Mathematics - 2019-06-21

12:02 AM

Also just about to apply for an apartment. Some stuff in the application I'll need clarification on and since it's past 5 my email won't likely get responded to until tomorrow but hopefully I'll be set to go then

Ted Shifrin

You might end up having to call.

Ultradark

He protec. he attac. But most of all he help you wit math

Daminark

Yeah that's pretty true. Hopefully by tomorrow this'll be over and I'll have an apartment in hand

Then + when I get a few days of sleep I'll finally get back to math

anakhro

12:25 AM

@Ultradark a few of those words are spelled wrong.

Secret

Hmm... in order to prove whether an actual infinity can only be formulated in a circular fashion, I guess proving the contrapositive may be easier:

To make this independent of foundations, we need to first find a predicate in first order logic that defines (not necessary uniquely) an actual infinite object

Infinite = not Finite

anakhro

@Secret what have you rigorously defined an "actual infinity" as?

Secret

I think P(x) where P = "is not finite" is sufficient. The challenge here is how to define "finite" in a foundationally independent manner under 1st order logic

In set theory alone there are already 5 notions of finite

anakhro

lol

Secret

What is observed so far is that things like natural numbers, dedekind finite sets, tarski finite sets and so on all shared some notion of limitation which make them different from infinity

If we can capture that limitation in first order logic than we may have a way to figure out whether defining actual infinity always require a self referential predicate

Karl Kronenfeld

12:44 AM

um what are you aiming for? what is a self-referential predicate?

Secret

I am not sure what the formal name is, but what I mean is some predicate P where one of its parameters is x itself. Thus x is defined to be what satisfies P(x)

What I rithaniel and many others are suspecting is that infinity is something that can only be defined wrt itself, and cannot be defined via a "constructive" mean

But the answer to this question remains painstakingly open

unlike e.g. even numbers y are those that satisfies Q(y) where Q = "is divisible by 2", where here there is no circularity in Q since "divisible" and "2" can be formulated without reference to each other or the notion of "even"

Karl Kronenfeld

Ah, I think I see what you mean.

anakhro

I don't follow, on the other hand.

Karl Kronenfeld

Yeah, it sounds like the only way forward would be to describe a predicate what distinguishes all infinities from all that is finite.

Otherwise, you would have non-self-referential predicates that merely reference other infinities.

Secret

Exactly

anakhro: To clarify. We knew we can define an even number y by having y to satisfy "y is divisible by 2". Here there is no self reference and no underlying circularity because none of the concepts "divisible", "2" will reference the concept "even". That is, if you follow down the predicates A,B that defines "divisible", "2", you can avoid the concept "even" to appear in any such A,B

anakhro

12:55 AM

Yes I get how that is not self-referential.

Secret

But the conjecture here is that, it is impossible for infinity to do so, meaning there is no way you can have a definition of infinity without it being referenced somewhere down the line

anakhro

Why do you conjecture that?

Secret

Because this question seemed to be still open. ZF and ZFC had not deal with this question directly because they have bypass it using the axiom of infinity (and large cardinal axioms when inaccessible cardinals were included), whereas Quine new foundations only push the thing higher into the universal set, which of course will contain infinity

Thus it will be interesting if the circularity of infinity can be rigorously proven independent of foundation systems under 1st order logic

anakhro

Sorry, what exactly is the question that is still open?

Secret

Whether infinity can be defined using a predicate that is neither self referential nor has no underlying circularity in the sense illustrated in the "is divisible by 2" example

That is, whether infinity can be defined without invoking other notions of infinities, including the universal class

anakhro

1:00 AM

What is "infinity" in that question?

Secret

anything that is not finite. As for finite, I am still thinking how to write out its predicate

anakhro

So the question isn't even well-defined?

Secret

Well, I have not fill in the blanks yet, but that Karl understood it seemed to suggest the sketch of the question is well defined

(rarely do people understood my queries on the first go)

Plan to do the rigorous shortly, reading that Rudyrucker book does give me more ideas on how to correctly think about infinity

anakhro

I question Karl's understanding.

Karl Kronenfeld

lol

anakhro

1:06 AM

You yourself have deferred the definition of "infinity" to the definition of "finite" to which you are still thinking how to define.

I don't think there is a question.

<3 u @KarlKronenfeld

Question, @Secret : can you define a "power set" analogously to how you want to define "infinity" (whatever that is)?

Because to me, it's equally valid to attack the axiom of infinity as the axiom of power set.

Secret

Not really, you can nuke the axiom of infinity and retain the axiom of power sets. That is because the power set of any finite set is always finite

anakhro

You can remove the axiom of power set.

I think some constructive set theorists get rid of it in favour for some other axiom.

Secret

Yes they do. Because power set does not control the existence of all infinities (only the uncountable ones), it is axiom of infinity that we need to ask question about its nature

anakhro

Why is that?

Secret

i mean, removing power set alone does not get us closer to the answer on whether infinity is non circular. Many constructivist still use some infinite sets such as the natural numbers and do not consider that circular

(also I am currently in seminar thus expect my response to delay)

anakhro

1:20 AM

My question, more succinctly, is what is wrong with the axiom of infinity?

Ryan Unger

1:31 AM

what is that

I doubt there's an infinity really

like I buy that there are arbitrarily large numbers

but idk if there's really an infinity

anakhro

Axiom of infinity

In axiomatic set theory and the branches of mathematics and philosophy that use it, the axiom of infinity is one of the axioms of Zermelo–Fraenkel set theory. It guarantees the existence of at least one infinite set, namely a set containing the natural numbers. It was first published by Ernst Zermelo as part of his set theory in 1908. == Formal statement == In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: ∃ I ( ∅ ∈ I ∧ ∀...

That's the axiom of infinity.

"if there's really an infinity" <--- ??? what do you mean ???

Ryan Unger

@anakhro well

that's how you get things like banach tarski

anakhro

Banach-Tarski is a math theorem.

Ryan Unger

>theorem

only if there's an infinity

anakhro

Yes, it has certain assumptions.

As does any theorem in mathematics.

I don't see the issue.

Karl Kronenfeld

1:40 AM

One of my dreams is to reference the enormous real-world potential of the banach-tarski paradox in a grant application for research on the axiom of choice.

anakhro

I can guess you are thinking that perhaps math ought to have some empirical credibility or something like that, @RyanUnger?

But if not, then I don't know what is "wrong" with the Banach-Tarski theorem.

Ryan Unger

@anakhro idk

I just don't buy certain stuff

anakhro

Yeah, I don't like cereal, so I don't buy it usually.

Ryan Unger

what, not even cheerios?

anakhro

No, cheerios are expensive, d00d.

Ryan Unger

1:48 AM

are they

I can say I've literally never thought about the price

anakhro

I like not spending too much on food.

Ryan Unger

food and alcohol is by far my biggest expenditure

then I guess it would be gas

anakhro

Rent is my biggest, then food.

Ryan Unger

ok discounting rent...

anakhro

How much do you spend on food a week?

Ryan Unger

1:53 AM

unclear

it varies

I spend about 300 a month I think. Maybe more. At most 400

there have been some outlying expensive meals

anakhro

Like burger king, si

Ryan Unger

how much do you spend

anakhro

In my undergrad, my goal was $20/week. I got closer to $30.

Now I spend a lot more, but I actually have a job.

Ryan Unger

holy fuck man

I spent 28 on a snack today

am I bougie

anakhro

The punchline is that my dollars is in canadian $$$.

$20\,\text{CAD}\approx 15\,\text{USD}$.

Ryan Unger

1:58 AM

dude that's insane

anakhro

It wasn't without malnutrition.

But life is better now.

Ryan Unger

good

anakhro

I think I spend close to $50/week

I think.

Ryan Unger

what job do you have

anakhro

grad student 4 lyfe

Rithaniel

2:04 AM

I'm looking forward to attaining an assistanceship

Though, I think the whole question about infinity is that the axiom of infinity is specifically there to say "let there be infinity." So the question is "Can you define infinity without axiomatically declaring it to exist and then prove that it does exist?"

I hold that you cannot, because using finite "tools" in a finite number of steps, you cannot attain an infinite object. There is a gap between the two and you can't cross it without making a jump.

anakhro

@Rithaniel @Rithaniel in the question you have in quotes there, what is the definition of "infinity"?

Ryan Unger

@anakhro why are you so against our denial of infinity

i'm not denying the existence of contact structures

anakhro

I don't think there is a denial of infinity. I think it's a whole lot of hot air.

Rithaniel

Well, I don't deny infinity. I'm fine with making that jump, I just think that there is a jump to be made.

anakhro

No more of a jump than any other axiom of ZFC.

Rithaniel

2:13 AM

Naturally. Axioms tend to be that way.

anakhro

ya that's the point

Rithaniel

Though other axioms of ZFC can be proven true as theorems in alternate systems of axioms.

anakhro

It's an axiom. You literally have to start somewhere, otherwise it's turtles all the way down.

Anything else is moving goal posts.

Rithaniel

The thing about the axiom of infinity that makes it unique is that all axioms which could potentially lead to the existence of infinity have to either directly or indirectly declare that infinity does exist.

Also, I agree entirely with your assessment, Anakhro. You do have to start somewhere.

(Though, I'm making pretty broad claims and I'm not sure I could back them up with rigor if it came to that.)

Perhaps I should say that it is a conjecture that the axiom of infinity has the property of not being derivable without an axiomatic declaration of the existence some infinite object or process.

anakhro

See, now there's a conjecture I can see as having some substance to it!

Thank you, Rithaniel.

Rithaniel

2:24 AM

No problem. It's still not quite as thorough as you would really want it to be.

anakhro

@Rithaniel on the subject of that, there is this: math.stackexchange.com/questions/779037/…

2

Rithaniel

Proofs at such a basic level can be difficult to follow.

I think I'm gonna take a nap and read that again tomorrow.

Secret

3:19 AM

@anakhro I see Rithaniel had clarified the conjecture to you

Secret

3:31 AM

But saying these in my own terms:

1. No, I do not think there is anything wrong about axiom of infinity (I am an infinity lover, otherwise I would not spent so much time studying it)

2. As Rithaniel said, the unique thing about axiom of infinity, or any theorem or results that involves infinity seemed to always rely on that infinity or some universal object is already exist

3. The conjecture is then: can we find a system of axioms, none of which axiomise the existence of infinity, such that the existence of infinity can be derived from them

4. A much stronger conjecture, will be the additional requirement that the system of axioms has a bounded number of axioms

Secret

4:07 AM

@anakhro @Rithaniel see below

> Work in 𝖹𝖥−Infinity. If 𝜔 is a set then we're done, so assume that 𝜔 is a proper class (in which case it is the class of all ordinals.) Then there is a definable surjection 𝜔→𝑉.

Why we can assume $\omega$ exists as a proper class in ZF-infinity?

we did not proved the existence of $\omega$ first before using it?

Ajay Mishra

4:40 AM

How to find the matrix of $\mathbb R^3 \to \mathbb R^2$ matrix, I have three unknowns and two variable?

I've checked the site they talk about, $\mathbb R^2 \to \mathbb R^3 $ which is easy because there are two unknowns and three equations for each row.

Secret

anarkhro, rithaniel: Some partial answers to the conjecture (Rithaniel's which is my (3)): Dedekind infinite sets are those that self inject to a subset. The existence of such requires at least countable choice in ZF. I am not sure if there is a non circular way to derive the existence of a set that self inject to a subset because countable choice is just one of the many forms of "axiomatic declaration of the existence some infinite object or process"

@AjayMishra Such matrix is 2x3. I am guessing you already have 2 linear equations in 3 unknowns. Then the matrix is just the coefficients

Ajay Mishra

@Secret "Then the matrix is just the coefficient" what does that meant?

Secret

ax+by+cz = d
gx+hy+iz = e

Then the matrix is just $\binom{a b c}{g h i}$

Ajay Mishra

I've to find T, if $T(1,0,-1) = (2,3)$ and $T(2,1,3) = (-1,0)$. It seems that impossible.

@Secret But that's not a real matrix( I mean that is full of variables)

Secret

hmm...

Ajay Mishra

4:49 AM

Isn't the third equation is necessary? And OP gave me just 2.

Secret

well, {(1,0,-1),(2,1,3)} does not form a basis in $\Bbb{R}^3$, I never seen such scenario before

Let me think

I don't think you have enough information to resolve T completely. You have 2 equations and 3 unknowns, essentially

Ajay Mishra

Yeah. I too think so. Thanks for your time.

Secret

Let T = (x,y,z) where x,y,z are vectors in $\Bbb{R}^2$. Then the two conditions means:
x-z=(2,3) and 2x+y+3z=(-1,0). Thus solving them you end up still with a vector equation in 2 unknowns, meaning your possible T can span a plane

Alessandro Codenotti

7:59 AM

@Rithaniel The thing about the axiom of union that makes it unique is that all axioms which could potentially lead to the existence of unions have to either directly or indirectly declare that unions exist. Where's the difference?

Secret

9:03 AM

@AlessandroCodenotti Really? I thought Union is a predicative notion?

(cause I never heard of any reverse mathematicians dropping out some notion of union, being at its base level basically string concatenation?)

Alessandro Codenotti

9:16 AM

I don't know what predicative means, I'm just talking about the ZFC axioms

Most of them assert the existence of some kind of set, which needs to be asserted by an axiom, in that regard I don't find infinity different from the rest

ÍgjøgnumMeg

quantamagazine.org/…

Alessandro Codenotti

formal verification of FLT's proof is not a number theory goal, it is a goal for proof checkers, FLT is just a "long, difficult and new" proof in this context (with the added bonus of being super famous I guess)

ÍgjøgnumMeg

10:04 AM

Was interesting to see that Kevin Buzzard commented saying that he's moved away from number theory and started working on formal proof verification

Balarka Sen

hott intensifies

s.harp

10:55 AM

@RyanUnger I was interested in seeing: If every bounded vector field is complete, the metric is complete

and also: a (bounded) $C^\infty$ linear combination of (bounded) complete vector fields is complete

specifically in my situation I have a global orthonormal frame consisting of complete vector fields, and I hope this makes the metric complete

Ultradark

11:09 AM

@Secret your main thing is chem right?

Secret

12:26 PM

@Ultradark yup

user193319

Theorem in Rudin's FA: Let $X$ be an infinite dimensional Banach space and let $T$ be a compact operator. Then $0$ is an eigenvalue of $T$. Here's Rudin's proof: if $0$ is not an eigenvalue, then $R(T) = X$.

...?

I still haven't figured out why $R(T) = X$ is a contradiction, but I'm not worried about that at the moment. Why does $0$ not being an eigenvalue imply $T$ is surjective?

I can see that it implies that $T$ is injective...

Thorgott

12:50 PM

Because $T=T-0\cdot I$ is invertible iff $0\not\in\sigma(T)$.

And $R(T)=X$ is a contradiction due to part $(b)$ of the Theorem, as far as I can tell.

user193319

1:23 PM

Thanks!

Quote from Rudin's FA: "An operator $T \in B(X)$ is said to be invertible if there exists $S \in B(X)$ such that $ST=I=TS$. In this case we write $S = T^{-1}$. By the open mapping theorem, this happens if and only if $N(T) = \{0\}$ and $R(T) = X$."...So, $T$ is invertible if and only if $N(T)=\{0\}$ and $R(T) = X$? Why do we need the open mapping theorem to know this?

So, does anyone have spikes on the back of their head from using smartphones? I don't: I don't really use my phone; besides, my neck would be killing me if I had to look down at it for more than 5 minutes, so I would probably just find a way of supporting my neck if I had to use it, thereby preventing any spikes.

Thorgott

I think you need the Open Mapping Theorem to guarantee that the inverse of a continuous linear operator is continuous?

user193319

Ah, you are right. That was kind of subtle.

So, not only does it have an inverse; it's inverse is also in $B(X)$.

Thorgott

Yeah, that $N(T)=\{0\}$ and $R(T)=X$ are equivalent to $T$ being bijective is elementary, so that is where the subtlety in the definition of invertible lies.

Rithaniel

2:42 PM

@AlessandroCodenotti Yes, my statement was poorly phrased. It was me reading back on it that particular comment which spurred me to say that I was perhaps making claims which were too broad.

Though, it is worth noting that the axiom of union doesn't directly state the existence of unions in the context we usually think of them. It needs to be used in tandem with the axiom of pairing in order to result in "traditional" unions.

I was quite tired last night, too. Perhaps I should not have been engaging in discussion of axioms and logic.

Secret

I had a feeling all axioms except infinity can be derivable by axioms that does not reference any of the respective concepts (union, powersets, relations with bounded variables, empty set etc.)

Meanwhile, I think the easier subquestion that can help on that conjecture may be to check whether one can derive the folllowing predicate:

"A set that self injects into one of its subsets"

without using countable choice. I only knew that is impossible in ZF, but what is harder to examine is all possible set theories

Rithaniel

Well, the tricky part is making sure your axioms are good axioms. Like, I could have an axiom that says "If you can define a list of elements, then there exists a set containing those elements" and then attain the concepts of union and intersection from that axiom, but it's sloppily worded (not that I can necessarily discern between sloppy wording and good wording, I'm just betting that it is sloppily written because I just now came up with it).

Secret

My current non rigorous thought on how one can model union using only strings is something like this:

Define string concatenation

Let s,t be strings. Then st is also a string

identify s,t,st as respective sets in the axiom of union

anakhro

@Secret the point is that you don't assume its existence as a set, I think.

Secret

ah right, I need to prove that st is a string

anakhro

2:56 PM

The axiom of infinity gives that you have an infinite set, not necessary for an object which is "infinite" in some philosophical sense. As far as I understand.

Secret

yes, it postulate there exists a dedekind infinite set

so classes can indeed be whatever

anakhro

Alessandro did have a point, though. The other axioms do assert explicitly the existence of a set.

So the axiom of infinity is not unique in this sense.

Secret

true

anakhro

But in any case, interested in what you think of that question/answer I posted.

Rithaniel

I still want to get a copy of the three volumes of the Principia Mathematica by Bertrand Russell to see what exactly they were doing that they took over 200 pages to prove 1+1=2

Secret

3:00 PM

However, objects postulated in the other axioms seemed to be finitistic, so it seems there is a way to find an axiomatic system that can derive them non circularly (in the sense Rithaniel used when he state the conjecture)

anakhro

@Rithaniel you can find it online relatively easily.

@Secret did you look at the question I posted?

Secret

Do you mean that link about real numbers, or a different question?

Rithaniel

I read that question/answer pair again. Still don't quite follow it, but I think there is an assumption that the real numbers are infinite.

skillpatrol

@Rithaniel 366 pages, actually :P

anakhro

@Secret math.stackexchange.com/questions/779037/…

Secret

3:01 PM

11 hours ago, by Secret

> Work in 𝖹𝖥−Infinity. If 𝜔 is a set then we're done, so assume that 𝜔 is a proper class (in which case it is the class of all ordinals.) Then there is a definable surjection 𝜔→𝑉.

11 hours ago, by Secret

Why we can assume $\omega$ exists as a proper class in ZF-infinity?

11 hours ago, by Secret

we did not proved the existence of $\omega$ first before using it?

anakhro

As I pointed out above, $\omega$ doesn't need the axiom of infinity.

Rithaniel

366? I'm going to actually try and remember that number for next time.

skillpatrol

leap year

Rithaniel

Hey, that's a good one.

Secret

@anakhro Ok I cannot quite follow on that. Can you show me the post you said about proving existence of $\omega$ without axiom of infinity, choice?

Because as far I am aware, at least in ZF, the existence of $\omega$ (as a set) is a consequence of axiom of infinity

anakhro

3:06 PM

Yes, precisely. "the existence of $\omega$" AS A SET.

They are not assuming it exists as a set.

Alessandro Codenotti

Would you consider an axiom saying "there is a function $f$ that maps dom(f) in dom(f) which is injective but not surjective" as directly referencing infinity?

anakhro

As you acknowledge in your question "Why we can assume $\omega$ exists as a proper class in ZF-infinity?"

Secret

anakhro: True, I was thinking a bit too far because when I am working on problems with my reverse mathematics hat on, I will not even declare something as a proper class if I cannot concretely construct it from something non circular, as otherwise it is easy to "cheat" by appealing to countable classes
Alessandro: That's what I am thinking about when trying to tackle the subcase of dedekind infinite sets. The problem here is whether there is a way to prove that such f exists without referencing f or any notion of infinity. But yes that is a sufficient condition for referencing infinity

That I still need to think about it as I literally started thinking about this question today

Alessandro Codenotti

$\omega$ is always at least a definable class (even a simple one, it's definable by a $\Sigma_0$ formula)

Secret

hmm... sure, peano axioms seemed to be enough to assert the existence of a collection of natural numbers (thus $\omega$ can be defined using just a bounded predicate)

Using countable class still feels like cheating to me, I might need to think about this...

Hmm... so, we have the natural numbers constructed from peano axioms

and because of that, we can define the class $\omega$ using some definable formula

thus the class $\omega$ must exists and is not of the same size as any natural number (As otherwise, the successor of the current largest number will lead to a contradiction), which means it can only be infinite (countable to be specific)

skillpatrol

3:25 PM

infinitely many

Secret

Hmm interesting, so a countable class is definable in 1st order logic

While now convinced that a countable class is definable without circularity, I do think using some axioms that derive "there is a function f that maps dom(f) in dom(f) which is injective but not surjective" seemed to be even better (because then not even countable is used), but that might be too harsh and I probably need to spend some time thinking about how to do that

Secret

Short summary of today's investigation on the nature of infinity:

It is not size nor invariance nor unboundness that characterise what is meant by something to be infinity. It is some notion of incompleteness and unreachability (that you cannot get the whole thing starting with any parts of it). Will investigate this more as I continue on the RudyRucker book as well investigating on the question on how to construct the aforementioned function f

Secret

Now to move on something that is not infinite for the rest of the night

Words I use are highlghted, and does not use them like pairs as shown

Konformist Liberal

4:11 PM

Hi! Does correspondence theorem for rings need axiom of choice?

Alessandro Codenotti

What's the correspondence theorem for rings?

Secret

11

Q: Correspondence theorem for rings.

Could someone provide a reference that includes a full and honest proof of the Correspondence Theorem for rings? Let $A$ be a multiplicative ring with identity and $I$ an ideal of $A$. There is a one-to-one correspondence between the ideals of $A$ that contain $I$ and the ideals of the quotie...

Alessandro Codenotti

Oh that

It doesn't look like it needs any choice to me

Konformist Liberal

It "feels" like we are choosing a $b$ for every $J$ (referring to the proof send above)

Alessandro Codenotti

I'm looking at the second proof

Leaky Nun

4:53 PM

facebook.com/HongKongPoliceForce/photos/a.964784386942859/…

please report as false news

Secret

done

Leaky Nun

5:19 PM

the police really have nothing to do

or rather they can't do anything

amnesty international claimed today that the police in June 12 abused their powers

Secret

This mess is going to go for a while

If you are going to protest, stay safe. For me, like all oversea buddies, we can only sign petitions and do our local protests

June 12, exploding that tear gas in the middle of the crowd...

Stagnation point

In fluid dynamics, a stagnation point is a point in a flow field where the local velocity of the fluid is zero. Stagnation points exist at the surface of objects in the flow field, where the fluid is brought to rest by the object. The Bernoulli equation shows that the static pressure is highest when the velocity is zero and hence static pressure is at its maximum value at stagnation points. This static pressure is called the stagnation pressure.The Bernoulli equation applicable to incompressible flow shows that the stagnation pressure is equal to the dynamic pressure plus static pressure. Total...

Fluid mechanics give extra points to play with

Michael.P

6:13 PM

Guys, what can $C(a,b)^2$ mean?

user193319

How can the Hahn-Banach extension theorem be applied when there is dominating sublinear functional? E.g., in a mere locally convex space.

@Michael.P What's the context?

Michael.P

@user193319 math.stackexchange.com/questions/3268706/…

The longer answer by user10354138

user193319

Why don't you ask him/her?

Michael.P

I am in the process of creating a chat room

user193319

6:38 PM

Ah, evidently "mere" local convexity is enough. See theorem 3.6 in Rudin's book on FA.

Alessandro Codenotti

7:11 PM

Hahn-Banach works on any vector space

Every real vector space let's say

MatheinBoulomenos

8:54 PM

Hahn-Banach also works over $\Bbb Q_p$

user193319

If $X$ is a (locally?) compact Hausdorff space, does the $C^*$-algebra $C(X)$ admit a trace?

Or $C_0(X)$, in the case that $X$ is locally compact Hausdorff?

Leaky Nun

you should be able to integrate $C_c(X)$ functions

user193319

Okay. So every locally compact Hausdorff admits a measure?

The radon measure?

Leaky Nun

oh you need a group to get a Haar measure

Tanner Swett

10:04 PM

Hey, is there a standard notation for specifying particular roots of particular polynomials?

I see that, for example, $x^6 - x - 1$ has a root that is approximately $1.13472$, but is there a more concise and standardized expression for this number than "the root of $x^6 - x - 1$ that is approximately $1.13472$"?

Semiclassical

@TannerSwett I doubt there's a systematic notation, given the sheer breadth of possibilities. But in your case I think the simplest one would be "the unique positive real root"

skullpetrol

10:27 PM

"...of the polynomial equation"

or the zeroes of the function

curio

good evening. Im a bit confused. Will a converging series of cosines, which involve non-integer spaced frequencies always yield a function that comes arbitrarily close back to its initial condition?

i guess this might be true since if it converges, at some point the next oscillations don't matter much anymore. So it basically reduces to a finite sum of oscillations, which will "rephase" like this

Tanner Swett

That's a really interesting question.

I'd say yes. There's guaranteed to be a point where the 10 largest terms all coincide to within 0.1 degrees, as well as a point where the 100 largest terms all coincide to within 0.01 degrees, and so forth.

curio

10:45 PM

yes. as long as there is a finite sum of oscillations, they will come arbitrarily close if you wait long enough. but what if you add infinitlly many? strange things can happen then right?

Tanner Swett

Right, but you can simply identify how close you want to come (say, you want to find a place that mirrors the initial condition to within 0.01%), and then you can just take enough terms that you've captured 99.99% of it.

curio

i guess you can choose an N large enough, so that the contribution of all terms after that become smaller then a certain value for all times. then the oscillations will coincide to a range of degrees, arbitrarily close to that vlaue

yeah exactly

that seems right

Tanner Swett

11:01 PM

You'll get a certain amount of error due to truncating the series, as well as a certain amount of error due to the phase not exactly matching, but both of those can be made arbitrarily small, at the same time.

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$Mathematics$ Mathematics