Mathematics - 2022-03-01 (page 1 of 2)

Jakobian

00:00

first thing I guess I can assume, is that $Z$ is a bounded rv

then by translating, I can say it's positive

or maybe mean zero would be better

then I could try defining $M_t = \int_0^t Z_s \mathrm{d}s$, where $E[M_t] = 0$

Xnero

Is this the correct explanation for why differentiating $y$ gives $\frac{dy}{dx}$?

$\frac{d(y)}{dx} = \frac{d(y)}{dy} * \frac{dy}{dx} = \frac{dy}{dx}$

Jakobian

I guess $E[M_tM_s] = \int_0^t\int_0^s E[Z_sZ_t]\mathrm{d}t\mathrm{d}s = 0$ for $s\neq t$ and $E[M_t^2] = Ct^2$

PenAndPaperMathematics

Yeah the EM fields, those will work :)

Just sound smaht here

Jakobian

00:16

I think I might be able to get that $M_t$ is a Wiener process here

wait I made a mistake

Ted Shifrin

@Xnero what is $y$, anyway?

It’s a function of $x$. So the derivative is the derivative. Bringing in a pretend chain rule is not helpful.

Jakobian

$E[M_tM_s] = \int_0^t \int_0^s E[Z_xZ_y]\mathrm{d}x\mathrm{d}y = \iint_{x = y, x\in [0, t], y\in [0, s]} E[Z_xZ_y]\mathrm{d}x\mathrm{d}y = 0$ I think, so it gives $E[M_t^2] = 0$ so $M_t = 0$ almost surely

Xnero

@TedShifrin I was trying to explain to myself why differentiating y (where y is a function of x) is just dy/dx. As differentiating 2y is 2 times dy/dx.

Ted Shifrin

That’s the definition of the derivative. Constant rule, product rule, sum rule, chain rule all come next.

Jakobian

Now $\frac{\mathrm{d}}{\mathrm{d}t}M_t = Z_t = 0$ almost surely

Xnero

00:24

@TedShifrin I know most of those rules. Technically, is my application of the chain rule is correct?

Ted Shifrin

You’re trying to bring in implicit differentiation, but it’s truly circular reasoning. What is $dy/dx$ in your second line?

It’s circular, as I just said.

Jakobian

I guess I solved it. Yay

Xnero

@TedShifrin What do you mean by "circular"?

I know it is pointless, but I was just trying to apply implicit differentiation which I learned recently to $y$ to see if I would get dy/dx so I was wondering if that is correct.

Ted Shifrin

If you don’t know what $dy/dx$ is, how can you use it to calculate anything?

Circular means assuming what you’re trying to do.

Xnero

@TedShifrin dy/dx is the gradient function of the rate of change of y with respect to x.

rb3652

00:31

I'm trying to understand matrix norms.

Xnero

@TedShifrin I guess it is just accepted that differentiating y is dy/dx then.

Ted Shifrin

That’s what the symbol means. Or you write $y=f(x)$ and the derivative is $f’(x)$.

leslie townes

it's what that notation means. it has no choice but to be 'accepted' as that.

oh, ted's on this.

rb3652

I understand that the $n$th norm of a vector is simply $(v_1^n+v_2^n+...+v_n^n)^(1/n)$

But I don't understand what the difference between the Frobenius norm and 2-norm is.

Ted Shifrin

@Xander Turns out it’s easy to google search textbooks by an author, but much harder to find research papers listed.

Xnero

00:33

@leslietownes It's not just notation though, it's a function.

leslie townes

i guess if you think of d/dx as something that can be applied to any function, however called, i guess de-bracketing of d/dx [y] as dy/dx is a notational convention. but it's not like dy/dx means something else and we're proving that it's equal to the derivative of y.

rb: the forbenius norm of a matrix is indeed the 2-norm of the matrix thought of as its list of entries (in any order).

Ted Shifrin

Matrix norm or vector norm?

Xnero

Then this is correct?

25 mins ago, by Xnero

$\frac{d(y)}{dx} = \frac{d(y)}{dy} * \frac{dy}{dx} = \frac{dy}{dx}$

leslie townes

all 'frobenius' tells you is that you're talking about something as a matrix.

Ted Shifrin

Xnero doesn’t listen, so I’m done.

All yours, mr lawyer.

Xnero

00:35

@TedShifrin It's a relatively simple yes or no.

Ted Shifrin

You made no effort to think about the things I explained. I’m done with you.

Circular reasoning is INCORRECT.

leslie townes

note that there are a number of choices of ways to think about a matrix as a one-dimensional array of numbers (and at least some treatments of vector norms require you to pick an ordering). so if there's a difference, it might be there.

Xnero

@TedShifrin What is that?

leslie townes

rb: it's traditional to write 'p' instead of n for that 'n-norm' above. it's sometimes helpful that you can choose p that isn't an integer.

Xnero

@TedShifrin Then how do you know that differentiating y does not give dy^2/dx?

anak

01:34

what

@Xnero what does "d/dx" mean to you?

love_sodam

I can't see the usefulness of metrization theorems stated in ch 6 in Munkres topology textbook.

leslie townes

01:54

love: i can't speak to all applications, but there are a lot of fields that do use topology but do not use metrization theorems (e.g. because the spaces of interest obviously are, or obviously are not, metrizable)

i can see why they would be included in a 'point set' topology book, i am not sure why they would be included/used in other settings

love_sodam

Yeah some professors skip Ch 6 metrization theorem but rather focus more on completion of topological space which I know the importance.

rb3652

Hi all

@leslietownes Thank you for the comment

But I'm still quite confused at why 2-norm is any different from the Frobenius norm.

My professor defines it as follows:

Well, I'll figure it out. This is a great community -- devoting free time to help others with their math problems. I think it should be more appreciated.

leslie townes

02:10

rb: oh, we were thinking different norms. i think it's not standard to call that the "2-norm" (at least when vector norms on ell^p are also being considered). that's not directly computable from matrix entries.

Ted Shifrin

SVD …. go SVD

leslie townes

consider upper triangular 2x2 matrices.

M17

What are the terms of conjecture?

What are the steps until something becomes a mathematical guess in specific steps

leslie townes

02:26

there's no formal process, M17. you could think of 'conjecture' as synonymous with 'educated guess.' people tend not to call wild guesses conjectures. most people use the term only to describe something if there's some evidence or heuristic reasoning to support it. but people do vary in how much evidence they want to see before they bother referring to something as a conjecture.

William John

I don't know how can I solve integral of tanx/(1+x^+x^4)

Should I expand tanx?

leslie townes

william: is there an exponent missing there? looks like it might be difficult and/or not have a 'nice' expression.

is this coming up in a context where it might be possible to answer the underlying question without a formula for an antiderivative of that thing?

M17

How l can send photo here?

PM 2Ring

@M17 Sorry, you need more rep to directly upload image files in Chat.

@M17 I case you missed it, leslie townes found the link to the question on the main site about this twin primes version of the Goldbach conjecture. math.stackexchange.com/questions/3134627/…

@M17 There is a way to work around that 100 rep limit. See meta.stackexchange.com/a/317856/334566 for details.

M17

Ok, thanks

PM 2Ring

02:39

I suspect that the twin primes version of Goldbach is at least as hard to prove as the plain prime version.

leslie townes

i do kind of like the idea of blending two famous conjectures into one. i'm wondering what else we might put in there.

PenAndPaperMathematics

That is a mistake. Generalize later

That's kind of like programmers should optimize lastly

But when coding sometimes we generalize

leslie townes

ok, to be clear, i was kidding.

PenAndPaperMathematics

but the first attack coding is a hardcoded prototype

leslie townes

i still want to know what else we might put in there.

maybe if we could find a way of associating something associatable with counterexamples to the conjecture with something on the critical line or critical strip.

PenAndPaperMathematics

02:43

@leslietownes $$f(n,k) := \sum_{c\mid d \mid n\# \\ \gcd(c,2k)=1} (-1)^{\omega(d)}\left(\lfloor \dfrac{b - x_{c,d}}{d} \rfloor + \lfloor \dfrac{x_{c,d} -a}{d} \rfloor \right) \\$$

PM 2Ring

Regarding that (6k-1)(6k+1)+26 pattern, I don't think it's very significant. I vaguely remember writing that code & discussing it with you a while ago. :)

PenAndPaperMathematics

That counts the twin primes in the interval $[a,b]$ for certain $x_{c,d}$ but an infinitude of them

Found by CRT

I invented this approach I think. One paper comes close

*discovered not invented

The idea is then now to tie together the different formulae for $2k$ and $2kj$-differenced prime pairs

That literally counts them in an exact way

o__o everyone talks twin primes, but no one cares

I'm working on a backward proof-by-induction from Zhang's result.

PM 2Ring

The proportion of those numbers which are primes gradually drops as k increases. And if we look at all pairs of the form [(6k-1), (6k+1)] in a given range starting from k=1, not just twin primes, we get roughly the same proportion. Sometimes the twins give a slightly higher ratio, but sometimes their ratio is lower.

PenAndPaperMathematics

Well, plug $k=1$ in to $f(n,k)$ and you get the exact number of them

Prove the formula is never eventually always zero after some point and you've got $1M

M17

@PM2Ring, I know it's not important

PenAndPaperMathematics

02:50

@M17 did you see the formula? I can teach you it - you know all of the involved operations

M17

I didn't say it was important or even about other styles

PM 2Ring

Oh, here's some code for the twin Goldbach thing.

sagecell.sagemath.org/…

Ajay

Hello

I just had a test on Biology

Vasoconstriction and Vasodilation are not fun too learn

PenAndPaperMathematics

Especially if you quit smoking and learn the hard way

Nicotine is a vasoconstrictor. When you quit, your face gets puffy

William John

@leslietownes nothing is missing. Professor gave answer as 0 for definite integral with some limit that I forgot. I was curious if someone can come up with the solution.

M17

02:55

9164/(E - R)= (E+R)

How i get, R and E?

Ajay

We have too learn about how vasoconstriction and vasodilation help too lower or increase body temperature.

Not much about smoking. Other than the fact than it causes lung cancer.

M17

@PM2Ring

PM 2Ring

@M17 Hey, there's nothing wrong with exploring number patterns. But it can suck up a lot of time. ;) The OEIS has lots of patterns.

Ajay

@M17 $(E-R)(E+R) = E^2 -R^2 = 9164$. Now you have a non-linear diophantine equation ;)

Actually not really, it's just some algebraic manipulation that gets you the values :(

High school stuff still

M17

PM, What's mean about OSIS?

@Ajay, Can it be resolved or not?

Ajay

03:01

It's an encyclopedia for integer sequences

yes it can

there should be integer values satisfying the equation

i think

PM 2Ring

Quadratic Diophantines are fun. I learned how to solve them efficiently a year or so ago, but my memory is a bit fuzzy, and I'd have to consult my notes for anything beyond simple Pell equations.

Ajay

I don't have any paper on me know so I can't check but I am very sure that it does

As Ted would say: "I would bet very seriously"

lol

PM 2Ring

@M17 The On-Line Encyclopedia of Integer Sequences

M17

I know the value of E and R and each of them has only one solution, I want the way to solve them if I don't know them

Ajay

Consider my answer to the following question

"Find the integer values of x and y in the equation $x^2 + y^2 = 77077$ where $x < y$ and $x,y\in{\mathbb{Z^{+}}}$."

M17

03:08

Is the solution to this equation available or not?

Ajay

"Find the integer values of x and y in the equation $x^2 + y^2 = 77077$ where $x < y$ and $x,y\in{\mathbb{Z^{+}}}$." Intuitively if we have not had experience with Diophantine equations we would think that x an y can take on infinitely many values or that we would require another equation with x and y variables. However, note that the equation has both variables squared, this is the key. With a square we have reduced the equation to simply having a unique solution for both variables x and y.

PM 2Ring

@M17 For equations of that form, you just need the factors of the RHS, then build integer pairs of those factors of the form E-R, E+R. The prime factors of 9164 are 2^2 * 29 * 79, so there aren't many solutions.

M17

To be clear, R and E are positive integers and there are no infinitely many solutions

Ajay

However, note that the equation has both variables squared, this is the key. With a square we have reduced the equation to simply having a unique solution for both variables x and y. Next we shall note what kind of number $77077$ is. It is not a prime number, therefore by the Fundamental Theorem of Arithmetic(FTA), we have $7^{2}\cdot11^{2}\cdot13$.

M17

@PM2Ring

Ajay

03:10

However this is not enough, we want this entire equation to be in its lowest prime factors. Therefore $13$ can be simplified further into $2^{2}+3^{2}$. Now we have that $x^2 + y^2 = 7^{2}\cdot11^{2}\cdot(2^{2} + 3^{2})$.

We have to be careful here, we will proceed slowly. First we take $7\cdot11$ yielding $77$, next we shall multiply $77\cdot2 = 154$ and $77\cdot3 = 231$, now the equation will be $x^2 + y^2 = 154\cdot231\cdot11\cdot7\cdot(2 + 3)$.

Without simplifying the quantity in the brackets we multiply it by $77$ since $11\cdot7 = 77$ giving us $x^2 + y^2 = 154\cdot231\cdot77(2 + 3)$ evaluating further we have $x^2 + y^2 = 154\cdot231\cdot154\cdot231$. Which is just $x^2 + y^2 = 154^{2}\cdot231^{2}$.

By direct comparison we have that $x = 154, y = 231$. And since $x < y$, indeed $x = 154, y = 231$.

PM 2Ring

@M17 Have you figured out how to view LaTeX in chat yet, or are you still seeing lots of $ signs?

Ajay

@M17 There is a much shorter way than this, but I feel this method is better for understanding how to evaluate these equations in the first place.

M17

I see dollar signs

Ajay

I wrote this for my 9-year old niece, so it should be very easy to understand.

Link for latex math.ucla.edu/~robjohn/math/mathjax.html

M17

@Ajay, It shows dollar signs, so your words may not be clear to me, I think it is important, but it is not apparent to me

Ajay

03:15

Use the link and then add the bookmark.

Then click on the bookmark and the latex will render

The signs will disappear.

M17

Can you send me a picture from latex?

9164/(E - R)= (E+R)

Ted Shifrin

@Ajay He would?

M17

E= one solution

R= one solution

Ajay

@TedShifrin Oh yes. The not so good chat we had earlier. Very sorry about that.

@M17 ok

Ted Shifrin

@Ajay LOL … I may have said it once :)

M17

03:20

@TedShifrin, can slove it?

Ajay

@M17 it's an answer to bigger question but the fact is that it evaluates a diophantine equation

PM 2Ring

@M17 This is just simple algebra. If E+R=p and E-R=q then 2E=p+q and 2R=p-q

Ajay

@TedShifrin I used your phrase in an english test and my teacher loved it ;)

@M17 I may have simplified the explanation on diophantine equations incorrectly but it still should be ok....He hopes.

Koro

I have $R=\{\begin{bmatrix}a&-b\\b&a\end{bmatrix}:a,b \in Z_p\}$. I have shown that R is commutative ring. I know that R will have $p^2$ elements.
Claim: If p=7, then R is a field.
Proof: every non zero element in R is a unit and has an inverse of the form: $(a^2+b^2)^{-1}\begin{bmatrix}a&b\\-b&a\end{bmatrix}$
Is this correct? Thanks.

M17

@Ajay, I use translation sometimes

Ajay

03:24

ok

M17

@PM2Ring do you know E andR?

Ajay

@M17 Why not try find it yourself with the explanation and steps we have given you. That way you learn!!

You cannot expect us too tell you the answer just like that.

M17

Unfortunately I use translation sometimes, so I can't translate the image

Ajay

What language are you fluent in?

M17

I know the answer, but I don't know the steps

Ajay

03:28

I'm confused now. What are you asking for? The values of E and R or the steps????

So you have the answer but are looking for the steps???

M17

AB=2291
A=? B=?
A,B are prime numbers
---------------------------------------
A+B=E
if A>B
A - B=R

E^2 - R^2=F
F/4=2291
F=9164
E^2 - R^2=9164
(E+R) (E - R)=9164
9164/(E - R)= (E+R)

Koro

discretion: if p=5 then I think that R is not a field. Because then $(2^2+1^2)\equiv 0$

PM 2Ring

@M17 I just told you the steps. E+R=2×79, E-R=2×29

Koro

the natural numbers which can be written as sum of squares of two natural numbers are called ...?

M17

@PM2Ring, yes

love_sodam

03:30

Need to show $a^2+b^2\neq 0$. This should be if and only if condition to be a field.

Koro

@love_sodam: I think so too. For Z_7, intuition suggests that $a^2+b^2$ is non zero.

M17

@PM2Ring, So it is easy to solve any equation in this way??

PM 2Ring

@Koro I just call them sums of two squares. ;) But Sloane calls them oeis.org/A001481

@M17 Any equation of the form E^2 - R^2 = n.

Koro

I'll also call them sum of squares.

love_sodam

$a^2+b^2 = 0\iff -1 = (a^{-1}b)^2$ in $\Bbb Z/p$.

M17

03:36

@PM2Ring, And through that I knew 29 and 79?

You know

Koro

@love_sodam Yeah, that makes sense :). Furthermore, $(a^{-1}b)^2=p-1$ and hence if p=7 then we have no solution to $x^2=p-1$ in Z_p. Thank you!

PM 2Ring

Sums of two squares. All natural numbers can be written as the sum of at most 4 squares. Fermat worked on that problem. There are lots of lovely patterns involved in the numbers that are the sum of 2 squares. An introduction to number theory (like Hardy & Wright) will generally have lots of info on this.

Koro

For p=5. $x^2=-1$ does have a solution though.

M17

@PM2Ring, Can you make a simple program for this equation, give it n and the program gives me R and E

love_sodam

@Koro $x= 2$.

Koro

03:40

yes

:)

PM 2Ring

FWIW, various medieval Indian mathematicians, like Brahmagupta, studied sums of two squares, and their application to Pell's equation.

Brahmagupta

Brahmagupta (c. 598 – c. 668 CE) was an Indian mathematician and astronomer. He is the author of two early works on mathematics and astronomy: the Brāhmasphuṭasiddhānta (BSS, "correctly established doctrine of Brahma", dated 628), a theoretical treatise, and the Khaṇḍakhādyaka ("edible bite", dated 665), a more practical text. Brahmagupta was the first to give rules to compute with zero. The texts composed by Brahmagupta were in elliptic verse in Sanskrit, as was common practice in Indian mathematics. As no proofs are given, it is not known how Brahmagupta's results were derived. == Life and... ==

M17

@PM2Ring

@Is it possible to create this program?

PM 2Ring

@M17 I think that would be a good project for you when you start to learn Python. ;)

M17

AB=2291
A=? B=?
A,B are prime numbers
---------------------------------------
A+B=E
if A>B
A - B=R

E^2 - R^2=F
F/4=2291
F=9164
E^2 - R^2=9164
(E+R) (E - R)=9164
9164/(E - R)= (E+R)

If we can easy to know E and R, you can solve pq=n

AB=2291

I have very simple projects in Python

PM 2Ring

Note that for 2E=p+q and 2R=p-q to have integer solutions, p and q must have the same parity. That is, they must both be odd or they must both be even. So if n=pq, then either n is odd, or its prime factorisation contains an even number of twos.

Koro

03:50

@M17: In general you can't!! Take n=16 for example.

Can you write 16 as product of two primes p and q?

What's correct however is that every n can be written as product of prime powers.

M17

P q prime numbers

Pq=n

Koro

@M17: It's not possible to solve it for every n as I said above.

M17

Ok

A=79, B=29

PM 2Ring

Eg, for 16, there's no natural solution. But there's the trivial integer solution $4^2-0^2$

M17

I multiply two prime numbers together and then take the result and assume that I don't know those two prime numbers

Koro

03:54

@PM2Ring: I understand that they want pq=16, where p and q are primes, which is not possible.

M17

This is the content of the idea

PM 2Ring

For $n=70$, we get $pq=2×5×7$, so no pairs $(p,q)$ can have the same parity, and so there are no integer solutions.

M17

@PM2Ring, But within the scope that I mentioned, there are solutions

PM 2Ring

@M17 This is the basis of Fermat's method of factorization. en.wikipedia.org/wiki/Fermat%27s_factorization_method

@M17 Yes, if n is odd, there are always solutions. And if n is prime, there's obviously exactly one solution (in natural numbers)

Koro

@PM2Ring n=81

M17

04:03

Is my idea correct about knowing A and B?

PM 2Ring

@Koro $81=9^2-0^2=15^2-12^2=41^2-40^2$

Koro

@PM2Ring in all the cases, 81 is not expressed as product of two primes.

I understand that they wanted 81=pq, where p and q are primes.

PM 2Ring

@Koro Sorry, I thought we were still talking about a difference of two squares.

Koro

@PM2Ring sorry, I misunderstood.

PM 2Ring

Hey, I might be misunderstanding too. :)

M17

04:11

@PM2Ring

You create programs when I don't ask you, but when I ask you don't:)

PM 2Ring

Oops. I made a mistake in chat.stackexchange.com/transcript/message/60544343#60544343 If n isn't odd, it must have at least 2 twos in its prime factorisation.

M17

It is better to make mathematical steps in an orderly manner without words in the text

9164/(E - R)= (E+R)

PM 2Ring

@M17 Maybe I'm trying to encourage you to learn Python. ;) Also, this is supposed to be a math chat. I don't think Ted & the other room owners and room regulars would like it much if this place turned into a free code writing service.

But I have written some code. :)

M17

04:27

True

PM 2Ring

sagecell.sagemath.org/…

M17

E^2-R^2=n

Where i write E and R

Sorry

Nn

n

E^2 - R^2=9164

PM 2Ring

BTW, you can type expressions into the m box, eg, 3*5. Even things like factorial(5)

M17

It shows me many results

PM 2Ring

That program uses m rather than n, because n already has a special meaning in Sage.

M17

04:33

I want to show me the value of R and E only, it shows many results

If R^2 - E^2=m

I will write m

And program give me R و E

It shows several results, including the correct answer

Thanks

Good program

PM 2Ring

@M17 If you give it 2291, it will show 54, 25. But it will also show 1146, 1145 because $1146^2 - 1145^2 = 2291$

M17

I writ 8164

9164 sorry

E+R=108
E - R=50

Is there an algebraic solution to that?

And another question, is the program based on a true algebra and not an algorithm?

PM 2Ring

1 hour ago, by PM 2Ring

@M17 This is just simple algebra. If E+R=p and E-R=q then 2E=p+q and 2R=p-q

M17

04:49

Ok

Nice

E+R=108
E - R=50

Is this possible algebraically?

E=?

R=?

PM 2Ring

@M17 Sage has a fast efficient algorithm for finding all the divisors of a number, based on finding the numbers prime factors. That's another good project for you when you start to learn Python. For small numbers, it doesn't matter much if you're efficient, you can just test every number under the square root.

@M17 You should be able to solve that yourself.

M17

I learned and watched a lot of Python, but I got bored because the course is about things other than ideas

PM 2Ring

@M17 If you can't solve that, you need to study basic algebra.

M17

I do slove now

E+R=108
E - R=50

2E=158

E=79

When I wasn't sleeping enough, sometimes very simple things seem unclear to me:)

Sorry, that's very easy, sorry for asking that question

MeltyButter

you know how people make wild speculations about how godel's incompleteness theorems could apply to all sorts of informal systems like utilitarianism or marxism or transubstantiation

and it's dismissed as crackpot because they're unspecified systems that don't even attempt to formalize a certain amount of arithmetic

but is it not true that all those informal systems I mentioned do in fact at least bud off from (or can at least with enormous effort can be shown to be contingent upon) a natural language that is capable of expressing that "certain amount of arithmetic"?

M17

05:05

AB=2291
A=? B=?
A,B are prime numbers
---------------------------------------
A+B=E
if A>B
A - B=R

E^2 - R^2=F
F/4=2291
F=9164
E^2 - R^2=9164
(E+R) (E - R)=9164
9164/(E - R)= (E+R)

A+B=108
A - B=50

2A=158
A=79
B=29

porridgemathematics

05:54

am I confused or is the following true? Let $E'$ be a smooth complex vector bundle over a differentiable manifold $X$, denote by $E$ the corresponding real-vector bundle (of twice the (real) dimension) obtained by restriction of scalars on each fiber, then multiplication by $i$ gives an isomorphism of real vector bundles $J : E \rightarrow E$, $J^2 = -Id$ and let $E_{\mathbb{C}}$ be the complexification of $E$. $E_{\mathbb{C}} = E^{(1,0)} \oplus E^{(0,1)}$ and $E' \equiv E^{(1,0)}$

here $E^{(1,0)},E^{(0,1)}$ are the eigensub-bundles of $J$ extended to the complexification corresponding to eigenvalues $i,-i$ respectively

so im not saying every complex vector bundle is the complexification of a real vector bundle, but is it true that every complex vector bundle can be realized as one of the eigenbundles of a real vector bundles with a $J$ - operator, of twice the dimension?

also being careful not to call this $J$ operator an almost complex structure since im not talking about tangent bundles necessarily

if all this is fine, can I go ahead and 'define' the conjugate bundle of $E$ as $E^{(0,1)}$?

and then show that this thing is isomorphic to the usual definition, defined by giving each fiber the conjugate action?

i seemed to have checked this myself yesteday, then came across a SE post asking about 'why the conjugate of every complex vector bundle is iso to the original' which didn't make sense to me because in the above if we started with a holomorphic vector bundle on a complex manifold $X$, my $E^{(1,0)}$ is holomorphic and my $E^{(0,1)}$ is anti-holomorphic in the sense its transition maps are killed by $\frac{\partial}{\partial z_i}$, and I think of $E^{(0,1)}$ as the conjugate normally

oh whoops, the SE post was referring to why the conjugate of the complexification of a real vector bundle is iso to its conjugate bundle..

but in any case - could anyone sanity check the original thing I asked?

copper.hat

06:11

sorry, not a manifold person

porridgemathematics

no worries

love_sodam

09:08

What is the naming invariant factor in finitely generated module over PID stands for? In linear algebra, invariant factor of a matrix is really invariant under the base field (containing coefficient of the given matrix).

M17

10:33

@Ajay

en.m.wikipedia.org/wiki/RSA_Factoring_Challenge$

Leave the page and look at the title Mathematics, is this knowledge of P and q among factor analysis problems?

Koro

12:05

If $I=\langle 2+2i\rangle$, then how many elements are there in $Z[i]/I$?

First of all, if a+bi +I is in $Z[i]/I$ then $a+bi+I=a+bi+2b+2bi+I=a+2b+i(3b)+I$.

4 is in I. But 2 is not in I.

I don't know how to get rid of i.

$2(a+bi)+I=2a+i(2b)+2b-2b+I=2(a-b)+I$.

Prithu biswas leftmse

12:25

For the integral $\int \frac{1}{3sinx+4cosx}dx$ , my textbook is telling me to let $a = r.cos(\theta)$ and $b = r.sin(\theta)$.Is there a name for this?

Lukas Heger

12:48

@porridgemathematics can you describe the map $E \to E^{(1,0)}$ that you have in mind?

M17

I need help with something in algebra

Yesterday, i send it

9164/(E - R)=(E+R)

And someone answered this

E+R=2×79, E-R=2×29

porridgemathematics

@LukasHeger basically it is the analogue of the map $\partial_{x} \rightarrow \partial_{z}$, $\partial_{y} \rightarrow i\partial_{z}$ adapted to the general case

M17

I want to explain how he got to this answer

To me

@XanderHenderson

porridgemathematics

uh sorry, I meant it is the analog of $\partial_{z} \rightarrow \frac{1}{2}(\partial_{x} - i \partial_{y})$

so i.e. , start with $E$ and some local section on it, $\sigma_1,...,\sigma_r$, restrict to $\mathbb{R}$ with corresponding local sectoin $\sigma_1, i\sigma_1, ... , \sigma_r, i\sigma_r$, complexify this, then $\frac{1}{2}(\sigma_j - i \sigma_j)$ forms a local frame for $E^{(1,0)}$, and the inclusion $E \rightarrow E^{(1,0)}$ will be $\sigma_j \rightarrow \frac{1}{2}(\sigma_j - i \sigma_j)$

Lukas Heger

@porridgemathematics so if I try express this in terms of elementary tensors on each fiber I get something like $v \mapsto \frac{1}{2}(1\otimes v - i \otimes iv)$? That seems to work

and the inverse is just $E^{(1,0)} \to E, \lambda \otimes v \mapsto \lambda v$?

seems correct to me, but I am very far from an expert on this manifold stuff

you should ask Ted when he's around, he's a complex geometer

porridgemathematics

13:07

hmm, for the first map why isn't it $v \rightarrow \frac{1}{2}(1 \otimes v - i \otimes v)$?

Lukas Heger

if we apply $J$ to $\frac{1}{2}(1 \otimes v - i \otimes v)$, I get $\frac{1}{2}(1 \otimes iv - i \otimes iv)$, but if I multiply with $i$ I get $\frac{1}{2}(i \otimes v + 1 \otimes v)$

porridgemathematics

oh yes

Lukas Heger

but the map I wrote works, it actually lands in the $i$-eigenspace of $J$

furthermore, it's actually an inverse to $\lambda \otimes v \mapsto \lambda v$ unlike $v\mapsto \frac{1}{2}(1 \otimes v - i \otimes v)$

porridgemathematics

yeah I actually meant your map! What I wanted was to mirror $\partial_{x} - i \partial_{y}$, but in this case my $\partial_{x} = \sigma_1$ and my $\partial_{y} = i\sigma_1$

so we would get your map

i dont think one half is necessary

Lukas Heger

@porridgemathematics I think it's somehow more canonical with the one half

porridgemathematics

13:13

its sort of meaningless addition here anyway unless we assume we are working on a tangent bundle of a complex manifold

yeah it did feel like that to me

Lukas Heger

because else you don't get $\lambda \otimes v \mapsto \lambda v$ as the inverse

porridgemathematics

oh I see

Lukas Heger

yeah, as I said, your reasoning seems to be correct, I just wanted to break it down to linear algebra as a sanity check

porridgemathematics

thanks a lot , it was really helpful

Ajay

13:41

@M17 I don't get what you are asking

You outlined the steps yourself so what is the problem?

Don't tag random people who weren't part of the original discussion for no reason

That is not going to bring attention to your problem.

Koro

0

Q: If $I=\langle 2+2i\rangle$, then how many elements are there in $Z[i]/I$?

If $I=\langle 2+2i\rangle$, then how many elements are there in $Z[i]/I$? First of all, if $a+bi +I$ is in $Z[i]/I$ then $a+bi+I=a+bi+2b+2bi+I=a+2b+i(3b)+I$. $4$ is in $I$ because $4=(2+2i)(1-i)\in I$. $2$ is not in $I$ else $2=(a+ib)(2+2i)$ for some integers $a$ and $b$. $2a-2b=2$ and $2a+2b=0\i...

can anyone please help me with this one? thanks.

porridgemathematics

@Koro what about using that this is iso to $\mathbb{Z}[i]/(2)$ and then that quotiented with the corresponding principal ideal of $1+i$ in that quotient?

i.e. quotient in succession

uh, actually the other way around will be easier

basically intuitively it should have only $0,1$, because quotienting by your ideal is like setting the relations $i = -1$ and $2 = 0$

Koro

@porridgemathematics I don't know that the quotient is isomorphic to Z/(2)

:(

porridgemathematics

oh nvm, it isnt

Koro

13:56

:(

Also, I can't cancel 2 from both sides: $2(a+ib)+I=2(a-b)+I$

porridgemathematics

ignore what I said, it only applies to things like $(2, (1+i))$

Koro

as R/I is not given to be an integral domain.

porridgemathematics

certainly not $( 2(1+i))$

Koro

the solution said: it will have 8 elements.

I don't know how.

porridgemathematics

do you know of smith normal form?

or the structure theorem for finitely generated modules over a PID?

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