Mathematics - 2024-01-03

psie

00:16

I'm working an exercise where I need to show every function, over a symmetric interval like $(-a,a)$, $a>0$, can be decomposed into a sum of an even and odd function. I've shown that, however, I also need to show the decomposition is unique. How do I do that?

the even part of $f$ is simply $g(x) = \frac {f(x) + f(-x)}2$

and the odd part of $f$ is $h(x) = \frac {f(x) - f(-x)}2$

leslie townes

depending on how you've argued that, you may only need to re order what is an assumption and what isn't, i.e. there may be no 'work' left to do. but it may help to at least think it through separately, if not write it separately. suppose f(x) = e(x) + o(x), with e even and o odd. can you show, using arguments that are probably very similar to ones you may have already come up with, that e(x) is given by the same formula as your "g"

psie

00:33

right, I would just argue the same way I already have 🙃

Jakobian

01:28

@psie this is a linear algebra problem

Thorgott

this is (not so) secretly an Eigenspace decomposition

ChoMedit

BibTex is easy, but are there more comfortable ways to manage the bibliography? Writing every information of papers is really boring.

Xander Henderson

02:09

@ChoMedit I don't understand. Using whatever system you use, you are going to have to put the data into some kind of file somewhere at some point in time.

Don't you just have a a single .bib file that you use over and over and over again?

Lucas Henrique

hi guys

Jakobian

hello

Lucas Henrique

is there any theorem that relates the tensor product of two finitely generated R-modules with the Tor functor?

Ted Shifrin

Tor is the higher derived functor of tensor product.

Lucas Henrique

yeah

I forgot to write down the rest of my question

Jakobian

02:19

Lucas Henrique

so what I meant is: it seems that the free parts of the modules should behave well, and the torsion parts should have some weird behavior that might depend on the Tor

so there might be some kind of relation like "let $M_1 = R^{I_1} \otimes T_1$, (...). then the following sequence is exact:" and then the middle guy is $M_1\otimes M_2$, $R^{I_1\times I_2}$ appears somewhere and maybe $\mathrm{Tor}(T_k)$ too

Ted Shifrin

This is gibberish to me. You need solid homological algebra. Maybe @Lukas can help.

ChoMedit

@XanderHenderson That's true, but like, to cite some non-journal-style papers(like lecture notes) or making a citation label are really annoying things. That's of course not a problem of bibtex itself. By the way, I don' have a single .bib file that I usually copy again, but I think writing it is not that hard. It's just a boring thing.

Ted Shifrin

It’s lucky for you that you were born after LaTeX. I wrote my PhD and my first several papers using typewriters.

Boring. Ha.

Lucas Henrique

@TedShifrin haha yeah

Ted Shifrin

02:31

I remember card catalogs and heavy tomes of Math Reviews.

Lucas Henrique

happily the homological algebra course will be offered in the following semester

Ted Shifrin

There are good books.

copper.hat

02:44

@PM2Ring I was at a nephew's wedding recently, and his contingent from Aberdeen came, but I must confess that it took a bit of listening to confirm (in my mind) that some were Scottish and not Irish. A bit sad given that a few decades ago I could make passable guesses at the county/shire of origin.

Xander Henderson

@ChoMedit My general process for most citations is "Google the paper -> Hit the "Cite" button -> Copy the bibtex code to my .bib file". It is relatively simple...

@TedShifrin My father was one of the very first people at University of Arizona to turn in a PhD dissertation as a computer printout (in... uh... 1982? 83?).

Lots of people hired folk to type up handwritten work.

Though that is harder to do in mathematics, where the symbols matter.

(I swear to g-d that the image in the "KayPro" article on Wikipedia could be me, except I didn't wear glasses until high school.)

Thorgott

@LucasHenrique so it sounds like you're working over a PID, so you have $M\cong R^r\oplus M_{tors}$ and $N\cong R^s\oplus N_{tors}$ for some $r,s\ge0$ and then $M\otimes N\cong R^{rs}\oplus M_{tors}^s\oplus N_{tors}^r\oplus(M_{tors}\otimes N_{tors})$. not sure if that's what you're getting.

Lucas Henrique

oh, i thought it would be harder lol. thanks Thorgott :)

Thorgott

np

you can be more precise if you decompose the torsion parts into cyclic summands and calculate $R/p^m\otimes R/q^n$ for primes $p,q$ and $m,n\ge0$

Dannyu NDos

07:22

I have a confusion about the terminogy of a "subring". Is it obliged to preserve the unity?

On the contrary, an ideal isn't obliged to preserve the unity. I presume that's why we never call them "normal subring".

leslie townes

this depends on your set of definitions. when the definition of "ring" includes a multiplicative identity element, the definition of "subring" usually requires that the "subring" have the exact same multiplicative identity element. it wouldn't surprise me if some weirdos somewhere don't do this, but that is the usual convention.

of course, for some people, a "ring" might not be required to have a multiplicative identity element, and in that case, a "subring" may not be required to have one either (even if the ambient "ring" does have one). my impression is that this convention is increasingly less common, although it is used in at least some popular algebra books that were written in the mid-late 20th century

there is enough variation in this from textbook to textbook (and certainly on the internet, and in this chat) that one should be cautious about any statement about what "we" do or do not require of "rings" :)

so e.g. under what i have termed the "usual convention," if Z is the integers (or more generally any nonzero ring with multiplicative identity), the subset Z x {0}, while an ideal in the product Z x Z, would not be a "subring" of Z x Z.

one potato two potato

09:18

@Jakobian @AlessandroCodenotti If $\alpha$ is an ordinal and $n\in\Bbb Z_+$, then is
$$(\omega^\alpha\cdot n+1)' = \begin{cases} \omega^\beta\cdot n+1 & \text{if }\beta+1 = \alpha\\
\omega^\alpha\cdot n+1 & \text{if }\alpha\text{ is a limit ordinal}\end{cases}$$
true?

here $'$ means the derived set

Mad

11:43

I am abit confused about Cartans closed Sub group theorem for Lie gorups. Are not Lie groups topological groups, is not an open subgroup of a Topological group automatically Closed? So by Cartans theorem, would this not mean any open or closed subgroup of a lie group a lie sub group

Thorgott

11:57

yes, open subgroups are automatically closed

but the fact that an open subgroup of a Lie group is itself a Lie group is a lot more trivial

Mad

Open subsets of a toplogical smooth manfiold are themselves such, they inherit also the group structure, and it remains to prove only that the multiplication and the inverse are smooth mappings

Which i guess is trivially done by taking the limitation on the subset?

Thorgott

yes, you are just restricting smooth maps to smooth submanifolds

the hard part is just proving that closed subgroups are automatically smooth submanifolds

but for open subgroups, this is trivial, cause all open subsets are automatically smooth submanifolds!

Mad

12:19

I see

Jakobian

12:59

@onepotatotwopotato no. For example the derived set won't contain any naturals

You seem to be concerned either by $\sup$ or ordinal type of the derived set ... but those are speculations by me

one potato two potato

@Jakobian why?

for example if $\alpha=2$ and $n=1$ then it's true

Jakobian

Because $\{n\}$ is open

Your set will consist of limit ordinals

one potato two potato

13:17

then what's $(\omega^\alpha\cdot n+1)'$ then?

Jakobian

The set of all limit ordinals $\leq \omega^\alpha\cdot n $

one potato two potato

can you provide a proof of that?

Jakobian

13:33

It pretty much follows from definitions

If $x$ is a successor ordinal and $y+1 = x$, then $\{x\} = (y, x+1)$ is open

If $x = 0$ then $\{0\} = [0, 1)$ is open

If $x$ is a limit ordinal and $y < x$ then $(y, x] = (y, x+1)$ is open and contains some $y < z < x$

It follows that $(y, x)$ is non-empty

If $U$ is any neighbourhood of $x$, then from definitions $U$ contains some $(y, x]$ for $y<x$

So $U\setminus \{x\}$ is non-empty

@onepotatotwopotato

one potato two potato

one confusion on basic: If $\alpha<\beta$ then $\alpha\subset\beta$ but containing $\beta$ does not imply $\alpha$ is also contained?

one potato two potato

14:10

I mean $\omega\in (\omega^\alpha\cdot n+1)'$ but this does not imply natural numbers are also contained?

Jakobian

14:21

@onepotatotwopotato There is no reason to believe that the derived set is an ordinal

Sha Vuklia

@leslietownes Thanks for the confirmation!

Jakobian

@onepotatotwopotato I'm not understanding you

one potato two potato

@Jakobian you said the derived set is the set of all limit ordinals $\leq \omega^\alpha\cdot n$, in particular $\omega$. But containing $\omega$ does not imply natural number $n$ is also contained?

Jakobian

Why would it

one potato two potato

$n\subset\omega$ isn't it? They are as sets but not as ordinal numbers?

Jakobian

14:25

$x\in y\in z\implies x\in z$ holds for transitive sets, but why would the derived set be transitive?

attack

is there notation involving $\nabla$ which is used for the hessian of a function?

one potato two potato

Hmm

attack

ive never really got what people mean by $\nabla^2$

Jakobian

@onepotatotwopotato but $\omega$ is not a subset of the derived set

@attack laplacian

$x\in y \implies x\subseteq y$ if $x, y$ are transitive

Why would the derived set be transitive or an ordinal?

one potato two potato

@Jakobian so you're saying $(\omega^\alpha\cdot n+1)' = \{\omega,\omega^2,\ldots\}\ni\omega$ and $\omega\supset n$ but this does not imply $n\in \{\omega,\omega^2,\ldots\}= (\omega^\alpha\cdot n+1)'$?

Jakobian

14:42

@onepotatotwopotato Yes

$n\in\omega$, but $n\notin\{\omega\}$ for example

one potato two potato

ok thank you

attack

@Jakobian is there a way to write hessian using $\nabla$?

Jakobian

@attack $(\nabla f_1, ..., \nabla f_n)$?

@attack Some authors seem to write $\nabla^2$ for Hessian, see:

Del squared

Del squared may refer to: Laplace operator, a differential operator often denoted by the symbol ∇2 Hessian matrix, sometimes denoted by ∇2 Aitken's delta-squared process, a numerical analysis technique used for accelerating the rate of convergence of a sequence == See also == DEL2, the second tier ice hockey league in Germany Del, a vector calculus differential operator Nabla symbol, the symbol used for the Del operator ∂, the partial derivative operator symbol Del (disambiguation)

But there might be lots of confusion between Laplacian and Hessian then...

attack

15:03

okok so thank you

Mad

15:49

Let $ M , N$ be Manifolds ,$ X,Y$ be vectorfields on those manifolds and $f$ some function between $M ,N$ we say the vectorfields are $f$ related if :
$ df \circ X= Y \circ f$
equivalently one reaches by applying some function $h$ on both sides for locally defined function on N that
$ X_p ( h\circ f) = Y_{f(p)}h $

i dont understand how the left side comes to place in the second definition.

Xander Henderson

16:05

Does anyone else find it kind of annoying when someone on Stack Exchange feels the need to include a title in their username? It seems to me that if you have to call yourself "Dr SuchAndSuch" on a silly internet website, you must be compensating for something. :/

Thorgott

16:16

I, BSc Thorgott, do not agree.

@Mad for $p\in M$, $(df\circ X)(p)=df_p(X_p)$ is the derivation that takes $h\mapsto X_p(h\circ f)$

Mad

should there be some sequence of steps and equalities between those two ? this is what i am struggling with

Frieren

I proved that $\lim_{x \to 0^+} \frac{1}{x} \int_0^x \frac{\arctan t}{x+t^2}dt = 1/2$ using inequalities. However, I was curious about how Hopital's rule behaves when the integrand depends on $x$ as well as the upper limit of integration. Indeed, it doesn't work: it leads to the wrong result $0$. Why Hopital's rule fails when the integrand depends on $x$?

Sine of the Time

How did you take the derivative of the integral? Did you use Leibniz rule?

you also should check if you have an indeterminate form

Thorgott

@Mad This is by the definition of the differential that we discussed recently

leslie townes

16:52

xander: [hands up]

although, in some parts of the world and/or in some fields it may be more expected and less of the attempted flex that it definitely looks like in US math

Xander Henderson

@leslietownes Yes! That is the word. "Flex".

leslie townes

i'm feeling confident today, maybe i'll edit Dr. Leslie, Ph.D. into my profile

Xander Henderson

17:12

@leslietownes Maybe I should edit my user name to read "Dr Alexander M Henderson, BA MS PhD"?

leslie townes

HSD [or GED]

Licensed Motorist

Two-Time Prospective Juror

Xander Henderson

@leslietownes Oh, I should put my teacher license number on there, too!

leslie townes

Selective Service Registrant

Ted Shifrin

18:19

@XanderHenderson How dare you not refer to I as Dr. Ted!

@attack The notation for $\nabla^2$ as the Laplacian comes from classical physics/vector analysis (short for $\nabla\cdot\nabla f$). The notation for $\nabla^2$ as the Hessian comes from differential geometry (applying the connection $\nabla$ twice to a function or other things).

Sahaj

What's the most elegant proof of irrationality of pi?

leslie townes

"elegant" is in the eye of the beholder, there's one often attributed to niven that is pretty slick that you see in textbooks a lot. it's probably on wikipedia.

Sahaj

Yes, know of that one. Indeed, elegant is quite subjective.

leslie townes

a sufficiently slick proof will often give rise to people writing very meta discussions about why the slick proof isn't "slick" at all, but is actually something any fool could have thought of, or is at least motivated by other considerations. i'm not sure how convincing these are, they tend to be less convincing as their lengths grow.

i just found pims.math.ca/~hoek/notes/dvi01/nivenpi2.pdf which is a very short and self contained discussion of a few of the high level ideas that might have motivated niven's proof, or at least "explain" the role of the functions considered in that proof in terms of abstract properties.

Sahaj

18:36

Very interesting, thanks

leslie townes

kinda cool, on the second page they explain how a similar idea can establish the irrationality of log(n), n > 1. i'm not sure that i had seen that before

Grasshopper

I am unable to solve the following double integral expression -
$$\int_{y_1}^{y_2}{
\int_{x_1}^{x_2}{\frac{2\pi\sqrt{
(X-x)^2 + (Y-y)^2 + Z^2
}}{\lambda}\space dx}
} \space dy$$

Can someone help?

Ted Shifrin

I do not see an equation.

Grasshopper

Ok, I fixed my vocabulary.

Ted Shifrin

18:51

This is not going to be pleasant. First, get rid of irrelevant constants (like $2\pi$ and $\lambda$). Can you do the one-dimensional integral $\int \sqrt{(a-x)^2+b^2}\,dx$ first?

Grasshopper

Ok

Ted Shifrin

But then integrating again is going to be horrendously horrible.

Grasshopper

Oh, I am sorry.

The equation is actually -
$$C_\text{total}=\int_{y_1}^{y_2}\int_{x_1}^{x_2}A\sin\left(\frac{
2\pi\sqrt{(X-x)^2 + (Y-y)^2 + Z^2}}{\lambda}\right) dx dy$$
where $x$ and $y$ are the variables and rest of the symbols are constants.

Actually, I am trying to create a simulation experiment (out of general interest).

Where I deduced that the magnitude at each pixel is given by the solution of this double integral equation.

Ted Shifrin

Almost surely you need to do numerical integration. No way to do this explicitly.

Grasshopper

I know enough about the theory of integrals to express myself through the equation, but, not enough to solve it. :(

Can you please elaborate?

Ted Shifrin

18:59

I guarantee you need to solve it by having a computer do the integrals numerically.

Of course, you'll need to put in the constants as actual numbers.

Grasshopper

Are the results precise in that case?

I can produce low precision images without integrals. I need integrals for high precision calculations.

Sahaj

Is there any known closed form for the sum $\sum_{n=1}^k \sin\left(\frac{x}{n}\right)$? Or perhaps some special cases such as $k=x$ etc.?

Ted Shifrin

"Precise"? What does that mean?

@Sahaj What does $k=x$ even mean?

You mean $k=x=N$ for some integer $N$?

Sahaj

Yes of course

$k,x$ etc are positive integers

Ted Shifrin

So you're asking for $\sin(N)+\sin(N/2)+\dots+\sin(1)$?

Sahaj

19:08

Yes, indeed

Ted Shifrin

Unless $N$ is of the form $m!$ for some $m$, I'd be surprised if there's a formula.

But I'm not particularly clever about these things.

Sahaj

Right

Regardless it'd be a trouble to evaluate

Grasshopper

@TedShifrin I need great precision in calculations. Upto the order of $10^{-10}$.

With the computation power I have, I can currently reach around $10^{-6}$

Ted Shifrin

I don't know how accurately your computer can do numerical integration.

I suspect that Mathematica can do it with that much accuracy if you tell it to.

Grasshopper

I am using a brute force algorithm now. Ideally, I can reach any precision, but, practically, there is a computation limit (amount of calculation I can do in a given time).

psie

19:13

If a function is neither even nor odd, can we know whether or not the complex Fourier coefficients $c_n$ will be purely imaginary, purely real, or complex (i.e. having both an imaginary part and a real part)?

Grasshopper

If that integral is solved, I will have an algorithm which is requires less computation.

Thats all.

leslie townes

psie: in general, the coefficient just is whatever it is, e.g. i don't know of any general condition that would be "simpler than," but equivalent to, a given c_n being real/imaginary/whatever

psie

ok

Grasshopper

@TedShifrin Anyway, thanks! I will try this too.

Ted Shifrin

Programs like Matlab and Mathematica are very sophisticated in doing numerical integration. I don't know what you mean by "brute force."

@psie Purely real should be equivalent to even and purely imaginary should be equivalent to odd (if you're talking about real-valued functions), I believe.

Xander Henderson

19:21

@Grasshopper For an integral like that, I doubt that computation time is all that relevant---unless you are evaluating that integral billions of times, I suppose.

psie

yeah

Grasshopper

@XanderHenderson Yes, I plan to integrate it billions of times actually if possible. But, I will try the suggestions first on smaller scale first.

Probably on a 1000x1000 array.

Xander Henderson

Why? What is the application?

Also, 1000x1000 is not billions. Only millions.

Grasshopper

Yes I am starting with that million.

@XanderHenderson Simulations related to light.

Xander Henderson

Well, that integral doesn't look like something that is going to be amenable to analytic approaches. However, you might be able to get some kind of "faster" expression out of it, e.g. some sort of local linearization, or a power series expansion, or something which can be computed faster.

But I don't see much reason to suspect that this integral itself will have a "nice" closed form, analytic representation.

Grasshopper

19:33

ok

Jakobian

@Sahaj For the closed form, I think we need to consider divisors of $n$ first, via some number theory-like procedure

Harmonic number

In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: Starting from n = 1, the sequence of harmonic numbers begins: Harmonic numbers are related to the harmonic mean in that the n-th harmonic number is also n times the reciprocal of the harmonic mean of the first n positive integers. Harmonic numbers have been studied since antiquity and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function, and appear in the expressions of various special functions...

looking at harmonic numbers, there is a formula $H_n = \frac{1}{n!}\cdot {n+1 \brack 2}$

The proof most likely boils down to $(n+1)!H_{n+1} = (n+1)\cdot n!H_n + n!$ and noting that ${n+1 \brack 2}$ satisfies the same recursion as $n!H_n$

Here $A_n = \sum_{n=1}^k \exp(z/n)$ satisfies $A_{n+1} = A_n + \exp(z/(n+1))$. By replacing $\exp(z/(n+1))$ by $c/(n+1)^k$, we might be able to obtain something if we first solve $B_{n+1} = B_n + c/(n+1)^k$

i.e. if we obtain formula for the generalized harmonic numbers $H_{n, k}$ for a fixed integer $k$

wikipedia lists formula $H_{n, k} = \zeta(k, 1)-\zeta(k, n+1)$ were $\zeta$ is the Hurwitz zeta function

So we can obtain a formula for $A_n$ in terms of a series of Hurwitz zeta functions at least

leslie townes

19:58

galaxy brain meme where the first one is "i want a closed form," the next one is "there is no nice closed form," the next one is "there is a closed form in terms of a series of zeta functions," and the last one is "i no longer want a closed form"

Jakobian

I mean, it's not completely hopeless until we calculate it

but I don't want to volunteer

robjohn

@Jakobian brackets meaning Stirling numbers?

Nvm, I looked at the Wikipedia article

Jakobian

yeah

I thought there might be similar formula for $A_n$ but I doubt it

robjohn

@Jakobian there is some index-variable confusion in the definition of $A_n$

Jakobian

$A_k = \sum_{n=1}^k \exp(z/n)$

robjohn

20:11

Yeah

Ted Shifrin

Happy new year, @robjohn. Still a green spider, I see :P

Jakobian

My idea is to write $\exp$ in terms of its power series, solve each of the recursions, and put it back together

hmm the formula for $H_{n, k}$ follows from definitions though... maybe its better to be more direct

$\sum_{k=1}^n\exp(z/k) = \sum_{k=0}^\infty \frac{z^k}{k!}\cdot H_{n, k}$

So its the exponential generating function for the sequence $H_{n, 0}, H_{n, 1}, ...$

Borel summation

In mathematics, Borel summation is a summation method for divergent series, introduced by Émile Borel (1899). It is particularly useful for summing divergent asymptotic series, and in some sense gives the best possible sum for such series. There are several variations of this method that are also called Borel summation, and a generalization of it called Mittag-Leffler summation. == Definition == There are (at least) three slightly different methods called Borel summation. They differ in which series they can sum, but are consistent, meaning that if two of the methods sum the same series they give...

there is a formula here to write it as normal generating function

$A(z) = \sum_{k=0}^\infty H_{n, k}z^k = \int_0^\infty e^{-t}(\sum_{k=1}^n \exp(tz/k))\mathrm{d}t$

no wait

robjohn

20:32

@TedShifrin Happy New Year! I’ll probably change back to the calendar when I get back home.

Ted Shifrin

Guess you guys are getting a lot of rain up that way. We had some serious rain just when I was out for my ophthalmology appointment after yesterday's cataract surgery (#1).

Jakobian

20:46

I retract all my attempts of calculating this sum, they're all circular

robjohn

We got about ⅙” this morning, but the forecast rain seems to have vanished. We got a bit over 4½” in December.

I hope your recovery goes well.

Ted Shifrin

21:06

Distance is already great, but close-up will take time ….

robjohn

21:20

Are there ocular calisthenics you need to do?

Xander Henderson

@robjohn We're supposed to get some of that rain from y'all, but it hasn't come yet. :(

Ted Shifrin

21:48

@robjohn No, no heavy lifting or bending.

Sha Vuklia

23:21

Let $X$ be a compact Hausdorff space, and let $\mu$ be a Radon measure on $X$. I don't see why the above equality is true. Here $\Vert \mu\Vert=\vert \mu\vert(X)$, where $\vert\mu\vert$ is the total variation of $\mu$. I can show that $\Vert h\mu\Vert\leq \int\vert h\vert d\vert\mu\vert$ by using the triangle inequality for measures (and the definition of total variation). Not sure how to go about showing equality. Any reference for this result is also welcome

Btw, I was thinking of maybe showing those for indicator functions, and then approximate $h$ by simple functions, but I am not sure about it

For indicator functions it works I believe

and for scalars times indicators too, and for sums of these functions if the measure is positive

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