Mathematics - 2021-08-18 (page 1 of 2)

Thorgott

00:00

well, constructing $\mathbb{Z}$ from $\mathbb{N}$ as a set is something that doesn't hurt doing once, though I'm not sure what kind of convoluted things you're doing

shintuku

i was doing this: pastebin.com/02h30nWG

Thorgott

oh lord

Derivative

00:25

@TedShifrin no, there was also talk about bureaucratic stuff

But like he was talking about fields like we know what they were, even though he didn't explain it

Thorgott

00:45

well, that's something that should be explained

Xander Henderson

@Derivative Fields are great.

Well, most of them.

I'm not really interested in soy fields. But corn and wheat are cool!

shintuku

as an econ student i can confirm

in sharp contrast, however, i think soy fields are remarkably interesting

I'm an alien Im an eagle alien

Soy berry fields forever

I prefer strawberries to soy beans myself. Imagine a world where there are no religions, countries, it's easy if you try.

^_^

I'm an alien Im an eagle alien

01:48

Coldplay - Adventure Of A Lifetime (Official Video)

►

I'm an alien Im an eagle alien

02:01

If you have an infnite intersection $\bigcap_{a,b,c \in A^3 \\ a,b,c \text{ distinct}} (U_a \cup U_b \cup U_c)$ then what can we conclude?

I'm an alien Im an eagle alien

02:18

-1

Q: If we have an infinite intersection $\bigcap_{\text{distinct } a,b,c \in X} (M_a \cup M_b \cup M_c)$, then how can we compute that as a union?

Let everything be in or a subset of $\Bbb{Z}$ through the discussion. Consider the problem: $$ \bigcap_{\text{distinct } a,b,c \in X} (M_a \cup M_b \cup M_c) $$ Where $X$ is an infinite set and $M_i$ is always a multiplicative submonoid of $\Bbb{Z}$ such that $\bigcap_{i\geq 1} M_{a_i} = \{1\}$ w...

I'm an alien Im an eagle alien

02:45

Thanks for the upvote, whoever did that

EricWofsey helped make it an okay post

Koro

02:59

Let C be the Cantor set and let f be a function continuous on $[0,1]\setminus C$. Then, is $f$ Riemann integrable on $[0,1]$?

Let $\epsilon \gt 0$ be given. I know that the Cantor set can’t contain any segment in it. But I have no idea how to proceed

I can’t take C as partition as partition by definition is a finite set.

robjohn

04:27

The complement of the Cantor set is a countable union of disjoint open intervals. The function is integrable on each of those intervals.

Koro

04:37

hmm

C is compact hence closed so $[0,1]\setminus C$ is open

in $[0,1]$

but that doesn't necessarily mean open in $\mathbb R$

but in this case the complement is open in $R$ also

as the construction of the Cantor set shows

$C$ is compact in $\mathbb R\implies C$ is closed in $\mathbb R\implies \mathbb R\setminus C$ is open in $\mathbb R$ and we know that every open set in $\mathbb R$ is a countable union of disjoint open intervals.

Semiclassical

comparing this question with the Chegg version (see link in comments) fills me with some sardonic amusement

like, if you're going to post two-third of a question text as an image, you might as well go the whole way

Koro

Hence $\mathbb R\setminus C$ is a countable collection of open intervals. $[0,1]-C\subset \mathbb R-C$ so yes $[0,1]\setminus C$ is a countable union of open intervals in $\mathbb R$.

hmm this explanation is wrong.

no this explanation is not required. simply by the construction of the Cantor set and using the fact that any union of open sets is open, we get the result that complement of $C$ in $[0,1]$ is a countable union of open intervals.

pritchard

05:00

Hi, I want to check, is it true that $f(V)\subseteq U\implies V\subseteq f^{-1}(U)$? where $f$ is any function.I believe it is true but want to confirm. Thank you!

Euler 2

greetings fellow tickians

why don't people simply use overmorrow

rather than day after tomorrow

lol

Semiclassical

@pritchard this assumes $f^{-1}$ exists in the first place, which isn't true for all functions

Koro

Hi semiclassical

Semiclassical

o/

Koro

I think they mean $f^{-1}$ in the sense of inverse image

Semiclassical

05:12

hmm

Koro

which exists for every function.

Semiclassical

fair

Koro

Do you have any suggestions for my integration question?

Semiclassical

nope

Koro

:(

I'm trying to follow the hint given in the book. :)

Nano Miratus

05:16

hi guys, i was wondering about a question, but i'm unsure if i'm allowed to post it on math.stackexchange

so, what i was wondering about, are there problems in mathematics which we know are solvable, but we just can't solve it at this time because of computational limitations?

Semiclassical

hmm

so something like: there's a finite amount of cases to test, but far too many to be computationally viable?

the problem is that this seems pretty broad

Nano Miratus

what i mean with that is, whenever i learn about some new unsolved problem, it's either logically unsolvable, or unsolved cause we are mathematically not ready for it (collatz conjecture), or unsolvable because the finite cases to test it are more than all atoms in the universe times 100

Semiclassical

well, that last case sounds like what you're talkingn about

Nano Miratus

@Semiclassical yeah, exactly...but specifically, is there a problem, where we have like idk, this many trillion cases to test, and our current supercomputers are not capable of doing that, but we know for a fact or we at least pretty sure that supercomputers in 10 years will be

Semiclassical

well

Nano Miratus

05:22

@Semiclassical i know, but i mean, is there an in between?

Semiclassical

the problem is that it's not so much a matter of "can it be done" but "how long"

like, if something would take 100 years of computer time to do

Nano Miratus

true...i have to point out i'm not a mathematician in any way, i'm just fascinated and try to learn stuff...so i would be equally satisfied with the answer: "no, such problems can't even exist, there is no in-between" because of some logical fact i'm not aware of

Semiclassical

if we take moore's law as a guide, then in 2 years that'll take 50 years, in 4 years it'd take 25, etc

Nano Miratus

@Semiclassical yeah

Semiclassical

i doubt it. practically the distinction is between problems people are willing to wait long enough to do, and those which they aren't

Nano Miratus

05:26

hm okay...then, what would be current problem people aren't willing to wait long enough?

Semiclassical

yeah

Nano Miratus

*a current problem

Koro

Let $\displaystyle C$ be the cantor set. Let $\displaystyle \epsilon >0$ be arbitrary.

Let $\displaystyle V_{\delta _{c}}( c)$ be an open interval around $\displaystyle c\in C$, where $\displaystyle \delta _{c} >0$ is chosen so small that $\displaystyle V_{\delta _{c}}( c) \cap (\mathbb{R} -C) =\emptyset $ and $\displaystyle \delta _{c} < \frac{\epsilon }{2}$.

Now $\displaystyle \cup _{c\in C} V_{\delta _{c}}( c)$ is an open cover of $\displaystyle C$ and \ must have a finite subcover as $\displaystyle C$ is compact.

2 hours ago, by Koro

Let C be the Cantor set and let f be a function continuous on $[0,1]\setminus C$. Then, is $f$ Riemann integrable on $[0,1]$?

Semiclassical

there is a tricky aspect there, of course. if you had a 1000 computers running in parallel on a problem, then that'd also change your answer

one obvious category of answer, though, which probably doesn't fit what you're looking for

Koro

Hint in the book says: Since $C$ is compact it can be covered by finitely many intervals whose sum of lengths is arbitrarily small. I think that's not possible.

because as shown above $n$ depends upon $\epsilon$

@robjohn

Nano Miratus

05:29

yeah, not really, but i'm fine if what i'm looking for doesn't exist lol

do you think i should even ask this question on this site or nah? don't wanna spam and if there's a better place, i'd like to know

Semiclassical

personally i wouldn't mind it, but i'm not sure this is the best SE for it

there's a bunch out there

the fact that it's directed towards mathematical instances of such, though, would help

the broad example I had in mind, btw, was cryptography

given a message encoded using, say, RSA with a 1000-bit key, is it practically possible to break it?

Nano Miratus

i'm not really familiar with any other stackexchange sites than the stackoverflow, superuser, math, and mathematica ones tbh :D

Semiclassical

fair

starting here seems reasonable

(looking up, the minimum standard for RSA is now 2048-bit keys. the question of "how long does it take to break it" has practical consequences for that)

Nano Miratus

alright, i'll try...any idea how i could phrase my question more concise or pack it all in one or two sentences? i usually am able to get my point across understandable but it often becomes a unnecessarily long post due to me not being a native english speaker

Semiclassical

hmmmm

Nano Miratus

05:37

and i wouldn't ask this if this wasn't exactly the critisism i received on my last question on here :D

"You should be able to ask this question in 1/4 the words. It will encourage people to help you." lol

they're not wrong

Semiclassical

lol

hmm

Nano Miratus

also, the research on this is pretty tough, i can only find countless lists of "unsolved" problems, but i just want the unsolvable ones :P

Semiclassical

yeah

Koro

one such problem could be: irrationality or rationality of $\pi+e$

Semiclassical

@Koro ehhh. that's not one where 'computer time' is currently relevant

which is the underlying context here

Koro

05:43

not sure if that comes under unsolvable one :)

okay @semi :)

Semiclassical

the phrase that comes to mind is "computationally inaccessible"

unfortunately trying to look up 'computationally hard problems' usually gets more into information about computational complexity

Nano Miratus

@Semiclassical yeah, that's mainly what i'm looking for, or more broadly "physically impossible (for now)"

Semiclassical

right

Nano Miratus

@Semiclassical yeahhh all that stuff about np=p ughhh

Semiclassical

lol

basically you're wanting to address what problems are currently hovering on the barrier of being computationally accessible

Nano Miratus

05:46

yep...i want to have the same feeling that i have following astronomy topics

basically xD

Semiclassical

the sort of thing where 10 years of computing advances alone could transition it from "would take a decade to do it" to "would take a year to do it"

robjohn

@Koro ??

Koro

that's clear to me now.

i think what i was missing was the sum of intervals in Cantor set approaches $0$

Ted Shifrin

@Koro Are you remembering that $C$ has measure $0$?

Nano Miratus

@Semiclassical yeah...i mean, there are examples of this in the past, aren't there?

robjohn

05:48

@Koro ??

Koro

To be more precise: $C=\cap E_n$ then length of $E_n$ is $(\frac 23)^n$

Semiclassical

presumably, yes

the difficulty is avoiding it being too broad

Koro

@robjohn: I am satisfied with this explanation:

58 mins ago, by Koro

no this explanation is not required. simply by the construction of the Cantor set and using the fact that any union of open sets is open, we get the result that complement of $C$ in $[0,1]$ is a countable union of open intervals.

Semiclassical

like, do you regard "breaking an encoded message with key length d" as a mathematical problem?

certainly it's a computational problem

but it feels too particular

Koro

Ted, I haven't yet been familiar with the term measure in the sense of measure theory. But I suppose you're referring to the fact that complement of $C$ is of length $1$.

Semiclassical

05:51

or you could frame some problems as "Is there a counterexample to this conjecture for any integer up to N"

robjohn

@Koro The Cantor set is closed, so its complement is open. Any open set in $\mathbb{R}$ is a countable union of disjoint open intervals.

Semiclassical

in which case progress amounts to showing that the answer is no, for larger and larger values of N

but again that doesn't seem like the right spirit

Koro

@robjohn yes i know that but got confused in whether the complement is open in R or in [0,1]

Nano Miratus

@Semiclassical well, i have another example in my mind, i'm not really sure what it was exactly, but i watched a video about a paper where the authors just tried and printed all hundreds of thousands of configurations of a specific graph to "prove" a conjecture...sure they didn't prove it the traditional way, but they showed the truth of it and sure, it's still a mystery on why exactly but yeah

Semiclassical

probably the 4-color theorem

in which case the main setup for the computer part of that was proving that there were "only" 1482 diagrams to check

Nano Miratus

05:54

yeah something like that...i think there are multiple problems that were just "solved" this way, and then, some even years later, "proven" in the traditional sense

Semiclassical

i wonder, hmm

in the case of the 4-color theorem, that's still basically the state of things. the main advances have been reducing the number of diagrams to 633 cases

Nano Miratus

idk the exact limits of supercomputers, but there just have to be some crazy multidimensional graph theory/geometry problems which are just too much for anything we have right now

@Semiclassical oh wow

Semiclassical

so maybe something in the realm of computer-assisted proofs

e.g., what are some open problems which admit computer-assisted proofs but currently require too much power

Nano Miratus

ughhh i'm trying to think of anything other than the 4-color theorem, anything where you can just say "alright let's try it in the 15th dimension" and it stays the same problem just with more iterations needed to solve it

Semiclassical

right

i wonder if there's other versions of the four-color theorem in that vein. say, on surfaces other than a plane or stronger notions of 'coloring'

Nano Miratus

06:01

i was also just thinking about that sums of three cubes "problem", but that's not really it, because either we know that all integers can be expressed this way or we don't, knowing that all < 100 are possible is not really the core problem

Semiclassical

right

Nano Miratus

@Semiclassical i'll look it up...but some of these problems are tricky in a way that they just exist in the second dimension or they just exist with pentagons or idk

Semiclassical

"is it true for all numbers up to 10^1000" is a mathematical problem to be sure, but it seems artificial

Nano Miratus

this en.wikipedia.org/wiki/Four_color_theorem#Generalizations talks about three-dimensional solid regions for example

but it seems we're limited mathematically here

alright imma go read this paper and understand like 10% of it, brb :D

discuss.wmie.uz.zgora.pl/php/…

Semiclassical

good luck

Nano Miratus

06:11

"If rooms in an office building are allowed to be any rectangular
solid, can we paint any configuration of rooms with a bounded number of
colors so that no two rooms sharing a wall or ceiling/floor get the same
color?
It turns out that the answer to this question is no"

wait what

also, the rest of that paper is to complex for me, i can't follow sets and set related symbols

oh, "bounded" just means "not as many regions there are", i get it

Nano Miratus

06:36

i mean, i guess the sudoku problems are something like that

seems like actually finding out what the maximum number of clues in a minimal sudoku is, is at least in the ballpark of super computers of the next decade...or maybe the current supercomputers are already capable of finding that out but they don't waste their time on stupid sudoku stuff :D

user400188

@NanoMiratus I am a bit confused by this. Cannot we just have alternating black and white rooms? Or is it that sharing an edge counts as sharing a floor/ceiling?

Nano Miratus

i was confused by this too but "bounded" in this context refers to the 4-color theorem, where you can say: it doesn't even **matter** how many regions, we only ever need 4 colors, period.

but it seems in three dimensions, they have proven that the number of colors is always at least equal to the number of regions - which kind of defeats the whole question, of course you can color every region with a different new color

@user400188 oh wait

i guess the emphasis is on can we paint any configuration?

now i'm confused again

but if you understand more than i do about this, have fun: discuss.wmie.uz.zgora.pl/php/…

i mean, it's saying "wall", "ceiling" and "floor", so i don't think edges touching matter

i think i got it: imagine a 3d chess pattern, just cubes of white and black...take one of those cubes, delete it, and stretch any adjecent cube to a rectangular solid to fill the space, does the number of colors still suffice?

and the answer is no, apparently...in two dimensions the answer is yes, but no matter how big, how many regions and how they are configured, you'll always need 4 colors max

user400188

07:06

@NanoMiratus I think the answer is still yes, you can do it with 54 colours. Label rooms in 3 by 3 grids as 1 to 9, and tile these with rooms labeled 10 to 18 in a plane (of cubes that is). Then produce a new plane above the first such that 18 has been added to ever number bellow it. Do it again and get a plane with 36 added to every number relative to the first plane. Above this one begin from the start (1 to 9 and 10 to 18) again.

Nano Miratus

yeah, but let's suppose i take one of those rooms out, and fill any of the free space with any of the other regions by stretching them

user400188

@NanoMiratus If we consider the first plane we made as the xy plane, and the z plane as up and down, then drilling downwards by one from the original will always put you in a plane where every number is a quantity above what was in the original plane, so it cannot be a duplicate. If we drill up, the same thing occurs. As for the matches on the original plane, you can just draw it and it will work by inspection.

Nano Miratus

do i still need just 54 colors?

well okay i need to visualize this somehow because i don't think i really get what you mean

user400188

@NanoMiratus This is the first plane. The one above it is the same but 18 has been added to every number. The one above that is the same but 36 is added to every number in the first plane. The one above this is the first plane again. The one bellow the first plane also has 36 added to everything in the first.

There can be no matches internal to a tile (a tile is a plane of numbers 1 to 9) once you stretch a cell in it by 1 space, because they are all different numbers. There can be no matches between neighbouring tiles because the set of numbers in one has no intersection with the one next to it, above it, or bellow it.

Nano Miratus

yeah but this is a specific configuration, isn't it?

user400188

07:19

Wasn't that the original question?

If it has to work for any configuration, then the answer is trivial, because if the number of colours was one less than the number of rooms, then you could have a configuration where two rooms sharing the same colour are touching to begin with.

Nano Miratus

the original question of the 4-color theorem at least, is, in any given scenario, how many buckets of paints do i need to bring with me, to be totally sure

and the fascinating thing is, only 4

user400188

@NanoMiratus I thought we were solving a 3D version, where every shape is also a cube, where the answer reported in a paper was supposed to be no.

Nano Miratus

i agree, the quote i used, and the whole paper is a bit ambigous about that in my opinion

user400188

Tbh, I just realised the problem I was trying to solve can be done with just 12 colours. Using the same method, and making the original 2D tiling using 4 colours.

Nano Miratus

well, the problem you are trying to solve can be "solved" using only 1 color

and 1 cube

you know

or 2 cubes and 2 colors

or not?

user400188

07:28

Not with deleting a cube and stretching another into its space.

@NanoMiratus I'm not sure if that is part of the original paper, I am just going off a quote you mentioned in the chat.

Nano Miratus

yeah sorry, the quote was a bit misleading i guess

user400188

In any case, I think I have found the problem. Rectangular solid, doesn't mean cube. It means... any rectangular solid.

Nano Miratus

yeah xD

still fascinating that there's no bounded number

it's just weird how this is a whole can of worms in 2 dimensions, and in 3 dimensions it all falls apart right at the start

user400188

I think it might still be easy to solve: Consider, for arguments sake, that the particular configuration is still cubes.
We know that for the case of cubes, it can be done with a finite number of colours. Call it X.
Then, consider a different configuration where just one of the offices is some finite number of multiples long compared to the rest, such that it intersects with another copy of itself.
This requires at least one more colour. We now need X plus 1 colours.

But then we could stretch the same rectangular solid again, and need one more colour. Ad infinitum.

Nano Miratus

does that explain intuitively why the number of colors is not bounded?

user400188

07:43

I don't think it needs to be intuitive. You could flesh that out into a proof pretty easily.

Nano Miratus

you can say roughly the same in 2 dimensions, and sure, you can always add one more color

but you don't need to, whatever you change, you can always somehow change the distribution of colors so you only need 4

user400188

If you can say it in 2D then I have made a mistake.

Nano Miratus

Consider, for arguments sake, that the particular configuration is still squares.
We know that for the case of squares, it can be done with a finite number of colours. Call it X.
Then, consider a different configuration where just one of the offices is some finite number of multiples long compared to the rest, such that it intersects with another copy of itself.
This requires at least one more colour. We now need X plus 1 colours. (which is false in 2d, we don't need 1 more color)

But then we could stretch the same rectangle again, and need one more colour. Ad infinitum. (no, we will never…

see what i mean?

user400188

Yes I noticed when you wrote "you can say roughly the same in 2 dimensions".

I forgot that you can just redistribute the same colours.

Nano Miratus

yeah that's the fascinating thing about 4-color in 2d...lets say you remove and stretch, it's mostly not at all obvious how on earth the colors need to be redistributed, but it can be done

so the statement "This requires at least one more colour" is a bold one

user400188

07:49

@NanoMiratus It wasn't bold, it was just wrong.

Nano Miratus

in some configurations you might not and in some configs this could even reduce the number of colors

íts just weird overall

user400188

Wait no, it wasn't wrong.

The mistake was actually "such that it intersects another copy of itself".

Nano Miratus

what? :D

that was fine in my opinion, just another way to look at it

user400188

Again I forgot that you can colour two identical shapes differently. Forget what I sad haha.

PM 2Ring

10:09

@NanoMiratus There's some good info about intractable problems on crypto.stackexchange.com Although in crypto, we want problems that are easy to solve if you have the key but require a brute-force search if you don't. And we want to be fairly confident that our analysis of the problem is correct. Otherwise, there could be a back-door solution that doesn't require a full brute-force search.

robjohn

@PM2Ring: $x\,\frac{1-\frac{53}{396}x^2+\frac{551}{166320}x^4}{1+\frac{13}{396}x^2+\frac{75}{166320}x^4}$ is extremely close to $\sin(x)$. At $\frac\pi2$, it is $1.000003$. Even at $\pi$, it is $0.004$.

PM 2Ring

A fun geometrical problem came up on Puzzling.SE a few months ago. You can arrange 6 identical cylinders (of sufficient length) so that each cylinder touches the other 5 (they have to touch on the curved surfaces). Solutions are easy to find by trial and error.

It was speculated that it's possible to do it with 7 cylinders, but it took 45 years before solutions were found. They searched 80 million paths in a space of ~121 billion paths, and found 2 solutions. See puzzling.stackexchange.com/q/109154/36040

@robjohn Nice!

robjohn

@PM2Ring when the cylinders are sliced, does that mean you can do this with 6 identical circles, or does it have to do with the extra dimension?

that is, the axes of the cylinders are not all parallel.

PM 2Ring

@robjohn Yes, it involves the extra dimension. There are diagrams on the Puzzling page, including an interactive one linked in my answer.

The cylinders are definitely not parallel. They're very skewed.

I suspect a 4D version of this would be even harder. But even in 3D, we don't know if 7 is the limit. Larger solutions might be possible. There might be a proof that it's impossible for some number n, but n>7.

robjohn

11:11

@PM2Ring: here is a comparison of several Padé approximants for $\sin(x)$

schn

@robjohn , thanks for the reply. Binding and scoping -- do you mean this as used in programming sort of?

I am familiar with the word binding appearing in binding constraints.

And the word scoping, or rather scope, as used in the first paragraph here.

PM 2Ring

11:34

@robjohn Pretty. They're all pretty good. And really, for practical calculation (eg, in a math library), you just need sin & cos on [0, pi/4]. In his implementation of the Standard C Library, P. J. Plauger uses a pair of 7th degree Chebyshev polys. He says they're adequate for IEEE 754 doubles, as well as a few other architectures requiring slightly more than 53 bits.

But he uses Padé (or Padé-like) expressions for most other trig functions. (I have a paper copy of the book, although I rarely write C these days).

schn

13:16

I am studying this article on Wikipedia about free and bound variables, in particular variable-binding operators. Does differentiation bind a variable to a set? Considering the definition of the derivative as a limit, it seems like differentiation should only bind the variable the limit is taken with respect to. I am commenting on the notation $D(x\mapsto x^2+2x+1)$ in the article.

robjohn

13:44

@schn in $\sum\limits_{n=1}^\infty\frac1{n^2}$, $n$ is bound to the sum. That is, outside the sum, it makes no sense to refer to $n$.

@schn in $\lim\limits_{h\to0}\frac{f(x+h)-f(x)}h$, $h$ is bound to the limit. $h$ is not mentioned outside this limit, so nothing is bound in $\frac{\mathrm{d}f}{\mathrm{d}x}$.

Flows.

14:06

When someone says a diffusion say $dX_t=b(X_t) dt + \sigma(X_t) dW_t$ is elliptic what do they mean exactly?

schn

@robjohn , right, so in the article the notation $D(x\mapsto x^2+2x+1)$ should apparently make the binding of the differentiation operator explicit, but there does not seem to be any binding of the variable that is being differentiated with respect to.

schn

14:30

Maybe $D(x\mapsto x^2+2x+1)$ does indeed make the binding explicit, but it would probably be even clearer to just write the definition of the derivative as a limit.

copper.hat

The operator $D$ operates on differentiable functions. $x \mapsto \text{blah}$ is just the function.

robjohn

@schn $x\mapsto x^2+2x+1$ is the function $f(x)=(x+1)^2$. $x$ is bound to the definition, but a liberty is taken to use that as the variable on which the derivative acts (context). The result is $2x+2$, and the $x$ is not bound, as that has meaning outside of the derivative.

schn

15:00

Thank you for clarifying this.

schn

15:20

Now, assuming this is what you meant with binding, @robjohn, how would you describe scoping?

robjohn

@schn scope is the extent of where a variable makes sense. If it is bound, then the scope would be inside the binding structure. If it is free, then the scope would be everywhere.

schn

Clear.

robjohn

it is the same in math as in comp sci

Shiranai

15:40

Any ideas how to find all $n \in \mathbb N$ for which $x^{2n}+x^n+1$ is divisible by $x^2+x+1$?

Koro

15:52

does $\int_0^1 \frac 1y |\sin \frac 1y| \,dy$ exist?

Actually I am looking for an example such that $\lim_{c\to 0}\int_c^1 f(x)\,dx$ exists but not for $\lim_{c\to 0}\int_c^1 |f(x)|\,dx$

leslie townes

16:33

interesting. i think you have the right idea, you want there to be positive-negative cancellation in f(x) that goes away when you slap on absolute values.

Koro

@leslie, this is an exercise problem in Rudin's book

leslie townes

in PMA? i must have a solution written up somewhere then.

it felt familiar.

Koro

yes, chapter 6 problem number 7 part b)

leslie townes

i'll look. i did most of chapters 1-8.

Koro

I posted solution to part a) of the exercise here today math.stackexchange.com/questions/4227355/…

cohomonoid

16:38

Hey can someone give a ruling on imbedding vs embedding? In the context of e.g. Sobolev spaces. Is it just preference?

I can't figure out how to do an n-gram on the arXiv corpus

leslie townes

i'd go with the e.

i was not a differential equations guy but saw e far more than i in functional analysis.

cohomonoid

thx that's how I was leaning, now i'm set for life

leslie townes

one thing i both like and dislike about math is that there is so much freedom to, uh, innovate, both in spelling and grammar, because it is an international field. this doesn't mean that the quality of the writing is bad, just that rigid rules are not in play.

i had a huge fight with a journal editor in a non math context about the use of the term 'data' in the singular. she had some stylebook that said 'data' was plural. i said, maybe somewhere it is, but here it just isn't. i lost.

Koro

datum

cohomonoid

I would have guess non-math academia was worse tbh

Koro

16:43

but nobody says datum even if they mean singular. I have always heard data

lone student

Hello. Can any software solve this problem? x^2+yz=9, y^2+xz=16, z^2+xy=25 find xy+yz+xz.

leslie townes

it's a lot of stylebooks and rigid rules.

cohomonoid

I'm still kind of torn about data/datum but I've accepted that I'm losing and should just get used to it

leslie townes

data is almost universally used in the singular in math, and it was a math context. it's like rice. you don't say the rice are cooked.

cohomonoid

related: xkcd 2039

leslie townes

16:44

or maybe you do if your stylebook tells you to.

cohomonoid

but rice is a mass noun, there are no rices

Koro

it feels bad when i can't find counterexamples :(

cohomonoid

data has become one but wasn't always

leslie townes

data kind of is too. the boundary data in the PDE context is not an aggregation of individual pieces of datums. it's one thing.

Koro

this one for example:

49 mins ago, by Koro

Actually I am looking for an example such that $\lim_{c\to 0}\int_c^1 f(x)\,dx$ exists but not for $\lim_{c\to 0}\int_c^1 |f(x)|\,dx$

cohomonoid

16:47

@leslietownes I agree I think its a mass noun to everyone now except for certain aging prescriptivist reactionaries

leslie townes

i don't feel too strongly about this but it was annoying. i do like how in math editors generally do not give edits at the line level, unless it is substantive. in other fields you sometimes get people who might be trained only as copy editors fiddling with word order, word choices, and things that are none of their business.

Koro

and what about "news"

that's always singular i think

cohomonoid

OMG have you heard the new

yeah I see what you're saying. Sort of like pant(s)

Koro

and scissors

leslie townes

xkcd 2039 is very on point, that made me laugh. thank you.

Koro

16:54

Leslie: I have one small question

0

Q: Showing that if $f$ is Riemann integrable on $[0,1]$ then it is true that $\int_0^1 f(x) \,dx=\lim_{c\to 0}\int_c^1 f(x)\, dx$

I want to prove that if $f$ is Riemann integrable on $[0,1]$ (i.e. $f\in\mathscr R$ on $[0,1]$) then it is true that $\int_0^1 f(x) \,dx=\lim_{c\to 0}\int_c^1 f(x)\, dx$. It is known to me that since $f\in\mathscr R$ on $[0,1]$ then Riemann integrals $\int_c^1f(x)\,dx$ and $\int_0^c f(x)\,dx$ bo...

real-analysis solution-verification riemann-integration

In the statement of this exercise in Rudin's PMA, it was mentioned for all $c\gt 0$

Shouldn't that have been "for all $c\in (0,1)$" instead?

Wolgwang

Doesn't the answer depend on the direction of rotation?

> A line has intercepts a and $\mathrm{b}$ on the coordinate axes. If keeping the origin fixed, the coordinate axes are rotated through 90$^\circ$ , the same line has intercepts $\mathrm{p}$ and $\mathrm{q}$, then

A $p=a, q=b$
B) $p=b, q=a$
C) $p=-b, q=-a$
D) $p=b, q=-a$

cohomonoid

its D) i think

because only one can change sign

but it could also be p=-b, q=a

because of what you said

Wolgwang

Thanks :-)

robjohn

17:17

I just tried to request the desktop site from my mobile phone and it would not go. Can anyone get the desktop site on a mobile device? It used to work a week or two ago.

Ted Shifrin

Works fine on my iPad, @robjohn.

Maybe clear cache, however one does that.

robjohn

Crap. It must be my phone’s browser.

cohomonoid

works fine on chrome on adroid

android 9 IIRC

Ted Shifrin

Time to reinstall, reboot.

Koro

@robjohn works on chrome on android, safari on ios mobile browsers

cohomonoid

17:21

though that's chrome 88 because google has just been making it worse and worse

robjohn

No, it works in chat. Let me try again on the main site.

cohomonoid

main site also works for me

leslie townes

it's working on chrome on my super modern samsung s5.

Ted Shifrin

Ultra-modern!

leslie townes

i had an IT problem at work last week and he asked what phone i had and did not believe me when i told him.

he said "normally people have newer phones."

Ted Shifrin

17:24

“Well, I’m no normal person.”

leslie townes

yeah. i think of the smartphone as a mechanism of control, so i avoid using it except for pictures of my cat and my daughter. the office put all this spyware on it. if i had a newer phone they'd probably watch me on a video stream.

cohomonoid

:thumbs: f the man

Ted Shifrin

If they want to put software on it, they’d better buy you an office cellphone.

cohomonoid

also s5 has replaceable battery! infinite lifetime

leslie townes

yes. i've replaced it only once.

ted we get 50 a month added to our paychecks to cover the cost of cell service. i think my bill is about $60, but whatever.

cohomonoid

17:27

you should pass it on to your kids, like this here s5 was my mothers phone and her mothers before her...

Ted Shifrin

Hmm …

leslie townes

it's a good phone but there is not a lot of storage space. there's probably an upgrade for that but i'm too cheap to upgrade a 7 year old phone.

schn

@robjohn , I guess then the substitution $v=hu$ is not motivated enough or is not needed in the problem, because it probably would not violate any bindings or scoping -- $hu$ only seems to appear under the integral sign.

cohomonoid

@leslietownes fwiw you can get a ~64gb microsd for like 10 bucks

leslie townes

gosh. i remember when storage was expensive. a dollar or more a megabyte.

Michael Albanese

17:45

Hi @TedShifrin

schn

@robjohn , the substitution $v=hu$ would probably be another change of variables, since $u$ is the integration variable. Thus it is not as straightforward as simply substituting $hu$ for $v$.

user10478

18:46

What am I doing wrong with these Fourier Series of $sin^2 t$ coefficients? wolframalpha.com/input/?i=%281%2Fpi%29+*+integral_0%5E%7B2+*+pi%7D+of+sin%5E2%28t%29+*+cos%28n+*+t%29+dt wolframalpha.com/input/?i=%281%2Fpi%29+*+integral_0%5E%7B2+*+pi%7D+of+sin%5E2%28t%29+*+sin%28n+*+t%29+dt It looks to me like both answers WA gives equal $0$ when $n$ is an integer, leaving only the $n = 0$ term (which obviously can't approximate a sinusoid by itself).

cohomonoid

the links didnt get posted well

user10478

:( I'll try a tinyurl

cohomonoid

So it's full Fourier series? like coeff of e^{-ix \xi}? Cause if it was Fourier sine series that wouldn't work obvi

When I do "Fourier series sin^2" on WA I get something reasonable

user10478

tinyurl.com/4sw6j787 tinyurl.com/9p48b7v

cohomonoid

yeah the fourier sin series shouldn't work because Sin^2 is even right?

thats not right..

user10478

18:52

I am not trying to do a sin series, those two links were the computations for sin and cos terms, and the problem is that they're both 0.

I'm cool with one of them being 0.

cohomonoid

Mathematica gives me:{0, 4/3, 0, -(4/15), 0, -(4/105)}

input Table[Integrate[Sin[x]^2 Sin[n x], {x, 0, [Pi]}], {n, 0, 5}]

Sayan Chattopadhyay

The leray serre spectral sequence works for fibrations that need not have homotopically equivalent fibres, right?

cohomonoid

output{0, 4/3, 0, -(4/15), 0, -(4/105)}

user10478

Is that for non-integer $n$ though?

cohomonoid

oops 2pi gives me all zero

Wait are you doing fourier series or fourier transform

user10478

18:54

Series

I'm doing Laplace's equation on a disk.

cohomonoid

so then n is an integer

user10478

yeah

cohomonoid

@SayanChattopadhyay I think you just need lifting property on CW complexes

or does it need path-connected base too?

cause that implies homo. eqv. fibers

user10478

I can't get an actually expression when I WA Fourier series sin^2

Sayan Chattopadhyay

Yeah it needs simply connected base, so yeah does not work

user10478

18:57

Which is why I entered the coefficient integrals individually.

cohomonoid

wolframalpha.com/input/?i=Fourier+series+sin%5E2

user10478

I see no expression :P

Only a table

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$Mathematics$ Mathematics