)$

Still figuring out how to formally prove this

Poincaré disk model

In geometry, the Poincaré disk model, also called the conformal disk model, is a model of 2-dimensional hyperbolic geometry in which the points of the geometry are inside the unit disk, and the straight lines consist of all segments of circles contained within that disk that are orthogonal to the boundary of the disk, plus all diameters of the disk. Along with the Klein model and the Poincaré half-space model, it was proposed by Eugenio Beltrami who used these models to show that hyperbolic geometry was equiconsistent with Euclidean geometry. It is named after Henri Poincaré, because his rediscovery...

So overall, it kinda reminds me of a one dimensional version of this

Silent

@LeakyNun, will you please confirm this:

12 mins ago, by Silent

@ÍgjøgnumMeg I showed that it is subgroup of group of roots of unity. $Z$ indeed contains identity, if $x,y$ are in $Z$, then $(xy)^{lcm (p^k,p^l)}=1$ and $x^{-1}$ lies indeed in $Z$.

Anjani Gupta

Soft question : Is there any way to remove the info bar on the right of this chat page?

user193319

If $A,B,C$ are sets, is it true that $(A \cap B) \setminus C = (A \setminus C) \cap (B \setminus C)$? I believe I proved that it is true, but I'm just wondering if it is obviously false and therefore I made some stupid mistake.

Semiclassical

12:59

Looks right

user193319

Thanks!

Semiclassical

a good way to check it is to draw a venn diagram

Anderson Felipe Viveiros

0

Q: The riemannian metric of a neighborhood of the boundary of a compact manifold

Let $M$ be a compact riemannian manifold with boundary $\partial M$. We have that $\partial M$ is also compact and I was able to show that there is some $a>0$ such that the map $F:[0,a]\times \partial M \to M$ given by $F(r,p)= \exp_p(r\nu(p))$ is a diffeomorphism onto its image, say $U$ (here $\...

riemannian-geometry manifolds-with-boundary riemannian-metric

Secret

13:13

Details

Rithaniel

So, given the integral $|\int_C\frac{dz}{z^4}|$ (where $C$ is the line from $i$ to $1$), is it true that this integral is less than or equal to $\frac{\pi}{2}$? (because that is the result I'm getting)

Semiclassical

well, you can compute that integral exactly using wolfram alpha

presumably, though, you've got some integral estimate/bound in mind

Rithaniel

What would be the syntax to get wolfram to handle that? I can't get it to spit out an answer. (Also, yes, the estimate I've been given is that it's less than or equal to $4\sqrt{2}$)

Unknown MathMan

@Silent Why you need $(xy)^{lcm\{p^k,p^l\}}=1$?

Semiclassical

well, you'd need to parametrize the integral appropriately

e.g. $z=t+(1-t)i$ from $t=0$ to $t=1$

in which case the integral becomes $\displaystyle \int_0^1 \frac{(1-i)}{(t+(1-t)i)^4}dt$, which I presume WA can do

Unknown MathMan

13:26

@Silent ooh, for closure. I think correct.

Silent

@UnknownMathMan yeah! you just saved me from latex typing :)

Semiclassical

yeah, WA gives an exact answer

Rithaniel

Yep, and it lines up with my result.

Semiclassical

neat

Silent

@UnknownMathMan, you may like to follow this.

Semiclassical

13:29

notably, the result is what you'd expect naively: $$\int_C \frac{dz}{z^4} = \frac{i-1}{3} = \left[\frac{-1}{3z^3}\right]_1^i$$

Rithaniel

My approach used $|\int_C f(z)dz|\leq\int_C |f(z)dz|=\int_C \frac{dz}{|z|^4}$

Semiclassical

or does it, hmm

Unknown MathMan

@Silent I think no issue if you take $p\in \mathbb{Z}$

Semiclassical

I mean, one just has $$\int_0^1 \frac{(1-i)}{(t+(1-t)i)^4}\,dt = \left[-\frac{1-i}{3}\frac{1}{(t+(1-t)i)^3}\right]_0^1$$

Unknown MathMan

@Silent Can you tell precisely which group of roots of unity contains this as a subgroup?

Semiclassical

13:33

so yeah, that's not quite the same. you pick up that factor of $(1-i)$ in addition

Silent

@UnknownMathMan i think the same. But, then perhaps $p$ prime version has some cool properties like being isomorphic to proper quotient of itself

@UnknownMathMan you mean like $\{z\in \Bbb C : z^{4^n}=1\}$ has $4th, 16th, 64th...$?

Unknown MathMan

yeah

Silent

I see, so, what's your point?

Unknown MathMan

you have said here that you are showing this is a subgroup of a group of (suppose $k$th) roots of unity. What is value of $k$ here you think?

Fargle

@Silent What would the square of the primitive 16th root of unity be? Would it be in the set?

Alucard

13:43

@AaronHall hi :) SE is small hehe

Silent

@Fargle By primitive root, you mean $exp(\frac{2\pi i}16})$, right?

Fargle

@Silent Yeah. Sorry, that was ambiguous---there are actually 8 primitive 16th roots of unity, but that's the one that I meant.

Although, hang on, I think I've messed something up here. Ignore me.

Silent

@Fargle i will come to u later asking what is primitive root, in fact, and how there are 8 here. right now i will work with your first comment.

oh! alright, i m halting

Fargle

@Silent A primitive nth root of unity is one for which $z^n = 1$, and for which $n$ is the smallest power that gives 1.

For example, -1 is a 4th root of unity since $(-1)^4 = 1$, but it is not a primitive 4th root of unity, because $(-1)^2 = 1$ too.

Silent

oh! that was cool!

Aaron Hall

13:49

@Alucard indeed

Silent

thanks

Fargle

Yeah. I don't think my line of logic necessarily works here...

That is, for showing why we don't talk about groups of the form you mentioned

Silent

its ok

i learned a new concept!

Fargle

It seems like part of the reason you wouldn't talk about it is that this group would actually be equal to the one with 2 in the power instead of 4.

But that's confusing, because then what if 6 is the power?

I'll have to think on this a while.

Semiclassical

I suspect that it's actually just $\{|z|=1\}$ in disguise

Fargle

13:56

@Semiclassical It doesn't seem like that should be the case--for example, a 5th root of unity wouldn't lie in $\{z \;|\;z^{6^{n}} = 1\}$, as far as I'm aware.

Semiclassical

my thinking is that, for any $z=e^{i\theta}$, we can write $\theta$ as a binary number

and therefore express $z$ as a possibly infinite product $\prod_{k=1}^\infty e^{i a_k/2^k}$ where $a_k=0,1$

Obviously, if you cut off $z^{2^n}$ at some finite $n$, this won't help

but if you allow arbitrarily large $n$?

Silent

group of roots of unity is isomorphic to $\Bbb Q/\Bbb Z$

Unknown MathMan

@Fargle Yeah

Fargle

But are you allowed to take infinite products in groups? Honest question, can't remember

Unknown MathMan

@Silent Simply to $\mathbb{Z}/n\mathbb{Z}$

Semiclassical

13:59

well, you'd better have closure

so any arbitrary product of elements should make sense

Fargle

But closure is only stipulated for finite products.

Semiclassical

hmm

Silent

@UnknownMathMan no! that is finite. the set i mentioned is infinite

Semiclassical

at the very least, you can definitely find a sequence of elements in $\{z|z^{2^n}=1\}$ converging to arbitrary $z$ on the unit circle

Unknown MathMan

@Silent ooh

Fargle

14:00

Absolutely, but the Prufer 2-group itself (which is what that is) is countable---those elements it "limits" to aren't included.

Silent

@Semiclassical i think i have seen this result somewhere in math se

Fargle

By analogy, you'd be saying that $\mathbb{Q} = \mathbb{R}$

Semiclassical

hrm.

true

Silent

@Fargle ?? is that meant to be considered as counterexample?

Semiclassical

though one difference is that Q is at least closed under infinite additions/multiplications

Fargle

14:02

@Semiclassical I disagree. $1 + 0.4 + 0.01 + 0.004 + \cdots = \sqrt{2}$

Semiclassical

derp

yeah

okay, yeah, that convinces me

Fargle

@Silent That was directed at Semi, sorry

@Semiclassical lol, I wasn't sure myself until I had that realization

Semiclassical

i guess the point is that the group will consist of every $e^{2\pi i x}$ with $x$ having a terminating base-2 expansion

Fargle

I was thinking by analogy with something in ring theory

Unknown MathMan

@Fargle But, to get $\sqrt{2}$ you need some numbers which will be irrational, $0.000000.....$ right?

Fargle

14:04

where you define the product $IJ$ of two ideals to be all finite sums $\sum a_ib_i$ with $a_i \in I$, $b_i \in J$

Semiclassical

ahhh

@UnknownMathMan no

Fargle

@UnknownMathMan Absolutely not. $\sqrt{2}$ is approximated by those partial sums, and the infinite sum is exactly equal to $\sqrt{2}$, but every single one of the infinite terms in that series is a rational number.

Semiclassical

take the sequence of decimal approximations to sqrt(2), truncated up to n digits

Fargle

There is no such real number as $0.000000...$ except for $0$.

Semiclassical

each thing you'd be adding is just 1/2^n

(in base 2)

Kari

14:06

Is this (even true and) a well known thing? $$ \int_{0}^{1} x^{-x} \text{d}x = \sum_{n=1}^{\infty} n^{-n} $$

Semiclassical

yep, sophomore's dream

(that's literally what it's called on WP)

Unknown MathMan

@Fargle I am showing it like that because otherwise it will seem like a rational number, I don't know how to write that number

Semiclassical

each term in Fargle's sum will be a set of finitely many zeros followed by one nonzero digit

Unknown MathMan

Ok got it. The number of summands are uncountable

Semiclassical

and that's a rational number a/10^k

Fargle

14:07

@UnknownMathMan But what we mean by a decimal representation $a.bcde...$ is just $a + b/10 + c/100 + d/1000 + e/10000 + \cdots$

Semiclassical

@UnknownMathMan no. it's still countable

the nth term will be of the form a/10^n

Unknown MathMan

Sorry, maybe countable, but infinitely many and non repeating

Fargle

So any number like "0.0000..." which has infinitely many zeroes and "eventually something" after those infinitely many zeroes would be $0 + 0/10 + 0/100 + \cdots$, and the sequence of partial sums is $0,0,0,0,0,\dots$, which just converges to $0$.

Unknown MathMan

and that makes root(2) irrational

Semiclassical

right

Unknown MathMan

14:09

am I correct?

Semiclassical

it's not a terminating or repeating expansion

so therefore not rational

Fargle

Indeed.

If you ever tried to sum uncountably many things conventionally, you'd always wind up diverging unless all but countably many of the things are zero.

There are, in some sense, "too many summands" in that case.

(That's why we use integration instead of summation on uncountable sets like intervals.)

Semiclassical

@Fargle and relatedly why we have to use probability densities for continuous r.v.s

Fargle

@Semiclassical Well noted, hadn't thought of that.

To put that a little more briefly, an infinite sum is (unless there's some weird, pathological instance where this isn't true, that I'm not aware of) always a sum of countably many things.

Semiclassical

countably many nonzero things, at least

Fargle

14:14

...right.

My kingdom for a cup of coffee...

Rithaniel

Hmmm, how would you directly prove Sophomore's Dream?

Semiclassical

i'd look at the WP page first to see if it's there

Unknown MathMan

Which of the following orders if a group have, then it is abelian: 49,6,15,35,24. I think 49(as it is $7^2$),24 and 15 are correct answers. are there any other true?

Fargle

@UnknownMathMan Check 24 again. There should be at least one obvious group of that order that isn't abelian.

Silent

oh!

Unknown MathMan

14:25

oops

$4!=24$ :p

So, $S_4$ is that group.

Silent

@UnknownMathMan how did you get 15 and 49 as answers? using sylow?

i have yet to touch that topic :/

Fargle

@Silent You don't need Sylow for 49 (and in fact it won't really help you). There's a result that any group of order $p^2$ is abelian for prime $p$.

Unknown MathMan

For 49, you can use a different thm. Any group of order $p^2$ with prime $p$ is abelian.

Silent

oh

for 15?

Fargle

@UnknownMathMan Try looking at 35 under Sylow, by analogy with how you did it for 15.

Unknown MathMan

14:30

@Fargle I have done for 15 and haven't used anything special for 3 and 5(except they are prime), so I think being product of 2 primes groups with order 35(=5x7) also abelian.

Semiclassical

can't be true as stated, since 2*3=6

might be correct for odd primes tho

Fargle

@UnknownMathMan That line of reasoning doesn't necessarily hold. For example, there are nonabelian groups of order $6 = 2\cdot 3$, $10 = 2 \cdot 5$, and $21 = 3 \cdot 7$.

Semiclassical

there we go

Fargle

Through Sylow you can discover that the condition is that, if $|G| = pq$ for $p$ and $q$ prime, $q < p$, and $q$ does not divide $p - 1$, then $G$ is cyclic.

2

Semiclassical

ah, nice

Fargle

14:33

So you are correct about 35, you just need a little more justification than you have.

Unknown MathMan

@Fargle nice

Fargle

(Exercise, provided you've done Sylow: prove the condition I just told you.)

Unknown MathMan

@Fargle I'll try :)

Fargle

Also, as a sidebar (though this doesn't help with your current problem), for any even number greater than 4, there is at least one nonabelian group of that order: the dihedral group.

2 is special that way I guess.

Unknown MathMan

14:56

For any group G with order $p^e,p\in\mathbb{P}$, then G has a normal subgroup of order strictly greater than $1$. My reasoning is - number of subgroups with order $p^e$ is $1$(from 3rd Sylow). From corollary of 2nd Sylow we can also say that, if number of Sylow $p$-subgrp is $1$ then the subgroup is normal. Hence, the statement is true. Am I correct?

Fargle

@UnknownMathMan All you've done is basically say, "$G$ is a normal subgroup of $G$, so we're done", but it seems like the problem wants you to find a proper normal subgroup.

$G$ is always a normal subgroup of $G$, for any group $G$.

Unknown MathMan

That's true.

Alessandro Codenotti

Hint: what do you know about the center of p-groups?

Fargle

^

Damn you, @Alessandro, you sniped me so hard I couldn't even finish typing.

Alessandro Codenotti

:P hi @Fargle, how are you?

Fargle

15:07

Pretty good! Just trying to get the blood flowing before I work on my homework, and study a bit for the math GRE. How about yourself?

rschwieb

Greetings

Fargle

Howdy @rschwieb

Alessandro Codenotti

Hi @rschwieb

Secret

You will need a cleverly defined net to define uncountable sums, but that is an overkill for reals because any reals are reachable already by a countable sum

Alessandro Codenotti

I'm doing well @Fargle, just a bit overworked with all the psets in uni

Fargle

15:09

@AlessandroCodenotti Ah yeah, I can imagine. It did look like you had a lot on your plate this semester. My condolences

Alessandro Codenotti

I was actually hoping to find @Mathei to bug him about algebraic geometry

Secret

hmm...

Rithaniel

Hmmm, I can't seem to find a pdf for "a classical introduction to modern number theory by Ireland and Rosen."

Secret

It is easy to see the difference between $\pi$ and $\pi - \frac{2}{3}$ is $\frac{2}{3}$ which is rational

But does there exist distinct irrationals $r,s$ such that $r+s=q$ where $q$ rational?

o wait, that is precisely what algebraic dependence is about

So if $\pi + e$ is rational, it means e.g. $e$ can be expressed as a sum of $\pi$ and a rational

hmm... what happens if we assume $e=\pi + q$ for some $q \in \Bbb{Q}$ and then compute:

$$e - \pi = \pi - (\pi + q)$$

Let $[x]_n$ be the nth truncation of the sequence of $x$

We are thus interested in:

$[e-\pi]_n = [\pi + q]_n - [\pi]_n$

If $e-\pi$ is rational, the difference on the RHS will eventually become constant after passing a finite $n$

Thus, we need to compare:

$[e]_n - [\pi]_n$ vs $[\pi + q]_n - [\pi]_n$

That is, if:

$$\exists q \in \Bbb{Q} \lim_{n\to \infty} [e]_n-[\pi]_n = q[\pi]$$ then we are done

Daminark

15:33

@Rithaniel it's mildly difficult to find just by Google searching, try libgen.io

Nick

something wrong with these notes: cs.cityu.edu.hk/~helena/cs31622000A/Notes02.pdf

Rithaniel

Hmm, my browser doesn't trust libgen, apparently.

Daminark

15:54

Rip

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$Mathematics$ Mathematics