Mathematics - 2022-01-08

Alex

00:00

Yes, I did not make a good choice of the term, I think. Well I don't have a beard :-(

however, I have some parts in my hair of color gray

copper.hat

i had some gray at 21

leslie townes

01:50

every silver lining's got a touch of grey

Ted Shifrin

So rules Munchkin?

leslie townes

since she was born i have white in my beard now. one hair per bphtpthbpthbhttht

Ted Shifrin

02:23

Uncountably many! Impressive!

leslie townes

it's hilbert's hotel down there

Ted Shifrin

I thought Hilbert was only countable!

dc3rd

how to fit 6 courses into a space for 4 courses?........hmmmm

Ted Shifrin

Go to Hilbert’s hotel. That is clear!

Even 4 hard courses can be much.

leslie townes

this is true. i have always advocated doing the bare minimum for courses.

Ted Shifrin

02:30

Well, you advocate laziness.

leslie townes

i graduated from the UC with 0.5 credits more than the bare minimum. with law school, 0.0 credits more than the minimum.

yes.

Ted Shifrin

I graduated with enough credits for a triple major.

Blah.

leslie townes

my wife did something like that, although only a double major.

dc3rd

I agree with the Hilbert Hotel and even 4 hard courses being much. I'm not doing them right now. Trying to plan my next 8 months and what I still feel I "require" to personally be sufficiently prepared to do the GRE

and well as we know with upper level math.....they are all hard.....no such thing as an easy course...

I got credits upon credits at this point.

leslie townes

i remember describing my course taking strategy to someone in law school and they thought i was joking. the brainwashing to do as much at once as possible is real.

dc3rd

02:35

@leslietownes fuuuuu******kkkk I've spent the last hour coming to peace with "not being able to do everything"

I have a full life to live I don't need to "know it all" within a small time span....that is what "life" is for...

leslie townes

the trouble is that if you work in an advising capacity you basically cannot tell people to do the bare minimum. it would be irresponsible to do it. god forbid they aim for 0.5 credits more than the minimum and fail to pass something, which they probably will, if they're in your office and asking how to do the minimum.

dc3rd

why would it be irresponsible? Provides one the opportunity to be able to truly synthesize the concepts

leslie townes

it's high risk to shoot for the minimum because you absolutely have to hit it.

dc3rd

true

leslie townes

i did a lot of stuff as a student that i discouraged people from doing as a prof. too high risk.

i did sometimes tell people to drop x course or skip y if it wasn't absolutely necessary.

but i did not turn my personal ethos of laziness into a policy of advising.

dc3rd

02:44

so my issue right now is determining what is and what isn't necessary...

123

02:58

Hi All...

Pls clear that what is the difference between method and algorithm in math, not computer science.

leslie townes

in most contexts, i don't think there is a difference. a central difficulty of learning math is learning which words are going to have precise definitions and which words don't. these don't.

AMDG

03:38

@robjohn I think I may have something.

This isn't like those other times, either. Turns out if you actually think, you can get results. Who would have thought?

desmos.com/calculator/clmg3zbkih

See cell 19.

This describes a quotient $\lfloor\frac{2^x}{y}\rfloor$ for two integers $x$ and $y$.

Notice that the second term has a power of two coefficient and there are no floors involved (ignore the $\lfloor x\rfloor$; that's just to make $x$ an integer due to Desmos' limitations).

Meaning it can be substituted with this whole expression but with that particular power of two.

So for any arbitrary amount of precision, you can reduce this to nested coefficients on the remainder and expand the polynomial.

Do this at worst 64 times and you can compute 64-bit divides.

But fast (tm)

Not sure why I didn't come up with this sooner :L

Anyways, will chat more about this in the morning.

(Alternatively, you can just use this to compute a reciprocal approximation and multiply by an integer in fixed-point)

Koro

05:44

Is it possible to write a power series for $\frac{3x^2}{5-2\sqrt[3] x}$ about $x=0$ ?

Supposing $\sqrt[3]{x}<\frac 52$, it follows that $\frac 1{1-\frac 25 \sqrt[3]x}=1+r+r^2+...$, where $r=\frac 25 \sqrt[3]x$.

I don't think that it is possible to write the said power series about $x=0$.

The expansion on right will always have non-integral powers of $x$.

Even if I try with Taylor's series expansion. Repeated application of Chain Rule should imply that there will be $x$ in the denominator after some time and the derivative would fail to exist after some time.

Odestheory12

06:22

@Koro well you can write it as a puiseux series

Puiseux series

In mathematics, Puiseux series are a generalization of power series that allow for negative and fractional exponents of the indeterminate. For example, the series x − 2 + 2 x − 1 /...

leslie townes

07:16

we don't use the p-word in this house

copper.hat

what the p...

leslie townes

your neck

copper.hat

pop, yeah, that's it, pop

oh i forgot, my evening entertainment was to solve one of Ted's puzzles.

pop my puiseux. sorry. i could not stop myself.

Mad Spaces

08:02

This seems to be obvious, but is there a difference between $\mathbb{R^{n+m}} $ and $\mathbb{R^n}\times \mathbb{R^m}$ it seems my professor use this as synonoms which looks wrong to me

Since one takes one element the other an ordered pair.

Koro

Of course, both the above are different vector spaces over field $\mathbb R$. However, they are isomorphic to each other as vector spaces.

@Odestheory12 but what about the 🔋 series?

🔋 is supposed to be emoji for ‘power’. :P

dtn

09:16

Hello. Help formalize the next task. I'm a little confused.

GoogleME

10:07

Is it true that if $f, g$ are coprime polynomials then the matrix $f(C(g))$ is invertible where $C(g)$ is the companion matrix?

PM 2Ring

11:05

@Koro Right. The 3rd derivative fails, and even if we use derivatives near 0 (like $10^{-6}$) they blow up pretty quickly, as this short Sage script shows:

sagecell.sagemath.org/…

But if you only need an approximate polynomial over a restricted domain, then you have a few options. Here are 4th degree Chebyshev & Minimax polys over [0, 1]. Chebyshev: $$-\frac{47}{1719} \, x^{4} + \frac{871}{3260} \, x^{3} + \frac{686}{893} \, x^{2} - \frac{15}{1826} \, x + \frac{1}{6963}$$ Minimax: $$-\frac{77}{2311} \, x^{4} + \frac{64}{229} \, x^{3} + \frac{355}{467} \, x^{2} - \frac{4}{617} \, x + \frac{1}{12951}$$

Here's a plot showing the difference of each of those polys from your function:

FuzzyPixelz

13:40

If any two elements of a join-semilattice are comparable by definition, then the set is not a poset anymore, so why define a join-semilattice in terms of a poset?

Lattice (order)

A lattice is an abstract structure studied in the mathematical subdisciplines of order theory and abstract algebra. It consists of a partially ordered set in which every pair of elements has a unique supremum (also called a least upper bound or join) and a unique infimum (also called a greatest lower bound or meet). An example is given by the power set of a set, partially ordered by inclusion, for which the supremum is the union and the infimum is the intersection. Another example is given by the natural numbers, partially ordered by divisibility, for which the supremum is the least common multiple...

FuzzyPixelz

13:57

For context, I was reading a Type Theory paper and the author used a join-semilattice only to make the join operator on it partial, which to my mind is just a partial order "with extra steps".

MethNoob

14:56

wtf is base e number lol

irrational base...? wtf

AMDG

16:04

Imagine not generalizing bases to all reals

Koro

@PM2Ring :)

AMDG

Imagine transforming functions of multiple arguments into functions of a single argument.

$e^{i\theta} \mapsto \lfloor 2^n \cos(\theta)\rfloor + \lfloor 2^n \sin(\theta)\rfloor 2^{-n}$

LearningCHelpMeV2

16:43

Hello, I would appreciate a hint if possible. I have tried to replicate the above technique by letting $qI = \int_0^{\frac{\pi}{2}} \frac{q\cos(x)}{p\sin(x) + q\cos(x)} dx$ and $pJ = \int_0^{\frac{\pi}{2}} \frac{p\sin(x)}{p\sin(x) + q\cos(x)} dx$. This results in $qI + pJ = \int_0^{\frac{\pi}{2}}\frac{q\cos(x) + p\sin(x)}{p\sin(x) + q\cos(x)}dx = \frac{\pi}{2}$.

However, I am having trouble on this integral $qI - pJ = \int_0^{\frac{\pi}{2}}\frac{q\cos(x) - p\sin(x)}{p\sin(x) + q\cos(x)}dx$

Koro

16:55

@LearningCHelpMeV2 Use this result $\int_0^a f(x) dx=\int_0^a f(a-x) dx$

1010011010

17:10

Hi there, I wish to improve on a Q&A by giving a more complete answer. (it's my own question) I'm reluctant to edit the answer that I accepted with a big explanation. Should I place a separate answer of my own instead? I will keep the check-mark and upvote in place as before

leslie townes

101: that's what i'd do. frankly i wish more people would answer their own questions.

it's pretty common to see a good answer that almost, but not quite, addresses the original question, with a tiny bit of stuff left out. i like to see the stuff that was left out. :)

1010011010

Yes indeed, and a lot of times a lot of basics are just assumed to be known while basics is a very ill-defined word, thus rendering some answers almost unusable to newcomers

PenAndPaperMathematics

Yes, that's true. Dem newbies got to study!

1010011010

Well let's just say, in the end new perspectives tend to find a way of being useful

PenAndPaperMathematics

A power object $PX$ in a topos is injective is what I'm trying to understand now from MacLane & Moerdijk

I've already got that $\Omega$ is injective

LearningCHelpMeV2

17:47

@Koro it didnt work

user193319

18:08

3

Q: If $f_n\rightarrow f$ uniformly on compact subsets $\Omega$ and $f$ is not constant, prove $f(\Omega)\subseteq \Omega$

Let $\Omega$ be open and connected, and let $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ such that $f_n(\Omega)\subseteq \Omega$. If $f_n\rightarrow f$ uniformly on compact subsets $\Omega$ and $f$ is not constant, prove $f(\Omega)\subseteq \Omega$ My attempt: Let $z_0\in \Omega...

complex-analysis

See Story123's answer. Why can we choose a small enough contour $\gamma$ around the point $z_0 \notin \Omega$ so that $\gamma$ doesn't intersect the open set $\Omega$?

user193319

18:19

Story123 claims that it's due to the fact that $\Omega$ is open. But if we can find such a contour for any point outside of $\Omega$, wouldn't that imply that the complement of $\Omega$ is open since any closed contour forms the boundary of a open set (right? or no?)

Koro

18:32

@LearningCHelpMeV2 Note that $p\sin x+q\cos x=\sqrt{p^2+q^2}\sin (x+b),$ where $\tan b=\frac qp$. Then note that the given integral is $I=\frac 1{\sqrt{p^2+q^2}}\int_0^{\frac \pi 2}\frac{\cos x\cos(x+b)}{\sin (2x+2b)}dx$

@Koro Then use this property.

@LearningCHelpMeV2 The idea is to keep the denominator of the integrand same so that we can add $I$'s to simplify the integral.

Ted Shifrin

@user193319 It's a badly written answer, but the idea is there. If $w_0\notin\Omega$, you want to show that $f(z)-w_0$ has no zeroes in $\Omega$. So show that there is no zero inside any closed contour contained in $\Omega$.

Koro

Oh, never mind @LearningCHelpMeV2: I miscalculated $I$. The above needs more work.

user193319

Ah, okay. I see how your rephrasing the argument. Let me write this up. Thanks!

Ted Shifrin

Sure thing. Maybe you should post a correct answer to that question when you're done :)

user193319

Oh, yeah. I suppose I could. I forgot I could answer questions, too lol

Koro

18:53

@LearningCHelpMeV2 First note that $pI-qJ=[\log(p\sin x+q\cos x)]_0^{\frac \pi 2}$. Then, note that $pI+qJ=\color{red}{\int_0^{\frac \pi 2}\frac{\sin (x+b)}{\cos (x-b)}\mathrm{dx}}$, where $\tan b=\frac pq$. To simply the integral in red, do the substitution $u=x-b$ and proceed.

PenAndPaperMathematics

23:10

@Shaun hey :)

Shaun

23:27

Hi @PenAndPaperMathematics. How are you?

PenAndPaperMathematics

hi

I mean hi, how are you

@Shaun, there's not much to finish off Prop 1, I think we can cover it next chat session

You should draw the diagrams on paper (with pen!) to really learn them

I have a notation that I'm using on those blog posts

Shaun

I'm a little concerned: my latest question is said to be a copyright infringement. I disagree but I'm also doubtful.

PenAndPaperMathematics

Show me

Shaun

4

Q: Showing ${\rm Aut}(Q_{2^n})\cong{\rm Hol}(\Bbb Z_{2^{n-1}})$ for $n>3$

This is part of Exercise 5.3.4 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE. The second part is here: Automorphism group of the quaternion group For previous questions of mine on generalised quaternion groups, see here: Find the...

group-theory group-isomorphism quaternions automorphism-group holomorph

Yeah, I think we could discuss Proposition 1 soon too.

PenAndPaperMathematics

@Shaun I upvoted all the positive comments and the post, hopefully we can resolve this. markvs has it out for you :o

You're just asking a question about exercises, as usual, I don't know why they're targeting you and not the whole site

Shaun

23:35

Thank you, @PenAndPaperMathematics. I don't take it personally. It seems like markvs is being sincere.

PenAndPaperMathematics

@Shaun, Quiz: is the product map $m \times m'$ of two monomorphisms also one?

I.e. $m : A \to B, m' : C \to D$.

I will accept an answer in $\textbf{Set}$ for now

Shaun

I'm tempted to say "yes" because monomorphisms are left cancellative by definition and that's something $m\times m'$ would inherit; however, I don't recall the definition of the product map. I'll look it up now . . .

PenAndPaperMathematics

See my recent post

enjoysmath.blogspot.com/2022/01/…

I teach that!

I actually just computed the fact in Set for myself, though you could probably derive it using general categories

$m \times m'$ is defined to be the unique map from a test object $A\times B$ into a product $C\times D$, by the UMP involving the projections $m\circ p_A, m' \circ p_B$. I.e. you literally draw the product diagram for $C \times D$, then draw in $m, m'$, see that they are a cone over $C, D$ and so conclude since a product is a limit, the UMP holds and there is a unique map from $A\times B \to C\times D$. We call that map $m \times m'$.

@Shaun thx for starring :)

Shaun

You're welcome :)

PenAndPaperMathematics

You might want to prove $m\times m'$ a monomorphism using the UMP, etc. If you do, you could teach me how it works, otherwise I'm assuming it's true because it is in Set.

It's true that $\text{id}$ is a monomorphism, right?

Shaun

23:46

Yes, it is mono.

PenAndPaperMathematics

Thus, as in corollary, given $m : S \rightarrowtail B$, we have that $\text{id} \times m :C \times S \rightarrowtail C\times B$ is also a monomorphism, by the above work up

The next part you need is the note at the bottom of the definition of Injective Object:

enjoysmath.blogspot.com/2022/01/…

Shaun

I'm quite tired now, @PenAndPaperMathematics. Let's chat some other time :)

PenAndPaperMathematics

Sounds good!

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$Mathematics$ Mathematics