12:00 AM
For that, $H$ and $K$ are both normal subgroups. You need at least one to be normal for the product to be a group.

however, why do we require a normal subgroup? isn't there a less restrictive condition?

Anyhow, I should not be talking algebra. It's forbidden.
How do you define the product of two elements of $HK$?

it's the product in $G$... (?)

Why does that make it an element of $HK$ again?

@LucasHenrique this is a notion that exists, it's called the Zappa-Szep product

12:02 AM
Never in my life heard of that!

there's also an external construction of the Zappa-Szep product, but it's rather unwieldy

(Nor in my death.)

yes Ted, I don't blame you
I just hoard useless knowledge

i'll be honest, i checked to see if that spelled something rude backwards

But @Lucas should appreciate that point of what I was saying, even though I'm forbidden to discuss algebra.
ROFL @Leslie ... Was it?

12:04 AM
not in any language that i speak.

More Z's and p's per length than any other name, I'm sure.

yeah, it's an important point

@TedShifrin ok, $HKHK = KHHK = KHK = KKH = KH = HK$ would be great

So doesn't that look a lot like normality for one of them?

having a normal subgroup is justifiable. thanks!

12:09 AM
So I looked at wiki, @Thor, and I don't see how one has a meaningful notion. "$G$ is said to be an internal Zappa=Szép product of $H$ and $K$." OK, but how do we multiply?
Oh, the external notion looks like a generalization of semidirect products.
I see, so it's stupid. You multiply elements of $HK$ in $G$ and use the fact that each element of $G$ is uniquely in $HK$ to say that's the answer. I do not like this.
I reject this out of hand.
But arcane reference award goes to Thor.

yeah, the external notion basically just comes by specifying maps that tell us how to switch the factors
but, I mean, what else would you expect

it worked for Zappa and Szep. i need to find something like that. bust my way back into academia.

12:24 AM
You'd be better off emulating Frank Zappa.

i think that's right.

12:48 AM
Out of curiosity, how long does it take one to write an undergraduate math textbook?

What should I do if I've noticed that a Wikipedia article states a formula several times, with a reference in which the formula isn't actually proved? Note that Prime-counting function is often used.

i have never edited wikipedia. maybe note it on the talk page? it may be howling into the void.
it would seem hard for random people in their basements (what i imagine most wikipedians to be) to analyze questions regarding what is or isn't proved in a textbook they might not have

*Correction: ...that the wikipedia article of the "prime counting function" is often visited.

i expect people might actually be reading the talk page. who knows. what a sordid corner of the internet.

12:59 AM
the most recent posts upon it appear to be from 2019 and discuss a book saying certain formulas do not have proofs in the literature. maybe uncited formulas are running rampant on that page.

I feel like writing it on the talk page results in absolutely nothing. I'll ask a few people if they also think that there is no proof (at least not in the references) and then I'll just delete everything

7 hours later…
7:43 AM
I have a doubt..
If 3/2 and 4 are the two roots of a quadratic equation ,then which one of the following is not the correct quadratic equation?
(A) 2x^2-11x+6=0 (B) 6x^2-33x+18=0 (C) -10x^2-55x-30=0 (D) 4x^2-41x+24=0
I tried using the analogy(x-3/2)(x-4) but it came to be of no use..
the equation comes out to have a constant of +12..
Ideally it should be +6 which I feel from the options any assitance about the same will be most welcome..

8:02 AM
Expand the product. Normalize all the equations by dividing by the coefficient of $x^2$.

I did..
This is what I got
$2x^2-11x+12=0$
What is meant by normalizing @robjohn

@RajorshiKoyal dividing by the coefficient of $x^2$ gives $x^2-\frac{11}2x+6=0$

ok
So how is that helpful here?

It allows you to compare the polynomials coefficient by coefficient.

ok..
So what is the answer here?
I do not find any match here?

8:16 AM
A: $x^2-\frac{11}2x+3=0$ B: $x^2-\frac{11}2x+3=0$ C: $x^2+\frac{11}2x+3=0$ D: $x^2-\frac{41}4x+6=0$

None match

maybe the question is wrong..

8:35 AM
I have another question so to get an equation as an identity for a quadratic equation (ax^2+bx+c=0) what do I need to do.
Are the terms a,b and c all of them independently going to represent the number 0?
@robjohn Please tell this to me..

yeet

8:54 AM
If $x \in \mathbb{C}^{m}$ and $A$ is a $m \times n$ matrix, prove the following inequalities:
(a) $\|x\|_{\infty} \leq\|x\|_{2}$.
(b) $\|x\|_{2} \leq \sqrt{m}\|x\|_{\infty}$.
(c) $\mid A\left\|_{\infty} \leq \sqrt{n}\right\| A \|_{2}$.
(d) $\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty}$.
How to do this last 2 inequalities I know some how I need to use (a) and (b) and use operator norm but vector product is not defined for this case
@LeakyNun Any idea or suggestion ?

@RajorshiKoyal I am not sure what you are asking.

9:18 AM
@mathsstudent Did you try to check some posts on the main site about those problems? Such as Matrix Norm Inequality $\lVert A\rVert_\infty \leq \sqrt{n} \lVert A\rVert_2$, Matrix norm inequalities for 2-norm and $\infty$-norm (And you'll be able to find many other posts for sure.)
This is what SearchOnMath return for (c) and for (d). (IIRC Approach0 will be available again after May 12.)

@MartinSleziak Is it unavailable now? I hadn't realized.
That's unfortunate. It would be nice if there were an announcement when you go to the site.

9:35 AM
in In the search of a question, May 5 at 18:48, by Wei Zhong
Announcement: Because we are using approach0 to attend a math search competition, and to the best of my team interests, approach0 will resume service after May 12.
in In the search of a question, May 6 at 9:18, by Wei Zhong
thanks, everybody. @MartinR, the message is changed now, refresh your browser to see. Thank you for your suggestion.

@MartinSleziak Thanks! I just looked at their Twitter post
I was just about to search for something, too! :-(

I missed that one:
I always forget that tweets are no longer oneboxed.
@robjohn Well, on the bright side, you have an opportunity to get more familiar with SearchOnMath (or some other search engine).

@MartinSleziak Indeed. I will look into them.

2 hours later…
11:52 AM
Is there a way that I can save a post of stack exchange Questions and then review them later ?
Also , I saw that you can clock on favourite but how do you review them afterwards

12:06 PM
you can bookmark questions and then access them through your profile
the bookmark is the symbol below the downvote button

12:23 PM
@Thorgott do you mean favourite ?

I don't think such an option exist
perhaps you're assuming that's what it is, because the bookmark symbol has a star on it, but it's a bookmark

12:42 PM
I think it used to be called favorites at some point

interesting

The question is: Define a first order language for trigonometry. State the arity of all the function and relation symbols you want your language to have.
My attempt: the language will be $$\mathcal{L} ~\text{is} ~ \{ 0, 1, +, \times, \sin, \cos , \tan , \leq \}$$

The arity of $+$ and $\times$ is two, and that of $\sin, \cos, \tan$ is one. The arity of $\leq$ is two.
Is my solution correct?

The question is : Is the field of Laurent series complete?
Is my solution correct?

12:57 PM
I have a finite nonempty set G with an associative binary operation. Cancellation also holds on G. Then G is to be proven to be a group.
I understand that since G is given to be non empty, there is no harm in assuming that let $a\in G$. Let $|G|=m$. And by closure $a,a^2,a^3,\cdots, a^{m-1}$ should also be in group. It is to be noted here that since $G$ is finite for every $i\in \mathbb N$ there must exist a $j$ such that $a^i=a^j$
Case 1: $a,a^2,\cdots, a^{m-1}$ are all distinct.
Then $a^m \in G$ by closure. It follows that $a^m$ can’t be equal to $a^j$ for any $j: 1\le j\lt m$.
I’m stuck here. Any hints on how to proceed? Thanks.
@leslietownes

1:28 PM
I'm not sure if this claim is correct
maybe I'm missing something
to be clear, you are not a priori assuming the existence of an identity element?

1:50 PM
Does anybody have any idea how do I find focus and directrix of this parabola :$9x^2 - 24xy+16y^2 -60y+100=0$

@Thorgott: Indeed. I’m not assuming the existence of identity.
But I somehow managed to solve it.

@Bhavay first convert it in standard form

@Rover it can't be . it contains xy . it is a tilted parabola.

@Bhavay: put $X=(3x-4y)$ and $Y=(y-\frac 53)$ and the equation becomes $X^2=60 Y$. Can you finish?

2:03 PM
@Bhavay Yes, it's tilted, so we can rotate axis by say thetha , then in standard form, i.e when it's axis are parallel to coordinate axes coefficient of xy =0.

@Koro Okay , so I plot $x^2 = 60y$ and after that , I am gonna need a little more help .
@Rover and how do I do that ?

By transforming axes and replace old coordinates by the relation of new ones .
By using this

2:20 PM
ah ok, now I see it too

Focus is at $(X,Y)=(0,15)=(3x-4y, y-\frac 53)$. Now solve for $x,y$ @Bhavay

@Rover and how do I know , by what angle do I need to tilt the axes ?
@Koro got it.

Similarly, find directrix using hint by @Rover

Is there anyone who can answer my question?

2:40 PM
If $f$ is a complex measurable function, what does $|f|$ mean?
Is it a real valued function which returns the absolute value of the complex output of $f$?

yes

ok ty

3:04 PM
Anyone here have more than 100k rep in math.SE

i have

3:48 PM
hello
is there anyone that can answer a very basic machine learning doubt?

I finally got 500 rep in math.SE yay

congrats.

4:30 PM
0

So I have just recently started exploring machine learning, and for a project I was required to train the YOLO v5 model. I first tried it on the coco128 dataset:https://www.kaggle.com/ultralytics/coco128.. repository of the yolo v5: https://github.com/ultralytics/yolov5 I followed this tutorial: ...

can anyone help with this question?
its a bit urgent
I dont want anything in depth, just a superficial expression

3 hours later…
7:15 PM
superficial expression?

Hi Ted

Heya, a Balarka

7:27 PM
@TedShifrin the expression that all the cheerleaders in high school have.

Among others … How does older age become you?

@TedShifrin the funny thing was, I was under the impression that I was already 62, then, when I recomputed, I was pleased to find that I had not aged.

7:42 PM
great news! or, in this case, non-news.

@robjohn I can't wait for next year's computation!

@TedShifrin yeah. I must have tripped while I was carrying the one...

With Common Core math, you can walk less carefully.

8:25 PM
Say we are integrating something as simple as $\frac{x}{e^{x^2}}$ and for some reason you mess up and u-sub. setting $u = e^{x^2}$.
You get something like $\frac{x}{u} \frac{du}{2x e^{x^2}}$ as the integrand. It kind of looks like you could "re sub" for that $e^{x^2}$ in the second fraction to get $\frac{1}{2} \frac{du}{u^2}$ as the integrand, which still integrates to the correct answer. Just wondering if you can "double substitute" like that or if this is some coincidence.
haven't done calc in a while

if $u = e^{x^2}$ then $du = 2x e^{x^2} \, dx$ and indeed there is nothing wrong with writing this as $du = 2x u \, dx$

yeah I see it, I guess I just forgot it could happen
pretty neat! forgot how nice calculus was

note it's not 'resubstitution' in the sense of a 'substitution' being a change of variable in the calculus sense. it's substitution in the sense of replacing an expression with an equivalent expression.
which is more general and more valuable. but yeah, a lot of people forget that.

yes, that's exactly how I meant it

most of the trig substitutions operate like that, which may be part of why people find them tricky.

2 hours later…
10:57 PM
Does anybody knows an online calculator which will give the equation of the conic in the new rotated coordinate axes ?

11:35 PM
im not sure what that would look like, but the problem is basically diagonalizing a $3\times 3$ matrix. is this coming from a linear algebra context? (you were asking about strangs linear algebra lectures, right?)

Why 3x3?