Mathematics - 2022-06-07

Ted Shifrin

00:15

@robjohn You concur that this be false?

Ted Shifrin

01:51

The room has died yet again. Do we have burial arrangements in place?

copper.hat

Let it RIP

Ted Shifrin

02:22

Cremation?

Koro

Hi

D.C. the III

$T$ is a linear operator on a complex inner product space. If $T$ is self adjoint, then $\langle T(x), x \rangle $ is real for all $x \in V$. I'm trying to prove this, but my problem is I don't know what it would mean for $\langle T(x), x \rangle $ to be real. I've seen a proof which is pretty straightforward, but I don't understand why that shows the value is real

copper.hat

I like having a place to visit, regardless of the state of remains.

D.C. the III

the proof just did some shifting of things while taking a conjugate....

copper.hat

@D.C.theIII $z$ is real iff $z =\bar{z}$.

Koro

02:24

Suppose that A is an n by n matrix with entries from C. Consider the set $W_A=\{A^n: n\ge 0\}$. What can be said about dimension of $W_A$?

Ted Shifrin

@Koro @copper Check my last query to robjohn.

copper.hat

@Koro Cayley Hamilton says something.

D.C. the III

@copper.hat and there you go....... seems like complex analysis/ complex variables will be next on the list of topics

Ted Shifrin

No, dc, this us rudimentary complex numbers..

Koro

@copper.hat I partially understand your point. But I'm having difficulty implementing it. I'll explain.

I think that dim W_A is less or equal to n^2.

@copper.hat I partially understand your point. But I'm having difficulty implementing it. I'll explain.

D.C. the III

02:25

so reviewing basic complex numbers first it is then....

but now having a math thinking background so it will actually stick with me this time

Ted Shifrin

Review the properties of a hermitian inner product.

Koro

I consider I, A, A^2, ..., A^n

copper.hat

@Koro If $A^n$ can be written in terms of $I,A,...,A^{n-1}$...

Koro

the list is linearly dependent because there are more than $n^2$ vectors.

Ted Shifrin

@Koro That’s correct, but CH does better.

Koro

02:27

So there is a minimum k such that $A^k= c_0I+c_1A+...+c_{k-1}A^{k-1}$.

with the convention that $A^0:=I$.

This in my understanding only shows that $k\le n$ (using CH).

But I don't know how that helps answer my question.

I meant $A^{n^2}$ in my earlier message where I mentioned the list.

Ted Shifrin

You’re worrying that $I$ is omitted?

Koro

no, I want to say that dim W_A is less or equal to n for any A.

copper.hat

Well, if $A=I$ the dimension is 1.

@Koro That is what Cayley Hamiliton gives....

@TedShifrin Is the OP you referenced a known quantity?

Koro

Suppose on the contrary that dim W_A=K>n. Let $A^{k_1}, A^{k_2},..., A^{k_K}$ be a basis of W_A.

Suppose that k_i is strictly increasing.

copper.hat

@Koro If $A^n \in \operatorname{sp} \{ I,..., A^{n-1} \}$ then so is any $A^{n+k}$.

Proof by constriction

Koro

02:36

great!

yeah. $A^{k_i}\in \rm span (I, A,..., A^{k-1})$ for any i.

where k is degree of the minimal polynomial

thanks a lot @copper.

copper.hat

indeed :-)

Koro

@TedShifrin the subspace question and orthogonal complement question?

Ted Shifrin

@Koro: No. The question about whether a function that is injective on a dense set must be injective on the whole domain.

Belu cat

02:55

Can someone tell how does the epsilon delta definition work actually I understood the definition But May I know what was the thought process of Cauchy when defining limits that way?

Can I claim that we can't find delta from the graph? Is such proofs allowed?

Nameless

03:24

The limit of a function as it approaches 'a' is 'L' if and only if as x approaches 'a' from both sides (values greater than or less than 'a') f(x) apporaches L

Calvin Khor

@Belucat I very often find delta from a graph. But when it comes time to write a proof, even the blind must be convinced by your argument

Belu cat

Ohh

Calvin Khor

@Belucat also mr-ideahamster.com/classes/assets/a_evepsilon.pdf

Belu cat

Thank you @CalvinKhor

Koro

Hi @Calvin!

Calvin Khor

03:31

@Koro hello, but Im going to have food! ttyl :)

@Belucat np!

Koro

@CalvinKhor bbl

robjohn

03:51

@TedShifrin yes.

If you mean that there is a continuous function so that it is injective on a dense subset yet not injective on the whole set.

I have a simple example

Ted Shifrin

Me too. :) Just wanted to confirm with you!

@copper I think that “pos def” Jacobian problem may be false without convexity.

Belu cat

04:11

What does this statement mean? If f is continuous on [a,b] then f is bounded above on [a,b] that is there is some number N such that f(x)<=N for all x in [a,b]

Ted Shifrin

No vertical asymptotes.

Belu cat

what does bounded mean?

Ted Shifrin

They told you in that sentence.

Belu cat

Oh Got it thank you

robjohn

bounded above means that there is some number $N$ so that $f(x)\le N$

Belu cat

04:15

Yeah Thank you!

What's the difference between theorem and definition?

Ted Shifrin

@robjohn: Here is another conundrum.

A theorem requires a proof.

A definition is telling you what a term means.

Belu cat

Oh Ok tq

Is y=3 graph continuous?

Ted Shifrin

You mean the function $f(x)=3$? What do you say?

Belu cat

04:30

Yeah

Oh Ok I got it tq

It is continuous

copper.hat

@TedShifrin Bummer. I believe you are correct.

Ted Shifrin

Counterexample seems subtle.

robjohn

@copper.hat bummer?

Ted Shifrin

Regarding the conundrum I just linked you, robjohn.

D.C. the III

@robjohn is there a way I can look at my full chat history?

and not just "recent"

Belu cat

04:54

So can I claim I can find a maximum value in a continuous function?

Belu cat

05:09

also a minimum value in a continuous function?

Spivak mentions when n is odd then f(x)=an x^n+an-1x^n-1.....+a0 has a root

Isn't it true when n is even too?

Why does he specify n being odd here?

leslie townes

presumably the a's are real and he wants a real root?

real polynomials of even degree do not need to have real roots (and even complex polynomials of degree zero do not need to have roots)

it's probably baked into the definition of 'root'

Belu cat

05:24

Yeah

Oh Ok tq

I got it

copper.hat

06:05

@robjohn I made a mistake in a comment to a question.

Abstract Space Crack

0

Q: Measuring the "maximum consecutive gap" in the setwise intersection of the usual "twin prime cosets" - how can we write a formula for it?

Use $(a) \subset \Bbb{Z}$ to mean ideal and $(a) + x$ a coset. $$ (6) \cap((5) \uplus ((5) + 2 )\uplus((5)+3)) \cap (5+1,7^2-1) $$ is non-empty because there exist the twin prime averages $12$ and $18$ in the set. Define $\mu(A) = \max \ \{x - y: x , y\in A, x \lt z \lt y \implies z \notin A\}$. ...

How do we measure the size of the "maximum consecutive gap" in the set above?

Wave

06:33

Hello, is the following correct: if I have a commutative ring R and denote it's maximal ideal p. Then the localization of R at p is R_p which is a local ring right? Because the maximal ideal of R_p is pR_p. But is it true that an element in pR_p is c*(r/b) where c is in p, r is in R and b is in R\p.

Karl Kroningfeld

Yes :)

Are there any reservations about what you said @wave ?

Wave

@KarlKroningfeld perfect thanks! No there are no reservations, but I only wanted to be sure since we haven't discussed this in the lecture. The prof only said that R_p is a local ring nothing more

Karl Kroningfeld

Also, one can take $p$ to be any prime ideal. Fun fact: the quotient $R_p/pR_p$ is the field of fractions of the integral domain $R/p$.

Wave

@KarlKroningfeld about your fun fact. So I know that since pR_p is a maximal ideal of the ring R_p then R_p/pR_p is a field. but how do we see that it's the field of fractions of the integral domain R/p?

I had the following definition of a field of fractions: If R is an integral domain and S=A\{0} then S^{-1}A is the field of fractions of A

Karl Kroningfeld

06:53

@Wave Briefly, you can localize the exact sequence of $R$-module $0\to p\to R \to R/p\to 0$, using the multiplicative set $S=R\setminus p$. That leads to $0\to pR_p\to R_p\to S^{-1}(R/p)\to 0$. Then, one shows that $S^{-1}(R/p)$ is the field of fractions of $R/p$.

Wave

@KarlKroningfeld sorry by localizing the exact sequence you mean tensoring by S so $S\otimes...$?

Karl Kroningfeld

You can localize $R$-modules, and that preserves exact sequences of $R$-modules. It is the same as tensoring with $S^{-1}R$.

Wave

@KarlKroningfeld sorry for the stupid questions I don't really understand tensoring yet. I mean I take the exact sequence you gave above and then tensor by S^{-1}R. for example I then get S^{-1}R\otimes p right? But how do I get to pR_p

Karl Kroningfeld

It's probably easier to see that $S^{-1}p$ is $pR_p$ and then identify $S^{-1}p\cong p\otimes S^{-1}R$.

I doubt I can fully explain the "fun fact", maybe I can find a decent reference.

Wave

okey thanks!

Karl Kroningfeld

07:13

@Wave The first answer here approaches it in a more direct way.

Atiyah-Macdonald say more about the approach I suggested (Corollary 3.4) though it may be equally hard to digest.

Wave

@KarlKroningfeld perfect thanks I will take a look!

robjohn

08:02

@copper.hat oh, no...

robjohn

08:16

@D.C.theIII not that I know of

Wave

Do we only need the field of fractions in the definition of an integrally closed domain since this is the smallest field R can be embedded in?

So I mean we could take any larger field K in which R can be embedded and then also say that R is integrally closed if R is an integral domain and the integral closure in K is R

user777

Hi,
I'm trying to find values for x and y such that
x-y <0
and
-x+y <0
for x, y \in R. when I sum these two equations up, I got 0 + 0 < 0 which is impossible, does this mean that I don't have a solution here by any real numbers?

Wave

@user777 Hi. I mean if you want x-y<0 then you can't have -x+y<0 since when you multiply x-y<0 with -1 you get -x+y>0.

@user777 does this help?

user777

08:32

@Wave yes, I understand!

Wave

Perfect!

user777

Does my conclusion right when I said 0 + 0 < 0 is impossible

Wave

@user777 sure 0+0<0 is impossible. In my opinion also your argument is correct because it's like a prove by contradiction which isn't really neccessary but it is correct in my opinion

user777

@Wave Thank you so much!

Wave

your welcome

abenthy

08:39

Does someone see an elegant way to show that for a real eigenvector $v$ of a real symmetric square matrix $A$ to an eigenvalue $\lambda$, the matrix $\begin{pmatrix}0 & v^\top \\ -v & A - \lambda I\end{pmatrix}$ is invertible? My only solution is to show with a rather ugly argument that the linear map induced by this matrix is injective.

Maybe there is a formula for the determinant here? I checked Wikipedia's article on block matrix determinants, but that didn't help.

Calvin Khor

09:20

@abenthy i think there's an assumption missing, A=0 is real symmetric, if we put v = (1,0,...0) there are zero rows

abenthy

09:46

Yeah right, thank you @Calvin. The eigenvalue has to be a simple eigenvalue.

robjohn

10:07

@abenthy yes, if $u$ is another eigenvector of $A$ with eigenvalue $\lambda$, then using Gram-Schmidt, we can find $u'$ perpendicular to $v$ which is still an eigenvector of $A$ with eigenvalue $\lambda$.

Then $\begin{pmatrix}0\\u'\end{pmatrix}$ is in the null space of the matrix.

Secret

11:14

How will you deal with a human who is declassifying all kinds of CIA shit across the planet and fucking up mathematical integrity and had no intent to remorse from his actions https://www.deviantart.com/galaxystarlights17/art/I-m-just-glow-up-and-relax-918424242 ?

Nasser

12:28

This is the second time I answer question, they read it then delete the question. All the time spend answering is wasted. math.stackexchange.com/questions/4467369/… something should be done about this. It is waste of time here to answer questions.

This is very rude. I will not bother answering questions any more on this forum.

Devansh Bhardwaj

13:06

But is deleting question not forbidden after it has been answered on stackexchange?

anak

@Nasser like I mean, no one is telling you that you have to answer questions. Maybe they don't like your answers? Maybe they are cheating on an exam? Who knows.

robjohn

13:30

@Nasser Rather than that, flag and the mods will undelete the post. We don't like people doing that.

2

one potato two potato

14:17

Is there anyone heard about 'topology of compact convergence'?

Calvin Khor

haven't heard of it, convergence on compacts maybe?

one potato two potato

But you heard about compact open topology right?

Wave

does someone know things about commutative algebra?

Govind75

14:53

If I have $X \sim MVN(\theta, I)$ and I want to calculate $E[f(||X||^2)]$, can I use the laplace approximation to say this is approximately $f(||\theta||^2)$

Belu cat

15:36

If someone asks what are numbers how would you guys answer him?

Shiki Ryougi

omg begula!!! :3 uwu

it depends on his level of knowledge

Belu cat

I was watching an IAS interview where he was questioned "What are numbers ?"

Shiki Ryougi

do the first two definitions satisfy you? wordnik.com/words/number

Belu cat

15:53

How can a number be derived from counting I mean only whole numbers could be counted right?

Belu cat

16:16

defined*

Wolgwang

818

A: What are imaginary numbers?

Let's go through some questions in order and see where it takes us. [Or skip to the bit about complex numbers below if you can't be bothered.] What are natural numbers? It took quite some evolution, but humans are blessed by their ability to notice that there is a similarity between the situati...

Abstract Space Crack

:math.stackexchange.com/questions/4467573/…

one potato two potato

16:32

Hausdorff metric problem in Munkres book is quite hard

Abstract Space Crack

That's what she said

Wolgwang

Hi!

How to find

$$\lim_{n\to\infty}\bigg[\frac1n+\frac1{n+1}+\frac1{n+2}\cdots+\frac1{3n}\bigg]$$

Ted Shifrin

Factor out $1/n$.

Wolgwang

$$\lim_{n\to\infty}\frac1n\bigg[1+\frac{n}{n+1}+\frac{n}{n+2}\cdots+\frac1{3}\bigg]$$

Ted Shifrin

Rewrite those middle terms,

Belu cat

16:41

If someone asks what are numbers how would you guys answer him?

Wolgwang

Something like $\frac{n}{n+1}=\frac{(n+1)-1}{n+1}=1-\frac{1}{n+1}$ ? >_<

Ted Shifrin

More basic than that.

Ssshhhh

Devansh Bhardwaj

oops

Ted Shifrin

:)

Wolgwang

@TedShifrin Hmm 🤔

Wave

16:50

sorry this is may be a stupid question but what are all the prime ideals of $\Bbb{C}[X]$? I mean it is a principal ideal domain so every ideal is generated by one element. I also know that all maximal ideals are $(P)$ where $P\in \Bbb{C}[X]$ is irreducible. Hence these are for sure also prime ideals. But then (0) is also a prime ideal. Now are there some further prime ideals?

Wolgwang

@TedShifrin IDK :(

Ted Shifrin

You didn't see what Devansh typed? There's only one other thing to do with $$\frac n{n+k}.$$

What are the irreducible polynomials, Wave?

Wave

@TedShifrin these are all the polynomials of the form (X+a) for some $a\in Bbb{C}$

Wolgwang

@TedShifrin I saw...the typical way to solve limits at infinity but how will that help here?

Ted Shifrin

Well, do it and look at the expression you have.

@Wave Yes. Right.

Wave

16:55

@TedShifrin so are all prime ideals of the form (X+a), and (0)? Because how do I know that there aren't futher?

Ted Shifrin

What else could there be?

Wave

I don't know but since this is really new to me I'm a bit unsure claiming that these are all

Ted Shifrin

Well, take any other polynomial and show that the ideal is not prime.

Wave

@TedShifrin Aha so you mean if for example I take the ideal (X^2-4), then I can take the polynomials [X+2] and [X-2] then clearly [X+2][X-2]\in (X^2-4) but neither [X+2] nor [X-2]\in (X^2-4).

Ted Shifrin

Right.

Wolgwang

17:05

@TedShifrin Do I have to distribute the limit inside the parenthesis?

Ted Shifrin

You cannot do that, Wolgwang. So what is your final expression after doing the algebra?

Wave

@TedShifrin perfect thanks!

Belu cat

Guys how do u define numbers?

Wolgwang

@TedShifrin Am I going in the right direction?

$$\lim_{n\to\infty}\frac1n\bigg[1+\frac{1}{1+\frac1n}+\frac{1}{1+\frac2n}\cdots+\frac12+\frac{1}{2+\frac1n}+\frac{1}{2+\frac2n}\cdots+\frac1{3}\bigg]$$

Ted Shifrin

That's good. It might be better just to write $1+k/n$ for $k=1,2,\dots,3n$. Doesn't this suggest something familiar to you from calculus?

Wolgwang

17:16

@TedShifrin Nope. They have just started teaching calculus.

Ted Shifrin

Hmm ... There's no way to do this without recognizing it as related to an integral.

Wolgwang

Is the answer going to be in terms of $\log$?

Ted Shifrin

Yes

Wolgwang

I know it is silly but can't I directly put $n=\infty$ in the initial expression? All terms will become zero?

Ted Shifrin

NO.

The number of terms keeps changing. You can do limit of a sum equals sum of the limits only when it's a (fixed) finite sum.

Wolgwang

17:24

Oh

Found a related question

We haven't studied Riemann sum. :-(

Wolgwang

$$\lim_{n→∞}\sum_n^k=\dfrac{k}{n^2+k^2}$$

Will this also require integrals?

Buddhini Angelika

Hi! When we are given a sum $\frac{26}{5} (\frac{-3}{10}) / (\frac{-67}{10}) \frac{1}{2}$, isn't it right that we take as $\frac{26}{5} (\frac{-3}{10}) (\frac{-10}{67}) \frac{1}{2}$ and simplify (according to BODMAS rule)?

Ted Shifrin

That looks like garbage, Wolgwang.

@Buddhini This is why one should never write / for division. Is the $\frac12$ in the numerator or in the denominator?

Buddhini Angelika

@TedShifrin I'm actually bit bothered by that. The bracket is only written for \frac{-67}{10}

Ted Shifrin

As I said, this is why writing / is horrible.

Buddhini Angelika

17:37

And instead of /, the otheer symbol with two dots on either side of - symbol

is present for division

Ted Shifrin

That also is horrible. Once one leaves primary school one should never write that.

$\div$

Buddhini Angelika

:)

Yeah

I found this question.

Ted Shifrin

If you want clarity using /, you should write $ab/(cd)$ to make sure both things are in the denominator.

BBIAB

Buddhini Angelika

But when simplifying I did as above by thinking that only $\frac{-67}{10}$ is in the denominator and the answer I got is 0.1164. But another friend of mine has thought that \frac{1}{2} is also in the denominator.

So I should probably explain to him that to be sure, we should put bracket to entire $\frac{-67}{10} \frac{1}{2}$?

Ted Shifrin

Yes, if you must use /.

Buddhini Angelika

17:51

@TedShifrin I'm confused again. Then, the answer is different based on the fact that we use / or $\div$, but not because of the bracket?

user777

Hi, I have a question: suppose I have A = frac{2^{n}}{10n}, Is it possible to write A as c^n for some c \in R?

leslie townes

for one fixed n? or to express the sequence of values like that, for all n, in that way?

yes in the first case (take c = the nth root of the right hand side). no in the second.

user777

@leslietownes for all n (as n approaches infinity)

Wave

sorry again, is an irreducible element of $\Bbb{C}[X,Y]$ of the form XY-a for some a in \Bbb{C}?

Ah no I saw that there are other irreducible elements but is there like an explicit form?

Ted Shifrin

@Buddhini If we write $$\frac{\frac{26}5\cdot\frac{-3}{10}}{\frac{-67}{10}\cdot\frac12},$$ then there is no ambiguity.

If you use / or $\div$, you must group numerator and denominator clearly.

Good morning, munchkin's pet.

Semiclassical

18:05

it may be tedious to put parantheses in places but it's very useful if you're worried about ambiguity

Ted Shifrin

Just better not to write / for fractions unless it's something simple like 2/3.

Buddhini Angelika

Okay, now its clear. :) Thank you very much @TedShifrin

:) :)

Have a nice day!

Ted Shifrin

You too.

Buddhini Angelika

Thanks!!

Semiclassical

@user777 if A(n) can be written as c^n, then A(n+1)/A(n) = A(n)/A(n-1) for all n

Buddhini Angelika

18:07

:)

Semiclassical

which is straightforward to test here

Wave

18:23

https://math.stackexchange.com/questions/4467661/how-do-i-find-all-prime-ideals-of-the-localization-s-1-bbbcx-y-where-s

Is there someone who can help me here?

copper.hat

sorry, i am not an ideal person.

user777

@Semiclassical Thank you, so A(n+1)/A(n) = n/(n+1) != 2(n-1)/n, therefore it cannot be written as c^n for some c in R.

Semiclassical

right

user777

Is this a theorem or where I can find this, I believe in analysis textbook

Semiclassical

can also pick small counterexamples, e.g. A(3)/A(2) != A(2)/A(1).

you don't need a theorem. if A(n) = b*c^n, then A(n+1)/A(n) = c = A(n)/A(n-1)

Wave

18:38

@copper.hat hahah;) no problem and nice word game.

Semiclassical

so A(n) ~ c^n would mean the ratio of consecutive terms is c, i.e., doesn't depend on n

user777

@Semiclassical I see the point!

Thank you

Semiclassical

np

Abstract Space Crack

19:37

My twin prime post is on fire - 3 upvotes!

I mean +3, that's a hyphen

I paved the road, now some smaht person just has to prove the inequality holds for each $n$.

The current conjecture checking code prints out:
1 1
< 4
2 5
< 19
3 11
< 41
4 29
< 111
5 41
< 155
6 65
< 273
7 107
< 341
8 149
< 507

as you can see, the inequality shouldn't be difficult to prove (it's not "very close" - it's actually a very large gap or leeway)

e.g. the last one reads that at $n = 8$ we have $\mu(A) = 149 \lt 507 = p_{n+1}^2 - p_n - 3$

From my experience looking at formula behavior, this doesn't seem to be one that's going to swoop back down; I think it will be forever growing

user193319

20:22

Is the Stone-Cech compactification of the natural numbers as a discrete topological space perfect?

Ted Shifrin

You saw my response to your f,g on a dense set question?

user193319

I did not! Let me take a look back.

leslie townes

21:11

what's a perfect space, again?

Ted Shifrin

22:00

Every point is a limit point.

leslie townes

22:34

are points of N isolated in Cech(N)?

i used to know that space pretty well for an analyst

to the extent anybody can know it, that is

Ted Shifrin

22:56

What I once knew is 50 years ago ….

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