Mathematics - 2023-01-23 (page 1 of 2)

ペガサス Seiya

00:00

@TedShifrin I'm the guy with the 3 hice, by the way.

Ted Shifrin

@Obliv It's not an accident two of my books have the subtitle "A Geometric Approach" — for which I am regularly roundly ridiculed on here.

@ペガサスSeiya You need to stop saying that. You sold them all.

ペガサス Seiya

@TedShifrin well, I used to be the 3 hice guy. Now I'm the guy with crippling debt

D.C. the III

00:16

How do people go about "creating" functions and transformations in the "adult world"? What I mean by this is in these courses I take and read up on we learn the properties of operators, transformations, functions, etc, but we don't actually create any. I could see how you might go about it when writing a textbook, but what about if you're modelling something?

Very nebulous question I know...

shintuku

you can use set theory to create an arbitrary function. in set theory you learn to prove whether what you defined exists or not

copper.hat

@robjohn i see you are milking that one...

D.C. the III

@shintuku I see.....well I haven't got to set theory yet. Topology and the likes is on the docket soon

shintuku

an alternative interpretation of your question is how do you learn to model phenomena with functions

then, whatever you're modeling can probably be represented by some sort of differential equation, and you learn to find appropriate differential equations by learning more about differential equations and getting intuition about their behaviour

PM 2Ring

@D.C.theIII If you're modelling something physical, you generally start with dimensional analysis. physics.stackexchange.com/q/98241/123208 That will (usually) give you a useful starting point. Then you try to come up with relevant differential equations. If you're lucky, some aspects of your system will be linear, and crude functions will give you useful approximations that you can refine. You can do a lot of useful stuff with Fourier / Laplace transforms, spherical harmonics, etc.

D.C. the III

00:31

@PM2Ring I see I have a bit of work left to do before I'm where I want to be.....

shintuku

00:50

Suppose the matrix $A$ has echelon form $U$ achieved with no row interchange. Prove $ L$ in $LU$ decomposition is a lower triangular matrix with 1's in the diagonal.

feels like this proof requires a huge amount of case work

is there no way of avoiding the case work?

Ted Shifrin

No. No cases because of the hypothesis. Work some examples and then do induction.

shintuku

that's what i thought at first, but consider the fact that $E_1A$ can zero not only the first entry of all other rows, but also the rest of their entries, and that if this doesn't happen due to the first pivot row and $E_1$, it could happen due to the second or third and $E_2$ or $E_3$, e.g., the second row could zero all the following rows instead of just their second entry

furthermore what if a set of subsequent pivots rows don't have their pivots placed in a perfect diagonal, meaning that they will have to zero more than one entry in the following rows

a rigorous proof of the most general case seems like a ton of case work

Ted Shifrin

If what you describe happens, wouldn’t you need row interchange to get to echelon form?

As I said, that hypothesis makes it case-free.

shintuku

no ted i chose those cases starting from a matrix that will need no row interchange to reduce

Ted Shifrin

The pivots don’t have to be in perfect diagonal form.

What you described is what makes you need a row interchange to move a zero row down to the end and another row up.

shintuku

01:03

1 0 0 0
0 0 1 2
0 0 1 2
0 0 1 2

you can make many such matrices, choosing an arbitrary row after which the rows start repeating

Ted Shifrin

That matrix has no problem unless you put another pivot on the bottom. $U$ has two rows of zeroes and $L$ is fine.

shintuku

i guess those are the cases then. first, a non problematic case with diagonal pivots. second, a case where there is a point where rows start repeating. third, the case where the rows are not in perfect diagonal, and fourth, one with both of these last two problems

Ted Shifrin

I don’t see why you need cases. The repetition means nothing.

shintuku

to give a rigorous proof, no? i will rethink this

Ted Shifrin

Where the pivots are affects $U$ and has no issue with $L$. No cases needed.

shintuku

01:10

thanks for the comments i will be thinking about this

Ted Shifrin

Remember that $L$ encodes the elementary row operations.

You must only see why every elementary matrix is lower triangular.

If you’re not used to thinking about products of elementary matrices, maybe my video on that topic will help.

shintuku

01:31

i'm seeing that every elementary matrix will have to be lower triangular. what i'm having a difficult time, however, is giving a rigorous proof for the statement: "in a given row, all entries before the pivot will be zeroed by a combination of rows above"

that's the proposition that translates the fact that lower triangular elementary matrices will produce the desired effect

Ted Shifrin

02:21

That’s the definition of pivot!

shintuku

hm... right

Josepsh

while you take care of creepy edgar. Teach that rich, spoiled brat what a real duel's all about ?. .

Obliv

02:45

what branch of math is the study of exponential functions and properties?

Semiclassical

03:02

precalculus? it's too small a subject to be of much interest in its own right

Obliv

03:28

What about knuth's up arrow notation or something in that vein?

Like "growth" functions? if that makes sense.

shintuku

suppose $E_1A = U_1$, $U_1$ is the first step towards the echelon form $U$ of $A$, and producing $U$ will require no row exchange. more specifically, $U_1$ has the first pivot column of $A$ made explicit, i.e., the first pivot column of $U_1$ (which may not be its first column) consists of a leading entry, and then has zero for all following entries

how would i even begin to prove that $E_1$ must be lower triangular? it's obvious if you do the computation

Obliv

03:50

I think they're called hyperoperators, idk what category this would be in but interesting.

robjohn

@AkivaWeinberger (99,175) seems to be another

Koro

@robjohn I did not understand the last equality. How does that follow?

i.e., how do we get $\sum 1/k$?

robjohn

$$\frac1{1-\frac1p}=1+\frac1p+\frac1{p^2}+\frac1{p^3}+\dots$$

correct?

Koro

yes

But the infinity product part?

robjohn

Take the product over all primes and you will find the reciprocal of any integer exactly once.

Koro

04:00

hmm, it seems true intuitively as every number is product of primes.

so it will appear somewhere.

robjohn

you can take finite products, and you will get all the integers up to at least the highest prime.

so as you take more terms in the product, you will get more terms in the harmonic series.

brb

Koro

we'll say that rearrangement is possible because of positivity of numbers and so...

1 will appear exactly once, 1/2 will appear after that, then 1/3, 1/4= (1/2)^2 and so on.

that's a cool method by the way :-).

Thanks a lot for sharing that.

Ted Shifrin

Thank Euler.

Obliv

Thank you Euler

Akiva Weinberger

@robjohn Here's a question

A while ago I was thinking a lot about the "threshold 3 infected squares" cellular automaton

in which a cell becomes 'infected' if three of its four edge-neighbors are infected, and you run it until it's stable

A 2x2 void would never get filled, nor would a cycle surrounded a disconnected region.

Thus, running that automaton on the coprime matrix would highlight voids and separated pieces

(by "coprime matrix" I mean the image from before - the set of pairs (x,y) with gcd(x,y)=1)

So, what do you get?

Koro

04:12

The following statement is plain wrong: Every nonzero integer can be written as a product of primes.

leslie townes

koro - are you angry about the sign thing, or the empty product thing, or both

Akiva Weinberger

For what it's worth, you need roughly 75% of the grid to have a chance of filling it all up; this is 6/pi^2=61% full, so we know there will be blanks left over. @robjohn

Obliv

Why is that wrong @Koro

Akiva Weinberger

I guess I don't know how to get -2?

Obliv

I'm literally about to write a program to do that very thing lol

Koro

04:14

1 is the counterexample.

Akiva Weinberger

Eh, empty product.

Same way empty sums are 0.

and 2^0=0!=1

Koro

We have the above theorem for non unit and non zero elements in UFD.

Obliv

Why is the identity element of addition different from that of multiplication, anyone know?

Akiva Weinberger

So -2 is considered "prime" because we're working with the ring theory definition of prime?

Then -1 would be the real counterexample, I guess

Koro

yes

@Obliv suppose they are the same, then it is basically the zero ring.

Obliv

04:17

For non zero, succession & addition identity element for natural numbers is 0 but for multiplication, exponentiation, tetration, etc it's 1.

Actually idk about the hyperoperators past exponentiation

Akiva Weinberger

Well, $a^1=a$ but $1^a\ne a$

shintuku

@Obliv the identity element $e$ is defined $e \circ x = x$, so you can see why it is different for multiplication and addition

Akiva Weinberger

You can also ask why addition and multiplication are commutative and exponentiation is the first time it fails

@Obliv I guess if we suppose the identities were the same - call it $e$, so that $e+x=ex=x$ - then evaluating $(e+e)x$ in two ways gives us $x=x+x$ for all $x$

or $e+x=x+x$, so using the cancellation law $e=x$ and so everything equals everything else

Koro

@leslietownes Ahh, so that is the spirit behind the statement.

empty product

Akiva Weinberger

-1 still is hard

Koro

04:23

:(

robjohn

$-1$ is a unit, so it is hard to include in a unique representation.

The unique representation only really applies to positive integers

$0$ has many factorizations

Koro

@Obliv suppose they are same- 0=1. Then for any x, x=1.x=0.x=0. So the whole ring is {0}. 0- the additive identity, 1- multiplicative identity of ring X. Is that what your question was?

@robjohn this is taken care of by saying that the representation is unique upto rearrangement and associates.

like: 6=2.3= (-2).(-3)=3.2=(-3).(-2)

for non unit and non zero elements.

I'll use the following definition: $p\in \mathbb Z$ is prime iff the ideal $(p)$ is a prime ideal.

this definition takes care of the $\pm 1$ issue.

robjohn

04:42

@Koro $(3)=(-3)$?

robjohn

04:55

@AkivaWeinberger not quite sure what you're asking. Did you want to run a cellular automaton on the result?

$(21,55),(99,175),(115,551),(369,575),(495,589)$ seem to be the ones with both numbers $\lt600$.

Koro

@robjohn yes, so 3 and -3 both are primes.

by this definition

Nasser

05:36

Can someone help me understand this about this pdf paper ? It is basic question. They have a paper basically saying "A New Approach for Solving Second Order Ordinary Differential Equations" where given second order ode, they use standard tradsformation v=-y'/y to convert it to Riccati. OK. But then they say given this particular solution to the Riccati ode, they can now obtain one solution to the second order ode. OK. But how did they find...

.. the solution to the Riccati ode in first place? They do not say. What Am I missing here? this is all on one page, so you do not need to read much. They just give the solution to the Riccati ode like this. But There is no known method to find particular solution to Riccati. I have no idea what I am missing about this paper.

So they are basically saying they found a method to solve any second order linear ode by doing this. But to do that they have to be able to find particular solution to Riccati, which there is no such method. I do not get it.

leslie townes

i don't know if it's a standard move here, but it's not uncommon to look for constant solutions to differential equations. so here, you're checking if there's any value of c for which 0 = 1/x + (1+x)/x c + c^2 holds identically. that leads to c = -1 pretty quickly.

whether there's always going to be a constant solution, or whether this is just some miracle that happens here, i dunno. it may be specific to whatever type of equation this is.

Ted Shifrin

It’s a miracle.

copper.hat

i managed to drop my new phone today. actually, it fell by itself.

upon urging of family members, i have ordered a case

a nice cab

not sure that's what they meant

Ted Shifrin

don’t take that paper too seriously. Lots of stuff gets published. If you happen to be able to guess a solution of the Riccati equation, then their method works.

sounds good to me, copper.

Nasser

05:51

@TedShifrin Yes ofcourse I know if one can find particular solution to Riccati, then this method works. But there is no known method to do that. And they do not even mention this fact. They claim this method can be used to solve second order ode without saying how to solve the Riccati itself.

Joe Shmo

this paper looks like it was written with micosoft word equations. Im calling BS

Ted Shifrin

As I said, plenty of mediocre stuff gets published. Not remotely a high quality journal.

I bet one of the authors was an undergrad.

Akiva Weinberger

@robjohn What do you mean by "the result"? I mean on this:

robjohn

yes, I just automated a check to look for the isolated points, and the ones I mentioned were what I found.

Akiva Weinberger

Sure, but running the automaton will reveal more than just isolated points.

Nasser

06:00

@TedShifrin I thought Journal papers are peer reviewed. so someone should have cought this and asked about it. Here it is at Journal page

Akiva Weinberger

For example, it will reveal the {20,21}x{14,15} square.

Ted Shifrin

My comments do not change. This is not a quality research journal.

if the only reference is a sophomore textbook, that tells you much.

Akiva Weinberger

(I'm a bit behind on the inside jokes in this group - but I wonder if Leslie hosts a journal…?)

Ted Shifrin

That would be truly high quality.

maybe a symbol-pushing approach.

Akiva Weinberger

"Linear algebra: a typographic approach" = instead of veracity, equations are written using whatever symbols look nicest

Someone somewhere has probably written a lament on how no-one typesets mathematics anymore, we just let Donald Knuth typeset it for us

Ted Shifrin

06:11

You’ve got to be kidding. Leslie can’t even be bothered to use ChatJax!

leslie townes

whoah, akiva? i thought we were friends!

Ted Shifrin

LOL

Akiva Weinberger

@leslietownes Which message specifically is this a response to

Ted Shifrin

As contrasted with us enemies?

Akiva Weinberger

You can't spell "friend" without "end". I mean, you could, quite easily, but not if you care about such things as correctness.

leslie townes

06:13

"i wonder if leslie hosts a journal" after a discussion of an article in a maybe not so great journal where peer review may have consisted of checking whether an electronic funds transfer went through

@TedShifrin yes, i expect it from you.

Ted Shifrin

You give a lot more abuse than you get from me!

leslie townes

$600, apparently, unless the authors are from one of a list of countries

Akiva Weinberger

@leslietownes No one with money has ever been wrong!

Thus peer review is not needed

leslie townes

this actually did give me an idea, which is to start a math journal. i could issue lesliecoin based on how many downloads a paper gets, so you can earn on your submissions. there is of course a small submission processing fee, currently payable only in US dollars.

hosting of the article itself is also done in distributed fashion, namely, we do not lock you down to any one website, such as anything that we maintain. you get to host it wherever you like. our license allows for this. we do provide a QR code that can be stamped on the front of your paper so everyone knows it was published in leslie j. m.

Ted Shifrin

The 21st century Annals of Math.

Akiva Weinberger

06:19

We don't sell products; we sell the marketplace. And by 'sell the marketplace' we mean 'play shooters, sometimes for upwards of 20 hours straight.'

robjohn

Is that meta-business?

Ted Shifrin

Very à propos, DogAteMy.

Akiva Weinberger

I just imagine the relevant question for pricing isn't "how much does it cost us to host it" but "what is hosting worth to you"

robjohn

@copper.hat I couldn't help but to churn out a response.

Ted Shifrin

No cream of the crop for @robjohn.

robjohn