Mathematics - 2024-09-10 (page 1 of 2)

Bml

12:53 AM

@SineoftheTime I did that. This is the question I posed on Math SE:

0

Q: How to solve $(a+b+c-x)^2 x \ge 4abc$, with $x \lt a+b+c$ and $x \in \mathbb{R}^{+}, \quad a,b,c \in \mathbb{R}^{+}$?

I'm interested in solving a non-trivial cubic inequality coming from some physical arguments. So, the first part of my question is dedicated to the creation of the (necessarily physical) context, the second part to the transition to mathematical language, since my goal here is to find a relation ...

Pizza

8:37 AM

👋👋👋

Sine of the Time

8:49 AM

@Bml well done

@Pizza hi

Pizza

@SineoftheTime In the exercise of the power series, I have to consider it as the thread $\sum^{\infty}_{n=0} a_n (x-x_0)^{n}$

Sine of the Time

this is the generic form of a power series

Jakobian

I had a nightmare where giant blue fish-like monsters were roaming outside of my window. There was 2 of them and each one different

Pizza

@SineoftheTime yes, however I think I have to use Leibniz, now I'll try to solve it for a moment

Sine of the Time

I too had a nightmare

Jakobian

8:59 AM

Same one?

Sine of the Time

no I got lost in the mountains after attending a funeral

the feeling of being lost is s dream is suffocating

@Pizza what is your strategy?

Bml

@SineoftheTime Did you understand the first part of the answer to my question? In my opinion there's a sort of typo, because it is said $y(1-y)^2-u\geq \frac{4abc}{(a+b+c)^3},\quad 0<y<1$, with $y=x/(a+b+c)$, but, according to the algebraic manipulation and according to what it is said immediately after, it should be $y(1-y)^2-u\geq 0$, with $u = \frac{4abc}{(a+b+c)^3}$. Am I wrong or is this really a typo?

Pizza

@SineoftheTime $\lim_{n\to\infty} \left(\frac{x^2}{3}\right)^n \cdot \frac{1}{n}$

Jakobian

@SineoftheTime I never experienced the feeling of being truly lost I suppose, neither in a dream nor real life

Pizza

And i have to see if $\lim_{n\to\infty} (a_n) = 0$ exists

I would do it like this

Sine of the Time

9:09 AM

note that $(-1)^n\frac{x^{2n}}{3^n(n+1)}=\left(-\frac{x^2}3\right)^n\cdot\frac1{n+1}$

if you let $y=-\frac{x^2}3$ you can study the convergence of $\sum \frac{y^n}{n+1}$ and then go back to $x$

Pizza

But $n + 1$ would be asymptotic to $n$ right?

Sine of the Time

yes but you have to be careful when applying asymptotic expansions

@Bml you can ask clarification in the comments

Pizza

@SineoftheTime so $\lim_{n\to\infty} \frac{y^n}{n} \Rightarrow |y|$

Using the Cauchy hadamard criterion

Jakobian

I do like my dreams, even the nightmares, because there is only a handful of them that I can remember, and I don't take them very seriously. The nightmares especially are more vivid than normal dreams

Sine of the Time

@Pizza ?

there are different cathegories of dream. If you think a lot about a specific thing during the day, you may also dream about it but you'll probably forget about it 5 minutes after waking up

Pizza

9:19 AM

@SineoftheTime I used the Cauchy-Hadamard criterion where the root n is used

So I'm left with only |y|

Because $\lim_{n\to\infty} n^{1/n} = 1$

Sine of the Time

what do you mean I'm left with $|y|$? Since the limit is $1$, the series converges for $|y|<1$

plus, since $b_n=\frac1{n+1}$, you have to compute the limit with $n+1$.

the limit is one but you can't write directly $\lim n^{1/n}$, expecially in an exam

Soumik Mukherjee

Many of my nightmares were about to not reaching the exam hall in time and failing the exam

Jam

If a function is constant on every rational and continuous then is constant on all R. Because i can approach every x with a sequence of rationals so the limit has to be that same constant. I think im right but i cannot write it in a more formal way.Can you help wrting it down?

Jakobian

9:36 AM

@Jam what does constant on every rational mean

Jam

f(q)=c for every q in Q

i think im just gonna write what i said.

Jakobian

If a function is constant on a dense set $D$ and continuous, then $f(\overline{D})\subseteq \overline{f(D)}$ where the latter is a singleton in any $T_1$ space

So, for continuous functions $f:X\to Y$ where $Y$ is a $T_1$ space, if $f$ is constant on a dense set, then $f$ is constant

@SineoftheTime never had that

@SoumikMukherjee never had that either

Contrast this with the theorem that if $f, g:X\to Y$ are continuous functions and $Y$ is a $T_2$ space, then if $f$ and $g$ are equal on a dense set, then $f=g$

Pizza

10:00 AM

@SineoftheTime I think I solved it

$a_n = (-1)^n \frac{1}{3^n(n+1)}$

$x^2 = y$ so i have $y^n$

$\lim_{n\to\infty} \sqrt[n]{|(-\frac{1}{3})^n \cdot \frac{1}{n+1}|} = 1/3$

$|y| < 3$ so $x < \sqrt{3}$ , $(-\sqrt{3}, \sqrt{3})$

Punctual convergence

Sine of the Time

don't forget the endpoints

Pizza

@SineoftheTime yes

Jakobian

You don't need to substitute. In fact a lot of the times such substitution will be impossible. Just consider limsup instead of lim

Pizza

At $x = -\sqrt{3}, \sqrt{3}$ the series the series simply converges so the interval is $[-\sqrt{3}, \sqrt{3}]$

Jakobian

E.g. consider the power series $\sum_{n=1}^\infty x^{n!}$. How would you determine convergence in this case?

What substitution will you do?

The answer is that no substitution is needed, and instead you observe that $\limsup_n \sqrt[n]{|a_n|}$ is equal to the limit along the subsequence $1!, 2!, ...$ that is $\lim_n |a_{n!}|^{1/n!}$

Pizza

10:19 AM

@Jakobian I would not know :(

Anyway thanks for this example, now I'm trying to understand what you wrote

Jakobian

Where $a_n = \begin{cases} 1 & n = k! \\ 0 & n\neq k!\end{cases}$

Where I'm being sloppy a bit on what I mean by $k!$

Sine of the Time

I don't think Pizza is familiar with limsup and liminf

Jakobian

@Pizza what allows you to determine radius of convergence of $\sum a_nx^n$ is $\limsup_n |a_n|^{1/n}$

You can think of this as comparison to geometric series, since this is basically what it is

psie

I have a basic question. Why does $\chi_{(n,n+1)}\to0$ pointwise? When I think about the graph of this function, all we're doing is shifting it to the right. So it'll always remain $1$ on some interval of length $1$ (assuming Lebesgue measure).

Maybe we're not shifting it to the right...

After all it's a function of $x$ and not of $n$...

Jakobian

@psie what is $\chi_{(n, n+1)}(x)$ for large enough $x$?

Thorgott

10:36 AM

your understanding of the graph is not wrong

psie

@Jakobian hmm, ok, for $x=10^{50}+1/2$, it'll be $0$, unless $n=10^{50}$?

I had to think long about your question for some reason...

Jakobian

For large enough $n$ I should have said

Sine of the Time

this is tricky at first

Jakobian

It really isn't

@psie substituting random numbers gets you nowhere

psie

I have to use the definition of pointwise convergence, meeh.

Sine of the Time

10:41 AM

for every $x_0\in \mathbb R$, you can find $n$ such that $x_0\notin]n,n+1[$

with $n$ large enough

psie

ah that makes sense!

Jakobian

does it make sense?

it doesn't to me, but if that makes sense to you then whatever

like of course I know what sine is trying to say, but that's not what they said

psie

@Jakobian well, for a fixed $x$, we can "shift" the interval $(n,n+1)$ so that it misses $x$ and so the sequence will be $0$ for infinitely many $n$. That's my understanding.

Jakobian

Not just for infinitely many $n$, but for all large enough $n$

psie

yeah, you're right

Jakobian

10:55 AM

You can just say $\chi_{(n, n+1)}(x) = 0$ for $n > x$

psie

ah great, then I don't have to unpack the definition of pointwise convergence :)

Jakobian

I don't know what you mean by that

of course you need to use definition of pointwise convergence

$\lim_{n\to\infty} \chi_{(n, n+1)}(x) = 0$ for all $x$, therefore $\chi_{(n, n+1)}$ converges pointwise to the function $f\equiv 0$

psie

ok 👍 yeah I was trying to be lazy, but probably doesn't work here :)

Jakobian

being lazy is fine, but being negligent is unacceptable

Pizza

11:10 AM

@Jakobian A geometric series will diverge for $|x|≥ 1$ , so from this for x^(n!) if |x| >= 1 It will diverge. For |x| < 1 It Will converge ?

Jakobian

11:33 AM

@Pizza A geometric series is a series of the form $\sum a^nx^n$

it will converge for $|x| < |a|^{-1}$ and diverge for $|x| > |a|^{-1}$

the point is that $\limsup_n \sqrt[n]{|a_n|}$ serves the role of "$a$" here

even if the coefficients aren't powers of $a$ and $\sum a_n x^n$ is your series

(interpret this as $|a|^{-1} = 0$ when $|a| = \infty$ and $|a|^{-1} = \infty$ when $|a| = 0$)

@Pizza the series $\sum x^{n!}$ is not a geometric series

Pizza

12:21 PM

@Jakobian No, i used direct comparison test for that reason

Sine of the Time

I don't think that series is relevant to you

Gian's Pizzeria

Hi

what are we talking about? anyway I have now become a machine

Sine of the Time

a machine?

Gian's Pizzeria

I'm kidding, what were you talking about anyway?

Sine of the Time

Pizza was solving an exercise on the convergence of a power series

Gian's Pizzeria

12:28 PM

Okay

@Pizza do this : Calculation of the numerical sum of $\sum^{\infty}_{n=0} \frac{n}{n^4+4}$

Pizza

I'll take a look, thanks anyway

Sine of the Time

12:43 PM

since you're still struggling with fundamental concepts, I suggest you not to do random exercises but to focus on the exam

Pizza

Yes

Gian's Pizzeria

1:09 PM

@SineoftheTime I had proposed it because he had written that exercises on sums could also be included

Obviously he is not obliged to do the exercise

Ryder Rude

1:42 PM

anyone interested in philosophies of consciousnes?

hi

Bml

@SineoftheTime I asked clarification and questions in comments. If you want, you can have a look and see what it looks like...

ephe

2:01 PM

@Thorgott @BenSteffan, thank you again for all your help regarding category theory stuff. The presentation went much better than I anticipated (youtube.com/watch?v=7a13k4S1eok), even though I had to throw a lot of content out and squeeze everything into 20 minutes (the maximum time allowed).

Mats Granvik

2:23 PM

Are the square free integers called a field?

ephe

@MatsGranvik Well, they are not closed under addition, so I don't think they even form a ring. Consider, for example, 10 and 6. 16 is not square-free.

Mats Granvik

@ephe Ok. What about the natural numbers, are they a ring?

ephe

@MatsGranvik No, because you don't have additive inverses for elements. If I remember correctly they do form a semiring (this needs checking I really don't remember).

Mats Granvik

Ok

Semiclassical

my real-analysis brain is being slow and i want to state an argument as simply as possible. Given a differentiable function $f$ and $a<b<c$ I've got $f(a)=f(c)<0$, $f(c)>0$. I also have $f'>0$ on $(a,b)$ and $f'<0$ on $(b,c)$.

by IVT there exists roots $x_1,x_2$ in the intervals $(a,b)$, $(b,c)$ respectively. by the signs of $f'$, i also know that $f$ is monotonic on these intervals and thus these roots are unique. i then want to conclude that $f(x)\geq 0$ on $[x_1,x_2]$

the last feels "obvious" to me but i'm forgetting how to spell it out. there's a proof by contradiction from the intermediate value theorem: if $f(x^*)$ for some $x^*\in (x_1,x_2)$ then IVT would yield another root between $x^*$ and $c$, which contradicts the uniqueness of the two roots already found.

Bml

2:46 PM

Hi @Semiclassical, thank you for the answer to my question. Have you read my comments (doubt and questions) under your answer?

Semiclassical

Yep. Was crafting a response before my laptop died @bml

The doubts should all be resolved

Mats Granvik

oeis.org/A366450

Semiclassical

The above is me trying to remember the cleanest way to give the argument for $f(y)\geq 0$ on one subinterval of $(0,1)$

Mats Granvik

1, -2, -1, -4, 1, 2, -2, -8, -3, -2, 1, 4, 4, 4, -1, -16, -2, 6, 0, -4, 2, -2, -1, 8, 5, -8, -9, 8, 0, 2, 7, -32, -1, 4, -2, 12, 3, 0, -4, -8, -8, -4, -6, -4, -3, 2, 8, 16, -14, -10, 2, -16, -6, 18, 1, 16, 0, 0, 5, 4, 12, -14, 6, -64, 4, 2, -7, 8, 1, 4, -3, 24, 4, -6, -5, 0, -2, 8, -10, -16, -27, 16, -6, -8

Bml

@Semiclassical Yes, I just saw your edit. Thank you.

Semiclassical

2:50 PM

As for the trigonometric solution, I was definitely being a bit lazy since at that point it’s more of a side argument

Sine of the Time

@Bml you ask a lot of questions on a post that is not too long. You should work out small details on your own first

Semiclassical

@SineoftheTime well, some of that was just typos that were indeed wrong

Bml

@Semiclassical Is this your response to my 1st and 2nd questions?

Semiclassical

It’s not my final response, but it’s what I’ll be adapting into the post (unless someone reminds me of a tidier way to make the argument)

Bml

@Semiclassical OK. Thank you very much anyway.

@Semiclassical As for my 3rd question, you changed "roots" into "branches". Could you explain this?

Semiclassical

2:59 PM

Sure. Roots was incorrect, so don’t linger on that word too much. There are three cases of Viete’s formula, corresponding to the index $k=0,1,2$, and each yielding a real root of the relevant depressed cubic

The word “branches” is just a synonym for those three cases

(This is consistent with the notion of branch points and branch cuts in complex analysis)

Of these branches, I checked numerically that the $k=0$ branch always gave a root in $(1,\infty)$. Since that root is physically irrelevant I ignored it

I was deliberately a bit lazy with that argument since it’s not really part of the main scope of the answer

Bml

@Semiclassical OK, but in what sense is one of these branches "negative"? How to understand what $k$ gives the negative/positive branch?

@Semiclassical Don't worry. This is not a deficiency in your response, which I think is excellent. I was just curious to go a little deeper into the question; it is a peculiar characteristic of mine. I hope it doesn't bother you.

Ben Steffan

Glad it went well! :>
20min sounds rough

so little time

ephe

@BenSteffan Yeah the my first trial run lasted 40 minutes and I was barely introducing simplicial sets. I may record the full version I had in mind later on using school equipment.

Pizza

@SineoftheTime Anyway I think I solved that exercise on the circulation, I used this: $\oint_\Gamma \mathbf{v} \cdot \mathrm{d}\Gamma = \iint_S \big(\nabla\times\mathbf{v}\big)\cdot \mathrm{d}S$

Ben Steffan

@ephe I had two talks where I had to leave material on the floor because I overestimated my speed last term, so I feel ya :/

very embarassing to barely get to the statement my bachelor thesis was about and not get to discuss the proof or any consequences :(

Jakobian

3:12 PM

Semiring

In abstract algebra, a semiring is an algebraic structure. Semirings are a generalization of rings, dropping the requirement that each element must have an additive inverse. At the same time, semirings are a generalization of bounded distributive lattices. The smallest semiring that is not a ring is the two-element Boolean algebra, for instance with logical disjunction ∨ {\displaystyle \lor } as addition. A motivating example that is neither a ring nor a lattice is the set of natural numbers N...

Sine of the Time

@Pizza ok, did you have trouble choosing $S$?

Pizza

In my case $S = x^2 + y^2 \leq 1$ and $z=3$ so I used $\vec{r}(r,\theta) = (r \cos\theta, r \sin\theta , 3) \quad r\in[0,1], \theta \in [0,2\pi]$

The normal vector was perpendicular to the plane so $\vec{n} = (0,0,1)$

ephe

@BenSteffan Yeah that was going to happen to me too but the day before the presentation my friend told me that no one was going to digest most of the stuff here in 5 minutes so I might as well throw it out, which I did. I think this was a very poor choice of subject for a 20 minute presentation. I thought I had 30 minutes at the start of the program but still even that is pretty short. To me it seems like 20 minutes is only appropriate to present a single thing, like Yoneda's embedding.

Ben Steffan

short talks are very difficult

Pizza

So I'm left with $(...,...,4z) \cdot (0,0,1) = 4z = 4 \cdot 3 = 12$

$\iint_S 12 dS = 12 \cdot \pi = 12\pi$

Ben Steffan

3:19 PM

there's very little you can do in 20 minutes, I agree

Pizza

Then it told me the curve was traveled in an anti-clockwise direction. If it had been clockwise I would have had to change $-\vec{n} = (0,0,-1)$

Semiclassical

@Bml one is associated with a plus, one is associated with the minus. That’s all that was signaling. Maybe just “sign” is sufficient

Sine of the Time

I always got confused by the convention of the sign

@MatsGranvik no

:D

Bml

@Semiclassical OK. As for the steps leading to the two roots $y_1, y_2$, do I try to sketch a solution or do you want to write it down in your answer? If you want, we can analyze it together...

nickbros123

Say $K$ and $H$ are groups, and I have a surjective homomorphism $\lambda: H \to K$ and a surjective homomorphism $\gamma:K \to H$, how can I show that $K$ and $H $ are isomorphic?

Ben Steffan

3:35 PM

Are $K$ and $H$ finite, perchance?

nickbros123

not necessarily

Sine of the Time

do you have any assumption on the cardinality of the groups?

Ben Steffan

then my hunch is that it's false

nickbros123

damn

Sine of the Time

when there's infinity, there's trouble

nickbros123

3:37 PM

well that sucks

i thought that was the last step in my proof of 1st isomorphism theorem

Sine of the Time

like abstract algebra :D

nickbros123

well we go again

Ben Steffan

@nickbros123 it's definitely false: consider $H = \prod_{i = 1}^\infty \mathbb{Z}$ and $K = \mathbb{Z} / 2 \times \prod_{i = 1}^\infty \mathbb{Z}$

Soumik Mukherjee

@nickbros123 It's false, take Z_2×G and G where G is infinite product of Z_4s

Ben Steffan

same idea :)

Soumik Mukherjee

3:43 PM

:)

nickbros123

though I have a proof for the theorem $$G/ker(f) \cong range(f) $$ where $f$ is a group homomorphism, (by making use of the map $\tilde{f}:G/ker(f) \to range(f)$ defined by $\tilde(f)([x])=f(x)$). I also had another approach where the last step was: we have a surjective homm from $G/ker(f)$ to range$(f)$ and vice versa. That this "CSB" esque result does not hold brings great pain to me :'(

Sine of the Time

I actually don't know the standard proof

Ben Steffan

don't you just write down the quotient projection and check the kernel is trivial

nickbros123

@BenSteffan yeah, basically the most straightforward map we can think of ought to prove the isomorphism :) I was just trying a different (failed) approach though :)

Sine of the Time

we've proved it in class using and undelying result

Ryder Rude

3:48 PM

There's 19 people in the chat

nickbros123

the map $\tilde{f}([x])=f(x) $ from $G/ \ker(f)$ to $image(f)$ is a straightforward bijection

Sine of the Time

pro tip: you can use \ker instead of ker

nickbros123

@SoumikMukherjee @BenSteffan cool counterexample tho

Ryder Rude

@nickbros123 this is also related to the dim(V)=dim(ker)+dim(range) for vector spaces

Ben Steffan

@RyderRude you don't say :^)

nickbros123

3:52 PM

@SineoftheTime \ker works but \image, \range etc doesnt lol

Sine of the Time

\Im works but it sucks

nickbros123

$\Im(f)$

Ben Steffan

nobody writes $\operatorname{image}$

nickbros123

huh?

Ben Steffan

that's for "imaginary part"

common pitfall

Sine of the Time

3:53 PM

yep

Ben Steffan

compare $\Re f$

\Re f

you hear the voice of Xander Henderson echoing in your mind: "Don't use mathfrak"

Semiclassical

@Bml I plan to edit my answer eventually but it’ll be a few hours at least. So I do encourage trying to work out your own solution for now

Bml

@Semiclassical OK, thanks. However, in your answer there's another typo: you wrote $u-4/27$ in place of $4/27-u$, for $0 < u < 4/27$. I submitted an edit to your answer.

Semiclassical

@BenSteffan mathfrak is great, so long as you never have to write it by hand

Ben Steffan

I feel ok about mathfrak

but some people in this chat have... differing opinions :^)

Semiclassical

4:02 PM

@Bml sounds right, I waffled on the choice of sign when defining the function f(y)

Ben Steffan

for handwriting there's Sütterlin or Kurrent, but this is a German secret :^)

Thorgott

4:22 PM

@ephe happy to hear it went well

I agree 20m is very little time for this topic, but your slides seem nice

Xander Henderson

@BenSteffan Honestly, I am sort of willing to accept $\Im(z)$ and $\Re(z)$. I usually change those to $\operatorname{Im}$ and $\operatorname{Re}$ , but I don't hate people who use $\Im(z)$.

Ben Steffan

@XanderHenderson can't believe that, coming from you :^)

I really dislike $\Im z$ and $\Re z$

Xander Henderson

@BenSteffan I don't like them, but I find them tolerable.

Ben Steffan

it's so out of place in analysis

Ryder Rude

do u think category theory is an alternative to formal logic

ncatlab.org/nlab/show/computational+trilogy

Thorgott

4:36 PM

if the nlab says it, it is gospel

Ben Steffan

congratulations, you have found the most nlab article, perhaps ever

the HoTT revolution is right around the corner, just an indefinite number of days away at this point

Thorgott

one of the funniest things I've ever experienced was when a family friend (who is a lawyer, not a mathematician at all) told me he somehow bought a copy of the HoTT book and tried reading it (only to give up very quickly, of course)

Ryder Rude

oh

@BenSteffan lol

@Thorgott category theory reads like alien language :P

Ben Steffan

4:58 PM

@Thorgott I bought a copy when I was 18, it's still sitting on my shelf

@RyderRude this is not about category theory per se, it's about homotopy type theory

I've read only a few pages of it, because I reads like a text written for programmers, not for mathematicians (in a sense)

maps in the HoTT book have names like $\mathrm{transportconst}$

you can sell this to programmers I'm sure but miss me with it

Ryder Rude

oh. I thought it sounded really cool because it is connecting logic and topology. Is this theory bad/overhyped?

@BenSteffan so it is hyped up to bring a revolution

Ben Steffan

it's a newfangled theory of foundations people have been developing and trying to sell for a few years now

it's connecting logic and homotopy theory in particular, I think

the thing is, as far as I can tell nobody outside of the HoTT community (which isn't that large) seems too convinced by it

there may be applications to formal verification; there's agda-unimath which I believe uses HoTT

the thing is, as far as I know nobody is really using agda-unimath much either

the playing field belongs to lean, and perhaps coq

takes this with a grain of salt because neither foundations nor formal verification are really my thing

Ryder Rude

Oh :P thanks anyway..

people r always looking to bring revolution. It may be a good theory if they stop pushing it for foundations

i also prefer math language over programmer stuff

Ben Steffan

it's foundations by definition

if it wasn't foundations it would probably just be homotopy theory

Ryder Rude

i just liked the idea of bridging logic and homotopy

but i cant say really. Beyond that overview, idk literally anything about what their bridge is

i first thought they would bridge logic and manifolds, which sadly isnt the case :P

Thorgott

6:05 PM

an implication is like a homotopy and that's where my understanding ends

ephe

6:55 PM

@Thorgott Thanks!

Vladimir Lysikov

7:43 PM

If I remeber correctly, it's not implication.
It's equality of two terms is like a path between two points, and equivalence between two equality proofs is like a homotopy of paths.

Ben Steffan

right, and I think you carry this upwards so that an equality type $x = y$ becomes an $\infty$-groupoid, and there's the homotopy theory

and the category theory, if you want to call it that

caveat emptor

Ryder is not Rude.

:: warning: heavy vocabulary work ahead ::

Xander Henderson

@RyderisnotRude. NO! I REFUSE!

Ryder is not Rude.

I respect your choice.

🙏

Xander Henderson

7:59 PM

NAMASTE, YOU JERK!

(putting "heavy vocab" in the chat... some people...)

Ben Steffan

should've said anima instead

absolutely spectacular choice of terminology by Clausen & Scholze there

bringing the fields of topology and proctology closer together :^)

Ryder is not Rude.

😆LoL 😂

psie

I have a basic question. Can you give an example of a function such that $\int f<\infty$, but $\int f^+$ or $\int f^-$ is not finite? I'm repeating some stuff on the notion of integrability (both $\int f^+$ and $\int f^-$ finite) and why its equivalent to $\int |f|<\infty$.

Ok, maybe this is silly. Any odd function would probably do.

The reason I was asking is because I do not quite understand why we neglect the condition $\int f<\infty$.

But I think we'd have problems with all the odd functions then.

leslie townes

8:16 PM

psie you need to say more about what you mean by 'integrable' here. this is where the choice of that notion begins to matter

in the lebesgue sense int f being finite, int |f| being finite, and int f^+ and int f^- being finite are all equivalent to what one usually has in mind by saying 'integrable'

the examples you're maybe thinking about where you get finiteness out of things canceling out in some kind of limit procedure are not lebesgue integrable

and the things you're computing when you compute those limits or whatever are not lebesgue integrals

psie

ah you're right

leslie townes

i think it's different in henstock land, or one of those other lands

or just certain 'improper' riemann integrals can be like that

i think folland explains this pretty well if you have access to that

Ryder is not Rude.

^

leslie townes

maybe a core example to keep in mind is if you do the usual measure theory treatment to positive integers with counting measure the notion of 'integrability' of a series you get out of that corresponds to absolute convergence, and not the convergence you might see in some limiting procedure that takes the ordering of N into account

psie

@leslietownes well, Folland says that we define the integral of a real-valued measurable function $f$ by $$\int f=\int f^+-\int f^-,$$provided one of $\int f^+$ or $\int f^-$ is finite (this makes sense). However, then he just states that $f$ is integrable if both $\int f^+$ and $\int f^-$ are finite and calls this integrable. He says this is equivalent to $\int |f|<\infty$, which also makes sense (I understand the equivalence).

leslie townes

8:27 PM

oh yeah, i forget sometimes he's fine with integrals being +oo or -oo

psie

What I don't understand is why $\int f<\infty$ is not enough to say something is integrable?

leslie townes

which does make sense in a lot of 'nonnegative' applications

Ben Steffan

@psie huh?

leslie townes

psie: he's willing e.g. to have int f take the value -oo, and that would certainly satisfy int f < infty and yet not meet his definition of 'integrable'

the key thing to remember is just that in this context 'integrable' is just frequently given this more restrictive meaning than 'the recipe defining the integral makes some kind of sense'

Ben Steffan

$\int f$ is the sum of two terms. If one of the summands is not finite $\int f$ won't be

leslie townes

8:32 PM

analysis books that use extended real numbers generally often have little hypotheses that allow for infinite integrals (or even the function itself taking infinite values) in at most one of the two infinities but not both, precisely to avoid having their hypotheses collide in odd ways with the usual conventions of the extended reals

e.g. "a < infty" maybe not being equivalent to "a is finite"

psie

ok, I have to meditate on this one

leslie townes

psie the key being -infty < +infty as it is usually set up, and he wants to reserve 'integrable' for finite things, not just ones less than +infty

psie

ok 👍

leslie townes

wow, rudin has a rare comment about a related issue "This terminology [about 'integrable'] may be a little confusing, if [int_E f+ - int_E f-] is +infty or -infty, then the integral of f over E is defined, although f is not integrable in the above sense of the word"

in his wonderful chapter 11 of PMA

Xander Henderson

"Wonderful"...

leslie townes

8:39 PM

rare moment when rudin somehow uses ink to acknowledge that something may be a little confusing

and yeah, just fulfilling our chat's contractual obligation for somebody to slam that part of the book today

Buddhini Angelika

Hi. When we speak about a module's aim and module's objectives is there a difference in Mathematics modules? I mean in Google the word "objectives" appear as a synonym for the word "aim".But some other descriptions say there's a difference in the two ...

Does it matter like that in Mathematics?

leslie townes

that kind of word choice might have some meaning within the context of a specific course of study at some school. there is no useful general distinction between the words that i am aware of

Buddhini Angelika

Hmm okay @leslietownes

There's no useful general distinction means we can use either word for the same place right? I just asked to double check whether I understood the exact meaning you meant?

leslie townes

i do not know of any general difference between those terms. if someone is using that word choice as if it matters, i would ask them specifically if they mean anything by it

Xander Henderson

@leslietownes Shocking. Rudin hordes ink as if it were a precious metal.

leslie townes

8:49 PM

i've worked in schools where sometimes for bureaucratic reasons a course would have a list of things that describe what the student is expected to know at the end of the class, and unlike most other course materials, this list was not something i was involved in preparing or any individual instructor had any say over. it was something the department used to track the course and talk about it with other departments.

Buddhini Angelika

Okay, thank you very much @leslietownes :) :)

leslie townes

i could see using a funny word to distinguish that particular list from anything that i had prepared, but i don't think there's a general choice that Everyone In Mathematics (tm) would make for this kind of purpose

Xander Henderson

@BuddhiniAngelika As words in vernacular English, they are indistinguishable. But in an official setting, there may be nuances that are important.

Ryder is not Rude.

"Objective" is specific.

Buddhini Angelika

@leslietownes okayy :)

@XanderHenderson okay thank you very much @XanderHenderson :)

Ryder is not Rude.

8:53 PM

"Aim", "Goal", etc are more general.

Buddhini Angelika

@RyderisnotRude. Hmm, do we use it in Mathematics too in that way?

Ryder is not Rude.

Depends on context.

Xander Henderson

I would understand the "aim" of a course to be a one or two sentence describing the overall goal of the course, eg "This course aims to introduce students to the basics of limits and differential calculus," while "objectives" are specific, testable things that students are supposed to know by the end of the course, eg "Students will be able to compute the derivative of a differentiable function."

This isn't about how the words are used in math; it is about how they are used in education.

"Objective" is a bit of education jargon.

Ryder is not Rude.

The object of the objective is what you will learn.

Buddhini Angelika

Thank you very much @RyderisnotRude. :)

Ryder is not Rude.

8:57 PM

np

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$Mathematics$ Mathematics