Mathematics - 2024-09-13

leslie townes

00:07

e.g. the functions chi_{E_n^k} are measurable by his definition of "measurable" (following corollary 2.2) and his definition of the borel sigma algebra (see proposition 1.2, which the discussion following corollary 2.2 refers to, and which even specifically addresses 'half open' intervals), linear combinations of measurable functions being measurable is stated (for sums of two functions and arbitrary measurable multiplications, not just scalar multiplications) in proposition 2.1

i'll leave you to find the statements and proofs about pointwise limits, etc :)

people sometimes have an expectation that if author labels a proposition with a number, then they can check wherever that proposition is being used simply by checking for express citations to that proposition by its number

that expectation is closest to reality for 'big' theorems and furthest away from reality for definitions and little technical results that get used so often that labeled citations to them would overwhelm the proofs of more complicated things

psie

ok 👍 thanks leslie for these pointers. I'm going to bed now.

Xander Henderson

00:52

@leslietownes Folland would never bother. That would use too much ink.

And if there is one thing Folland HATES, it is using ink.

Ben Steffan

does bourbaki not have a treatise on the topic

they love using ink

leslie townes

01:48

xander he did publish his book during The Ink Wars, when ink was in short supply and you had to go through mad max situations just to get one barrel of it

Bob

03:24

Hi

is anybody around?

Soumik Mukherjee

Yup

Bob

I am in the United States

Soumik Mukherjee

04:02

Where is alice?

nickbros123

04:14

Lol

Soumik Mukherjee

Hey @nickbros123 do you know python?

nickbros123

04:39

@SoumikMukherjee little bit, why?

I am not a programmer though, I just use python to do numerical stuff

Soumik Mukherjee

Do you know how to compute svd in python?

(singular value decomposition)

nickbros123

I think it's in numpy

Perhaps u should check out the geeksforgeeks page on this

Soumik Mukherjee

Okay, thanks

PM 2Ring

05:09

@SoumikMukherjee There's a good example of SVD in Numpy here math.stackexchange.com/a/99317/207316

Here's a demo I did a little while ago, derived from that code. It's mostly plain Python, but it uses Sage for the 3D plotting.

sagecell.sagemath.org/…

Soumik Mukherjee

08:11

@PM2Ring Thanks!

one potato two potato

08:22

Recently interested in Finsler geometry

Soumik Mukherjee

10:10

How to compute the Jacobi symbol (123456789/1111111111) without any factoring other than dividing out powers of two?

psie

12:57

Basic question. If I have a set $A\subset \mathbb R$ that is a finite union of open intervals, is it always possible to make this into a disjoint union, preserving the finiteness and the open intervals?

I think I know how to make the sets disjoint. Suppose $A=\bigcup_1^n A_k$ where $A_n$ are open intervals. Then define $E_1=A_1$ and $$E_k=A_k\setminus\left[\bigcup_1^{k-1}A_j\right],\quad k=2,\ldots,n.$$These sets would do it I guess, but would they be open?

Vladimir Lysikov

@psie No, these would not be open, for example, take just two intersecting intervals, you will get one open interval and one half-closed

psie

ok, I thought so

Vladimir Lysikov

13:14

@psie A union of two intersecting open intervals is an open interval, so if our union involves two intersecting intervals, we can join them and get a union such that the number of intervals is one less

the number of intervals is finite, so by repeating this we either get a disjoint union, or zero intervals

*one interval, not zero

psie

@VladimirLysikov I see, makes sense.

nickbros123

13:50

this is also how to prove that open sets in $R$ are utmost countable union of disjoint intervals. Take all the points in the set, and for a particular $x$, take the largest interval that contains it and call it $I_x$. For any two $x,y$ either $I_x=I_y$ or they are disjoint (for if not u can union them). So u have created disjoint intervals. now since $R$ is separable (or rationals is dense in R), each interval contains a rational number, so there are utmost countable intervals

cool property of R used here is union of two open intervals remains an open interval whereas for R^n or something union of balls dont result in bals

Jakobian

14:57

Take maximal convex subsets

Soumik Mukherjee

@nickbros123 all intervals are not balls in R though

Soham Saha

15:23

For geometry problems, I prefer to make a rough sketch of the situation with all the given info by hand. But circles always trip me up. Anyone has ideas how someone can roughly sketch a circle? (for example: given 3 points, approximately draw a circle that passes through them?)

Ryder Rude

do u lean more toward formalism or Platonism and why

hi

Jakobian

15:39

@RyderRude formalism. But I don't think it's interesting to discuss

Ryder Rude

15:55

@Jakobian oh

i too lean toward formalism

Platonism is kinda meaningless, as all consistent universes exist in platonism

e.g. Platonism doesn't say whether continuum hypothesis should be true or false, as both universes exist

Sahaj

16:15

hi

Sine of the Time

Hi

Sahaj

Does someone know how i can find the possible values of an expression like $3x + 2y + 5z$ where the variables are in $\{0,1,2,3,\dots, 10\}$? Using a computer code I've verified that $\{0,2,3,4,\dots,100\}$ is the answer for this specific case, but any idea how to do it generally via hand?

Soham Saha

Expression is always linear in $x$, $y$ and $z$?

Sahaj

yes, a multivariable expression linear in all variables, where all the variables are bounded and integers

Soham Saha

Same bound for all variables?

Jakobian

16:20

@Sahaj this should be borderline impossible to do.

Sahaj

why do you think so?

Ryder Rude

I think one can work out the maximum and minimum

but exact values is hard

Jakobian

because I know its a problem in number theory to find when such sums will eventually be every natural number

Sahaj

@RyderRude Indeed, and for this expression its pretty trivial

Ryder Rude

unless u do 10^3 computations

Sahaj

16:21

@Jakobian Could you point to some reference on the same? Seems quite related

Jakobian

Numerical semigroup

In mathematics, a numerical semigroup is a special kind of a semigroup. Its underlying set is the set of all nonnegative integers except a finite number and the binary operation is the operation of addition of integers. Also, the integer 0 must be an element of the semigroup. For example, while the set {0, 2, 3, 4, 5, 6, ...} is a numerical semigroup, the set {0, 1, 3, 5, 6, ...} is not because 1 is in the set and 1 + 1 = 2 is not in the set. Numerical semigroups are commutative monoids and are also known as numerical monoids. The definition of numerical semigroup is intimately related to the...

Sahaj

Thanks

Ryder Rude

maybe we should work with the 1D case first. I think slopes can be utilised in the general case

y=5x x\in (0,10)

Sahaj

@SohamSaha sorry didn't see this earlier; yes

Ryder Rude

so this is just maximum and minimum with jumps of 5

in general, maybe we want to utilise the slopes to get the jumps

Soham Saha

16:23

@Sahaj ok, no problem

Ryder Rude

nvm that method is too naive..

Jakobian

I've learned about this from my introduction to number theory classes

Mats Granvik

Speaking of number theory, do you know other functions that have a similar property like this:
$$\underset{\text{factor}}{-2}\left(\underset{\text{non-trivial zeros}}{\sum_{\rho} \frac{1}{\rho (1{-}\rho)}} + \underset{\text{trivial zeros}}{\sum_{k \geq 1} \frac{1}{-2k(1-(-2k))}} - \underset{\text{the pole}}{1}\right)= \underset{s\to 1}{\text{lim}}\left(\underset{\text{the whole matrix}}{\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}}+\underset{\text{main diagonal}}{\frac{\zeta (s)}{\zeta (s-1)}}\right)$$

I tried briefly looking in the NIST Handbook of Mathematical Functions, but I did not find any.

The pole could be in Czechoslovakia, I don't know.

Difficult word to spell by the way, Czechoslovakia.

Soham Saha

@Jakobian @Sahaj Naive brute-forcing should be $O(n^l)$ where bound length=$n$, number of variable=$l$, I think…

Which is too bad for large $l$

Perhaps it would be possible to reduce the complexity by some more efficient algorithm, idk…

Vladimir Lysikov

16:54

Maybe look up quantifier elimination for Presburger arithmetic.
What you want is basically eliminate quantifiers from the formula of teh form $(\exists xyz)(x \leq 10 \wedge y \leq 10 \wedge z \leq 10 \wedge u = 2x + 3y + 5z)$

Why do you need these sets of values?

Sahaj

@VladimirLysikov Really only out of curiosity. I was trying to figure out what scores can be possibly achieved by a student on an exam which has 10 questions each of 2 points, 3 points and 5 points.

Vladimir Lysikov

Ah, then nevermind, I think brute force would work for any reasonable exam :)

Bml

17:13

Hi everyone. How could I prove $f(x) = \frac{2}{3} [1 + cos (\frac{1}{3} \arccos (27/2 x -1))] > 1 \forall x \in (0, 4/27)$?

Jakobian

17:34

@Bml I believe this will be strictly monotone. So all you have to show is $f(0)$ and $f(4/27)$ are $\geq 1$

Yeah since $\arccos$ is strictly monotone and $\cos$ on the range of $\frac{1}{3}\arccos$ is also strictly monotone

Soham Saha

18:08

@Sahaj Just a wild guess, but is this about the IOQM?

Bml

@Jakobian OK. Usually, to see if a function is strictly monotone over an interval I calculate the first derivative of the function and determined the intervals in which it is positive or negative. But why, if a function is strictly monotone, is it sufficient to check that the function calculated at the extremes of the definition set is $\geq 1$ (rather than $> 1$) to determine that $f(x) > 1$ on that interval? Could you please explain?

Jakobian

18:44

@Bml "Usually, to see if a function is strictly monotone over an interval I calculate the first derivative of the function and determined the intervals in which it is positive or negative.", that's a waste of time

at least in this example, you often can just determine it by functions being compositions of increasing functions

You defined your function on $(0, 4/27)$ but your function is defined on $[0, 4/27]$ as well

and since it's strictly monotone, its global minimum will be either at $0$ or at $4/27$

on the rest of the interval it will be strictly greater than the global minimum

@Jakobian monotone/strictly monotone functions

user10478

Say I solve a problem, maybe with some branching computations as opposed to each step following linearly from the previous step, and in the end I choose a set of steps within the computations which seem important. So I have some steps I'll label A, B, C, D, E, F..., and I want to understand which pairs of these steps are "bijection-like."
For example, I would want to represent whether knowing B lets you compute D and knowing D lets you compute B. If you can't, I would want to represent the direction of "information loss." I guess for a linear computation, this could be represented for al…

Bml

19:03

@Jakobian Why can't it be a global maximum? We do not know whether it is increasing or decreasing, only that it is monotone.

@Jakobian It is a restriction of the function to the interval (0, 4/27), isn't it?

@Jakobian You said "$\cos$ on the range of $\frac{1}{3}\arccos$ is also strictly monotone": why is it obvious?

Jakobian

@Bml one will be a global maximum, the other a global minimum

@Bml Yes. And? It's not like you can't consider it on $[0, 4/27]$ because of that

@Bml $\cos$ on range of $\arccos$ is strictly monotone - it's bijective and continuous there. The range of $\frac{1}{3}\arccos$ is an even smaller interval.

Bml

@Jakobian Ah, OK. I thought that since it is defined only on $(0, 4/27)$, the values $0$ and $4/27$ should not be considered. I still don't understand why, since I want $f(x) > 1 \forall x \in (0, 4/27)$, I have to check that $f(0) \geq 1$ and $f(4/27) \geq 1$. The equal sign is not included...

Jakobian

Okay, suppose that one of those values will be equal to $1$. Does that matter?

Bml

@Jakobian "$\cos$ on range of $\arccos$ is strictly monotone": where can I find an analysis of this, even on Math SE? It doesn't seem like an obvious fact.

Jakobian

@Bml most texts about real analysis

being a continuous injection $f:I\to \mathbb{R}$ where $I$ is an interval implies that $f$ must be strictly monotone

this is just intermediate value theorem

Bml

19:18

@Jakobian We have $f(0) = 1$ and $f(4/27) = 4/3 > 1$. $f(x)$, however, must be strictly greater than 1, not greater (or equal).

Jakobian

@Bml $f$ is injective. So how can it be $1$ on anywhere else than for $x = 0$

and you want it to be $> 1$ for $x\in (0, 4/27)$ anyway

I don't understand the objections you raise

Bml

@Jakobian Can that be helpful?

1

Q: Monotonicity of $f$ and $f^{-1}$

Let $f$ be a strictly monotonic function on an interval $I$. Then, is $f^{-1}$ of the same monotonicity? I believe that the answer is YES. Here is my attempt: Since $f$ is monotonic in an interval, then it is invertible. If $f$ is strictly increasing and $f^{-1}$ is strictly decreasing on $I$, th...

Jakobian

@Bml I don't want to read whatever is in this post. What exactly do you want to find helpful here?

Bml

19:35

@Jakobian Something like, " Let $f: I \to U$ be an injective function, therefore strictly monotonic. Let $g: U \to I$ be an injective function, hence strictly monotone, such that $g = f^{-1} (x)$. Then, $f(f^{-1}(x))$ is injective, hence strictly monotonic in $I$. Is there such a theorem?

Such a theorem would make $cos(arccos(x))$ an injective function, thus strictly monotonic.

Sine of the Time

$f(f^{-1}(x))$ is the identity

plus the composition of injective functions is injective

@CroCo a couple of days ago I saw this question, maybe you find it useful to that problem you were solving

Bml

@SineoftheTime OK. How to generalize this to the case $f(k f^{-1}(x))$, with $0 < k < 1$?

Sine of the Time

33 mins ago, by Jakobian

@Bml $\cos$ on range of $\arccos$ is strictly monotone - it's bijective and continuous there. The range of $\frac{1}{3}\arccos$ is an even smaller interval.

besides, if $k\neq 0$, $kf^{-1}$ is still injective

Bml

20:01

@SineoftheTime Does this occur because it is a combination of a real scalar and an injective function?

Thorgott

multiplication by a non-zero number is an injective function, composition of injective functions is injective

Sine of the Time

did you try to prove it?

Jakobian

@SineoftheTime it's still under the composition of strictly increasing monotone. This is $x\mapsto kx$ and $\arccos$

@Bml you can't really

I mean you can, in some sense, but I wouldn't recommend to look at this in some kind of general way here, at least not this part of the argument

@Bml continuity is really important here. You can't conclude that injective functions are monotone.

@Bml $\cos(\arccos(x)) = x$

all the confusion here really should have been answered before we even define what $\arccos$ is

so if there is no confusion about what $\arccos$ is, then I don't get whats confusing here

@leslietownes ever seen this notation? math.stackexchange.com/questions/4970931/…

Bml

20:36

@Jakobian Yes. $\cos(x)$ is a continuous injection in $[0, \pi]$, hence strictly monotone on that interval. $\arccos(x)$ is a continuous injection in $[-1, 1]$, so it is $k \arccos x$ with $k \neq 0$, hence it is strictly monotone on that interval. So, $\cos(k \arccos x), k \neq 0$ is continuous injection in $[0, \pi]$, hence strictly monotone on that interval. Is it correct now?

Jakobian

@Bml Well not quite, since if you take $k\arccos x$ then it can have range outside of $[0, \pi]$

that's why it's important that it was a small number like $1/3$

Bml

@Jakobian I modify my previous statement, replacing $k \neq 0$ with $0 < k < 1$. Is this now correct?

Jakobian

yes

naturallyInconsistent

@Bml Is that $\dfrac{27}{2x}-1$ or $\dfrac{27}{2x-1}$ because if it is the former, the $\lim$ that $x\to0$ makes the argument going into the $\arccos$ tend to infinity, and $\arccos$ is only uniquely real-valued when its argument is between $-1$ and $+1$

Jakobian

It's clearly $\frac{27}{2}x-1$ just not written correctly

look at the domain

naturallyInconsistent

20:48

@Jakobian uugh, this is so bad

I was also looking at the domain and the latter choice is clearly also not tolerable

Jakobian

latter choice as in what I wrote, or what you wrote?

naturallyInconsistent

no, obviously what I wrote. Yours make too much more sense

Bml

@Jakobian Yes.

mick

hi all

Sine of the Time

hi

mick

20:54

i asked a question theoretically hard i guess , but LHF for numerical checks for those who can

Sahaj

@SohamSaha It was the inspiration for the question since the exam has a fairly neat marking scheme.

mick

0

Q: the zero's of $f(s,a) = \sum_{n=1}^{a} (\frac{n^2 + n}{2})^{-s} $

I was looking at the zero's of $$f(s,a) = \sum_{n=1}^{a} (\frac{n^2 + n}{2})^{-s} $$ for integer $a>3$ in the strip $0 < \operatorname{Re}(s) < \frac{1}{2}$. And ofcourse the limiting case " triangular zeta function " : $$f(s) = \sum_{n=1}^{\infty} (\frac{n^2 + n}{2})^{-s} $$ in the strip $0 < \o...

complex-analysis roots dirichlet-series truncation-error

Bml

21:10

@naturallyInconsistent This math question stems from a physically-oriented interest of mine. If you are interested, read this question of mine and the corresponding answer.

2

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 ...

real-analysis inequality physics

CroCo

@SineoftheTime thank you so much. Indeed, I did show my supervisor and he approved it as a correct answer.

naturallyInconsistent

@Bml All I'd like to say is that the answer is very nice and Callen is a wonderful book.

Ryder is not Rude.

M I A O ~

naturallyInconsistent

M I A O ~

Ryder is not Rude.

21:27

:)

Bml

21:40

@naturallyInconsistent There is another inequality I would like to resolve, again related to the same physically-oriented question. The inequality is $\frac{2}{3}-\frac{2}{3} \sin\left(\frac{pi}{6}- \frac{\arccos \left(\frac{54 abc}{(a+b+c)^3} - 1\right)}{3}\right) > (\sqrt{a} - \sqrt {b})^2 + c$. Is this inequality valid $\forall a, b, c > 0$? Does it have a trivial solution? If you like, I could ask a new question bringing in the physical argument that leads to this inequality.

naturallyInconsistent

Isnt that clearly an offshoot of that same question and should thus be handled in similar ways?

Bml

@naturallyInconsistent It is not really a consequence; there is a related but different physical argument to support it. The problem is, do you have any idea what the ways might be to approach such an inequality involving a nontrivial trigonometric expression? I really wouldn't know how to define a new function, as Semiclassical did who answered the question...

Bml

22:12

@naturallyInconsistent I thought a little bit and concluded that $\frac{2}{3}-\frac{2}{3} \sin\left(\frac{pi}{6}- \frac{\arccos \left(\frac{54 abc}{(a+b+c)^3} - 1\right)}{3}\right) > (\sqrt{a} - \sqrt {b})^2 + c$ isn't really an inequality, as we don't have to solve for a variable. It only reduces to proving that LHS is strictly greater than RHS for all values $a, b, c > 0$. The problem is that I have no input in mind to do this.

psie

22:30

In Folland's theorem 2.27b, on the differentiation under the integral sign, he has $h_n(x)=\frac{f(x,t_n)-f(x,t_0)}{t_n-t_0}$, where $f:X\times[a,b]\to\mathbb C$ and $\partial f/\partial t$ is assumed to exist. He then applies the mean value theorem as follows: $$|h_n(x)|\leq\sup_{t\in[a,b]}\left|\frac{\partial f}{\partial t}(x,t)\right|.$$

I'm a bit rusty with multivariable things, but we can apply the mean value theorem because $f$ is continuous in $t$, right? Moreover, do you know a reference for where I can find this version of the mean value theorem?

Jakobian

@psie Let $F:[a, b]\to\mathbb{R}^n$ be differentiable on $(a, b)$ and continuous on $[a, b]$ then $(b-a)^{-1}|F(b)-F(a)|\leq |F'(c)|$ for some $c\in (a, b)$.

Mean value theorem

In mathematics, the mean value or Lagrange theorem from the branch of mathematical analysis states, roughly, that for a "smooth" arc between two endpoints on a plane, there is at least one point on the arc whose tangent line is parallel to the line connecting its endpoints (its secant, see figure). The MVT's importance lies in its serving as the basis for the proof of other important mathematical theorems, including the fundamental theorem of calculus, and, an implication—that at some point, a curve's slope must be equal to its average slope—that supports its use in developing various types of...

Sine of the Time

Rudin PMA, th 5.19

psie

ok 👍

Jakobian

Feels like there is no reason to go for $\mathbb{C}$ here

$\mathbb{R}^n$ should do

psie

yeah, I don't understand why $\mathbb C$ either. But what you wrote above for $\mathbb R^n$ applies to $\mathbb C$ as well, right? (we identify it with $\mathbb R^2$ I guess)

Jakobian

22:44

@psie yes, for the topics Folland is talking about, there is no reason to distinguish between $\mathbb{R}^2$ and $\mathbb{C}$

psie

ok

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