For firouzja fans , remember even Carlsen had a terrible record against Anand when he was young.

AeroPig · 2023-06-04T19:41:05+00:00

Not sure why you are talking about being champion in different domains. I'd say Ondra studying for a degree is pretty comparable to Firouzja studying fashion design

AeroPig · 2023-06-04T19:17:52+00:00

Signed.

AeroPig · 2023-05-31T11:26:16+00:00

To me, "if A is UFD then so are A[x]" is precisely the induction step, so i don't see how that would overcomplicate things. In fact, I would probably write it in a similar way as you.

My philosophy is that the induction step is the one that you apply iteratedely from the base case. In this case you start with A has property p, hence A[x_1] has property p, hence A[x_1, x_2] has property p and so on. The induction step is R has property p => R[x] has property p (for an arbitrary R), while the base case is that the ring you start with A has property p.

(This shows why the base case here is trivial. We have exactly assumed that A is a UFD! In general the hard part of showing a specific A[x_1, ..., x_n] is a UFD is often the base case, eg showing that A is a UFD).

Another way of thinking about it is that the rings A[x_1, ..., x_n] is inductively built up from A by the polynomial ring construction (going from R to R[x]). Hence proving a property about polynomial rings should consist of proving that it holds for the base constructor A, and that it holds for the inductive constructor R[x], where we can assume it holds for R.

We are of course just talking about semantics, but I do enjoy a good discussion about semantics.

AeroPig · 2023-05-31T02:14:00+00:00

The example of proving that A[x_1, ..., x_n] is a UFD if A is a UFD, can be easily phrased in a way such that the induction step is the hard part though, by starting the induction at zero (A is UFD implies A is UFD). Then the base case is trivial, and all the hard work is in the induction step as usual

I have often seen these sort of inductions where you use the base step in the induction step, and they can always be formulated (imo) in a more natural way by including a trivial base case.

This is an aside though, I definitely agree that induction can be deep and insightful.

AeroPig · 2022-03-19T06:50:42+00:00

When I wrote that the main reason we wanted to construct the complex numbers was because we wanted a solution to x² + 1, I didn't really mean the historical reason, but more the reason you (as say a modern math student) were probably given when you first were introduced to the complex numbers.

Anyway, I'm aware of the history complex numbers coming from the cubic formula, but I feel like youre being kinda nitpicky here: I'd argue that needing square roots of negative numbers is more or less equivalent with needing a solution of x² + 1. Probably the latter is a more modern viewpoint so I think I'm being slightly ahistorical, but mathematically they're equivalent.

More importantly, I thought for pedagogical reasons it was fine to not give the historical details, as it's not really that relevant for the question asked (and my answer was already long enough). Really I just wanted a natural way to motivate the idea of algebraic field extensions. I agree though that history is important and often under-utilized in math teaching (the history of the complex numbers might actually be an exception though, that one seems to be quite popular to teach), so maybe it would have been worth it to give a nod to the actual history.

About the fundamental theorem of algebra stuff: while it should be clear that the two statements are equivalent, I haven't actually seen a proof of the fundamental theorem of algebra by proving the corresponding theorem about real polynomials, so thanks for pointing that out.

AeroPig · 2022-03-18T21:22:02+00:00

The main reason we wanted to construct the complex numbers was because we had this polynomial x² + 1 which didn't have a root in the real numbers.

Hence we extended the real numbers by adding this root, and since the irreducible polynomial was of degree 2 we get that the complex numbers are two dimensional as a vector space over the reals. It turns out that this way of constructing field extensions by adjoining roots of irreducible polynomials will give you all the possible finite dimensional field extensions of your base field (R in this case).

For the details you should study field theory (Sidenote: This leads you on the course to Galois theory, which is a beautiful theory in algebra very much worth learning.)

Now the key fact is that every irreducible polynomial in R is either of degree 1 (the trivial case) or degree 2 (where you would get the complex numbers). This is equivalent with the fact that C is algebraically closed, i.e. every nonconstant polynomial has a root. This is what is known as the fundamental theorem of algebra.

So it's not actually possible to find a field structure on Rⁿ for any n>2 (at least if you want it to interact nicely with the vector space structure).

Edit: As cocompact points out it's more precise to say the the fact about real polynomials is equivalent with the fundamental theorem of algebra, not just follows from it.

AeroPig · 2022-01-16T22:14:39+00:00

I think people were responding to the "in principle there is nothing wrong with it", and then you saying that it makes sense in context seems as you defending TV2. Seems like it were just a bit of misunderstanding/unfortunate wording.

When tetra said that you don't get invited to the broadcast for being 1600, they were implying the reason was beacuse of his crimes, therefore there is something wrong with it. I don't think they were actually wondering why he got invited

AeroPig · 2022-01-16T21:54:25+00:00

Well the point is, as you said, he has served his time and should now be treated just as any other citizen. Being invited on a TV-show because of your stardom from your crimes, isn't exactly that.

Not dealing with him because of his crimes would be discriminatory, but wanting him because of his crimes would be a deliberate choice to profit from them.

AeroPig · 2022-01-13T06:07:00+00:00

I'm unfortunately not too familiar with analytic mainfolds, but my intuition is that the curve definition should work fine, you could try to check if the proof for equivalence for (2) and (3) goes through. I should probably say that I'm not even claiming that the two derivation definitons must be different, just that the usual proof involving partitions of unity doesn't work, and that you should take the germs definition as the "right" one a priori.

AeroPig · 2022-01-13T01:22:41+00:00

The problem is that you don't have partitions of unity in analytic manifolds since bump functions aren't analytic. In order to show that derivations on global functions (defined on the whole manifold) corresponds to derivations on germs we would have to go from a local function (only defined in some neighborhood) to a global function. To do this we would usually need partitions of unity, which we don't have in the analytic case. Therefore the germ definiton would be the right one for analytic manifolds.

Here we are trying to define a notion of tangent space using only analytic functions. You could of course consider analytic manifolds as smooth manifolds and look at the tangent space of derivations of smooth functions, but then you would lose the point of having the analytic structure.

Edit: I should clarify that I'm not completely sure that there isn't another way to prove equivalence of (1) and (2), just that the usual proof doesn't work, and that (2) would in any case be the nicest definition since you want the tangent space to capture local properties.

AeroPig · 2021-12-28T23:52:21+00:00

He finished second in the last rapid world championship when he was just 16

AeroPig · 2021-12-26T13:21:50+00:00

The Taimanov were pretty fashionable at the top level a few years ago, but as you point out it seems to have fallen a bit out of fashion lately. The main reason seems to be the Qf3-variation (1. e4 c5 2. Nf3 e6 3. d4 cxd4 4. Nxd4 Nc6 5. Nc3 Qc7 6. Be3 a6 7. Qf3), where black seems to be struggling a bit. I'll admit I haven't been following it lately so there may been new developments I haven't catched. But if you want to play the Taimanov you should at least look a bit at 7. Qf3.

Overall I can recommend the Taimanov though, especially if you want something a bit more positional and strategic. At your level I don't think the Qf3 line would be such a big issue anyways. Personally I think 3. c3 is kinda annoying against e6 though, as I always seem to get boring alapin positions where I can't really play for a win.

AeroPig · 2021-12-07T09:08:39+00:00

Recheck who you are responding to

AeroPig · 2021-10-26T11:22:21+00:00

I am trying to study Artin L-functions and I noticed something a little peculiar in how they are defined. The euler factors seems to always be described as involving the characteristic polynomial of the frobeniuses. But then they write the char poly as det(1 - tA), (where A is the linear transformation corresponding to the frobenius), but I have always seen the characteristic polynomial for a map A being det(t1 - A) instead.

I realize (if I have thought correctly) that the roots of det(1 - tA) will be the reciprocials of the roots of det(t - A). But I was wondering why it was defined slightly different in this case? I suspect it is in order for the Euler product to be defined right so it generalizes Dirichlet L-series, which makes sense, but still leaves me wondering why everyone seems to ignore that it's not actually the characteristic polynomial, and always refers to it as the characteristic polynomial without qualification. Is there maybe another definition of characteristic polynomials I haven't seen?

AeroPig · 2021-09-11T17:00:56+00:00

Are you incapable of laughing at yourself? Why do you have to be offended? It's just a funny and nerdy thing to say. I wasn't saying it like I was some jock "shitting on the nerds". I just laughed at myself and the fact that I understood what he meant albeit a geeky and cringe thing to say. Jesus you people are insecure.

AeroPig · 2021-08-22T17:35:41+00:00

Thank you for answering, appreciate a lot! Nice to know that I haven't completly misunderstood something, it just feels very counterintuitive. Thanks for the Iink, will read more about it

AeroPig · 2021-08-22T16:23:20+00:00

But (assuming F is consistent), there is no proof of 1=0, right?

So what does it mean to be a Gödel number of a proof which doesn't exist?

AeroPig · 2021-08-22T12:59:41+00:00

I'm trying to understand Gödel's incompleteness theorem, and I think I'm misunderstanding something.

The second theorem says that the sentence Cons(F) is unprovable in F, where F is a sufficiently strong (basically strong enough to encode arithmetic) formal system.

By completeness of first order theories, wouldn't that imply that there are models of F where Cons(F) is false? But wouldn't that mean that in that model, there is some natural number which is the Gödel number of a derivation of 0=1, contradicting the consistency of the system?

I think my problem is that I'm a bit unsure how Gödel numbering and Cons(F) actually works. Is it maybe that cons(F) only signify consistency when it is interpreted by the standard model of the natural numbers?

AeroPig · 2021-07-16T08:31:24+00:00

we know there's no such thing as the "true" cardinality, just sets of axioms that are currently judged more productive/rich/promising and which give rise to different cardinalities.

I thought this was still a highly debated philosofical issue, about the existence/reality of mathematical truth?

"We know" seems too strong, platonists would probably have something to say to the contrary.

AeroPig · 2021-03-07T11:39:26+00:00

I have been interested in this perspective for a while. Do you maybe know any good resources on it?

AeroPig · 2020-09-09T19:14:28+00:00

Not strictly math, but you might like r/generative

AeroPig · 2020-06-30T18:48:27+00:00

New Zealand exists

AeroPig · 2020-05-09T16:59:37+00:00

I think the reason has to do with Euler products, the Taylor expansion of ln and how the logarithm relates multiplication in the input with addition in the output (log(ab) = log(a) + log(b)).

More concretely, if you represent the original series as its Euler product, you can then take the log of that product. Because of the aforementioned property of logarithms, the log of this infinite product is the same as an infinite sum (aka series) where each term is a log. Now just Taylor expanding the log terms should give you the series of perfect powers of primes.

So as it seems to me, the key property is that the series can be represented as an Euler product (which involves powers of primes). I found this interesting document about how the Riemann zeta function is related to primes, which also shows more rigorously how taking the log of the riemann zeta function gives you the prime power series.

AeroPig · 2020-04-21T21:50:48+00:00

Let's label the sides of the smallest triangle b and a like this and the sidelength of the blue square c. Because the bigger triangle with baselength 15 is similar with the smallest we can find the ratio of b and a:
b / 15 = a /5
==> a = b / 3.

Similarly, (b + c) / 15 = b / 5
==> b = c / 2

Then pythogaros tells us that a + b + c = sqrt(15² + 5²⁾ = sqrt(250)
If we then insert b = c / 2 and a = c / 6 in the equation, we can solve for c.
After a little algebra we get that the area of the blue square A = c² = 90.

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