Mathematics - 2023-04-09 (page 1 of 2)

user 85795

00:04

🙉

shintuku

why would i be interested in an Lp space with p = 10 over, say, p = 20? i.e. what are we trying to do when we try higher values of $p$

AMDG

I'm just looking for a path of reasonable effort forward given what I should know for my profession, and what I need to know for finding optimal solutions within my profession.

I just need a list of options to decide on

user 85795

00:20

@leslietownes your point is well taken, sir. But, for over a hundred years this is what the nearest library had

12 hours ago, by user 85795

AMDG

Bruh, there's nothing wrong with wanting to find the best available option one can reasonably find within say a few months at most.

With the internet, the library of the world is available.

user 85795

Have you read any of Concrete Mathematics by Knuth.

AMDG

No, unironically. I haven't read much of anything. Most of what I've done, if I didn't recall it from what I learned in public school, I figured out for myself because myself is all I (thought) I had.

user 85795

Try it.

AMDG

Though if I were to read Knuth now, it'd probably be quite pleasant.

shintuku

00:23

@user85795 how do you quote yourself in chat?

AMDG

:<message_id>

user 85795

Just like you quote anyone else.

shintuku

testing :63339289

:63339289

AMDG

lol no without spaces and also I just realized that's not the quote you're talking about

shintuku

great success

AMDG

00:25

Doing that creates a reply message

shintuku

@user85795 how do you quote people at all

AMDG

This is the question of time. It has plagued man since the dawn of time.

user 85795

Top left corner of the message is an arrow to click on.

AMDG

?????

user 85795

"reply to message" will pop up

shintuku

00:27

that only makes a reply, no? not the quote

@user85795 test

yeah it only does a reply

AMDG

Nor does it show up for your own messages because yes, no one would ever want to reply to himself.

user 85795

Above that is a 'permalink'

AMDG

said a programmer at the dawn of time

Yes but vat iz dee seentaks

user 85795

Copy the 'Permalink'

AMDG

In discord, it's straightforward because it uses markdown which... is funny because it varies slightly in this chat from markdown that we're accustomed to on there.

> quote

> quote

This shouldn't be quoted, but it is

shintuku

00:30

2 mins ago, by shintuku

yeah it only does a reply

yesss!!

thanks

AMDG

For some reason the newlines do not delimit quotes like in Discord.

52 secs ago, by AMDG

> quote

This shouldn't be quoted, but it is

user 85795

chat.stackexchange.com/faq#permalink

AMDG

The best interface in the world. You have to post a message link explicitly and then for some reason it gets turned into a quote.

Which would unironically be ideal in discord

Completely opposite problem

Would be better if it did just give the link verbatim followed by an embed like Discord:

user 85795

The Art Of Computer Programming by Knuth is a classic.

AMDG

I've never read the classics. Or played them for that matter.

user 85795

00:35

12 mins ago, by user 85795

Try it.

AMDG

@user85795 Well as you said here... "[n][...][o]"

Important example of course. Always gotta have more than one.

The Hobbit but just Tauriel's lines from the book

►

^ really interesting watch. Love it. Very informative. Even provides page numbers and everything.

robjohn

01:22

@AMDG Must be short

user 85795

Yup, as short as this chapter.

Apr 1 at 6:42, by user 726941

AMDG

@robjohn Spoiler alert: it's 14 seconds

Have some common sense algebra guys.

Akiva Weinberger

01:38

It's wild conjecture time

0

Q: Does there exist a sequence of antiderivatives $f_0,f_1,f_2,\dots$ with $f_i(0),f_i(1)$ integers?

Does there exist an infinite sequence of differentiable functions $f_0,f_1,f_2,\ldots:[0,1]\to\Bbb R$, not all the zero function, such that, for all $i$, $f_{i+1}'=f_i$, and in addition $f_i(0)$ and $f_i(1)$ are both integers for all $i$? I conjecture the answer is no, though I'm unsure of how to...

AMDG

Yes. Absolutely. Without a doubt. By the Nah'yeev-May-D'tup Theorem.

Akiva Weinberger

Ah yes, I know his nephew

AMDG

You mean their nephews. And I don't mean that in a woke sense either. Those three people have been instrumental... I mean, without them, how would anything in mathematics be proven? Their laughable objections--a devil's advocate if you will--are the only reason why rigor exists in mathematics!

user 85795

@AMDG if you replace "garbage" with ♾️ some of them approach "sensible."

Just like physicists used to write the speed of light as c = ♾️.

AMDG

@user85795 LIsten. Bud. I know you're new to this, so I'll go easy on you... but... we all know you can't sense abstract entities, let alone fathom infinity in all its being, so that makes it even less sensible.

Akiva Weinberger

01:48

Infinity smells like a raspberry

AMDG

That's not true. It smells like bananas, the objectively worst-smelling vegetable of all time.

user 85795

Can you use "true" with ♾️

AMDG

I don't know, can you?

user 85795

Does it equal itself.

AMDG

The correct way to phrase such questions is "May I use 'true' with ♾️?"

user 85795

01:51

What is "true" then.

Akiva Weinberger

Let $\kappa$ be the cardinality of the set of all grammatically incorrect sentences.

user 85795

Light travelled at ♾️

c = ♾️

AMDG

@AkivaWeinberger Then κ is me every time I try to speak another language.

user 85795

No experiment said otherwise.

Albeit they believed in the aether wind :P

AMDG

That must mean that $c = -\frac{1}{12}$

user 85795

01:54

lol

So the aether wind created the minus sign in -1/12

Or was it Ramanujan

AMDG

Now simply put, if you $\infty\exp(\frac{90}{360}\pi i)$, then clearly we get $8$.

user 85795

@AMDG thanks for taking it easy on me, bud 🙏

AMDG

Egg. Tough crowd I see.

@user85795 Any time, young apprentice.

AMDG

02:10

Naysayers will say this actually equals -1 chat.stackexchange.com/transcript/message/63339590#63339590

Unfortunately I forgot the factor of 2 so it's all downhill from here

user 85795

@AkivaWeinberger we'll have to try and get professor Hamkins into the chatroom to confirm that.

Perhaps, invite him from Twitter?

Ajay

02:25

Does anybody know why the conjugate of $a + b$ is $a - b$ where $a$ and $b$ are real numbers?

Ted Shifrin

That actually doesn’t make sense in general.

The general terminology comes from field theory — things like $\sqrt 2 \pm\sqrt 3$.

leslie townes

yeah. maybe more helpful to think of that as an operation on algebraic expressions (e.g. the difference of squares identity, used e.g. in 'rationalizing the denominator' when a and b are sqrts of stuff) than on real numbers.

totally makes sense to say, flip the sign in a - b to a + b. makes less sense to say, like, what do you get from the real number 1 when you flip the sign. is 1 = 1 - 0 for this purpose, or 0 - (-1), or some other thing.

user 858770

02:44

@Ajay did you forget the "i" in "bi"?

Ted Shifrin

It generalizes to things with cube roots, too.

Ajay

03:13

@user858770 no.

@leslietownes I've never thought of it like that

I always just think flip the sign

Ted Shifrin

Only for square roots.

The conjugate of $3+\root3\of 2$ is surprising.

Akiva Weinberger

If $a+bi$ is the root of a polynomial with real coefficients, then $a-bi$ is also a root of the same polynomial

Ajay

Conjugate pairs

👍🏽

Ajay

03:36

@TedShifrin Why is the conjugate of that surprising?

wouldn't the plus just become minus?

Ted Shifrin

Nope.

Ajay

Why?

Ted Shifrin

What is point of the conjugate? You multiply and end up in the rational numbers.

Ajay

03:59

Say we have a sequence $(a_n)^{\infty}_{n=1}$ of real numbers. Is the following implication always true? $$\sum_{n=1}^{\infty} a_n \quad \textrm{converges} \implies \sum_{n=1}^{\infty} a^{3}_n \quad \textrm{converges}$$

Ted Shifrin

I’ve assigned that question before.

What have you figured out?

Ajay

I tried the comparison test

Ted Shifrin

Assuming what?

Ajay

$|a_n|$ is convergent

Ted Shifrin

OK, so then we’re done.

Ajay

04:12

We know that $|a_n| \to 0$ whenever $n \to \infty$

Huh, how?

Ted Shifrin

If abs conv, then it is true. But what otherwise?

Ajay

Wait, so then the implication is false?

Wait i'm being stupid

leslie townes

huh? thats a normal question to ask. it's an interesting problem, if sum a_n is not assumed to converge absolutely.

at least, i don't see it as a "look at it and just 'see the answer'" kind of thing.

exactly the kind of thing a prof like ted would assign.

Ted Shifrin

04:32

I even put this into Spivak as an exercise in the 3rd edition, I think. It’s surprising.

@Ajay Did I say that?

Ajay

@TedShifrin Say what?

Ted Shifrin

See what I linked to.

leslie townes

ted invited you to think about what happened if sum |a_n| is not assumed to be convergent. he didn't tell you what might or might not happen in that case. it's a separate thing, not resolved one way or the other by addressing only the absolutely convergent case.

it's certainly worth thinking about why it might not be true in general, or why it ought to be. i don't know that the absolutely convergent case is much of a help.

Franklin

06:30

Can anyone please help me with this:

0

Q: Find a particular integral of $y=\frac{1}{D^2+4D+5}x.$

Find a particular integral of $y=\frac{1}{D^2+4D+5}x.$ I tried solving it like this: Given, $y=\frac{1}{D^2+4D+5}x.$ So, we can say, $$y=\frac{1}{D^2+4D+5}x\implies y=\frac{1}{4D}(\frac D4+\frac{5}{4D}+1)^{-1}x=\frac{1}{4D}(1-\frac D4-\frac{5}{4D}+\frac{D^2}{16}+\frac{25}{16D^2}+....$$ But this...

calculus ordinary-differential-equations

Ajay

Never mind, I didn't see the date

robjohn

07:08

The minor edit was not really needed.

Prateek Mourya

07:26

how to show

@robjohn

robjohn

What does $\lt$ mean in this context?

leslie townes

08:02

what indeed

the "for any linear transformation T" at the end of that is so stylistically awful.

Koro

08:44

@robjohn this usually denotes subspace.

subgroup also depending upon the context.

leslie townes

sheesh, koro. spoiler alert.

one potato two potato

nobody denotes like that

robjohn

09:53

@PrateekMourya Pretty much, you need to verify the properties of a subspace. If $x,y\in S$ and $a,b\in\mathbb{R}\text{ (or }\mathbb{C}\text{)}$, then $ax+by\in S$, etc.

Sourav Ghosh

10:26

@PrateekMourya Do you the definition of a subspace or linear subspace?

As robjohn said, all you need to check for all u, v in \ker T and \alpha in K such that T(u+\alpha v) \in \ker T

T(u+\alpha v) =Tu+\alpha Tv=0

PNDas

10:54

We found solution of wave equation in $\mathbb{R, R}_+$. For $\mathbb R_+$, we have two cases when $0\leq t\leq x$ and $0\leq x\leq t$. When we find solution in $\mathbb R^3$ using Euler-Poisson-Darboux equation,we convert the equation to $\mathbb R_+$.

But we only consider $0\leq r\leq t$. why?

users.math.msu.edu/users/yanb/849-full-note.pdf. section 4.3.2

Franklin

11:47

I was struggling with this problem a lot and doing enormous amount of bashing but all my travails ended up as a failure!

It was then I found this above solution.

It's a pretty clever approach. But I dont the motivation behind it?

Is it just heuristics ?

To me, this approach seems uncanny in a way, that there was indeed a dilly of a simplification step involved in the solution( I mean considering the two new variables x and y and proceeding like that : this was witty in my opinion)

Koro

12:05

x^2 is prime element of k(x^2), k is field.

Ajay

12:25

Hey @robjohn Could you permanently remove one of my answers?

Can you delete my answer to this question: math.stackexchange.com/questions/4675611/…

I skimmed past the post description and didn't see that the OP had not tried anything on their own

PNDas

12:49

@Franklin To me it's pretty obvious. The only part that is not obvious is why $x,y$ are integers.

This also became clear after thinking for a few seconds.

robjohn

13:03

@Ajay there is no way to remove an answer completely. People with enough reputation will still be able to see it. However, since you've deleted it, no one can flag or downvote it.

Happy Passover/Easter

Arthur

13:20

1

Q: The average of 10 scores is $25,$ and the lowest score is $20.$ So, the highest score must be...

The average score of $10$ students in a test is $25.$ The lowest score is $20.$ Then the highest score is: $$A.100,$$ $$B.70,$$ $$C.30,$$ $$D.75$$ The answer key suggests option $B$ as the answer. I don't understand why this differs from my solution: Let the marks obtained by each student be $x...

algebra-precalculus statistics solution-verification average means

Can anyone please help me with this strange question

?

robjohn

It can be at most $70$

it must be at least $25\frac59$

I see that answer is already there.

PNDas

13:50

In my book it is written that for any $\xi\in\mathbb R^n$, $|\widehat{f_n}(\xi)-\widehat{f}(\xi)|\leq ||f_n-f||_{L^1(\mathbb R^n)}$. What I want to ask is shouldn't there be a $||\xi||$ term?

Is it true that $|f(x)|\leq ||f||_{L^1}$?

Franklin

@PNDas so is that a standard way, of solving these problems?

robjohn

@PNDas There is no $\|\xi\|$ there. The $\|\hat f\|_\infty\le\|f\|_1$

If $f\in L^1$, then $\hat f$ is continuous, therefore, we can get the pointwise maximum.

PNDas

Ahh yes

I forgot about that inequality.

Thank you.

@Franklin Well the straight forward way would be $b-d=(c+9)^{3/2}-c^{5/4}$, but then I don't know how to proceed from here.

Otherway would be $9=a-c=(b^{1/3})^2-(d^{2/5})^2$.

Franklin

14:14

@PNDas That's the point. I am not saying this impossible of course. But this appears to be some far out of the box, way of bashing things out. I guess this should be kept in mind as an important strategy for solving these problems!

Sourav Ghosh

How to count the number of matrices similar to a 3×3 daigonal matrix diag(1,1,2) over some finite field, say F_3 ?

Obliv

$\overline{X}\to \mu$ as $n$ increases?

or the mean of the $\overline{X}$s approaches $\mu$ as the number of samples increases? is that more accurate

Sourav Ghosh

A~B iff char(A) =char(B) & min. poly. (A) =min poly(B) as matrices are of order 3.

A~B iff trace(A) =trace(B) & det(A) =det(B)

A~B iff char(A) =char(B) & min. poly. (A) =min poly(B) as matrices are of order 3.

Obliv

Nvm it's clear to me that the set of $\overline{X}$ has mean of $\mu$ for any given size sample

or $$\mu = \frac{\sum_i \overline{X}_i}{(N/n)}$$ I believe

Sourav Ghosh

So the problem is reduced to find number of matrices of size 3×3 over F_3 having trace 4=1 and det 2

Franklin

14:46

Can anyone please help me how to ascertain the number of terms in the expression $(x_1+x_2+...+x_r)^n$ ?

My reasoning goes like this: If we have, $(x_1+x_2+...+x_r)^n$ , then all the terms in this expansion, will be of the form $(K)(x_1^{a_1}x_2^{a_2}...x_r^{a_r}$, where, $a_1+a_2+...+a_r=n$, and hence all possible terms are of the number of non-negative solutions of $a_1+a_2+...+a_r=n$, i.e $\binom{n+r-1}{r}. (K is a real constant ).

Are my reasonings correct/valid ?

I am sorry, there's a typo in the statement: "hence all possible terms are of the number of non-negative solutions of $a_1+a_2+...+a_r=n$, i.e $\binom{n+r-1}{r}. (K is a real constant )." This should be hence all possible terms are of the number of non-negative solutions of $a_1+a_2+...+a_r=n$, i.e $\binom{n+r-1}{r-1}. (K is a real constant ).

Arthur

15:25

The correct link is

0

Q: What is the percentage of students required atleast equal to?

If atleast 90 percent students are good at sports, atleast 80 percent students are good at music and atleast 70 percent students are good at studies, then the percentage of students who are good in all three is atleast equal to? My approach: I couldn't quite proceed in the the proper way, howeve...

probability

I need some help in this...

The answers given are incomplete in my opinion

[I did a repost as my prev messages were faulty]

Sorry for all the confusion...

Ted Shifrin

17:10

@Arthur So let's see your attempt at a more complete solution. The second one posted is quite correct (especially with the comment after).

Franklin

17:20

Thanks everyone! I got it verified and I feel that my approach is correct.

Can anyone please help me with this:

0

Q: Let $f:[0,1]\to (1,\infty)$ be a continuous function. let $g(x)=\frac 1x$ for $x>0.$ then, the equation $f(x)=g(x)$ has

Let $f:[0,1]\to (1,\infty)$ be a continuous function. let $g(x)=\frac 1x$ for $x>0.$ then, the equation $f(x)=g(x)$ has $$A.\text{no solution},\quad \quad B.\text{all points in (0,1]},\quad \quad C. \text{atleast one solution}, \quad \quad D.\text{none of the above}$$ My solution goes like this...

functions solution-verification

Arthur

@TedShifrin the solution posted by drhab is correct, I see. I find sense in it. Is that answer you were talking about?

Ted Shifrin

That is terribly written, @Franklin. I don't see a proof of anything.

@Arthur: Is that the second posted solution? No, there's no reason it should be correct.

Franklin

@TedShifrin Was there something to prove ? 😂😂😂

Ted Shifrin

If you're going to post on a mathematics site, why not post actual mathematics?

We're sick of your non-stop multiple choice stuff.

Hippalectryon

@Franklin Your question doesn't even have a finished title ^^"

Ted Shifrin

17:28

@Hippa :D Not surprising :)

Just writing out the entire question is NOT a title.

Hippalectryon

@TedShifrin Are you coming back to France anytime soon by any chance ?

Franklin

@TedShifrin 😂😂😂 Now, that you say it. Yeah, lately, I am posting soooo much mcqs. But I actually created a post on mcqs after a long while. Mostly, I do discuss here....

Ted Shifrin

@Hippalectryon I haven't traveled anywhere since Covid. I would like to, but it may have to wait until I have surgery on my lower back. Why do you ask?

@Franklin The point is that this is a good example of an exercise where you can practice using actual mathematics from your basic calculus/analysis course.

Hippalectryon

@TedShifrin Just wondering :) Sorry to hear about your back

Ted Shifrin

We need a fancier meal next time, @Hippa :)

Franklin

17:32

@TedShifrin I am not that creative writing titles, Idk. I will let some community members suggest it...🥲🥲🥲

Hippalectryon

@TedShifrin Let me know next time you come and I'll invite you :D

@Franklin It has nothing to do with being creative, it has to do with writing a title that makes sense without reading your question

Franklin

@TedShifrin I hope the surgery goes well

Ted Shifrin

I just last night got an email from a Danish woman who, as an exchange student at UGA in 1985-86, took my Calculus with Theory course. She, along with a German guy and a number of the plain old American students, was so impressive. Anyhow, she just emailed out of the blue to thank me for being a good teacher :P

Hippalectryon

@Franklin Also I don't know if you realize, but the way you ask your "question" feels less like a genuine math question and more like a "I have homework to do and i'd like to get good grades so please verify my answers", which is not the spirit of MSE at all

Ted Shifrin

Oh oh, I feel @Hippa's meme identity returning.

Franklin

17:34

@Hippalectryon "with writing a title that makes sense without reading your question" -Now, that is what creativity is!

@Hippalectryon to be honest:

Hippalectryon

@TedShifrin That's so nice !

Ted Shifrin

How about something like A less-than-obvious application of the Intermediate Value Theorem?

Franklin

@Hippalectryon If this sort of questions, were given on homework, then the students would throw eggs or even worse stones on him out of frustration!

Ted Shifrin

No, without the multiple choice, this is actually a good basic analysis question.

Franklin

@TedShifrin But I feel, the options made it easier...

Ted Shifrin

17:37

Franklin, you have to remember that those of us in the US and Europe do not come from a culture of MCQ once we are adults.

We're not interested in your easiest MCQ answer. We're interested in your learning mathematics and making efforts to learn it.

Hippalectryon

What's MCQ btw ?

Ted Shifrin

multiple choice question

Tu aurais dû deviner :P

Hippalectryon

> those of us in the US and Europe do not come from a culture of MCQ

Aah though that was some kind of university / ... !!

Franklin

@TedShifrin I will have to think about it, without mcqs. You know, that's a nice one, indeed...

Ted Shifrin

LOL @Hippa

Franklin

17:39

@TedShifrin Just for curiousity: Are all the tests there of, subjective type?

Ted Shifrin

There are lots and lots of totally standard applications of the IVT. The fact that one function here is defined only on $(0,1]$ makes this one a tiny bit more interesting.

Franklin

@TedShifrin true...can't agree more...

Ted Shifrin

In real mathematics courses (not high school level), some lazy teachers give short-answer tests, but most require work to be written out and evaluated.

Franklin

@TedShifrin @Hippalectryon Now, that's a real homework what @TedShifrin suggests me to do, i.e making it a non-mcq question!!!!!

Ted Shifrin

Of course, for things like most math competitions we have to write MCQ to get things machine graded quickly.

Franklin

17:41

@TedShifrin in undergraduate entrances? Are they all subjective?

Ted Shifrin

This is what most MSE questions have become since you left, @Hippa.

On the advanced placement exams (which earn college credit) there is a third or so which is actual written responses. The rest is MCQ.

The fact that there are MCQ in the world doesn't mean that MSE should be overrun with them. IMHO.

Hippalectryon

@Franklin Sure, but MSE isn't a platform where people verify your homework !

@TedShifrin :/ I mean there was always those questions, but you mean it's become much worse / more frequent ?

Ted Shifrin

MCQ and L'Hôpital's rule. I want them ALL banned.

Franklin

@Hippalectryon come on, come on, come on...that's not a homework. How should I explain...the answer's not given there. Now, you are making me uncomfortable!!! I feel embarassed...🥲🥲🥲

Ted Shifrin

Yes, worse and more frequent and now there's a negligible probability of interesting questions.

So, in one paragraph or less, what's your thesis on, @Hippa?

Hippalectryon

17:52

> I feel embarassed...🥲🥲🥲

I mean... https://i.imgur.com/6JUro5p.png MSE is a place where people help *each other*, not where people take their precious time to check your answers. If you start contributing to the website by answering questions yourself, you'll maybe realize why we're so turned off by this kind of question :)

@TedShifrin I'm working on new "mathematical" models to perform numerical simulations of fluid equations on very decimated spaces (hence at a very reduced numerical cost).

Ted Shifrin

What are "very decimated spaces"?

Hippalectryon

Think of a "normal" numerical simulation. Usually, it's performed on a spatial grid of a given resolution, where every "pixel" has the same size. Hence to simulate a system with a biggest size L and a pixel size l, you need N=(L/l)^D points in D dimensions.

We're working on a "subset" of the grid (in spectral space), such that N~log(L/l)^D

Ted Shifrin

Ah, sparser sampling.

Hippalectryon

Yep. We call that "log-grids".

Sourav Ghosh

@Franklin g(x) =1/x has no continuous extension over [0, 1] :)

Ted Shifrin

17:57

@SouravGhosh Why did you comment this?

Sourav Ghosh

@TedShifrin Is there anything wrong?

Ted Shifrin

I think we all know that, and that's why my chosen title for the question was "a less-than-obvious application of IVT."

Sourav Ghosh

Continuity is weaker than IVT :)

Ted Shifrin

18:12

I do not get your point.

Sourav Ghosh

18:28

How to count the number of matrices similar to a 3×3 diagonal matrix diag(1,1,2) over some finite field, say F_3 ?

A~B iff char(A) =char(B) & min. poly. (A) =min poly(B) as matrices are of order 3.
A~B iff trace(A) =trace(B) & det(A) =det(B)

Ted Shifrin

The word similar has a well-defined meaning.

Now you're defining $\sim$ with weaker notions.

Sourav Ghosh

Yes. Two matrices are similar iff both represents the same linear map (possibly under two different bases)

Ted Shifrin

So what are the last two lines?

Sourav Ghosh

Well. For matrices of small size( upto 3×3) last two lines are equivalent but not in general.

Ted Shifrin

I just cannot make sense of things you say. I'm out.

Sourav Ghosh

18:34

Notation A~B : A, B are similar.

schn

18:48

I am studying the following series: $$\sum_{k=1}^\infty \frac{1+(-1)^{k}k}{k^{2}}.$$ Does it converge absolutely? I am not allowed to use the alternating series test.

I have tried to use the traingle inequality to get an upper bound, but that only diverges.

(I know that the above series converges...)

Ted Shifrin

What do the terms look like for large $k$?

schn

$ \frac{1}{k^2}$ ?

Ted Shifrin

@SouravGhosh what about $\begin{bmatrix} 1\\&1\\&&2\end{bmatrix}$ and $begin{bmatrix} 1&1\\&1\\&&2\end{bmatrix}$?6

@schn Really?

You need work on your basic intuition.

D Left Adjoint to U

Koro

@DLeftAdjointtoU cool!

D Left Adjoint to U

18:59

I have hacked the infamous Quiver editor

into my own django template

Ted Shifrin

Ugh. stupid \ missing.

D Left Adjoint to U

I'm actually using BSS to visually edit a lot like a WYSIWYG web editor. Then hand coding the Django side

Koro

@TedShifrin :(

D Left Adjoint to U

@Koro want to see the code?

Would you like to see the ruins my friend

schn

@TedShifrin maybe $\frac{(-1)^k}{k^2}$?

Koro

19:00

I won't be able to understand it atm. :(

Ted Shifrin

We need more than 10 milliseconds to edit on mobile.

Koro

I'm doing field theory.

D Left Adjoint to U

@Koro nice. What theorem are you on?

Ted Shifrin

@schn don’t worry about sign. Worry about magnitude.

Koro

@DLeftAdjointtoU So I have done: algebraic extensions, splitting fields.

I know definitions of separable and normal fields.

D Left Adjoint to U

19:02

Nice.

Want to study together on that?

Ted Shifrin

Extensions, not fields

Koro

yeah we can.

D Left Adjoint to U

K, I'll create a room

Koro

sure

Ted Shifrin

@schn what does $\dfrac{k+k^2}{2k+k^3}$ look like? Does the series converge or diverge?

schn

19:06

let me think

D.C. the III

Good day to the peanut gallery.

Ted. I had some questions on understanding with regards to showing $f(x) + g() = \frac{\pi}{4}$.

(i) To do the first part I only was successful in working it out because it was hinted to look at $f(x) + g(x)$ as a constant and to compute the value of $f(x) + g(x)$ at $x = 0$. From this everythign worked out. But where did the "idea" for thinking about things like this come from? I wouldn't have thought about looking at that quantity as a constant and I would not have thought of computing it at $x = 0$.

Ted Shifrin

I respond only to hazelnuts.

D.C. the III

Nutella?

I have a feeling you don't like nutella

Ted Shifrin

No, way too sweet.

D.C. the III

if there was a more mellow version?

Koro

19:18

Have you ever used jaggery powder instead of white sugar?

D.C. the III

or thin layer sparsley spread over a baguette...surely this would be good

Ted Shifrin

Any time you integrate you need a point (initial condition) to get the constant. $0$ is the only computable point. Of course you used (a).

schn

@TedShifrin $$\frac{k+k^2}{2k+k^3}\geq \frac{k^2}{2k^3+k^3}=\frac{1}{3k}$$, we expect it to diverge :)

D.C. the III

How was (a) used?

Ted Shifrin

You should use limit comparison and take 2 seconds, not 20 minutes.

D.C. the III

19:21

the mere fact that $f'(x) + g'(x) = 0$ implying that my value is a constant?

Ted Shifrin

To know the function is constant? How else is the value at $0$ telling you anything?

D.C. the III

Answered my own question when I wrote that last question.

Ted Shifrin

@schn the dominating terms tell you it looks like $k^2/k^3 = 1/k$ for large $k$.

schn

yes

Ted Shifrin

In your question, what are the dominating terms?

schn

19:25

for large $k$, we have $\frac{1}{k}$, no?

Ted Shifrin

Yes, ignoring sign.

You wanted to look at absolute values.

Koro

Let $f(x)\in k[x]$ be irreducible of degree p, p prime. Let L be a splitting field of f(x) over k. If $g(x)\in k[x]$ is a polynomial of degree q, q>p, such that there is a in L with g(a)=0. How to show that g(x) is not irreducible in k[x]?

schn

Yes.

Ted Shifrin

@Koro Surely $f$ is relevant.

Koro

yeah, I don't understand how to bring that into mix.

Suppose g(x) is irred. Then I know that k[x]/(g(x)) $\cong k(a)$.

Ted Shifrin

19:31

No contradiction. Think about two polynomials.

Koro

I still don't understand how to prove it though.

examples for verifying the situation are fine.

Ted Shifrin

Given any two polynomials, what is natural to consider?

Koro

product?

Ted Shifrin

Try again.

Koro

ah, another thing: [k(a): k]= deg g(x)=q

(if g is assumed irred.)

Ted Shifrin

19:37

What is the most important notion in ring theory going back to the integers?

Koro

I have no idea.

in groups (finite), order. But in rings, I don't know.

Ted Shifrin

Seriously? Two integers? Two polynomials? What is their ….

Koro

ah, gcd

Ted Shifrin

Now go finish.

Koro

I'm afraid I don't understand how to use this information.

:(

if f has a root b in L, then [k(b):k]=p

If I could somehow show here that $k(a)\subset k(b),$ subfield or $k(b)\subset k(a)$, subfield, then I'll get a contradiction as follows: $[k(a):k]=[k(a):k(b)][k(b):k]$

But the containment does not seem true and it may not be true considering the fact I nowhere used the fact that L is a splitting field.

Ted Shifrin

19:52

So you totally ignored what I suggested. Your typical behavior, I might add.

Can they be relatively prime?

Koro

I didn't ignore. I thought on but didn't understand how to go ahead as I said earlier.

Ted Shifrin

🤷‍♂️🤷‍♂️

Koro

I think they can be rel prime.

$a_1(x) f(x)+a_2(x) g(x)=1$, $a_i(x)\in k[x]$.

It is not obvious from here why such $a_i$'s should not exist.

The above is from the fact that: if (f(x), g(x))= k[x], then...

Ted Shifrin

Plug in $x=a\in L$.

Koro

We get that a is not a root of f as g(a)=0.

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$Mathematics$ Mathematics