Mathematics - 2023-10-31 (page 1 of 2)

copper.hat

00:43

is that a "well regulated" integral?

Ted Shifrin

00:55

A well-regulated militia?

one potato two potato

01:07

What is the type of this kind of pde?: For given $H:\Bbb R^3\to\Bbb R$: smooth function, $L:\Bbb R^3\to\Bbb R^3$: smooth vector field, $v\in S^2$, find a smooth function $f$ on $\Bbb R^3$ such that $L\cdot v = D_vf+Hf$

Semiclassical

01:23

trying to decide whether it's worth posting this to the main site

suppose i want to find a sequence of rotations that takes me from one sphere orientation to another. in mechanics we usually do this via Euler angles, to the misery of all physics students

the standard decomposition is to look for a Z-X-Z sequence of rotations

and that makes sense, b/c in mechanics you've typically got the Z-axis picked out (by the direction of gravity, say)

but aside from that this sequence is arbitrary, e.g., one could instead look for a X-Z-X decomposition

suppose i know the Euler angles for that decomposition. how does one go from these angles to those for the Z-X-Z decomposition?

i imagine the answer is something something quaternions

looks like the Wikipedia page covers this well enough. oh well

Horiatiki

01:42

A little further up, in chat, I published the link to a question with BOUNTY+50 on the Gauss-Newton algorithm. Can anyone help me?

Ted Shifrin

@onepotatotwopotato First-order linear (inhomogeneous).

Jakobian

02:18

I'm thinking, whats the example of a function $f$ such that $|f|$ is H-K integrable, but $f$ isn't

$2\cdot 1_A -1$ where $A$ is not Lebesgue meaasurable

all H-K integrable functions are Lebesgue measurable

maybe there's something simpler, I don't know

one potato two potato

03:37

@TedShifrin I was expecting some kind of the name of pde like heat equation

copper.hat

@TedShifrin i always wonder why few make any noise about the "well regulated" part. Seems pretty clear to me.

copper.hat

04:16

i'll never get my jump suit.

Ted Shifrin

@copper.hat You still owe me one.

copper.hat

@TedShifrin Then I will need to get two suits...

Ted Shifrin

@onepotatotwopotato Look up the classic equations. For starters, you’ll find they’re second order.

@copper.hat I never got one for 100K, so why do you deserve two?

copper.hat

@TedShifrin As a reward for my suspension.

Ted Shifrin

Suspended animation?

copper.hat

04:26

I walk as such at the moment. A fun christmas coming up, will have (insurance permitting) my hip replaced in the last week of December.

Ted Shifrin

Exciting! I don’t yet have neck/back surgeries scheduled.

copper.hat

Urgh. I'm doing 'prehab' at the moment. Not sure it helps, but the PT chap is interesting to talk to. He probably breathes a sigh of relief when I leave.

Ted Shifrin

I go to PT, chiro, and massage each twice a month.

copper.hat

The exercises help in some way that is hard to be precise about. Having someone to talk to who cannot escape is also good.

Ted Shifrin

04:45

LOL ... yes, you do love a captive audience.

Jakobian

06:10

My example above is not Lebesgue measurable. I wonder if this holds for them that $|f|$ is integrable then so is $f$

copper.hat

06:34

If $A$ is a non measurable set then $2\cdot 1_A -1$ is not measurable but its absolute value is.

Koro

07:00

How was multiplication formula applied in the last line?

Is it legal to apply it in this case?

\hat f 1 is the usual Fourier transformation of f and \hat f2 is Fourier transformation defined differently.

Also, multiplication formula holds for L1 functions.

copper.hat

Anyone see something wrong with this answer ode question)? math.stackexchange.com/q/4797419/27978

@Koro what space is $\phi$ in, it's a bit hard to read.

Koro

07:20

Hi :).
$\phi$ is in Schwartz space $S(R^n)$.

And I know that the multiplication formula holds for L^1 functions $f_1$ and $f_2$.

copper.hat

07:34

I'm not sure what you mean by the multiplication formula, but the result is true for two members of $S$, so I guess I would proceed from there?

Koro

By the formula, I mean that if f, g are in L^1(R^n), then $(f,\hat g)= (\hat f, g)$, where the notation (f,g)= $\int f(x) g(x) dx$

@copper.hat yes, but the problem is Fourier transformation may not exist for an L^2 function.

So I suppose by taking members from S, you mean approximating ‘Fourier transformation of L^2 function’.

But then what guarantees that this approximated Fourier transformation is indeed the Fourier transformation?

leslie townes

koro, you (or maybe your source) are using language a little imprecisely here. the integral formula that is often used to define the fourier transform on L^1 may not converge for elements of L^2. there is nevertheless a unique extension of the map defined by the integral formula (regarded e.g. as a map from S(R^n) into L^2(R^n) for example) to a unitary map on L^2(R^n). this extension, which is often/always also called "the fourier transform," does exist for L^2 functions.

unitary, or scalar multiple of a unitary, i guess, depending on how/whether you normalize.

so this thing you call "this approximated fourier transformation" may be just the fourier transform.

this is not something you can easily piece together across different references (whcih may use different definitions) but wherever the details come up, they would be in verifying that such an extension exists and is unique.

it might even be possible to set definitions up so that it holds more or less by definition, in which case it would probably be a theorem that such definition is implemented by an integral formula for L^1 functions or compactly supported smooth functions or whatever.

copper's vibe of "use some kind of approximation argument" is the right vibe, but the details of turning the vibe into a theorem might depend on exactly how the fourier transform is defined on L^2

some MSE questions that address stuff in this area( which may be proceeding from different definitions than your notes above) include math.stackexchange.com/questions/560316/… and math.stackexchange.com/questions/1429086/…

Koro

07:50

Thanks Leslie for the message.

@leslietownes it turns out to be. That’s what the proof in my notes seem to be doing.

@leslietownes it turns out to be. That’s what the proof in my notes seems to be doing.

Message duplicated on its own.

copper.hat

I think rudins functional analysis has it? Been a while and I am away from my library atm.

leslie townes

yeah, it can look like cheating, unless the fourier transform is simply defined that way :) but it does take some thought to check that the "approximation" recipe makes sense.

the first MSE link above is partly about that.

Koro

Thanks for references (links and the book), I’ll look into these.

copper.hat

The close it quickly gang is bugging me again

leslie townes

rudin would definitely do it, the only question might be whether rudin uses some weird functional analysis definition by which the identities are trivial and the hard part is proving that they relate to the thing that people who aren't rudin would define via integral formulas.

i don't think he does anything that weird in his functional analysis book, but functional analysts sometimes do that.

copper.hat

07:54

Definitely at the border of my comfort zone.

Koro

Here is what I meant by “approximated FT”: I know that L^2 functions can be approximated by functions in S. Fourier transformation of functions in S also lie in S. S is a subset of L^2.

So we define the limit of Fourier transforms of approximation functions in S to be the FT of L^2.

@leslietownes these links have enough details. This will do.

Thanks a lot.

copper.hat

09:15

Anyone have any idea what is wrong with my answer here math.stackexchange.com/q/4797419/27978?

Two downvotes in a short period with no explanation?

leslie townes

copper, some people downvote answers to PSQs, in an attempt to deter people from answering them before they are closed

its the next level beyond just voting to close

copper.hat

:-) that jumpsuit is just getting further and further away

Koro

09:36

@copper.hat tried to reduce the downvotes impact by +1

The question probably got closed because no attempt was shown in the post.

Thomas Finley

Show that any ring of order 6 is commutative.

Till now, I have only come across the term "order" in Group Theory. What does it mean in the context of Ring Theory?

leslie townes

it would be good practice to specify the context if you are using the term yourself, and if an author is using it, they should ideally supply the context. but a ring is (among other things) a group with respect to its addition operation, and without further context other than knowing that x is an element of a ring, i would start with a presumption that "the order of x" means the order of x in that additive group

math.stackexchange.com/questions/1510957/… might be of interest

Gwyn

09:55

Problem: let $X$ be a topological space. Prove that if $U$ is open, then $\partial U =\overline{U}\setminus U$. My work: since $U$ is open, $X\setminus U$ is closed and so $X\setminus U =\overline{X \setminus U}$. Hence, by definition of boundary, we have $\partial U = \overline{U} \cap \overline{X\setminus U} = \overline{U} \cap (X\setminus U)$.

Now, I would like to conclude by saying that $\overline{U} \cap (X\setminus U)=\overline{U} \setminus U$; I think it's true because $\overline{U} \cap (X\setminus U)$ means "being in $\overline{U}$ and being in $X$ and not being in $U$" and, since $X$ the universe set, saying "being in $\overline{U}$ and in $X$" is equivalent to be in $\overline{U}$. But I'm not sure that this argument is correct. Can someone confirm it or explain why is wrong?

Thomas Finley

@leslietownes This question is posed in an excercise of the book, "Topics in Algebra" by IN Herstein.

Koro

order means cardinality for all finite objects.

Jakobian

10:49

@copper.hat why did you copy my example

Thorgott

11:30

@ThomasFinley the same thing

the order of a group is the cardinality of its underlying set. the order of a ring is the order of its underlying group, i.e. the cardinality of its underlying set.

@Gwyn yes, this is correct

Jakobian

11:42

@Gwyn $\partial A = \overline{A}\cap \overline{X\setminus A} = \overline{A}\setminus\text{int}(A)$

@Thorgott really? I don't think I ever heard of it in that context

user 726941

now you have :P

Jakobian

Order (ring theory)

In mathematics, an order in the sense of ring theory is a subring O {\displaystyle {\mathcal {O}}} of a ring A {\displaystyle A} , such that A {\displaystyle A} is a finite-dimensional algebra over the field Q {\displaystyle \mathbb {Q} } of rational numbers O {\displaystyle {\mathcal...

this somehow reinforces me that we shouldn't use the word order when dealing with rings?

Gwyn

Thanks to both Thorgott and Jakobian for the answers)

X4J

13:09

Suppose $f: [a, b] \rightarrow \mathbb{R}$ is monotonic.
Prove that there exists $c \in [a, b]$ such that $\int_a^b f(x) dx$ = $f(a)(c-a)$ + $f(b)(b-c)$

I could prove it, but now I am trying to think if we can always choose c such that $c \in (a, b)$

Xander Henderson

13:23

@copper.hat Addressing your comment that people are demanding "pro forma rubbish": I don't think that anyone is asking for that. They are asking for context, which will help to explain what the asker knows, and what kind of answer they are expecting.

This is the policy on the site---it is part of a compromise, which is intended to reduce the number of homework problems dumped on the site. If you don't like the policy, lobby folk on meta to change it.

For what it is worth, I am not a big fan of the policy, because it still permits too much homework dumping.

@X4J If the function is constant, then you can choose $c=a$ or $c=b$. The proof I have in mind guarantees a $c \in (a,b)$, but perhaps the author has something in mind for some special cases in which either $a$ or $b$ can be chosen.

Koro

13:40

@X4J no, take f(x)=x^2, a=-1,b=1. There is no c that does the job.

Xander Henderson

@Koro Is that function monotonic?

Koro

@XanderHenderson they already know the answer for monotonic.

Xander Henderson

@Koro That is not how I understood the question.

I understood it as the difference between $c \in [a,b]$ and $c \in (a,b)$.

Koro

"if we can always choose..." I interpreted that as in general because before this, they said 'I could prove it,'.

Xander Henderson

But I suppose that it is up to @X4J to clarify.

Yeah, they have not asked a clear question.

Koro

13:43

let's downvote them.

:-)

X4J

Sorry Xander, I meant that if we can claim the same argument as above but for $c \in (a, b)$ and that would be true.

As Koro showed it is not true

Xander Henderson

The argument does work for $c \in (a,b)$, assuming monotonicity.

I am still confused about your question.

@Koro ;)

Koro

(removed)

X4J

This is the original one

Suppose $f: [a, b] \rightarrow \mathbb{R}$ is monotonic. <br>
Prove that there exists $c \in [a, b]$ such that $\int_a^b f(x) dx$ = $f(a)(c-a)$ + $f(b)(b-c)$

And it is true

The second is to determine if the following statement is true:
Suppose $f: [a, b] \rightarrow \mathbb{R}$ is monotonic. <br>
Prove that there exists $c \in (a, b)$ such that $\int_a^b f(x) dx$ = $f(a)(c-a)$ + $f(b)(b-c)$

Xander Henderson

I don't understand which part of the argument you are trying to change: are you looking for $f$ not monotonic, or for $c$ explicitly in $(a,b)$, rather than $[a,b]$?

X4J

13:49

for $c$ explicitly in $(a,b)$, rather than $[a,b]$

Xander Henderson

@Koro Ha ha! I win! :P

X4J

But wait I think it isnt true

Koro

the outcome of the battle got revised. You win. 😁👍

X4J

taking $f: [0, 1] \rightarrow \mathbb{R}$ with f(0) = 0 and else f(x) = 1

Xander Henderson

@Koro (Not that it matters---I'm being childish.)

Koro

13:50

me 2

Xander Henderson

@X4J Yes, if your function is not continuous, bad things can happen.

X4J

even if it is strictly monotonic?

Xander Henderson

@X4J The only way that $c$ can be one of the endpoints is if the function is "essentially" constant.

Koro

say a functions f is in L^1 (R) and L^2(R) both. It is known that there exists a sequence of functions {g_n} in S(R) -Schwartz space that L^1 converges to f, and there also exists a sequence of functions {h_n} in S(R) -Schwartz space that L^2 converges to f. Question is: does there exist a sequence $\phi_n$ of functions in S(R) -Schwartz space, that converges to f in L^1 and L^2 both ways?

Xander Henderson

Note that if $f$ is monotonic, then $\int_{a}^{x} f(t)\,\mathrm{d}t$ is continuous, and so you can make some kind of intermediate value theorem argument.

Koro

14:00

here is an excellent answer math.stackexchange.com/a/560454/266435

Jakobian

@Koro doesn't convolution do the job? I think what you do is consider molifiers $\phi_n$, and then $f* \phi_n\to f$ in $L^p$ for all $p$ that $f$ is in

Koro

but I'm not sure how they get smoothness via convolution at the end of the answer.

Jakobian

its standard that convolution preserves all of those nice properties

34

Q: Why convolution regularize functions?

There is a tool in mathematics that I have used a lot of times and I'm still not confortable with. In fact I can't figure out (by this I mean that I cannot understand it geometrically) why does convolution regularize things. It is know for example that if $u\in L_{loc}^1(\mathbb{R}^N)$ and $f\in ...

convolution

convolution "commutes" with the derivative

Koro

but note that f_m's are not necessarily smooth (terminology of the answer) so convoluting them does give us compact support but smoothness?

Jakobian

since molifiers are smooth, you really just deriving the molifiers

Koro

14:03

I don't yet know molifiers...

Jakobian

@Koro the compactly supported approximation has to be by smooth functions

Koro

I know kernels/approximate identities though

Jakobian

@Koro molifiers are just those things that let you approximate the identity

Mollifier

In mathematics, mollifiers (also known as approximations to the identity) are smooth functions with special properties, used for example in distribution theory to create sequences of smooth functions approximating nonsmooth (generalized) functions, via convolution. Intuitively, given a function which is rather irregular, by convolving it with a mollifier the function gets "mollified", that is, its sharp features are smoothed, while still remaining close to the original nonsmooth (generalized) function.They are also known as Friedrichs mollifiers after Kurt Otto Friedrichs, who introduced them....

Koro

@Jakobian I see. That's what seems to be happening in the answer.

Jakobian

it is whats happening

Koro

14:05

I'll get to mollifiers soon.

also, mollifier sounds like a chemical's name

not sure how the word made its way to...

Jakobian

it sounds like moles to me

Thomas Finley

14:19

show that a ring with 6 elements is commutative.

I don't know how to proceed with this problem.

Xander Henderson

@Koro "to mollify" is a standard English word, meaning something like "to soften, or smooth out, or to reduce the severity of". The usage in mathematics makes perfect sense to me---a mollifier "smooths out" a function, in the sense that convolution with a mollifier gives you something which is in $C^{\infty}$.

Koro

@XanderHenderson yes, when I looked at the wiki page that Jakobian linked, I noted that 'Mollifiers' was not hyperlinked so thought it's a mathematical word. But digging deeper into the history of the word, it appears to have come from French.

Xander Henderson

@Koro And Latin before that. :D

Koro

mollifier

Jakobian

14:35

@ThomasFinley we might try to start with the additive group, and worry about multiplication later. The only abelian with $6$ elements is $\mathbb{Z}/6\mathbb{Z}$, so we know we need to add some structure to it

Now we can consider all possibilities for multiplication, and rule out the case of non-commutativity step by step

it might be a little trickier if you consider non-unital rings as well

trickier in the sense, more cases to consider

Thomas Finley

14:53

Thanks! Meanwhile, I was able to prove that every abelian group of order 6 is cyclic and this makes the result more or less immediate.

Yes, if we consider non-unital rings, the things seem to get more lengthy

Jakobian

15:05

@ThomasFinley oh, I know, this should be easy actually, non-unital or not

just write $x = 1+...+1$, $y = 1+...+1$ and use distributivity for both $xy$ and $yx$

$1$ meaning the additive generator of the cyclic group, not necessarily the unit of the ring

If the addivitive group of a ring $R$ is generated by just one element $a$, then writing $x, y\in R$ as $x = na$ and $y = ma$, we'd have $xy = yx = nma^2$

so $R$ must be commutative

where $n, m\in\mathbb{Z}$

This automatically gives us commutativity of rings of square-free size, like $6$

Thomas Finley

15:32

@Jakobian Yes, that's the apparent idea. The main thing is to write elements of R $x,y$ in the form $ma,na$ where $R=<a>$ and $m,n\in\Bbb Z^+\cup \{0\}.$ The commutativity follows from the equality: $xy=(ma)(na)=(mn)(a^2)=nm(a^2)=(na)(ma)=yx,$ and we are done as $x,y$ are arbitrary elements of R.

This is precisely the argument that we have to do for this particular problem.

Thomas Finley

15:46

0

Q: Prove that a ring of order $6$ can never be an integral domain.

Prove that a ring of order $6$ can never be an integral domain. My solution: Let $R$ be a ring of order $6$ which is an integral domain. This means, that $1+1\neq 0\in R$ and we note that, $(1+1)(1+1+1)=0,$ a contradiction as $1+1,1+1+1\neq 0.$ So, $R$ is not an integral domain. However, I fee...

ring-theory solution-verification fake-proofs integral-domain

Hey guys! Need some help with this!

Jakobian

@ThomasFinley is $1$ the additive generator of $R$?

For this it works, $1+1$ and $1+1+1$ are both non-zero

The justification that $(1+1)(1+1+1)$ is non-zero seems to not be there though

You can justify it by saying that its $6\cdot (1*1)$, where $\cdot$ means added $6$ times

so its $0$

Sine of the Time

@TedShifrin I found the transform of to that "famous" function

Ted Shifrin

@Sine Oh, robjohn said one word which gave it away. Convolution.

Silly of us not to have noticed.

Sine of the Time

we still did not introduce the convolution product :(

Ted Shifrin

Product? I'm talking integral.

$f*g(x) = \int f(y)g(x-y)dy$

It plays very nicely with Fourier transform.

leslie townes

15:57

munchkin just got her kindergarten report card, they had two behavioral remarks, one was "often rude to instructors," the other was "plays nicely with the fourier transform"

Ted Shifrin

Rude only to instructors?

Thomas Finley

@Jakobian No, in this case, I assumed $1$ is a unit element of $R$, i.e $r.1=1.r=r,\forall r\in R.$

Lucky Chouhan

16:26

Hello @TedShifrin

Ted Shifrin

Hi, Lucky.

Lucky Chouhan

May I ask you 'what are you doing now?'?

Ted Shifrin

No.

Lucky Chouhan

Uno Ted $k^3 = \left( \sum_{i=1}^k i \right)^2 - \left( \sum_{i=1}^{k-1} i \right)^2$

@TedShifrin why?😭

Ted Shifrin

Because $k^2(k+1)^2 - k^2(k-1)^2=4k^3$. I don't know "why."

Lucky Chouhan

16:32

@TedShifrin Yeah, I was trying to tell you that we can prove that $(1^3 + \cdots +n^3) = (1+2+\cdots + n)^2$ using that relation.

Ted Shifrin

Of course.

I think you can find a "proof with pictures" for this somewhere if you look.

Lucky Chouhan

"I don't know why". Just like that! You're retired professor.

Ted Shifrin

You are very annoying.

Lucky Chouhan

Squared triangular number

In number theory, the sum of the first n cubes is the square of the nth triangular number. That is, 1 3 + 2 3 + 3 3 + ⋯ + n 3 = ( 1 + 2 + 3 + ⋯...

@TedShifrin Oh don't say so. If I am curious that doesn't mean I am annoying. But sometimes people get annoyed by me :( Do you read novels?

@TedShifrin you know I really know very few intellectual people. There is lot to learn from you, but you don't understand. Btw how is life??

X4J

Your curiosity is embodied as disrespect I must say.

Lucky Chouhan

16:38

@X4J Wtf... I never disrespected Ted.

Ted Shifrin

You made some uncalled for snide remarks a few days ago when I wasn't in the room. You think you're funny, but calling me out for being impatient when I was being extraordinarily patient with someone does piss me off. Just quit.

X4J

Nah you just keep tagging him over and over again although it implicitly seems uncomfortable

Horiatiki

Hi everyone :) A few days ago I posted a question with BOUNTY+50 on the Gauss-Newton algorithm, starting from a nonlinear system. Is there anyone kind and talented enough to help me please?

Ted Shifrin

I don't know what Gauss-Newton refers to. Is that Newton's method?

Horiatiki

Gauss–Newton algorithm

The Gauss–Newton algorithm is used to solve non-linear least squares problems, which is equivalent to minimizing a sum of squared function values. It is an extension of Newton's method for finding a minimum of a non-linear function. Since a sum of squares must be nonnegative, the algorithm can be viewed as using Newton's method to iteratively approximate zeroes of the components of the sum, and thus minimizing the sum. In this sense, the algorithm is also an effective method for solving overdetermined systems of equations. It has the advantage that second derivatives, which can be challenging to...

Ted Shifrin

16:46

OK, so this looks like using linear algebra to implement the least squares solution of an inconsistent linear system. Why are you trying to use this for a nonlinear system?

Horiatiki

I won't republish the question link out of respect for everyone, to avoid spam, because I already published it yesterday in the group. You can find it in my profile. The question is titled "Gauss-Newton algorithm to solve a non-linear system?"

@TedShifrin I need a parameter to use in Poisson. I want to calculate this parameter using the Gauss-Newton algorithm, starting from a nonlinear system.

@TedShifrin My goal is to use Poisson (which is not relevant to this question, because with this question i just want to get the only parameter of Gauss-Netwon to use in Poisson) to individually calculate the probabilities of a football club scoring 0 goals, 1 goal, 2 goals, 3 goals, 4 goals, etc. etc. So I need to find a parameter using the Gauss-Newton algorithm.

Jakobian

Since it was yesterday, imo it'd be okay if you were to repost it, especially since someone is interested in your question

Horiatiki

@TedShifrin In an experimental test, i tried to use the Arithmetic Mean in Poisson, then i use the parameter 1.6 to calculate the probability that a football team individually scores goals, 1 goal, 2 goals, 3 goals, 4 goals, etc. etc. The results are decent, but I would like to get something more precise. So I thought of using another parameter, which is what i would like to find with the Gauss-Newton algorithm and the nonlinear system.

copper.hat

@Koro Thanks! But no need, I was just whining.

Ted Shifrin

Just link your post with the []() procedure here. Then it isn't spam.

copper.hat

16:48

@Jakobian I wasn't copying anything, I was giving an example???

Ted Shifrin

Copper loves to whine.

Jakobian

@copper.hat of what

Horiatiki

@TedShifrin Just link your post with the []() procedure? What do you want to say? I was thinking of simply copying and pasting the link

Jakobian

I gave an example of a function such that $|f|$ is H-K integrable but $f$ isn't, you posted the exact same example

Koro

@copper.hat you put an effort in answering and that deserves +1 from me.

copper.hat

16:49

I don't remember, you messaged me asking why I copied something of yours?

Thanks @Koro

Ted Shifrin

You can link in here by putting what you want to refer to it as in [], followed immediately by the link in ().

Jakobian

I literally posted $2\cdot 1_A-1$, the exact same notation, before you

Ted Shifrin

I know nothing about the H-K integral, but if everyone comes up with the same example of why it's important, I'm not impressed.

Jakobian

its very close to what you posted (time-wise)

Koro

@ThomasFinley I posted an answer to your post.

Horiatiki

16:51

@TedShifrin Can you give me a practical example please? Sorry, but I've never done this. Thank you

copper.hat

@Jakobian Sorry, I didn't notice, I thought you were asking for an example.

Jakobian

@copper.hat yes, I was asking for an example, but I also gave myself one

but I was also asking for another example

Ted Shifrin

For what it's worth, this is an example.

I'm not sure how you're going to view that to see what I did.

Jakobian

a function $f$ which would be Lebesgue measurable, $|f|$ is H-K integrable, but $f$ not H-K integrable

I suspect that if $f$ is Lebesgue measurable, there is no such function (is what I meant)

Ted Shifrin

Oh, anyhow, reading more carefully in your Gauss-Newton link, I see that it is Newton's method in the multivariable setting. So I don't know what you expect us to do.

Jakobian

16:53

I think it should imply that $f$ is Lebesgue integrable, hence $f$ as well.

copper.hat

Gauss Newton is basically Newton with the Hessain replaced by $I$.

Ted Shifrin

Oh, that seems like a horrid idea.

Soumik Mukherjee

@Horiatiki [text here](link here)

Ted Shifrin

But now that copper's here, perhaps he can help if you make it clear.

@Soumik I already did that much, although not quite as explicitly. +1 to you if it works.

Horiatiki

@SoumikMukherjee Thank you :)

Ted Shifrin

16:55

@Soumik to wit.

copper.hat

@Jakobian A little late to ask, but what is H-K?

Horiatiki

BOUNTY+50: Gauss-Newton algorithm to solve a non-linear system?

Ted Shifrin

ROFL

Jakobian

@copper.hat gauge integral

Xander Henderson

@copper.hat Henstock-Kurtzweil.

Ted Shifrin

16:56

There you go, @Horiatiki.

Koro

what is gauge integral?

Jakobian

@XanderHenderson I thought it was Kurzweil

copper.hat

Thanks both of you!

Horiatiki

@TedShifrin :)

I hope someone is kind and talented enough to help me :'(

Koro

is it Henstock-Kurtzweil integral?

Soumik Mukherjee

16:57

@TedShifrin Ooh

Xander Henderson

@Jakobian I don't spell German.

Spell it however you like. Google will fix it for you.

Jakobian

@Koro one of the most general integrals you can have for the real line

Koro

Why not Lebesgue?

Jakobian

it allows you to integrate things, even if they're not absolutely convergent

Ted Shifrin

@Koro Don't get him started again!

Jakobian

16:58

@TedShifrin its you guys that are getting started

Ted Shifrin

Not I.

Jakobian

I just want to learn integrals

Koro

anyways, I have a question:

copper.hat

@XanderHenderson I was cranky last night. However, I think posting half-hearted efforts or regurgitating theorems does not constitute effort, and many reasonable questions get closed. Furthermore, it happens without allowing a reasonable amount of time for the OP to even respond.

Koro

I hope there is an easy answer to this.

Xander Henderson

16:59

@copper.hat Honestly, I have long advocated for quick closure, very slow deletion.

Koro

Tell me one practical usage of the statement "$\mathbb R^n$ is contractible."

Jakobian

@XanderHenderson Kurzweil is a Czech

though surname might be German, who knows

Xander Henderson

Close the question quickly so that it can be improved, before someone gives an answer which might end up being made irrelevant by edits.

@Jakobian I don't spell in that language, either.

Again, spell it however you like. Google will fix it for you.

Ted Shifrin

@Horiatiki I find it very hard to decipher, and your handwriting doesn't help. Can you typeset the relevant material? So $F$ is the function with the parameter you're trying to solve for, and the givens are the $f_e$ values?

copper.hat

@XanderHenderson The closure is hard to undo, so it effectively removes the question. Absolutely makes sense for dups or near dups, but the requirement of 'effort' should at least allow some response time.

Jakobian

17:01

@XanderHenderson its a matter of spelling in general it seems

Ted Shifrin

@Koro Any maps whatsoever to $\Bbb R^n$ are homotopic.

Koro

like if I tell a kid this fact and suppose the kid understands and still says "whatever". How would you counter that?

2 mins ago, by Koro

Tell me one practical usage of the statement "$\mathbb R^n$ is contractible."

copper.hat

My name Joe is notoriously difficult to spell.

Xander Henderson

@copper.hat The requirement is not for "effort". The requirement is for "context".

Jakobian

If someone doesn't care then why should you

Ted Shifrin

17:03

This one is very important: If $\Bbb R^n$ is the universal covering space of $X$, then the higher homotopy groups of $X$ all vanish.

The site is messing up and requiring infinitely many efforts to post.

Koro

of course, I'll tell homotope in layman terms.

I can say "take two loops at a point and you can 'homotope' one onto the other."

Ted Shifrin

Why are you talking to "a kid"?

Xander Henderson

Oh, now it seems to be working for me.

Koro

of course, I'll tell homotope in layman terms.

Xander Henderson

I wonder what that was all about...

Koro

17:04

I can say "take two loops at a point and you can 'homotope' one onto the other."

Ted Shifrin

The site is messed up. It took me 8 efforts to post what finally showed up above. Koro, did you see it?

Koro

my messages multiplied.

Ted Shifrin

It doesn't have to be loops. It can be maps from any $X$ to $\Bbb R^n$.

Xander Henderson

Yeah, chat was momentarily b0rked.

Seems to be working now, I think.

Sine of the Time

@Koro you can compress a cotton candy and eat it in one bite

Horiatiki

17:05

@TedShifrin The sheet is an extra attachment added, but it doesn't have much importance. I also wrote the things written on the paper in the application. Exactly, it's as you say

Ted Shifrin

I gather you don't want a serious mathematical answer to your question.

Koro

the chat seems fixed now.

no, I am expecting a practical usage of the fact like Fourier transformations can be used in segregating specific frequencies from a mix of frequencies.

something like that...

Ted Shifrin

NO CLUE.

They're Fourier transforms, by the way.

Jakobian

first we were at homotopies, now we're at Fourier transforms somehow?

copper.hat

@XanderHenderson Yes, but context is a bit vague. I mean a question about ordering of solutions of an ode really provides its own context.

Lukas Heger

17:09

@Koro because $\Bbb R^n$ is contractible, a surface of genus $\geq 1$ has zero higher homotopy groups (as it has universal cover $\Bbb R^2$), thus group cohomology of surface groups may be computed topologically/geometrically

Ted Shifrin

I just made the homotopy point, @Lukas. But this is far, far afield.

Horiatiki

@TedShifrin What do you mean sorry?

冥王 Hades

Is it Halloween yet?

Ted Shifrin

I wasn't talking to you, Horiatiki.

Koro

If there is such a usage, then I can motivate a kid to learn more about this.

copper.hat

17:10

I may be missing context here, but Fourier transforms and their ilk have myriad 'practical' applications. (Not clear that faster communication of snapchat messages is practical.)

Jakobian

thats funny

Ted Shifrin

I was about to ask you, though. Can you just boil this down to an explicit question here. You have an overdetermined system. ONE variable only? The function with the one variable is the Poisson distribution with parameter (variable) $\mu$? @Horiatiki

copper.hat

However, motivating like that can be difficult, for example, the FFT is useful for computing convolutions, but then you have to explain why convolutions are useful and that gets you back to square one.

Ted Shifrin

@copper.hat The question is why contractibility of $\Bbb R^n$ is germane.

copper.hat

I'm transforming myself.

leslie townes

17:12

you'll go blind, copper

copper.hat

reminds me of a movie, Kentucky fried chicken

Ted Shifrin

@copper isn't Gauss-Newton with just one variable just regular Newton?

冥王 Hades

Reminds me of a food chain called Kentucky Fried Chicken

Horiatiki

@TedShifrin Sorry, I didn't understand your question

Ted Shifrin

Why not LSD, @Hades?

Koro

17:14

@copper.hat like integration has many practical uses and one of them is designing buildings etc., so does fourier analysis. What is "contractibility of R^n" good for practically?

Ted Shifrin

@Horiatiki I'm trying to distill the essential elements of your questions to make it understandable.

冥王 Hades

@TedShifrin Lysergic acid diethylamide?

Horiatiki

@TedShifrin The basic, initial idea, is to start from List_Goals: [1, 2, 2, 1, 2] which i organize in the non-linear system, building it with the goals [1, 2, 2, 1, 2] and the number of times the event occurs

Jakobian

how is a random fact supposed to be practical though

integration is a whole field

Horiatiki

If a football club, in the last 5 matches,scores [1, 2, 2, 1, 2], it means that:

0 goals, they were scored 0 times ( );
1 or less goals, scored 2 times (1, 1);
2 or less goals, scored 5 times (1, 2, 2, 1, 2);

Ted Shifrin

17:16

I have $f_\mu(i)=b_i$ for several values of $i$. This is an inconsistent (overdetermined) system of equations.

Koro

anyways, such questions are considered 'philosophical' and are dismissed in class.

Horiatiki

If the goals scored are <= 0 events, then i will have 0/5 = 0;
If the goals scored are <= 1 events, then i will have 2/5 = 0.4;
If the goals scored are <= 2 events, then i will have 5/5 = 1;

f(0) = 0/5 = 0;
f(1) = 2/5= 0.4;
f(2) = 5/5= 1;

System: {f(0) = F(0) } therefore 0/5= 0
{f(1) = F(1) } therefore 2/5= 0.4
{f(2) = F(2) } therefore 5/5= 1

Ted Shifrin

I don't want any of that. It just obscures the math.

Koro

But I think these are genuine questions!!

Jakobian

everything is practical if it has its uses, but how is this question different than "what are applications of $(x^2)' = 2x$"

Lukas Heger

17:17

Contractibility of $\Bbb R^n$ implies manifolds are locally contractible, which implies they have a well-behaved covering theory. I'm no expert, but I'm fairly sure covering spaces of manifolds have applications in physics

Ted Shifrin

@Koro Every closed $k$-form is exact. This settles classical questions in greater generality, generalizing every curl-free vector field has a potential, and every div-free vector field is a curl.

That's very real world if you deal with vector fields.

copper.hat

@TedShifrin With one variable, you would have the Hessian $2(f(x)f''(x) + f'(x)^2)$ and Newton's method would use this Hessian. Gauss Newton uses $2(f(x) \cdot 1 + f'(x)^2)$.

Ted Shifrin

@Copper But all he wants to do is get the optimal solution of a simple overdetermined system.

$f_\mu(0)=a_0$, $f_\mu(1)=a_1$, $f_\mu(2)=a_2$ (maybe more). How do we solve for $\mu$?

Koro

@TedShifrin curl divergence are practical, yes.

Ted Shifrin

Yeah, this really has nothing to do with Newton's method. It is a discrete system.

Koro

17:20

they won't have a place in physics otherwise.

Ted Shifrin

I'm sticking with that example, Koro. Stick with $n=2$ and $n=3$.

You need $H^1=0$ for the curl=0 question, $H^2=0$ for the div=0 question. Contractible gives both.

copper.hat

miimise least squares, GN would be appropriate.

Ted Shifrin

OK. Can you interpret it explicitly enough to tell Horiatiki what to do? His $f_\mu$ is essentially the Poisson distribution.

copper.hat

wow, i got 3 downvotes on my ode answer. must have bothered someone.

Ted Shifrin

What answer?

Horiatiki

17:23

I'll go now and close the PC. I greet you and point out the link again in case it is needed. BOUNTY+50: Gauss-Newton algorithm to solve a non-linear system? I know it's a bit of a difficult question, I apologize for the difficulty. Thanks anyway for your attention :)

Ted Shifrin

It's not that difficult. But it's overcomplicated by the details of your particular situation, which are not relevant to the math question.

copper.hat

@TedShifrin my whine from late last night math.stackexchange.com/q/4797419/27978

Ted Shifrin

You probably got downvotes for answering a PSQ.

copper.hat

not looking for upvotes, just a reason for the downvotes. i'm being punitive for reasons unrelated to the answer is a fine explanation.

Ted Shifrin

Yeah, there is a whole brigade of militant anti-PSQers who will penalize anyone who answers.

Xander Henderson

17:25

@copper.hat I have no idea why you are getting downvotes, as I am not one of the people who cast any downvotes. I would guess that @TedShifrin is correct, and that the downvoters believe that you shouldn't have answered the question.

Ted Shifrin

This question was more sophisticated, but still a PSQ.

Xander Henderson

But I also don't think that it is productive to complain or demand explanation.

Ted Shifrin

I have been punished for similarly answering advanced diff geo questions with no effort.

For once, I concur with @Xander.

copper.hat

@XanderHenderson i need an outlet

Ted Shifrin

Answer Horiatiki's question and get the bounty, @copper.

Xander Henderson

17:26

@copper.hat I mean, if all you need is to rant, and have people say "there there", we're here for you in chat.

Ted Shifrin

I rephrased it simply for you just above :)

copper.hat

sometimes folks have replied with reasonable objections and mostly they were addressed.

Xander Henderson

But ranty comments about downvoters on the main site will be deleted.

copper.hat

@XanderHenderson i was covering my bases, in chat & the question

i thought my pro forma bs comment was very polite :-)

Ted Shifrin

Ultimately, the fact that PSQs are proliferating and the brigade is more militant will deter many long-term members like us from participating in the site.

I get tired of telling people to show effort.

copper.hat

17:28

frankly if it were not for chat i would probably have flown a long time ago

Horiatiki

@TedShifrin I know you are right. I'm sorry. This is the best I can show you. I have never used the Gauss-Newton algorithm. I'm just a math enthusiast. I just wanted to find a new parameter for Poisson. Then a Norwegian mathematician recommended Gauss-Newton to me. He gave me some initial advice and wrote me that paper. I am studying Gauss-Newton algorithm hard and I understand the theory of him, but I have difficulty solving it with practice

Ted Shifrin

@copper Focus. Be productive and take 5 minutes to help Horiatiki instead of ranting.

copper.hat

i'm trying to find their question

Ted Shifrin

I phrased it explicitly above, but he linked to it too.

copper.hat

17:56

@Horiatiki Can you elaborate where you are having difficulty?

As an aside, the term Gauss Newton is used if different ways. But that is not relevant here.

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$Mathematics$ Mathematics