Mathematics - 2023-03-06

Unknown x

01:13

Let L be the line with parametric equations$ [ \begin{array}{l} x=5+2 t \\ y=-6+t \\ z=-7 \end{array} ]$ Find the vector equation for a line that passes through the point $P=(−8,−5,2)$ and intersects L at a point that is distance 3 from the point $Q (5,−6,−7)$

Our Line intersect $L$ at $R=(5+2*3,-6+3,-7)=(11,-3,-7)$. right?

The direction of the line is $\vec{OR}-\vec{OP}$. right? where $O$ is the origin.

Hence, the direction of the given line is $(11,-3,-7)-(-8,-5,2)=(19,2,-9)$. Am I correct?

Ted Shifrin

01:40

@Unknownx Wrong.

Unknown x

@TedShifrin yes, I understood sir, I need to find the unit vector direction. right?

Ted Shifrin

Right. And there are two points on the line that are a distance $3$ from $Q$. So I guess there can be two answers.

Unknown x

@TedShifrin yes sir. Thank you very much :)

冥王 Hades

02:17

Which is hotter? The core of the sun or the ramen that just burned my tongue

Silly Goose

03:41

is it true that every neighborhood is bounded?

Franklin

Can anyone please help me plot the graph of the function f(x)=sin(\pi/4(x-[x])) if [x] is odd ,x\geq 0 and f(x)=cos(\pi/4(1-x+[x]))$ if [x] is even ,x\geq 0 analytically ?

For starters, how to plot y=sin(x-[x]) ?

Koro

@leslietownes Thanks a lot :-). It seems that there is no programming in that course.

@mick probably not

@SillyGoose what definition of neighborhood are you using? The answer will depend upon that.

@Franklin Note that $\sin (x-[x])$ is periodic.

Silly Goose

(from baby Rudin)

Given metric space $(X, d)$ a neighborhood of $p \in X$ is defined $N_r(x) = \{y \in X : d(x, y) < r\}$

i was thinking choose point $p$ and real number $r + 1$ and the neighborhood is hence bounded

here is rudin's def of bounded as well

This is all to ultimately understand why we know $\bar{V}_n$ is bounded. I get why it is closed (definition of closure).

Franklin

03:58

@Koro yes, that only implies we must plot it between [0,2\pi] but what else ?

Koro

@Franklin No. I mean period of $\sin (x-[x])$, not of $\sin x$.

@SillyGoose bounded then.

leslie townes

@Koro now i wonder what the point is of a course like that, without programming. :)

Silly Goose

okay great

Koro

@SillyGoose $V_1$ is open ball of radius r centered at $x_1$.

Silly Goose

right and $\bar{V}_n$ is simply including the "edge" points. Hence the point $x_n$ and the radius $r + 1$ still means it is bounded :D

Franklin

04:12

@Koro I see, but what's the period of sin(x+[x]) ?

I dont get this at all

Koro

x-[x] = fractional part of {x}, which has period 1.

Ted Shifrin

@SillyGoose No.

Koro

sin (x+1-[x+1])=sin(x+1-[x]-1)=sin(x-[x]). :-)

Ted Shifrin

@Franklin Plot on $[0,1]$.

Koro

so plot sin x on [0,1].

Then, {x}=x on [0,1). So you get sin{x} in [0,1]. Repeat this business throughout by periodicity.

@leslietownes 😅

Franklin

04:18

@Koro Ohh... so we get sinx in [0,1] right ?

Koro

sin (x)= sin {x} in [0,1].

now use periodicity of sin{x} to extend.

Franklin

@Koro Yeah! And thus, since y=sin(x-[x]) is periodic, we just repeat this pattern in [1,2] , [2,4],... , right?

Thanks a lot!

Koro

right. Note that you can do this for any f(x), not just for sine x.

Franklin

@Koro Thank you so much!

I get it now...

Koro

gr8 !

copper.hat

04:34

i use desmos

Silly Goose

If $V_n = \{y \in \mathbb{R}^k : |x_n - y| \leq r\} how is $V_n$ not bounded when considering the point $x_n$ and real number $r+1$?

since by construction $|x_n - y| \leq r < r + 1$ for all $y \in \V_n$

leslie townes

see, this is why i don't use chatjax

any one of those sets V_n is bounded, but perhaps there is some argument that would like to make use of a bound on the V_n that does not depend upon n, and in such an argument, the fact that the x_n's could 'escape to infinity' could prevent that from happening

Silly Goose

oh i see hm

copper.hat

-{1 \over 12}

Silly Goose

04:56

hm so how would i more generally see these $V_n$ as being bounded

leslie townes

for any y in V_n, |y| = |y - x_n + x_n| <= |y - x_n| + |x_n| <= r + |x_n|

the extreme right hand side of this is independent of y

Silly Goose

05:11

hm i guess i am having trouble seeing how this makes our boundedness independent of n

leslie townes

it doesn't

who said that it does

Silly Goose

oh i thought you were saying we'd like to show the boundedness in a way independent of n

leslie townes

i did say, maybe somebody would like to do that, and i did observe that none of the above arguments do that, but this was mostly as an attempt to guess at an answer to the question "how is V_n not bounded" when each set V_n is in fact bounded

there might or might not be a bound on the sets V_n that is independent of n, depending on the properties of the sequence x_n

but the issue of whether there is or is not such a bound that is independent of n, is formally unrelated to the question of whether each set V_n is individually bounded

Silly Goose

oh i see so you were provided a proof of some form of boundedness

leslie townes

yeah

Silly Goose

05:17

wait so would be not need one more step to get a strictly less than in your chain of inequalities

or maybe im misunderstanding

leslie townes

what are you trying to do, where it would be necessary to have a strict inequality

i was operating from the assumption that "S is bounded" if and only if "there is some number p such that |x| <= p for all x in S"

Silly Goose

ah i see

rudin's def uses a strictly less than for the def of boundedness i think

leslie townes

if S is bounded in this sense, you can arrange for strict inequality by choosing a slightly larger p, but, i'm just unsure about the purpose of that

what a goofball

Silly Goose

xD

Silly Goose

05:54

bleh these theorems in rudin are quite hard to understand for me

Koro

I got 50% marks in my algebraic topology midsem.

which is fine

@SillyGoose Drawing pictures will help.

Silly Goose

is the idea with the segment business on the lower half of the image this: that every segment contains this special segment (24) for some $m$. But we also know that $P$ does not share any points with any such special segment (24) by construction (we removed those points or they weren't included in the first place). Hence, because $P$ does not contain any points of the segments which are subsets of every single arbitrary segment, $P$ does not contain any arbitrary segment?

I think the one bit of uncertainty i have is about: I think every single arbitrary segment contains (24) for some $m$, which some stack post said you can prove with archimedean property, which I can see as plausible. The rest i think I can follow from there, though.

Koro

yeah, something like that

Silly Goose

Maybe something like establishing that $3^{-m} < 3m$ for $m > 0$. Then use archimedean to make m as small as need be to be less than $\frac{\beta - \alpha}{6}$

Koro

the special segment is of length 3^{-m} but in P look at $E_{m+1}$ for example, and it is contained in P.

Silly Goose

06:01

so each of the $E_{m+1}$ have at maximum segments of length $3^-{m+1}$?

Koro

yeah. And so P can't contain the segment (24).

why do people take green tea?

I don't quite like the taste.

Silly Goose

oh i see so P cannot contain any arbitrary (24). And since every arbitrary segment contains some special case of (24), P cannot contain any arbitrary segment then

how are you brewing it? and what green tea?

Koro

indeed

Silly Goose

so is my thinking for why any arbitrary segment must contain a segment of the form (24) along the right lines?

Koro

I follow the instruction on its box, which says to add hot water and let it be for some 2 mins.

Silly Goose

06:04

is it a particular brand of tea?

i have heard and experienced that green tea is very easy to brew such that it tastes not so great

e.g. if you use boiling water it will taste not so great

Koro

oh, I am using boiling water.

Silly Goose

75-85 celcius will make it taste how it is meant to heh

and apperently you are supposed to steep for very short amounts of time depending on what green tea, but i like more flavor :P so i steep for longer

Koro

oh I see, thanks. I'll remember that.

Silly Goose

but yeah 75 - 85 celcius and 1-2 mins is probably a rough range of how to brew green tea.

Koro

I do it 2 mins but with boiling water.

Silly Goose

06:07

i personally like sencha green tea a lot it is quite vegetal but light and fresh tasting

Koro

That's why probably it didn't taste good.

Silly Goose

heh perhaps

it could also be the tea. e.g. popular american brands like lipton and pure leaf green tea taste quite different than say if you get green tea from anywhere else

and i do not like lipton/pure leaf green tea :P

Koro

about your question: do you understand why every segment should contain segment (24)?

I guess liking green tea takes some getting used to it. The flavour as you're about to finish the tea is good. :)

there are so many other varieties of tea. There is orange tea for god sake.

lemon tea tastes great by the way.

Silly Goose

and there are different varieties of each variety :) jasmine green tea is my favorite right now i am hoping to order some soon

sencha and jasmine green teas

let me try writing my reasoning for why every segment should contain (24)

black tea is much more resilliant to improper brewing--you just throw some boiling water on and let it sit for however long you wish XD

okay this is what i got :P

can probably prove the ($3^n > n$) by induction or what not

Koro

07:33

@SillyGoose Almost ok. You should also add why $\gamma_n\in (\alpha,\beta)$.

Silly Goose

also a separate question: is this proof by contraposition? since we are proving not A implies not B; hence, B implies A. Then we are doing the converse?

ah i see. I was taking that for granted via Rudin's statement of why :) ty @Koro

Koro

07:58

I meant $\gamma_n\color{red}{\subset }(\alpha, \beta)$ earlier. $\in$ in place of $\subset $ was indeed a typo.

@SillyGoose I think it is by contradiction.

But I think that this depends upon how one writes it: If one writes in the beginning 'suppose on the contrary that negation of "if x in E, y in E,..." is true, then it becomes a proof by contradiction.

Here since it is not written there, so I think that you may treat the first part of the proof as a proof by contraposition.

Koro

08:18

My algebraic topology teacher said: don't study from book as there is no time.

Study only what I do in class and do only the exercises that I assign.

CroCo

09:26

Good morning everyone,
could you please take a look at this question
https://math.stackexchange.com/questions/4653020/is-m-1-frac14-m-3i-yes-no

Why $M^{-1}=(1/4)(M+3I)$?

Obviously $MM^{-1} \neq I$.

I'm aware of Cayley–Hamilton theorem though.

which says every square matrix satisfies its own characteristic equation.

CroCo

09:42

please check my answer there. thank you

Not Tfue

10:06

To prove that f is injective using the condition f(x) = f(y) implies x = y, we can proceed as follows:

Suppose that f(x) = f(y) for some real numbers x and y in (0,1]. Then we have:

f(x) = {n | x_n = 2}
f(y) = {n | y_n = 2}

where x_n and y_n denote the decimal expansions of x and y, respectively.

Since f(x) = f(y), it follows that the sets of natural numbers corresponding to x and y are identical. Therefore, we have:

x_n = y_n

for all n in the set of natural numbers. This implies that x = y, since the decimal expansions of x and y are identical.

Is my proof of why f(x) is injection looks okay?

Does*

CroCo

10:21

Good morning Sir @TedShifrin, could you please take a look at this question
https://math.stackexchange.com/questions/4653020/is-m-1-frac14-m-3i-yes-no
Specifically, why $M^{-1}=(1/4)(M+3I)$? I feel the accepted answer is incorrect.

Am I wrong? My answer got busted :)

robjohn

11:10

@冥王Hades you asked this question and another. Did you ever get an answer to either? You had once said you would post the eariler question, but it doesn't appear that you have posted either. I have answers to both.

@冥王Hades of what?

冥王 Hades

11:58

@robjohn yeah I'm sorry, I've been meaning to post those questions but I just couldn't find the time. I'm going to post one of them in a few hours today and I'll ping you as soon as I do

@robjohn its from another question I saw on here a while ago

Its notoriously difficult without a regular n-gon approach but I'm just insane enough to derive one without it

Franklin

12:23

Find the positive integers $x$ such that $[\frac x7]+[\frac x7]=1.$ -Can anyone please help me with this?

Not Tfue

12:58

@PM2Ring unfortunately I changed my phone and still won't work :( replying to chatjax problem

I am using Chrome currently

robjohn

13:51

@NotTfue I assume you’ve tried all the applicable suggestions lower down on the page?

The installation page

Franklin

14:08

@robjohn Sorry if this is silly, but what's chatjax ?

Is the discussion going about somekind of an app?

robjohn

@Franklin if you look in the room description (on a non-phone) or here, you will see a link to the installation page for a JavaScript bookmarklet that enables MathJax in chat.

I’m going to UCLA to proctor an exam. I may be back intermittently, but out quite a bit.

Koro

14:26

@robjohn ohh nice :)

Franklin

@robjohn ohh! Thanks 😊

ILikeMathematics

@robjohn "I buy physical versions of most books."
When the book you are looking for is priced at 600$ or more, you would still lend it out the library, right?

PM 2Ring

15:26

@NotTfue Chatjax works in the Samsung Internet browser on my Galaxy S9 phone, but the bookmark has to be in the Bookmarks bar (or a folder in that bar). If it's just in the general Bookmarks then javascript: URLs are blocked in recent versions of Samsung Internet. On Chrome, you have to follow the specific Samsung instructions at the end of the Chatjax info page. At least, those instructions worked last time I tested them, but I think I've updated Chrome a few times since then.

Rajarshi_Rit

hi

PM 2Ring

@NotTfue In Samsung Internet, do you have the Bookmark bar displayed? If not, open Settings, go to Layout and menus, and Enable bookmark bar.

Obliv

15:44

how is $s = \frac{\sigma}{\sqrt{n}}$ derived, is it by $\sigma = \frac{\sqrt{\sum (x-\mu)^2}}{N}$ ?

and is $\sigma = \sqrt{npq}$ also from that same definition

as in $s = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{\sum (x-\mu)^2}}{N} \frac{1}{\sqrt{n}} = \frac{\sqrt{\sum (x-\bar{x})^2}}{n-1}$

PM 2Ring

@NotTfue Ok. I just tested Chatjax in Chrome. It works. It's rather tedious, though. It's much nicer in the Samsung browser.

Obliv

and $s = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{npq}}{\sqrt{n}}$ for a binomial distribution?

i'm also curious if this is a parameter used in statistics $\frac{|{x-\mu}|}{N}$

PM 2Ring

@Obliv Basically, yes. It might be easier to start from the variance. en.wikipedia.org/wiki/Variance Generally, Wikipedia can be an unpleasant place to learn new mathematics, but that page isn't too bad.

@Obliv Sure, but it's painful to do calculus with absolute value.

Obliv

16:01

just to clarify, this is only for samples taken through "replacement"?

what does this mean exactly. Say I have $\{0,1,2,3,4,5,6,7\}$ and I take a sample, is it restricted in any way

like $\{0,0\}$ isn't allowed or something

PM 2Ring

Square roots are much easier to manipulate, but more importantly, the square root of the sum of a bunch of squares is a natural way to measure distance.

Obliv

Gotcha

PM 2Ring

@Obliv Sampling with replacement means you take a sample, record it, and throw it back into the pool. So yes, you can get (0, 0).

It's probably not a good idea to use set notation for sequences that may contain repeated items. ;)

Obliv

for population size $N$ and sample size $n$, we have $N^n$ samples

by replacement*

PM 2Ring

Yes. Assuming we're sampling them one at a time.

Obliv

16:07

so $s = \frac{\sigma}{\sqrt{n}}$ is the standard deviation of the samples within this context?

Rajarshi_Rit

Amazed to see the [music theory] tag

Obliv

and you can derive this by just using $\sigma = \frac{\sqrt{\sum (x-\mu)^2}}{N}$ but for the samples and $\mu$ is the mean of the sample means

Rajarshi_Rit

Please recommend me book on that

Obliv

with the denominator being $N^n$

Rajarshi_Rit

math in music theory

PM 2Ring

16:10

It gets more complicated if, eg, we take 5 cards from a deck of 52, record them, then shuffle them back into the deck & then take another 5 cards. So each individual group of 5 cards is done without replacement, but the overall experiment is done with replacement. But I Am Not A Statistician!

Franklin

math.stackexchange.com/questions/4653342/… Can anyone please help me with this ?

PM 2Ring

@Rajarshi_Rit That gets discussed a bit on the Music Theory & Practice stack. There's some great info there on the mathematics underlying different scale temperaments. There's no perfect solution because $\log_2 3$ is irrational. I briefly mention that here: math.stackexchange.com/a/3983124/207316

Obliv

I think i've been using the wrong definition.. should be $\sqrt{\frac{\sum(x-\mu)^2}{N}}$ yikes

so $s = \frac{\sigma}{\sqrt{n}} = \sqrt{\frac{\sum(x-\mu)^2}{N}} \frac{1}{\sqrt{n}}$

Rajarshi_Rit

@PM2Ring thanks so is music theory totally off topic here??

only curious that a tag exists so i thought there might be some different kind of music theory discussed by mathematicians

PM 2Ring

@Rajarshi_Rit Huh? We have a tag for it, with a bunch of good questions. music-theory

parz

16:22

I have arrived. Much wow.

Rajarshi_Rit

@PM2Ring then there must be some good book from a math perspective on music theory

but im afraid to ask in main thread

Ted Shifrin

Check out David Benson’s book on math and music.

Rajarshi_Rit

16:53

thank

@TedShifrin thank you so much

user10478

17:30

Suppose there is an item that deteriorates with use, its condition expressed as a number between $0$ (completely broken condition) and $1$ (perfect condition), for which I want to assign a monetary value. To make the units nicer, let's think of the condition as a monetary value; someone has guaranteed to pay me $T$ dollars at any time to sell the item, where $T$ is its condition. $M(T)$ is the multiplicative markup on $T$ I assign to the item, and the final monetary value is $V(T) = M(T) * T$.

The first intuition (which I'm treating as an axiom) is that $V(T)$ should be monotonically increasing in $T$. The second is that $M(T)$ should be monotonically decreasing in $T$. I can also ask (and answer) myself what I feel an approximate $V$ should be at a handful of points in $T$.

So far, I have tried Lagrange Polynomial Interpolation on such a handful of points to output a candidate $V(T)$, but when I divide this by $T$ to produce the corresponding $M(T)$, the result violates the second axiom; namely, polynomials produce non-decreasing regions in $M(T)$ around the LPI input points.

Are there ways to construct a $V(T)$, in elementary but perhaps non-polynomial functions, so that the two axioms are satisfied, and it also passes through a handful of input points (the points can be trusted to themselves satisfy the axioms)?

Ted Shifrin

18:27

@leslie @robjohn @Thor Croco's question to me above led me to think about something I've never considered before. We all know that if $p(t)$ is the characteristic polynomial of $A$, then $p(t)-\det A = tq(t)$ and $q(A)/\det A$ gives $A^{-1}$. Of course, the same must work with the minimal polynomial. What's the sleight of hand that says we get the same inverse matrix? Obviously, this is nothing about matrices, but just a fact about quotient rings by polynomial.

Thorgott

shouldn't it be $-q(A)/\det A$

Ted Shifrin

I don't think so, but that's not the main point :P

Thorgott

I think the sleight of hand might be this: if $R$ is a $k$-algebra, $0\neq x\in R$ and $p\in k[t]$ is an irreducible polynomial such that $p(x)=0$, then we can write $p(t)-p(0)=tq(t)$ for some polynomial $q\in k[t]$. Note $p(0)=0$ implies that $p=t$ by irreducibility, hence $x=0$, so $p(0)\neq0$. Then, evaluating this polynomial identity in $t=x$ yields $-p(0)=p(x)-p(0)=xq(x)$. Note also that $x$ and $q(x)$ necessarily commute. Thus, $-q(x)/p(0)$ is a multiplicative inverse for $x$.

This should be basically the same as the argument that the subring generated by an algebraic element in a field extension is itself a field

Ted Shifrin

No one says $p$ is irreducible, of course. You're missing — I think — the point of my question. I understand all this. Why does the same manipulation with the minimal polynomial give you the same inverse? The polynomial expressions you get for the inverse are entirely different when the minimal polynomial is not the characteristic polynomial.

Thorgott

18:42

well, the minimal polynomial is irreducible

but if the emphasis is why you get the same inverse, the answer is just that inverses are unique, no?

there is no reason for the polynomial expressions to be the same, I think

Ted Shifrin

No, the minimal polynomial needn't be irreducible, either.

No, but the polynomial expressions must be related in such a way that $tq_1(t)=1\pmod{p(t)}$ and $tq_2(t)=1\pmod{m(t)}$, so $t(q_1(t)-q_2(t))=0$ mod ?

Thorgott

sorry, you're right, I was getting mixed up

Ted Shifrin

There's got to be something not entirely coincidental going on in $R/I$ versus $R/J$ when $I\subset J$.

I'm sort of surprised/ashamed that I've never contemplated this before. I've always just used the characteristic polynomial and not given it another thought.

PNDas

A few days ago, my teacher mentioned that with some extra conditions we can write a $L^p$ function to be convolution of one $L^1$ and one $L^p$ function

Does anyone have any reference?

Ted Shifrin

Your sentence is confused, I think.

You mean one $L^1$ function and one $L^p$ function?

PNDas

18:50

Yes sorry

$f\in L^p$ is given then there exists functions $g\in L^1$ and $h\in L^p$ such that $f=g\ast h$

with some conditions on $f$

Thorgott

the kernel of the evaluation hom $k[x]\rightarrow M(n,k)$ is generated by $m(t)$, the fact that $Aq_1(A)=1$ implies that $tq_1(t)=1\mod m(t)$ too, no?

Ted Shifrin

Have you googled and/or searched on the site?

Thorgott

so $t(q_1(t)-q_2(t))=0\mod m(t)$

Ted Shifrin

Yeah, that sounds right, Thor. But it's weird. Of course $m(t)|p(t)$, but what goes on when we subtract the respective constant terms?

PNDas

What to google? I googled "for any L^p function does there exist L^1 and L^p functions such that it is their convolution" but didn't found anything useful

Lemme search chatgpt

Ted Shifrin

18:58

Too longwinded for searching

Try when is a function a convolution?

PNDas

Chatgpt says; take $p=2$, For instance, consider $f(x) = \frac{1}{\sqrt{x}}$ on $(0,1)$, which is in $L^2((0,1))$ but not in $L^1((0,1))$. Suppose there exists $g,h$ such that $f=g*h$. Then, taking the Fourier transform, we have $\hat{f}=\hat{g}\cdot \hat{h}$, where $\hat{f},\hat{g},\hat{h}$ are the Fourier transforms of $f,g,h$, respectively.

However, since $\hat{f}$ is not in $L^1(\mathbb{R})$, it cannot be the product of two functions in $L^1(\mathbb{R})$, which implies that such $g$ and $h$ do not exist.

Ted Shifrin

I certainly have zero trust in ChatGPT.

Thorgott

we can write $t(q_1(t)-q_2(t))=m(t)n(t)$ for some polynomial $n$. the LHS has no constant term, so the RHS doesn't either. since we assume that $A$ is invertible, $m(t)$ has a constant term, so necessarily $n(t)=tn'(t)$ for some $n'(t)$. then, cancelling, we get $q_1(t)-q_2(t)=m(t)n'(t)$, so $q_1(t)=q_2(t)\mod m(t)$.

PNDas

Ha ha

Ted Shifrin

That seems OK, @Thor.

Maybe my question wasn't interesting after all. But if we have two polynomial expressions for $A^{-1}$, I guess they must differ by something in the ideal generated by the minimal polynomial. End of story. ?

PNDas

19:05

I'll just ask my teacher about this again.

Ted Shifrin

@PNDas T/F The product of $L^1$ functions is always $L^1$. That sounds like garbage to me.

PNDas

F, $1/\sqrt x$

Thorgott

this reminds me of the Jordan-Chevalley decomposition

PNDas

I'm the only student in that class.

Ted Shifrin

So, as usual, ChatGPT wrote crap.

I think this is well-understood, @PNDas. Obviously $f = f\star\delta$, but the restriction makes it interesting.

PNDas

19:08

@TedShifrin Yes if you are thinking of them as distributions.

Ted Shifrin

I no longer own my real analysis library, but I'm pretty sure there is theory answering your question. Surely @leslie knows the result.

PNDas

This is a proper course. But except me nobody took it.

@TedShifrin Do you have any reference on the range of convolution operation? I know few easy ones like $L^p\ast L^q\subset L^r$, (where $\frac1p+\frac1q=1+\frac1r$) and $L^2\ast L^2\not\subset L^2$.

Ted Shifrin

No, I know there are results, but I no longer have much of a library. I suggest asking @leslie.

PNDas

Okay thanks

Koro

pathetic macbook. Forever stuck on waiting while using 'airdrop'.

Ted Shifrin

19:18

The standard homework problem, of course, is $L^1*L^*1\subset L^1$.

Koro

using we transfer to transfer file from one device to another

Ted Shifrin

@PNDas. Ah, the usual Hölder/Fubini games show that $\|f*g\|_p \le \|f\|_1\|g\|_p$.

PNDas

I stated a more general version.

@TedShifrin Hmm that is true. So?

Ted Shifrin

Wasn't that your question?

Oh, no, you're asking for the image of the convolution map.

Seems like every $L^p$ function should be the $L^p$-limit of convolutions with approximate identities.

PNDas

@TedShifrin Actually, I was doing the proof of this and the teacher made that remark.

@TedShifrin $\rho_{\varepsilon}\ast f\to f$ in $L^p$ for all $1\leq p<\infty$

Ted Shifrin

19:24

Right.

Well, if $f\in L^p$ to start with.

PNDas

Well let's take the simplest case. Is $L^1\ast L^1=L^1$?

Ted Shifrin

Can you show you get everything in $L^1\cap L^2$?

冥王 Hades

21:24

confused screaming

Koro

is this cyclic?

冥王 Hades

Nope

20° and 40° are produced by the same arc. Safe to say it isn't cyclic

冥王 Hades

21:43

Found a proof

Akiva Weinberger

22:57

Ask ChatGPT to prove the square root of 4 is irrational

Amit

When taking the wedge product of two forms of a Grasmann algebra $P \wedge Q$ let's say $P$ is a $k$-form and $Q$ is an $m$-form, if $m+k > dimM$ is the result 0?

冥王 Hades

92% score on my Real Analysis exam.

Lower than before

Akiva Weinberger

That's not too bad imo

冥王 Hades

23:15

Yeah its good but it also means I'm slacking

Amit

I think I got it, it becomes linearly dependent so it must be zero due to anti commutation of the basis forms

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$Mathematics$ Mathematics