Mathematics - 2017-08-22 (page 4 of 4)

SteamyRoot

19:01

Renaming historical terminology is a nightmare

Tobias Kildetoft

Even things named in the last few decades can cause trouble if not consistent (ahem...tilting modules)

Semiclassical

What it seems to come down to is that "algebraic operation" isn't used consistently with "algebraic number" or "algebraic function"

And, sure. I'm okay (in principle) with renaming "algebraic operation", since it's not used with anywhere near as high of frequency (or with anywhere near as relevance) as "algebraic number"

But I see no reason at all to rename "algebraic number." That usage is accepted and understood. If that makes "algebraic operation" bad, so much the worse for "algebraic operation".

Will Hunting

19:15

Well, you can always create your own definitions. If you are famous enough, enough people will follow you.

Semiclassical

On the other hand, acting as though you're in a position to dictate usage of certain terminology is a good way to get dismissed as a crank.

Secret

Jun 21 at 18:51, by user107952

Since I am the most rigorous, I consider it my life mission to teach people rigor. I will tell you all a secret very, very few humans know. The argument "I am the most rigorous human. H is a human. Therefore, I am more rigorous than H." is invalid. You need the additional premise, "H is not me." Consider yourself all honored to learn this secret from the most rigorous human.

1 hour ago, by Wietlol

im closer to being an expert than an idiot

2/3

[Rambles]

Tobias Kildetoft

@Semiclassical But I like the good old "proof by redefining things"

Will Hunting

@Secret That user is a crank.

Secret

hopefully we won't met a 3rd one...

Semiclassical

19:19

@TobiasKildetoft smbc-comics.com/comic/compatibilism

Will Hunting

I am not even on that graph, go figure, LOL.

Secret

well, most of us are probably somewhere in the middle

Will Hunting

There was once I said the following: I am in this world but not of this world, and someone flagged me, LOL.

Secret

Well, I am here and not here at the same time, as I am a wavefunction

Tobias Kildetoft

What graph?

Secret

19:22

anyway, must head to sleep 5:21 after listened to the seminar

Will Hunting

1 hour ago, by Leaky Nun

9 mins ago, by Leaky Nun

Tobias Kildetoft

Ahh, that one

Will Hunting

Well, technically, the graph of the function is just the function which is just the set of ordered pairs you see there, LOL.

Balarka Sen

@Semiclassical Good comic!

This is appropriate in so many contexts

Semiclassical

a pity I don't actually know enough about Dennett to assess its validity

Tobias Kildetoft

19:25

@Semiclassical Who is Dennett?

Semiclassical

Daniel Dennett

Daniel Clement Dennett III (born March 28, 1942) is an American philosopher, writer, and cognitive scientist whose research centers on the philosophy of mind, philosophy of science, and philosophy of biology, particularly as those fields relate to evolutionary biology and cognitive science. As of 2017, he is the co-director of the Center for Cognitive Studies and the Austin B. Fletcher Professor of Philosophy at Tufts University. Dennett is an atheist and secularist, a member of the Secular Coalition for America advisory board, and a member of the Committee for Skeptical Inquiry, as well as an...

I can say that he's got an awesome beard, though.

Will Hunting

What a nice beard, LOL.

Semiclassical

I have no idea whether or not the comic is a good summary of Dennett's views on free will / compatibalism, but it is a good representation of his beard.

Akiva Weinberger

Random thing I was reading

> If we had no intonation, our speech would be - in the literal sense of the word - monotonous.

Your humor kills me /s

Semiclassical

I always find it hard to take philosophizing against free will seriously, though.

Will Hunting

19:29

Well, we just say monotone or monotonic function, LOL.

Semiclassical

"For not believing in free will, you're investing a lot of effort into convincing me."

Will Hunting

Does anyone call it monotonous function ever?

Akiva Weinberger

I'd say the identity function is pretty monotonous

Luigi M

I'm asked to exhibit a n-1 form w on a compact orientable n-manifold with boundary for which dw never vanishes. Does the form ydx on the unit ball in R2 works as an example?

DaneJoe

@Semiclassical You realize literally all of mathematics is just an arbitrary human construct right? But anyway, even though it is an arbitrary stance, it is considerable for purely human purposes that algebraic number be kept over algebraic operation.

Luigi M

19:37

d(ydx)=dydx and it should be the volume form on the unitary ball

DaneJoe

I think I will do that

But anyway, regarding sum(ln(k))=ln(product(k)), you didn't transform the boundaries at all. I agree that for this very specific case, that it happens to work out. But if you were trying to prove anything, you at least would have to show you can arbitrarily substitute variables of summation.

this isn't like substituting x in a taylor series

x is a constant with respect to the sum you can do anything to it

but there's no known formulas that let you the variable of summation into any other arbitrary function of k

substitute the*

Tobias Kildetoft

@DaneJoe I still don't get what your objection is here

Could you give an example where this fails?

Leaky Nun

@TobiasKildetoft it stems from a misunderstanding of what sum and product means

i.e. symbols with no meanings

DaneJoe

@TobiasKildetoft Easily. Let's say you substitute sum(e^x) for u and it goes to n. Well, the sum of u to n is (n^2-n)/2. If you back substituted, you would get e^(2x)/2+e^x/2. This is false. It's actually (e^(n+1)-1)/(e-1)

Tobias Kildetoft

@DaneJoe substitute what in what? Be precise

DaneJoe

19:45

sorry I said e^x but meant e^n

@TobiasKildetoft Um, the sum, which I already said...

Leaky Nun

@DaneJoe it will be clear if you include the summation variable.

DaneJoe

if this chat recognized any of the latex that was typed in it then I would

Leaky Nun

@DaneJoe it does

Tobias Kildetoft

@DaneJoe It does, you just need to enable it

DaneJoe

$x$

still no

where is it enabled?

Leaky Nun

19:47

$\displaystyle \sum_{x=0}^n e^x \color{red}{\ne} \sum_{u=0}^n u = \frac12n(n+1) \color{red}{\ne} \frac12e^x(e^x+1)$

@DaneJoe tinyurl.com/cfqcvpc

DaneJoe

yeah that all looks like raw latex to me

not like an actual equation

Leaky Nun

click the link

DaneJoe

yeah

i don't have a "bookmark bar"

but ill see if I can get it to work

Leaky Nun

that'll do for now

DaneJoe

that's closer to what I'm saying

Leaky Nun

19:50

ya, that's two wrong substitutions

DaneJoe

right

Leaky Nun

so your objection is that we haven't written down sum(ln(k)) = ln(product(k)) clearly

DaneJoe

the objection is that semi or someone said sum(ln(f(k)))=ln(product(f(k)))

that it's not just true for k

but fur any function of k

and i would like to see the proof of that if it's anywhere

for*

Tobias Kildetoft

@DaneJoe Proof is the same.

DaneJoe

@TobiasKildetoft That sentence doesn't make grammatical sense to me.

Tobias Kildetoft

19:52

@DaneJoe The proof does not change in any meaningful way when you replace $k$ by $f(k)$

DaneJoe

okay, so you said the proof does not change, but you haven't given the proof itself

Leaky Nun

@DaneJoe we already mentioned induction

Tobias Kildetoft

@DaneJoe Right, because I distinctly recall that being discussed ad infinitum yesterday

DaneJoe

It couldn't have been for ad-infinitum because the universe would have already been destroyed which means we couldn't be discussing it.

but anyway so you can show it's true for any f(k)+1

you're 100% confident in that

Tobias Kildetoft

that does not make sense

DaneJoe

19:54

okay so here we go

does this specific formula have a name so I can just look it up?

Leaky Nun

@DaneJoe do you know what induction is?

DaneJoe

where you start out assuming a formula is true

and then show it's true for any k+1?

Leaky Nun

close

of course you don't assume that it's true; that would be begging the question

DaneJoe

no I explicitly remember multiple people using the word "assume"

and then I just looked it up

and found "assume"

Leaky Nun

@DaneJoe yes, but not just the formula itself

you are missing out an important part

DaneJoe

19:57

okay but do you have the proof or not?

because if you don't this is a waste of time

Leaky Nun

@DaneJoe of course I have

but you need to understand induction first

as it is the main tool of the proof

DaneJoe

I have confidence that I can interpret it properly if applied to the ln formula

Leaky Nun

@DaneJoe then tell me what induction is

DaneJoe

this is taking too long

im just going to try and look it up

Leaky Nun

@DaneJoe then make sure not to miss out important terms

DaneJoe

19:59

it doesn't matter because the goal isn't to explain induction, it's to use it

it doesn't matter if I get it right or not as long as the proof is provided

the burden of understanding it is my own

here we go, a khan academy video that explains it faster than you

Leaky Nun

@DaneJoe good

DaneJoe

20:26

Anyway, still doesn't address the general case where you can't arbitrarily substitute for any variable in any summation and then back substitute

Tobias Kildetoft

@DaneJoe What general case?

You can usually substitute without problems as long as you do it correctly

Leaky Nun

@DaneJoe so you've looked up induction?

Mike Miller

@Tobias Remind me the flavors of representation theory you work with?

Tobias Kildetoft

@MikeMiller algebraic groups in positive characteristic, semisimple complex Lie algebras, and 2-representations of fiat 2-categories

Mike Miller

fiat?

Tobias Kildetoft

20:32

@MikeMiller "Finitary with Involution and Adjunction Two-morphisms"

Mike Miller

what's the interest in those?

Tobias Kildetoft

They are very well-behaved and yet tend to show up in many places naturally

The main examples being Soergel bimodules

Mike Miller

Aha, I can understand the interest. Presumably the extra structure also gives you extra structure on various kinds of cohomology.

Tobias Kildetoft

@MikeMiller Hmm, I am not sure if there is a particularly well-developed theory of cohomology for these. Even extensions is a quite recent thing for them, and it looks a lot different than one is used to.

Mike Miller

Aha, I'm confusing two unrelated things. In any case I can trust the interest :)

Will Hunting

20:40

Hello @MikeMiller are you very busy these days?

Mike Miller

Aye. Taking a break for a bit.

Will Hunting

Maybe I will send you an email later.

Vrouvrou

hello

I have always this problem, if $F(x)=\int_0^x f(t) dt$ then $\int_{x_0}^{x}f(t)dt=F(x)-F(x_0)$ or it is equal to F(x) ?

if $F(x_0)\neq0$

Will Hunting

@Vrouvrou Did you type correctly?

Vrouvrou

yes why ?

Will Hunting

20:54

Oh I guess you did. Just many symbols there.

Leaky Nun

@Vrouvrou the former option is correct

by... the fundamental theorem of calculus

Tobias Kildetoft

@MikeMiller I have the impression that most people felt the general theory of fiat 2-categories and their representations might be a bit too much generalization for its own sake. But it has turned out to be useful after all.

Vrouvrou

@LeakyNun you mean $\int_{x_0}^{x}f(t)dt=F(x)-F(x_0)$ is correct

Leaky Nun

@Vrouvrou yes

Vrouvrou

thank you

DaneJoe

20:56

@TobiasKildetoft Right but neither you nor anyone from any paper I've read has ever proven any correct method at all. As I already said, the substitution in sum_{k=1}^{n}(e^k) does not yield e^(2n)/2+e^n/2 when back substituted.

at least for the cause of when you substitute for e^k

Tobias Kildetoft

That is because the substitution was not done correctly, as the limits were left unchanged.

DaneJoe

right, that's the exact point I brought up before

Akiva Weinberger

Is that your real name

or is it a pun on Jane Doe

DaneJoe

just because it happens to work out nicely in the case sum(log)=log(product) doesn't always mean it will work for any sum

Tobias Kildetoft

Of course, there will not always be a meaningful way to do that

DaneJoe

20:59

well meaning is relative

so if there's any way, it's worth considering

Tobias Kildetoft

I am really still not sure what sort of statement you are objecting to here. The examples seem to have essentially nothing in common

DaneJoe

that's the intention

that's what shows that the general case isn't always true

it just happens to be true in the instance of the log product formula

Akiva Weinberger

So the answer is "no comment"

DaneJoe

it seems like the answer is "no one actually knows"

Akiva Weinberger

To my earlier question

Tobias Kildetoft

21:01

But that formula is not a substitution. So why compare them?

DaneJoe

because we were talking about an instance of a substitution in that formula

that it was true not just for k

but for any f(k)

but that's not true for any sum relationship

just for that specific formula

Akiva Weinberger

So I'll assume it's the second one

DaneJoe

@AkivaWeinberger It doesn't matter what answer I give because any answer could be a lie or the truth with no way of you knowing.

Semiclassical

Mu.

Here's a possible question. Suppose $\sum_{k=1}^n f(k)=f(n!)$ for all $n$. What can one say about $f(k)$?

Tobias Kildetoft

Yes, of course you cannot in general replace any function with any other function.

Semiclassical

21:03

certainly $f(x)=\log x$ works.

DaneJoe

@TobiasKildetoft Right, so what is the substitution method that allows you to do that, like with integration?

Semiclassical

Is this still about why $\log (\prod_{k=1}^n f(k))=\sum_{k=1}^n \log f(k)$?

DaneJoe

@Semiclassical Not specifically, its about a general substitution method for any given summation.

Tobias Kildetoft

Ahh, we are back to that. No idea if there is one.

Semiclassical

When it comes to sums, the allowed 'changes of variables' are a lot more limited as far as I understand it

DaneJoe

21:06

right...

that's why I was asking because that's what it seemed like to me

but

Semiclassical

There's no substitution going on in the above, is my point.

DaneJoe

if there's any method at all

well...you substitute k f(k)...the formula works

but only for that sum formula

k for f(k)*

Semiclassical

Well, sure. $f(k)=k$ is just a specific instance of $f(k)$.

It'd also be true if $f(k)=k^2$.

DaneJoe

right

but lets say I had a different summation like sum(e^k)

how would you substitute for u-e^k and get the right answer?

u=e^k*

Semiclassical

You're making an apples to oranges comparison as far as I can tell.

DaneJoe

21:09

I'm not, but what you're noticing is a fault in the method that would lead a sum to be bounded at something other than integers

which is why it would be important if a substitution method worked

Semiclassical

Sure, but the logic in these two cases is very different. Yours has to do with the range of values of $k$.

The other one really doesn't care a whit about the range of $k$.

DaneJoe

And again, it happens to work nicely that way for that other formula

but as far as I can tell

as other people have said

there's still no known substitution method for the general case of any sum

as there is with an integral

Semiclassical

Sure. I'm not sure why one would expect there to be so.

I mean, there's some obvious cases. For instance, $\sum_{j=1}^n f(j)=\sum_{j'=0}^n f(j'+1)$.

I suppose the point is really that there's not as many nice-looking bijections on the integers as there are on the real line.

So, for instance, suppose $g$ is a bijection from the positive integers to the positive integers. Then $\sum_{j=1}^\infty f(j)=\sum_{j=1}^\infty f(g(j))$.

Vrouvrou

@LeakyNun i found $F(x)=\int_{a}^x f(t)dt $ without saying that $F(a)=0$ wjhat about this ? please

Leaky Nun

@Vrouvrou still ok

Semiclassical

21:17

I don't think this ends up being terribly useful in practice, though.

Leaky Nun

@Semiclassical nope.

you can only do finitely many swaps with guarantee that the limit is the same

Semiclassical

bah, analysis

yeah, that's fair.

Vrouvrou

@LeakyNun the two means that $F$ is the premitive of $f$ ?

Leaky Nun

@Vrouvrou primitive.

Semiclassical

What I have said amounts to this: If $\pi$ is a permutation of $\{1,2,3,\ldots ,n\}$, then $\sum_{j=1}^n f(j)=\sum_{j=1}^n f(\pi(j))$

Vrouvrou

21:19

yes sorry

@LeakyNun

Semiclassical

Better?

Leaky Nun

@Vrouvrou we usually use "primitive" to describe indefinite integrals

@Semiclassical yes

Semiclassical

mmkay.

yeah, just noticed/fixed that

Vrouvrou

@LeakyNun what do you mean by "indefinite"

Leaky Nun

@Vrouvrou without the limits

Vrouvrou

21:23

any limits ?

Astyx

A primitive, not the primitive

Leaky Nun

@Vrouvrou $\int x \ \mathrm dx$ is indefinite integral; $\int_a^b x \ \mathrm dx$ is definite integral

Vrouvrou

ok thank's

so what is $F$ when $F(t)=\int_0^t f(x)dx$ ?

@LeakyNun

Leaky Nun

eh

no idea

@Astyx how do you call it?

Vrouvrou

21:47

good night for all

bonne nuit @Astyx

thank's for your help

Will Hunting

@Vrouvrou It should be thanks, not thank's.

ALannister

22:15

2

Q: Please help me check my proof that the transcendental numbers are dense in $\mathbb{R}$

I need to prove that the set of all transcendental numbers is dense in $\mathbb{R}$, and to that end, I have written the following proof: Let $\mathbb{T}$ denote the set of transcendental numbers in $\mathbb{R}$. First, notice that $\mathbb{T} \subset \mathbb{R}\setminus \mathbb{A}$, where $\...

ALannister

22:26

@TedShifrin if you show up, could you help? <3

anon

23:21

only one dragon is missing

ALannister

23:40

LOL!

So, you read the proof @anon

anon

no

Will Hunting

The great anon!

ALannister

Oh, never mind then.

anon

ok, now I have. is your definition of dense "has an element in every interval"?

ALannister

Yes @anon

anon

23:44

then yeah that's a proof

hello JL

ALannister

Thanks, bud

Will Hunting

Budweiser!

ALannister

Why does His Lordness the Shark seem to think I should do q + pi then?

anon

I mean, you'll want to say "there exists a q, besides 0" in the proof, but it's not much of a difference.

Will Hunting

He's just a shark.

Akiva Weinberger

23:51

Q+pi and Q*pi\{0} are both things that are dense and made of transcendentals

which is all you need

Well, besides love

Secret

Last night dream there is a weird workshop which taught you how to prioritise layout of information in your writing. Basically the abstract came first, then 3 statements that you labelled as important and then countable infinite number of statements. In particular, their will be a followed up workshop called The Magical Controller that will teach you what to do with those countable emotions as it implies the order on how they are laid out in the writing is important.

Akiva Weinberger

I uh

Maybe that's your subconscious telling you to go to a writing course

Not very subtle to be honest

Will Hunting

@AkivaWeinberger Some people don't need love, but I do.

Akiva Weinberger

$\heartsuit$

Will Hunting

My favourite love song is 'Nothing's gonna change my love for you'.

Secret

23:58

I suspect it might be an allusion to either the proposal or my program script. Or that somehow the infinity discussion yesterday get warped somehow

Transcript for

$Mathematics$ Mathematics