Mathematics - 2021-10-15 (page 2 of 3)

i saw a halloween costume in a store the other day that i think i could turn into a scary strawberry.

is ruby chocolate a scam or 100% legit and op?

they put the vaccine in it. that and salad dressing.

ha, of course there's a patent. patents.google.com/patent/US9107430B2

it expires 2029?

does that mean it will get cheaper then?

user178758

5:08 PM

\o @BalarkaSen

how'd u find that so fast

hi

it might or might not. i haven't checked to see if they have other patents. in pharmaceuticals this is often an issue. you can certainly dye white chocolate red without paying a royalty to these guys.

but it does taste different

it has a bit of a berry flavor

or maybe I just want to believe it does

it's the steak in the matrix. same as cotton candy grapes and several flavors of white claw.

5:11 PM

is white claw good?

I have never bought it

but I found a very good 6.8 marzenbier for 40 cents

it's OK, but it's overpriced

no patents on white claw, as the market saturation can attest.

The product that i've never understood is energy drinks

don't people like coffee

it's way cheaper and imho tastes way better

and you can get a shit-ton of caffeine out of it if you want

@leslietownes A scary strawberry with sharp Olivia claws?

@Holee Plenty of people do not like coffee.

How do you know

did you meet one

my wife drinks way more coffee than i do, and hates the taste.

if you think of an energy drink as a mixer with other drinks, they have very different flavor profiles from coffee.

5:15 PM

has she ever tasted a coffee she likes

i don't think so.

The only time I drank energy drinks was in jager bombs

but I'm done with those

@TedShifrin yes. i should send you the photo of the cat monster. it's something else.

user178758

Why not post it here?

Q: Is it really bad style to write proofs by contrapositive as proofs by contradiction?

Can someone explain this to me

0

I have seen it said on Mathematics Stack Exchange that proofs by contrapositive are generally preferred over proofs by contradiction (for instance here and here). In other words, it is bad style to prove an implication $P\to Q$ by assuming $P\land\neg Q$, and then giving a direct proof of $\neg Q...

Q: Help in understanding proof of theorem 2.14 in Friedberg's linear algebra

5:19 PM

0

The theorem is: Let $V$ and $W$ be finite dimensional vector spaces with ordered basis $B=\{v_1,v_2,...,v_m\}$ and $C=\{w_1,...,w_n\}$ respectively and let $T:V\to W$ be linear.Then, for each $u\in V$, we have $$[T(u)]_C=[T]_B^C[u]_B \tag 1$$ The proof is this: Fix $u\in V$ and define linear maps...

linear-algebra linear-transformations

Can someone please explain this to me? Thanks.

@HoleeCannoli Explain what? I disagree with the answer and was called ignorant for it.

@Koro Explain what precisely?

I am trying to understand all of it

to it's entirety if possible

so first we have a user called joe

@TedShifrin in equation $(2)$, second equality.

this equality: $[g(1)]_C=[g]_\alpha^C$

koro, not to discourage you from studying this, but as a background fact, this result seems structured by wanting to leverage something he must have already done elsewhere with linear maps from F into F-vector spaces. it's not how a normal human being would structure this proof.

I totally hate that proof.

5:24 PM

Leslie, honestly I know one very natural proof of this.

Just write down the definition of the matrix $[T]^C_B$. This is just the definition of the matrix and linearity.

ted and i can't stop agreeing.

In fact, this result follows by definition of matrix of linear maps.

@TedShifrin nvm it seems like too much work

The whole point of the definition of matrix multiplication times column vector is to make this equation hold.

Thorgott

5:25 PM

I don't like that notation

@Holee If you ask a specific question, we can try to help, but just saying entirety gets us nowhere.

I guess I might be able to figure out the brackets with super and sub scripts, but that is not explained there.

Today, I had a question in the morning about dual spaces. In the search of understanding, I saw Friedberg's LA and I understood a new concept double dual space. Using that, I could prove the result.

I wanted to understand the context that led to that memish situation

the chat's immune system has thoroughly rejected friedberg's proof

5:26 PM

Yes, I was using $V\cong V^{**}$ in my sketched argument.

Then I started studying linear transformation again and came across theorem 2.14 and got stuck at the step I mentioned earlier.

@TedShifrin “it’s your understanding of formal proof that is flawed.” Quite a bit of gumption on that one

I think these discussions convince me that I like my own linear algebra book better than most of the others.

In my topologica data analysis we had to come up with a proof that gluing mobius bands through their boundary gives a klein bottle by giviing suitable triangulations

@TedShifrin yes, Ted. This makes sense to me now :)

5:27 PM

but it seemed hard

@Semiclassic Well, I have never written a formal proof in that sense except in a mathematical logic class.

so a teammate gave a fake proof that actually shows its a tetrahedra but I wanted to go home so I rolled with it

It might not be totally condescending, but I still am very bothered by that notion of proof by contrapositive. I wanted @Alessandro's help.

@HoleeCannoli proof by exhaustion

haha literally

hopefully the grader also gets exhausted

Alessandro Codenotti

5:29 PM

Sorry, I'm busy at the moment

The best way to do that is to draw rectangles with identifications, yes, @Holee. WTF does this have to do with proof by contradiction/contrapositive?

nothing to do with it

Thankfully.

I just remembered that I'm going to have to do it and tell them kindly that they were shitposting during our meeting

yeah but if you identify too much your simplicial complex collapses too much

I'm going to have to do it carefully tomorrow and check munkres and stuff

it's been a while since I worked with that stuff

I think you're making this too difficult, @Holee. Just draw the two rectangles with appropriate identifications. And add the additional identification for gluing along the boundary.

5:33 PM

Oh ok

So I show it's a klein bottle first without using simplicial complex?

and then I provide my own simplicial complex for the Klein Bottle?

because I am asked to calculate homology stuff after

that makes sense

although after that we have to glue another surface to it that touches it weirdly , which makes us get some weird topology that isn't a manifold, and we also gotta turn it into a simplicial complex

You can always triangulate if you insist, but there are much better ways of computing homology (cellularly).

Too complicated for me.

same :/

Maybe we'll just go with the "its a tetrahedra" meme

A tetrahedrON, many tetrahedrA

5:39 PM

oh :/

think i've found the most disgusting integral i'll see all week

Mazltov.

i'm not sure whether to be impressed that it has an explicit antiderivative or appalled

i heard Mazltov in some tv-show. I forgot what it meant.

it means gl in jewish

5:44 PM

@HoleeCannoli The language is "Hebrew", not "jewish".

I meant jewish culture

I was also baiting tbh

And "mazel tov" is generally more "congratulations" than "good luck" (though it can certainly be translated that way, it has a congratulatory connotation).

antiderivative:

this better be good.

what better be?

5:46 PM

$$u^{3/2} - 2 x^3 - u \sqrt{u + x^2} + 2 x^2 \sqrt{u + x^2}$$

as a function of $x$

that doesn't look so bad in that form but

imagine differentiating that and seeing that it has an antiderivative

@Xander But a lot of it is more Yiddish than Hebrew :P

it's not as bad as i realized once i get rid of irrelevant parameters tho

i could have guessed the -2x^3 part. is u also a function of x?

nah

it's not that cruel

main weirdness is that it converges to a constant as $x\to\infty$

well, if that's ugly to you, you should hear what formulas like that say about you behind your back.

5:49 PM

lol

@TedShifrin I am considering commenting there: "Indeed, the contrapositive of $A\to B$ is $\lnot B\to\lnot A$. Proving $\lnot B\to\lnot A$ is done by assuming $\lnot B$ and then proving $\lnot A$. One way of proving $\lnot A$ (by contradiction) is to assume $\lnot\lnot A$ (which is equivalent to $A$) and then arrive at a contradiction. However, proving $\lnot A$ does not mean assuming $A$ and arriving at a contradiction."

that would be a sick burn ngl

@TedShifrin Sure enough. Though my understanding is that it has been incorporated into modern Hebrew, even if it not originally of Hebrew origin, so it isn't exactly wrong to say that it is Hebrew. :P

original had $x=k\sqrt{4m}$ and $u=n\lambda$, which made it look uglier

(understood as a function of $k$)

@robjohn The question is closed (not sure why, honestly). But I think that if "formal proof" means that every proof is a proof by contradiction, then I'm just stupid

Anyhow, since I'm not a student of "formal logic" or "formal proof," I gave up. That's why I wanted Alessandro's help.

5:51 PM

semi seems silly not to toss planck's constant in there somewhere.

@TedShifrin That is not what formal proof means, but it seems that they think so.

@XanderHenderson i remember now. Thanks :)

Several theys.

I couldn't ascertain their qualifications

yes

it was closed because it is primarily opinion based

5:52 PM

oh. original version of the integrand wasn't rationalized

and we have very high standards in this site

I don't think the difference between contrapositive and contradiction is a matter of opinion.

When I taught (possibly incompetently) intro to higher math courses, I tried to explain this very clearly.

the person that critisized you has 3k downvotes though, I wouldn't worry too much

Agh, if I ever have to use Inkscape (I was going to draw the Klein bottle as identification space), it's going to take some learning. I was so competent with Illustrator.

$\lnot Q\to\lnot P$ is the contrapositive of $P\to Q$. contradiction is a method of proof: Prove $P$ by assuming $\lnot P$ and arriving at a contradiction.

5:54 PM

You mean that person has downvoted 3k times? Or has he/she been downvoted 3k times?

yes

proof by contradiction for $P\implies Q$: show that $P\wedge \neg Q$ is false
proof by contrapositive: show that $\neg Q\implies \neg P$

Oh wait no

We all know this, Semiclassic. But apparently not those other experts.

@TedShifrin They have an extremely high downvote given to upvote given ratio.

5:55 PM

yeah

i mean, they're logically equivalent but

not equivalent as proof strats

That person has downvoted 30k times, which is about 3% of all downvotes on the site is what I should have said :)

jeeze

@robjohn how do you find that ratio

n/m, i think i figured it out

i'm 4,434 upvotes / 122 downvotes

though some of my downvotes are on deleted questions and therefore no shown

Holy cow. Now I'm curious to know what competence that person actually has.

At least said user is a professional in the art of the downvote

Well, the person who wrote the dubious answer is a student "specializing" in logic, blah blah.

6:00 PM

@Semiclassical go to the bottom of the activity page under "Votes cast".

gotta go under activity for that

I just defended myself to the horde of "professional student logicians."

@Semiclassical Yes, that is what I get when I just look at the page without the specifier.

might be in my preferences.

huh

so that is what you see next to the "nuke this account" button ?

6:04 PM

good to know regardless

yikes, all of my business is just out here on this profile page

dang you have 1 downvote what was that about

it must have been pretty bad, whatever it was

or i'd eaten something that upset my stomach

@Semiclassical now the link takes you directly to the proper part of the proper page.

Hmm, I have 1100+ upvotes and fewer than 200 downvotes. Meh.

6:08 PM

thx

@HoleeCannoli who has one downvote?

not u

@robjohn Can we see how many times other people have downvoted me?

I know that...

the CTO of lesliecoin

6:09 PM

apparently i have only done one downvote

@TedShifrin I think you can by making an SQL query on data.stackexchange.com

Way too hard for me :P

I'll go back to my fuzzy logic.

today all profiles look more colourful.

i'm calling for a formal forensic analysis and recount of all statements made in ted's mathematical work. clearly there's some kind of fraud going on.

orange looks more orange

6:12 PM

@TedShifrin It is not made easy.

@TedShifrin I don't think that the difference between contraposition and contradiction is a matter of opinion, but it is also not clear that this is the crux of the question, which is also about whether it is good style to use contradiction.

Lack of clarity might also have been a reasonable close reason.

But the question ultimately reads:

> In this example, does it really matter that the proof by contrapositive above is phrased as a proof by contradiction? And if not, why are the objections to proof by contradiction raised above not relevant here? When is it important to care about whether a proof is by contrapositive, or proof by contradiction?

Isn't that a matter of opinion?

@Semiclassical: I think there should be some celebration when someone reaches 30000 downvotes given ;-p

i'm not sure 'celebration' is the right word for it

They should replace their downvote button with a honorary gold downvote button that doesn't work

3

hmm. be productive or take a Friday nap

dilemmas

6:19 PM

I certainly discouraged my students from doing proofs by unnecessary contradiction, but sometimes one just needs to do it. :P

yeah

at least we're not doing 'don't use law of excluded middle' shenanigans

i can appreciate why one would do that, but it usually just gives me a headache

6:36 PM

I think i called my ugliest integral of the week too early

b/c this one is a bit of an arse:

$\int_0^1 \frac{x'}{x}\ln\left|\frac{x+x'}{x-x'}\right|\,dx'$

Are these coming up in physics homework?

yeah

this one is basically part (a) of an exercise

First, obvious change of variables.

$y=x'/x$?

yup

6:44 PM

yeah, tho it makes the upper endpoint gross.

this is easy parts.

Gross? Nah.

i mean, it's $\int_0^{1/x} x y \ln\left|\frac{1+y}{1-y}\right|dy $

not clear to me that's really any nicer

Easy peasy.

break the log up.

@TedShifrin I don't have an axe.

i mean, you can do that, so that it's now $\int_0^{1/x}x y \ln |1\pm y|\,dy$

6:47 PM

a hatchet will suffice !

@TedShifrin All I've got is a pocket knife.

the $x$ is irrelephant. Integrate by parts.

@TedShifrin What do you get if you cross an elephant and a rhinoceros?

Elephino...

shrug.

fair. i guess i don't see why it's useful to split up the logs for that. $\frac{d}{dy}\left|\frac{1+y}{1-y}\right|= \frac{2}{1-y^2}$ is nice enough by itself

6:49 PM

@XanderHenderson here

I give up.

@robjohn Yay!

I just knew I could do it in my head if I split them.

fair

6:51 PM

@Ted: I wonder if those logicians would consider any two proofs of the same theorem logically equivalent.

I sure hope not.

But they aren’t impressing me.

Nor, clearly, I them.

same input, same output, same proof.

Voting someone out of office is logically equivalent to going back in time and killing their parents before they were born.

If someone had found a simple way of proving Fermat's Last Theorem right off, they might be equivalent, but just think of all the other math that was discovered along the way.

hi, for showing that the sequence $\frac{n!}{n^n}$ converges to $0$, is it fine if I do this: $\frac{n!}{n^n} = \frac{n\cdot(n-1)\cdot(n-2) \cdots 2\cdot1}{n\cdot n \cdot n \cdots n \cdot n} = (1 - \frac{1}{n})\cdot(1 - \frac{2}{n})\cdots (1 - \frac{n-2}{n})\cdot(1 - \frac{n-1}{n})$. Then since the multiplication of convergent sequences also converges and converges to the multiplication of the limits, I get $1\cdot1\cdots0\cdot0 = 0$. Is this fine?

No

If it were a fixed size product, that would be fine.

Consider $(1+1/n)^n$.

7:02 PM

@athing Of course, you could say that it is no greater than $\frac1n$ (the last term in the product) and that would work.

also, it's pretty suspicious that you seemingly end up with a product of 0's and 1's, without any logic for when $1-k/n$ goes to $0$ vs. $1$

@Semiclassical Yes that's I felt too :p

robjohn: thanks! What do you exactly mean by "fixed size product", can you elaborate please?

the number of terms in the product is growing with $n$

n not going to \infty?

It is not?

7:05 PM

so you're not in a situation where you have $A_n B_n C_n\to ABC$

$$\prod_{k=1}^{100}\left(1-\frac kn\right)$$ as $n\to\infty$, that product goes to $1$

Ted: Yes, yes; I meant "did you mean it would be fixed size product if n were not going to \infty".

no

the point is that, if you had a product like $A^1_n A^2_n\cdots A^k_n$, and each factor converged as $n\to \infty$, then the product of sequences would converge to the product of their limits

Well, sure, if $n$ is FIXED.

but here you have more and more factors as $n$ increases, so that doesn't work

7:08 PM

No limits anywhere.

@TedShifrin It was broken?

Neutered.

on an amusing note, i'm pretty sure we have the same nickname for our family cat

our official name for her is Gabby

my nickname for her is Screech :P

@Semiclassical thanks; but why doesn't it work?

4 mins ago, by Semiclassical

but here you have more and more factors as $n$ increases, so that doesn't work

7:13 PM

In my case, no nickname :)

the limit rule you're appealing to is not for arbitrarily many factors

it's for exactly two at a time

@athing Consider the $e$ limit I gave you above.

and that means you need to have a scenario $A_n^1 A_n^2 \cdots A_n^k$ if you're going to take $n\to\infty$ and expect that to converge to $A^1 A^2\cdots A^k$

fixed number of factors

Thanks Semiclass and @TedShifrin! You wrote "FIXED" above in all capitals, does that have a special meaning or were you shouting only :p

speaking with emphasis :P

7:15 PM

emphasizing, yes.

$(1-1/n)^k\to 1$ as $n\to\infty$ for any finite $k$

nevertheless, $(1-1/n)^n\to 1/e$ as $n\to\infty$

no

$1/e$ in your case

yeah, fixed

fixed … again?

derp

too late to edit the first message, so i fixed the second

7:22 PM

\o @LukasHeger

@TedShifrin at the end of the day, what i get from integration by parts is $$\int_0^1 \frac{x'}{x}\ln\left|\frac{x'+x}{x'-x}\right|=\frac12+\frac{1-x^2}{4x}\ln\left|\frac{1+x}{1-x}\right|$$

which i know is correct

but it's still an integral that feels like pulling teeth for me

I have taught integral calculus a lot of times! Shrug.

though the next integral is...to integrate that thing

not interesting

(that's a slight lie, there's a factor of $x^2$ that enters as well. but mostly that)

yeah, i hate it

it's slightly less nausea inducing for me if i write it as $\int_0^1\int_0^1 xx' \ln\left| \dfrac{x'+x}{x'-x}\right|dxdx'$

but only slightly

7:32 PM

change order of integration? 🙄😳🤣

lol

pretty sure that wouldn't help here :P

Duh

lol

actually, that integrates out to $1/2$

which is sorta interestging

incredibly bored and need some math to think about

i have ran out of interesting problems for now

Perhaps an obvious symmetry argument.

7:33 PM

yeah, i wonder now

Poor Balarka!

the problem is very much set up as "do the horrible integral over $x'$, and then do the even more horrible integral over $x$"

but that doesn't mean it's the -smart- thing to do

have you tried applying formal logic

man, why didn't i think of that

There you go. Contradiction is universal!

7:36 PM

hmm, actually

apparently $\int_0^x x'\ln\frac{x'+x}{x-x'}\,dx'=x^2$

and then $\int_0^1 x^3 \,dx=\frac14$ is trivial

which gets doubled b/c i've only done the $x'<x$ part

moreover, my qualms about $u=x'/x$ go away in that case, since now the upper limit would be $u=1$

@BalarkaSen hey

Hi @Lukas

Hi @Ted

$$\int_0^1\int_0^1 xx'\ln\left|\frac{x'+x}{x'-x}\right|\,dxdx'=2\int_0^1 \int_0^x xx' \ln\left(\frac{x'+x}{x-x'}\right)dx'dx$$

Maybe clever rewrite + COV in the double integral. I dunno.

7:44 PM

ya

there's something fancy going on

but...ehhh

assume the right hand side, and also not the right hand side, and try to deduce not the left hand side.

The obvious rotational change doesn’t help.

yeah

LOL, Leslie. Tell the logician experts!

what is something that you can grab with your left hand but not with your right hand?

7:46 PM

actually, the substitution $u=x'/x$ now just splits the integral in two

your right hand?

how did you know

idk, seems obvious to me

that's not exactly a sphinx-level riddle, is it

you must be a category theorist or some othe kind of genius

7:47 PM

It’s worthy of munchkin.

@HoleeCannoli this is true on so many levels

I do some category theory, yeah

$$2\int_0^1 \int_0^x xx' \ln\left(\frac{x'+x}{x-x'}\right)dx'dx=2\int_0^1 \int_0^1 x^2 u^2 \ln\left(\frac{1+u}{1-u}\right)\,du\,dx=2\int_0^1 x^2 \,dx \int_0^1 u^2 \ln\left(\frac{1+u}{1-u}\right)\,du$$

munchkin is right handed. we're all a little disappointed in her

that's a ltitle cute

or it would be, if it gave the right answer...hrm

7:52 PM

If you have left-right confusion, just remember that right is the side where the tensor product is exact

lol

hmm

it's where the inner product is conjugate-linear instead of linear

i think i have to be careful and write $u=x'/x$, $v=x$, otherwise i think i'm missing a factor from the Jacobian

@leslietownes that's not very helpful lol

7:55 PM

i love mnemonics that don't help, or actively harm

s.harp

In german you say that the left side is where the thumb is right

everything's placid when you add the water to the acid

$P_n(z)=\sum_{m \geq 1}\left[\delta_{m,n} + 2\pi \cdot (-1)^{\frac k2} \cdot \left(\frac mn\right)^{\frac {k-1}2} \cdot \sum_{c \geq 1} \left( \frac 1c \cdot \left(\sum_{\substack{d (\operatorname{mod} c) \\ (c,d)=1} \\ d \bar{d} \equiv 1 \pmod c} e^{2\pi i \frac{md + n\bar{d}}{c}} \right)\cdot \left(\frac{2\pi \sqrt{mn}}c\right) ^{k-1} \sum_{\ell \geq 0} \frac{\left( -\left(\frac{2\pi \sqrt{mn}}c\right) \right)^\ell}{\ell! (k-1+\ell)!} \right)\right]e^{2\pi i m z}$

do yu have a mnemonic that helps with that?

no, but i do have drinks

also, if that's just the Legendre polynomials, i'mma be deeply disturbed by this rewriting

no, that's the Poincaré series

7:58 PM

good

what drinks u got?

mostly water tbh

always a classic

i've got a lot of green tea

actually, better joke: if i see a formula like that, i want lethe not mnemonic

7:59 PM

that reminds me of this brilliand question on main