Mathematics - 2020-01-15 (page 1 of 2)

Akiva Weinberger

00:05

@TedShifrin On a degenerate torus where the innermost longitude is collapsed to a point, what's the (improper) integral of the Gaussian curvature everywhere away from that point?

Ordinarily you'd expect 0 from Gauss-Bonnet but the curvature at the bad point blows up so I think that throws things off

Besides, it's also basically a sphere where you've pushed the north and south poles together

@BalarkaSen ^

Ted Shifrin

00:33

DogAteMy: What a fascinating question. I think that by symmetry the curvature integral is still $0$. (Yes, $K$ blows up as you approach the point, but there's a compensating zero in $dA$ at that point.) Alternatively, the image of the Gauss map is probably still net $0$. But you're certainly right that $\chi = 1$, but we don't expect Gauss-Bonnet to work. Presumably we can cut out a little disk around the point and apply the boundary version and you get geodesic curvature there.

Akiva Weinberger

01:10

Hm you're probably right

Kinda disappointing

What that means, though, is that if you slice it in half (perpendicular to the axis of rotation) you get a topological disc, whose boundary is geodesic, such that the integral of the curvature is 0

(Compare that to a hemisphere, which is also a topological disc with geodesic boundary, but has $\int KdA=2\pi$)

What that means is that, to make this topological disc satisfy Gauss–Bonnet, we need to give the bad point a curvature of $2\pi\delta(0)$

(where that's the Dirac delta function)

Ultradark

01:47

@AkivaWeinberger I can't remember what balarka and you said several weeks ago about that space I posed which was either isometric to minkowski plane or isometric

I can't remember whether it was isomorphic or isometric

If distances are shrunken it can't be isometric, correct?

So then it was probably isomorphic

Akiva Weinberger

01:59

@TedShifrin In other words, as a metric on the torus, the metric blows up on a ring which contributes $0$ to the integral of the curvature. But as a metric on the sphere, the metric blows up at two points, which would have to contribute $4\pi$ to the integral of the curvature (i.e. each point should have curvature $2\pi\cdot\infty$, in the sense of the Dirac delta) if we want to preserve Gauss–Bonnet.

Ultradark

oh actually I think it's isometric, if, the metric is preserved

and balarka said it was I believe. time to try to prove that

Akiva Weinberger

(That would be made rigorous by viewing it as the limit of metrics on the sphere.)

yh05

I need help with the following proof. Why is the underlined statement true?

Akiva Weinberger

@yh05 On the left, we have some multiple of $g(x)$, so it has degree at least $\deg(g)=m$ (or it's the zero polynomial)

On the right, we have the difference of two polynomials of degree less than $m$, so the difference must also be less than $m$

The only way for these to be equal is if they're zero

yh05

Why?

Akiva Weinberger

02:09

If they're not zero, we have "degree at least m" = "degree less than m"

which is impossible

yh05

Why?

Akiva Weinberger

Which part @yh05

ÍgjøgnumMeg

I miss 24hr libraries

studying at night time at home is 90% procrastination

anakhro

02:25

@AkivaWeinberger you can do better. :)

Akiva Weinberger

Dammit!

@TedShifrin ^

anakhro

The very cool thing is that the 3D analogue to the problem---they have absolutely no idea.

Akiva Weinberger

I have a new idea

Inscribe the circle in a square

Head towards one corner ($\sqrt2$)

Head tangentially towards the circle ($1$)

Half-circle ($\pi$)

Tangent out one ($1$)

anakhro

An improvement.

Akiva Weinberger

This gives 6.556

What did I have before

I had 6.712

anakhro

02:29

You said 6.7 or something

Yeah

(hint: you can do better)

Akiva Weinberger

What if I inscribe the circle in a triangle and do the same thing

Hm… intuition says no

anakhro

@AkivaWeinberger are you at all a fan of L^p spaces?

Akiva Weinberger

This isn't related to the problem, is it?

anakhro

No. I am afraid not. :P

Akiva Weinberger

@anakhro A little

anakhro

02:32

I am just stuck on seeing how the two different versions of Young's inequality are equivalent: en.wikipedia.org/wiki/Young's_convolution_inequality

From "equivalently, if p,q,r>=1 and ..."

Akiva Weinberger

Young's convolution inequality

In mathematics, Young's convolution inequality is a mathematical inequality about the convolution of two functions, named after William Henry Young. == Statement == === Euclidean Space === In real analysis, the following result is called Young's convolution inequality:Suppose f is in Lp(Rd) and g is in Lq(Rd) and 1 p + 1 q = 1 r + 1 {\...

anakhro

The convolution one, and the one with the 3 p,q,r norms on the right and the left is the big double integral

Just in R^n, not the generalization

In the integrand I see the convolution of g and h

And I feel like Holder's inequality might come in handy

But I can't figure out a way to massage out an equivalence.

And despite it being a common equivalence, no textbook I have seen explains it. >: (

Akiva Weinberger

What happens if we set $h=1$ in the second one

anakhro

Maybe I am silly or tired, but wouldn't that mean the right hand side blows up, since 1 is not L^r?

Akiva Weinberger

Oh

Whoops yeah

anakhro

02:41

Glad it wasn't that easy.

wipes sweat off brow

Akiva Weinberger

I'm still a beginner here

What's the identity for convolution?

$\delta$?

yh05

@AkivaWeinberger I haven't proven that if polynomial p=q then deg of p = deg of q. So I can't use this to obtain a contradiction. To prove this, don't I need to first prove that p=q iff the coefficients p_n = q_n for every n?

Akiva Weinberger

Oh hold on

I didn't even realize that $p,q,r$ are different in each one @anakhro

anakhro

Perhaps you have the right idea and you can cook up an h.

Akiva Weinberger

@yh05 That's the definition of polynomial equivalence

for arbitrary fields, anyway

If we're working in the field of real numbers, then polynomials are equal iff they're equal as functions

Lemma: A real polynomial is zero for all values of x iff it's the zero polynomial

yh05

02:46

Instead of taking it as a definition, I want to prove it.

https://math.stackexchange.com/questions/586611/proof-two-polynomials-px-and-qx-attain-same-value-for-every-x-in-mat

But the non-analytic proof relies on collaries of theorem E.1. So the whole argument becomes circular.

Akiva Weinberger

Proof: Suppose otherwise. Let its degree be m. Then divide by x^m. You end up with a+b/x+c/x^2+d/x^3+… = 0 for all x

Now take the limit as x->infty

You get a=0

Contradiction

This is basically the second answer in your link

except I divided by x^degree first

@yh05 I should point out that this is false for finite fields

Do you know modular arithmetic?

Consider x^p+x in the field Z/pZ

For example x^3+x in the integers mod 3

If you plug in 0,1,2 you can check that x^3+x is 0 for each of them

So, when F=Z/3Z, x^3+x is equal to 0 as a function, even though (by definition) it's not equal to 0 as a polynomial

Sorry: I meant x^3-x

Not x^3+x

So 0^3-0=0, 1^3-1=0, and 2^3-2=6=0 in Z/3Z

yh05

03:18

@AkivaWeinberger I see. I'm trying to prove for polynomial over $\mathbb{R}$.

anakhro

04:09

@yh05 what are you trying to prove exactly?

Akiva Weinberger

04:23

@anakhro A polynomial is zero everywhere iff its coefficients are all zero

Equivalently, two polynomials are equal everywhere iff their coefficients are equal to each other

@yh05 Then what's wrong with the analytic proofs?

Stupid Questions Inc

@TedShifrin Apparently one can show that $\sum_{n=1}^m c_n=\sum_{n=1}^m a_n\sum_{n=1}^m b_n$ and from there the result follows immediately so for $c_n=\sum_{k=1}^n a_kb_n+\sum_{k=1}^{n-1} a_nb_k$, $\sum_{n=1}^\infty c_n$ is convergent

I don't know how one was supposed to eagle eye that $\sum_{n=1}^m c_n=\sum_{n=1}^m a_n\sum_{n=1}^m b_n$

A Perspicaciously Curious Mind

08:31

Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos)

6

Nick

11:47

How to calculate the angle for the zigzag

►

A Perspicaciously Curious Mind

Perturbative

13:22

Hey @AlessandroCodenotti

Alessandro Codenotti

Hi

Perturbative

Quick question about Bonn, are you currently staying in accommodation provided by Bonn or on your own?

Alessandro Codenotti

I'm on my own in a flatshare with other students. The university dorm was already full

Perturbative

Ah, so you're paying the expenses for the flatshare on your own?

And was it a struggle to get accommodation?

Alessandro Codenotti

@Perturbative yes

@Perturbative It wasn't too bad, but I got lucky. The city centre is very expensive so I'm living in a village nearby

And it's very hard to find a flatshare if you're not physically in Bonn since you're supposed to visit them first

user

13:48

Do any of you know resources for exercises on Picard's theorem differential equations, autonomous systems of differential equations?

Lucas Henrique

Hey chat

Thorgott

Hey

A Perspicaciously Curious Mind

Hi

Silent

14:28

Let $f:\Bbb R^2\to\Bbb R$. Let $(a,b)\in\Bbb R^2$. Suppose that $\lim\limits_{(x,y)\to(a,b)}f(x,y)$ exists and that $\lim\limits_{x\to a}f(x,y)$ and $\lim\limits_{y\to b}f(x,y)$ exist in a neighborhood of $(a,b)$ for each $y$ and each $x$ respectively. Then Apostol's exercise says $\lim\limits_{(x,y)\to(a,b)}f(x,y)=\lim\limits_{y\to b}\left(\lim\limits_{x\to a}f(x,y)\right)=\lim\limits_{x\to a}\left(\lim\limits_{y\to b}f(x,y)\right)$.

Do above conditions imply that $f$ has to be continuous in a deleted neighborhood of $(a,b)$?

Thorgott

14:42

Is this saying that $\lim_{x\rightarrow a}f(x,y)$ exists for all $y$ (in the neighborhood)?

Silent

yes

Perturbative

@AlessandroCodenotti Okay thanks for the info I'll keep that in mind!

Akiva Weinberger

14:57

@Silent Let $f(x,y)$ be $1$ iff $x=y$ and $0$ otherwise

and look at $(a,b)=(0,0)$

MyWrathAcademia

Is there a rule to find the divisor of a cubic polynomial? Wikipedia says that for a cubic polynomial ax^3 + bx^2 + cx + d that one of the products of factorization is a one degree polynomial of the form qx - p and that q must be a divisor of a and p must be a divisor of d. Does this mean that a must always be exactly divisible by q and d must always be exactly divisible by p in order for a divisor to divide a polynomial into a quotient and remainder?

Thorgott

The limit doesn't exist then, maybe you want $f(x,y)=x$ if $x=y$?

Akiva Weinberger

Oh I misread

Didn't see the bit saying the limit exists

@Thorgott That should work actually yeah

$f(x,y)=\begin{cases}x\ (=y),&x=y\\0,&x\ne y\end{cases}$ @Silent

Thorgott

Yeah, that's nice

Akiva Weinberger

@MyWrathAcademia Yes (if all the coefficients involved are whole numbers)

Thorgott

15:05

As far as a rule to find the divisor is concerned, the Cardano formula gives you all roots, i.e. a factorization into irreducibles (over C).

Akiva Weinberger

There's a neat theorem that says, a polynomial is reducible over the integers iff it's reducible over the rationals

In other words, if you have a polynomial with integer coefficients, if you can factor it into smaller polynomials with rational coefficients, you can factor it into smaller polynomials with integer coefficients

This is called Gauss's lemma

Thorgott

It's a very useful result

MyWrathAcademia

@AkivaWeinberger thanks for confirming but why does the divisor x - 3 divide the cubic polynomial x^3 - 2x^2 + 0x - 4 into the quotient x^2 + x + 3 and remainder -13 when d in this cubic polynomial is 4 which is not divisible by the divisor's second term p which is 3?

Akiva Weinberger

Because it has remainder

so it's not a factor

If qx-p is a factor of ax^3+bx^2+cx+d (i.e. it divides it without remainder) then q's a divisor of a and p's a divisor of d

MyWrathAcademia

@AkivaWeinberger thanks that clears it up, that's the piece of information I was missing from the Wikipedia article.

I was trying to find p and q for any cubic polynomial based on thinking that for any cubic polynomial ax^3+bx^2+cx+d a must always be exactly divisible by q and d must always be exactly divisible by p. Now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder, how can I find p and q of a divisor (i.e. finding qx - p) when given only the dividend (i.e. cubic polynomial)?

Akiva Weinberger

15:21

This is equivalent to finding a root p/q of the polynomial

You can do trial and error, but note that there's not always a solution (the polynomial might not have a rational root)

There's also an exact formula for the root, Cardano's formula (the cubic equivalent of the quadratic formula), and numerical methods like Newton's method

What Wikipedia page were you looking at? This one? https://en.wikipedia.org/wiki/Rational_root_theorem

MyWrathAcademia

This one: en.wikipedia.org/wiki/Cubic_equation#In_mathematics

Ultradark

Does the Minkowski metric $ds^2=c^2dt^2-dx^2$ remain invariant upon a change of scale?

let's say you plot the speed of light $c^2$ on a different scale than a linear one. It shouldn't change the metric right?

in principle it seems correct that the metrics should remain equivalent, but I am not quite sure how to prove this

user193319

15:39

Let $K$ be a convex set in a vector space such that $0$ is an internal point of $K$, let $p_K$ be the associated Minkowski functional, and let $z \notin K$. Why must $p_K(z) \ge 1$ hold?

MyWrathAcademia

@AkivaWeinberger I planned to do trial and error if p and q satisfied some criteria such as always being a factor of a and d respectively but now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder. Do you know of any condition/criteria for p and q when qx -p divides ax^3+bx^2+cx+d with a remainder, or in other words do you know of any condition/criteria for p and q regardless of whether qx -p divides ax^3+bx^2+cx+d with or without a remainder?

user193319

If it were true that $z \in p_K(z) \cdot K$, then if $p_K(z) < 1$ were the case, I could write $z = p_K(z) \cdot \frac{z}{p_K(z)} + (1 - p_K(z)) \cdot 0 \in K$, which would be a contradiction...

However, I don't know if it's true that $z \in p_K(z) \cdot K$.

MyWrathAcademia

I am also not completely familiar with the terminology here so please clarify that when you say "there's not always a solution" you mean that there's not always a root when you divide a polynomial by its divisor and that this happens when the polynomial might not have a root that is an integer (i.e. what you mean by a rational root)?

Silent

@AkivaWeinberger Wow! thanks a lot!

Thorgott

Any polynomial divides any other polynomial with a remainder, though "dividing with a remainder" is a somewhat oxymoronic phrasing.

Silent

15:50

@Thorgott Thank you :)

Lucas Henrique

@Semi, look at what I've found:

4

A: Prove that $x^3 + y^3 = 9$ has infinitely many rational solutions

Here is the 'tangent-chord' construction of infinitely many rational points from a single such point. Suppose we have a point $p_0=(x_0,y_0)$ on the cubic curve $f(x,y)=x^3+y^3=9$. The tangent vector at this point is given by $$\left(\frac{\partial f}{\partial y},-\frac{\partial f}{\partial x}\r...

Thorgott

np

Semiclassical

Ah, nice

I never did bother trying to plug the hole, lol

And I’m not gonna do so now :P

Lucas Henrique

I didn't understand your solution 2 years ago, but now (finally) I do

Semiclassical

Nice

Lucas Henrique

15:58

I have a question, tho: why did you choose the tangent as $v \perp \nabla f$ and not $v \parallel \nabla f$?

Leaky Nun

@LucasHenrique do you play chess?

Semiclassical

Because it wouldn’t be a tangent line in the second case? The cubic is a level set, not a parametric curve

Lucas Henrique

Not always, but yeah

Semiclassical

You can parametrize the curve, of course, and if you did then you’d need to construct the tangent vector

Leaky Nun

@LucasHenrique wanna play?

Semiclassical

16:04

But you’d find that it’s still perpendicular to the gradient of f

Lucas Henrique

@Semiclassical so $\gamma(s) = (s^3, 9-s^3)$?

Abhas Kumar Sinha

@Semiclassical Good evening sir.

Semiclassical

No. You still need it to satisfy x^3+y^3=9

Lucas Henrique

I wouldn't say I suck at vector calculus, but I'm not even close to good lol. But I would say I suck at chess, @Leaky :p

Leaky Nun

me too

Lucas Henrique

16:06

oh, lol. that's a line

I'm a bit tired, sorry

@LeakyNun let's play!

Leaky Nun

@LucasHenrique what time control?

Semiclassical

A rational parametrization of this is probably best, eg stereographic projection

Leaky Nun

or no time limit?

Semiclassical

Mind if I spectate?

Leaky Nun

@Semiclassical sure

Lucas Henrique

16:08

you choose

Leaky Nun

@LucasHenrique how much time do you have?

Lucas Henrique

Plenty of it

Leaky Nun

@LucasHenrique lichess.org/1eHbI0ij

@Semiclassical same link to spectate

Lucas Henrique

16:25

told ya

hahahaha

Semiclassical

Yeah, I watched

Lucas Henrique

I gave my queen away

that was so sad

Semiclassical

My problem with chess is a sense of decision paralysis

Too many options and too little sense of their long-term implications

Leaky Nun

@Semiclassical want a game?

Semiclassical

Yeah, sure. 15 mins is fine

Leaky Nun

16:28

@Semiclassical with 15 sec increments?

Semiclassical

I meant whatever time control you had in the last game

Leaky Nun

ok

@Semiclassical lichess.org/ziXiIVM0

MyWrathAcademia

Is the product of factorizing a reducible polynomial always in the form (qx - p) (ax^2 + bx + c)? that is, is the coefficient of the first term of the lower degree polynomial always positive and the coefficient of the second term always negative?

For some reason I thought for a cubic polynomial ax^3+bx^2+cx+d only one specific divisor can be used to obtain all the roots of the cubic polynomial but since any polynomial less than degree 3 can divide a cubic polynomial into a quotient does that mean that the roots of a cubic equation can be infinite?

Can someone clarify this. May be I should start thinking of a root as simply a solution to a polynomial therefore any first degree or second degree polynomial should have a solution (i.e. should be able to be solved by either factorization or quadratic formula)

So I'm just realizing that a cubic polynomial cannot just be divided by a single polynomial of the form qx -p, any polynomial of the form qx - p can be the divisor

Thorgott

16:46

To your first question, that coefficient is only a matter of convention; after all, qx-p=qx+(-p). To your second question, it is not true that any polynomial can divide a cubic (if that's what you mean) and the number of roots of a cubic is at most 3. To your third statement, a root is by definition a solution to the equation "polynomial = 0". Whether polynomials have a solution depends on the field you're working on (x^2+1 has no real solution). The fourth statement is wrong, like the second.

For clarity, a polnomial P(x) divides a polynomial Q(x) if there is another polynomial R(x) such that Q(x)=P(x)R(x).

adesh mishra

@Thorgott I want to know: are you an academic student? I’m asking it beacuse your replies to me are very proficient (just like sir @Semiclassical and I know he is not a student) . If you don’t mind much then please just tell me only at what stage of learning are you in ?

Thorgott

I'm a math undergrad

adesh mishra

Really? I doubt it

user21820

@Thorgott Why don't you write out a sketch of what you think the proof is? I can't criticize a vague handwaving proof. But I can say for sure that a proof that merely claims some ring to be a field cannot be correct. I don't know what you mean by "plain induction". It should be clear from the proof I gave that all you need is a well-ordering of the initial field. If it is countable then obviously you don't need any transfinite recursion.

MyWrathAcademia

17:01

@Thorgott thanks a lot, that helped. However in my second question I meant that a polynomial can reduce (i.e. divide) another polynomial into a quotient and no remainder or a quotient and a remainder. For example for the polynomial x^3 - 2x^2 + 0x - 4 the polynomial x - 3 can divide it into the quotient x^2 + x + 3 and the remainder 13 and also the polynomial x - 5 can divide the same cubic polynomial into the quotient x^2 + 3x + 15 and the remainder 71.

@Thorgott From these two quotients you can obtain four solutions (i.e. 4 roots). I think that your statement "For clarity, a polnomial P(x) divides a polynomial Q(x) if there is another polynomial R(x) such that Q(x)=P(x)R(x)." is true only when polynomial P(x) divides polynomial Q(x) without a remainder.

user21820

@Thorgott: And to make it easier to follow the discussion, please post your proof in the other chat-room.

Semiclassical

@LeakyNun lol, stockfish evidently disagrees strongly that I was in any way playing well :P

Leaky Nun

@Semiclassical the evaluation isn't everything I guess

@Semiclassical eg SF can't tell you the sharpness of a position (how easy it is for me to blunder)

Thorgott

@MyWrathAcademia What I gave you is the definition of what it means for one polynomial to divide the other. Of course, this means there is no remainder. But saying "a polynomial divides another polynomial with a remainder" is vapid, because that's always true (except for the zero polynomial). I don't see how dividing by a polynomial that leaves a remainder gives you a root.

Leaky Nun

@Semiclassical I would say you avoided trades quite well

Semiclassical

17:10

@LeakyNun true

Leaky Nun

but losing those two pawns are kind of unfortunate

Semiclassical

At the end I was aiming to finally attack that pawn holding your defense down in the back

But somehow missed that putting the queen there was a very silly idea

Alessandro Codenotti

@LeakyNun SF can be tuned to play with contempt against human players

user21820

@AlessandroCodenotti Oh, is there a setting on Lichess for that?

Leaky Nun

@Semiclassical but those pawns aren't used for defense

Alessandro Codenotti

17:13

@user21820 I don't think so

Semiclassical

Hmm

Yeah, I guess not

Leaky Nun

since you castled queenside, my a and b pawns are used for attack I think

that's why I sort of gave up my e pawn and snatched your g and h pawns

(which you didn't take because you wanted to avoid trading)

Semiclassical

Well, I mostly had in mind that the b pawn was supporting your knight which was supporting your e pawn

Leaky Nun

ah

Semiclassical

I was also doing my best to keep my rooks active and linked

Leaky Nun

17:16

yeah I can see that

Semiclassical

To the detriment of my pawns

Ted Shifrin

Agh. Rookie chess talk.

Semiclassical

Lol

Ted Shifrin

hi @Lukas

Semiclassical

Anyways

Lukas Heger

17:17

hi @Ted

Leaky Nun

@Semiclassical after Nxg7 maybe you could play Rg8?

I feel like rooks are best on the back rank

and also Rg8 asks me a question

Semiclassical

yeah, they’ve got a lot of freedom back there

You can probably sense I’m not a very analytical player tho

Leaky Nun

maybe I would put my rooks and queen on a single file

or mostly just stack them on the g and h files (so 2+1)

instead of aligning them on the 5th rank

Semiclassical

Right

Leaky Nun

but what do I know

Semiclassical

17:20

In other news, my laptop is a bit borked

Leaky Nun

in what sense?

Semiclassical

One of the two hinges came apart partially

I probably pulled on that side of the case in a bad way

No issue of hard drive or battery or anything, so it’s easy to back up

But not really much good as a laptop if you can’t close/open it

Leaky Nun

instead of that dubious queen move I would probably have played (and I was calculating for you) 20... Nxe4 21. Nxe4 Ref5

Semiclassical

So off to the shop to see if it can be repaired

Leaky Nun

oh right I calculated that it doesn't work

also SF can't tell you about strategy

maybe it can if you ask it nicely

MyWrathAcademia

17:24

@Thorgott This is how dividing by a polynomial that leaves a remainder gives you a root: Dividing the polynomial x^3 - 2x^2 + 0x - 4 by the polynomial x - 5 gives the quadratic x^2 + 3x + 15 and the remainder 71. The quadratic x^2 + 3x + 15 via the quadratic formula has the solutions x = 2.653311931459 or x = -5.653311931459, and the root of the polynomial x - 5 is of course 5. So the cubic ploynomial x^3 - 2x^2 + 0x - 4 has these three solutions as its three roots. Feel free to correct me.

@Thorgott My idea of what it means to divide a polynomial and what a remainder is differs from yours, I believe that divide does not mean to exactly divide something leaving no remainder, and I didn't consider 0 to be a remainder so may be once you consider 0 to be a remainder then may be you are right that it is always true that a polynomial divides another polynomial with a remainder.

Leaky Nun

@Semiclassical do you have any comments for me?

btw I allowed SF to play for a while on the position before that dubious queen move, and in 4 moves it moved the king to b8, rook to g8, and queen to e8

Semiclassical

That knight was quite annoying :P

That’s my main observation

@LeakyNun heh, so SF really wanted everything on the back rank

Leaky Nun

also SF traded down to a RQ vs RR endgame

Semiclassical

Hmm

Leaky Nun

20... Kb8 21. Rfd1 Rg8 22. Rac1 Qf8 23. g3 Qe8 24. Nd4 Qe7 25. Nd5 Nxd5 26. exd5 a6 27. d6 Qxd6 28. Nb5 Qxd1+ 29. Rxd1 axb5 30. Rxd7

but I've been told that if SF is losing then SF just sees every move as losing and doesn't try to complicate the position

@Semiclassical are there discrepancies between what you thought I would play and what I played?

Semiclassical

17:42

Nah

Thorgott

@MyWrathAcademia try plugging 5 into x^3-2x^2-4 and then tell me whether you still think it's a root

Semiclassical

I’ll note that, if one number isn’t divisible by another, then long division can either be used to find the remainder or to find the decimal expansion for the ratio

MyWrathAcademia

@Thorgott plugging 5 into x^3-2x^2-4 gives 71. Are you hinting that x - 5 is not a root because the result of plugging 5 into x^3-2x^2-4 is not 0?

Semiclassical

So, in the former procedure, it does make sense to talk about $a\quotient b$ regardless of whether $b$ is divisible by $a$

Ted Shifrin

@MyWrathAcademia: Be careful with language here. $5$ is a root means $x-5$ is a factor.

Thorgott

17:56

I'm suggesting 5 is not a root, because plugging 5 into the polynomial is not 0. x-5 is a polynomial, roots are numbers.

MyWrathAcademia

@TedShifrin thanks for catching that

Leaky Nun

@Semiclassical this is interesting

Ted Shifrin

@Leaky: I think you need a separate chess chatroom.

@StupidQuestionsInc Oh, my apologies. I was too obsessed with breaking it into two pieces. Interestingly, following my suggested idea leads to an example where one of the pieces need not converge, but presumably the other piece (which would be a mess) cancels out the bad stuff. Think about what they're doing with a picture (like setting up a double integral).

One of the sums is getting the $(k,\ell)$ with $k\le \ell$ and the other is getting the terms with $k\ge\ell$. (Change the order of summation on one.) I guess the diagonal is appearing twice. Not sure if I'm missing a small detail or not.

MyWrathAcademia

@TedShifrin @Thorgott thanks a lot, I think I'm understanding how to know whether a divisor of a cubic polynomial can get distinct roots for that ploynomial. So a factors result must always be a root. Does that mean that only a polynomial qx - p that is a factor (x is a root) can be used to find the other roots of the cubic polynomial of the form ax^3 + bx^2 + cx + d where these other roots are the solutions of the quadratic formula?

@Thorgott since you confirmed that one of the factors of the cubic polynomial ax^3 + bx^2 +cx + d is always in the form qx - p (or in other words, 1x + (-p)) I think I'm finally understanding how to find which values for p and q in px - q to use to divide the polynomial ax^3 + bx^2 + cx + d in order to find three roots.

Ted Shifrin

18:11

@StupidQuestionsInc: I guess I didn't pay enough attention to the indices in your original question. It looks like they did make sure not to include the diagonal in the second sum.

MyWrathAcademia

I meant *qx + (-p)*

Lukas Heger

@Ted would you mind reading over my CV?

Ted Shifrin

@Lukas: I can do that, but I honestly don't know what the European standards are. If you're applying to graduate school, I think you're worrying way too much about it.

Lukas Heger

I'm just applying for masters

Balarka Sen

Hi @TedShifrin

Ted Shifrin

18:13

Hi, a @Balarka.

Lukas Heger

Hi @Balarka

Ted Shifrin

@Lukas: In this country they care about your courses, your grades, and most of all letters of recommendation. All the seminar talks and conferences you've attended should be a bonus. You should certainly mention those in whatever little essay you write.

Lukas Heger

they can see my grades in my transcript

Thorgott

If $qx-p$ is a factor, the corresponding root is $p/q$. Dividing the cubic by this linear factor gives a quadratic whose roots can be calculated using the quadratic formula, yes. However, it isn't at all easy to find such a linear factor. The conditions that q divides d and p divides a discussed earlier are necessary, but not sufficient.

Balarka Sen

@LukasHeger Wait what? Your real name is not Mathein??

Or is this the fake name

Lukas Heger

18:15

@TedShifrin I have to submit a separate motivation letter, I still need to write that

@Balarka yeah, mathein boulomenos just means "someone who wants to learn"

Balarka Sen

Lol

I see

@TedShifrin What do you suppose is the best way to begin to learn French

Lukas Heger

read EGA

/jk

Balarka Sen

I would rather read Thom

:3

Ted Shifrin

@Lukas: As I said, in this country I don't think we even ask for CVs. You can list all relevant academic experiences somewhere, I guess, but that essay gives you an opportunity to impress people.

@Balarka: Learn for math or learn for actual grammar and literature?

Balarka Sen

Both. Former has slight priority.

I would like to be able to read French literature

Leaky Nun

18:19

wait till you get to the "passé simple" lmao

which is anything but simple

Ted Shifrin

Yes, French literature is lovely (à mon avis). For that you probably should just take a few courses. For reading math, get a French math dictionary and just start reading. If you read math you sort of know a bit, it'll be pretty easy.

I can send you my little paper in French (if I didn't already).

Well, you won't find passé simple in mathematics, but you will find it in older literature for sure.

Lukas Heger

@TedShifrin should I mention that I knew calculus at 14 or does that sound too ... I don't know a good adjective

Balarka Sen

My uni unfortunately does not offer language/literature courses, I was thinking of self-learning

Ted Shifrin

No, don't do that.

That's crazy ... stupid uni.

Balarka Sen

I know

Lukas Heger

18:20

and that I went from grade 8 straight to grade 11?

Balarka Sen

I would rather take a lang/lit course than a CS or stat course

Ted Shifrin

@Lukas: A lot of us who study math in university successfully learned calculus very young.

@Lukas: All of that is hardly relevant. All they care about is your mathematical talent.

The fact that you TAed for uni algebra courses, etc., is far more significant.

Lukas Heger

I'm not sure what I should write, I don't want to praise myself too much

Ted Shifrin

Oh, these seminar talks are all undergraduates, or is it mostly graduate students?

Lukas Heger

all of them with one exception were masters level which I guess equals early grad student

Ted Shifrin

18:23

Talk about some specific mathematics you've studied or worked on and why it is so interesting to you. I mean specifics. They hear general bullshit from everyone.

So when you say "student seminar" you should make it clear that it's a "graduate student seminar."

Balarka Sen

@TedShifrin Let me know if you know of a good textbook. I am too old fashioned for duolingo or whatever

Lukas Heger

is it fair to call it graduate student seminar? technically masters students are not graduate students here, I think

or is graduate student anyone who already has a degree?

Ted Shifrin

I don't know texts other than what are standard college texts, and I already was fluent in French when I went to coleege, @Balarka.

Lukas Heger

I thought it was PhD students

Ted Shifrin

@Lukas, in the US graduate includes both masters and PhD. I think that's universal.

Lukas Heger

18:25

okay

Balarka Sen

@TedShifrin Gotcha. Thanks anyhow

Lukas Heger

should I indicate which lectures are graduate level as well?

Ted Shifrin

Yes, sure.

@Balarka: Literature I can recommend.

I wouldn't recommend that you start by reading Elie Cartan. René Thom is more words than symbols, which makes it harder going.

Balarka Sen

I was thinking of diving into Le Petit Prince directly, given I have read the English translation. But I wonder if I know enough French for that

Ted Shifrin

Oh, sure.

Balarka Sen

18:27

Yeah that's why I had Thom in mind

Ted Shifrin

Or Winnie the Pooh in French :P

Balarka Sen

Oh could be good!

There's a book by I. A. Richards which is like a bunch of comic strips of stick figures talking in French but I can't find it on the internet anymore

Lukas Heger

@Ted I adjusted it. You wouldn't mistake "graduate student seminar" for "research seminar" (which is where most PhD students give talks here), right?

Balarka Sen

Oh fuck I could read Tintin in French

How hard can that be

Why didn't I think of this

luchonacho

hi there. Trivial question. What is the notation used to express the inverse of a multivariate function? Do you specify the variable with which the inverse is being computed? Say y = f(x,w). x=f^{-1}(y,w) is one inverse, and the other is w=f^{-1}(y,x). The f is the same. Does it matter, or is it evident by the variables in parenthesis which is the one being inverted?

In my particular case, I am evaluating the inverse. So, actually I have f^{-1}(3,4). I thought if I use f_x^{-1}(3,4) it would be clear, but subscript usually means derivative.

Ted Shifrin

18:37

@Lukas: Again, I don't know European grad culture so well. If you are worried, say "masters student seminar", I dunno.

Thorgott

It sounds as if you don't want to invert the multivariate function, but rather a partial evaluation obtained from it

luchonacho

but you evaluate after inverting, of course

so...no (?)

Thorgott

Here's what I can offer (if I understand correctly). You have a function $f\colon A\times B\rightarrow C$; then you can define functions $f_a\colon B\rightarrow C, b\mapsto f(a,b)$ for $a\in A$ and $f^b\colon A\rightarrow C,a\mapsto f(a,b)$ for $b\in B$ and in case these are bijective, you can properly talk about their inverses.

luchonacho

So notationwise, you are using subscripts to define inverse functions? Is that conventional? That is my question.

$f_a\colon B\rightarrow C

Thorgott

No, these are in no way inverse functions. They are a kind of partial evaluation and as far as that is concerned, this notation is fairly standard.

luchonacho

18:50

let me rephrase. Assume $f\colon A\times B\rightarrow C$. How would you, notationwise, differentiate between $f^{-1}\colon C\times B\rightarrow A$ and $f^{-1}\colon A\times C\rightarrow B$?

perhaps using $f_C^{-1}$ and $f_B^{-1}$ respectively?

Thorgott

I don't know what these are supposed to be. The inverse of $f$ (in case it is bijective) is a map $C\rightarrow A\times B$.

luchonacho

Example. y=xw. One "inverse" is x=y/w, another is w=y/x.

If, y = f(x,w), is notation x=f^{-1}(y,w) and w=f^{-1}(y,x), respectively, standard?

Semiclassical

you've got $f(x,w)=xw$, $g(y,w)=y/w$, $h(x,y)=y/x$, and thus $f(g(y,w),w) = y$, $f(x,h(x,y)) = x$

luchonacho

@Semiclassical so you just use different names?

because it is not, streactly speaking, an inverse?

Semiclassical

all I'm doing is saying what you'd do absent a better notation

there may be a more standard one

luchonacho

18:57

(so inverse notation misleading)

Thorgott

These notations are not standard, they are not even permissible. $f^{-1}$ already has a specific meaning different from what you are talking about. I'm afraid I don't know a standard notation for what you're looking for.

Just giving them a name is probably a good idea.

Transcript for

$Mathematics$ Mathematics