@DHMO That's exactly how I want to use this notation (as the closure) but I do not want to go against conventions for obvious reasons

@Astyx don't ask me then, lol

@Ramanujan Firstly, we have $a \cdot b = \Vert a \Vert \Vert b \Vert \cos (a,b)$

@DHMO Usually one starts at S(1), because N does not contain 0.

The extended real line is a compactification of R, not the closure (the closure in which space? R is already closed in R)

@BalarkaSen your country is strange

@Alessandro Yes, but there is no equivalent for $\Bbb Z$ is there ?

@BalarkaSen but induction need not start at $n = 0$...

@DHMO my teacher also takes first n=1

@meow-mix It has to start somewhere.

yes... it starts at $n$

$S(n)$ is assumed and it's proven that $S(k) \implies S(k+1)$

@DHMO so?

Huh?

I'd guess adding $\pm\infty$ to Z and using the order topology would work

No, that's not sufficient.

You need that S(1) holds.

@Balarka Sure it includes 0.

@BalarkaSen no, he's starting from S(n)

Mmm right

Thanks

e.g. the theorem holds for all integer i >= 4, then n = 4 @BalarkaSen

@Ramanujan Then, we have $\Vert a \times b \Vert = \Vert a \Vert \Vert b \Vert \sin (a,b)$

@DHMO He says "S(n) is assumed".

hi chat

@BalarkaSen by that i meant

He should use some other letter then.

it's given; already proven

Hi @Semiclassical

@Astyx To me, there are a few things which make Chebyshev polynomials (first and second kind) interesting.

@Ramanujan well, we have that $-1 \le \sin \le 1$ and $-1 \le \cos \le 1$

Tell me more

one is the minimax property, but I don't do a lot of approximation stuff so I can't say a lot directly on that.

also, when people were arguing about $\mathbb{Z}$ and how to define it, you could just define it up to isomorphism as the free group of a singleton

@Ramanujan So if their product is $1$, then they must both individually be $1$.

@DHMO you mean cos?

The more interesting route to me starts from their relation to sines/cosines.

@Ramanujan never mind that

the vectors are not relevant

Definition of Z ain't a problem. The conventions about N is usually confusing.

i was talking about

Whether it includes 0, or not. Anyway induction works in any case.

i think it was @dhmo, not sure

minimax property ?

What do you mean by that ?

@DHMO a.b=abcos(a,b)

That's the approximation stuff.

(or both $-1$)

what does "." mean?

That's the name I remember seeing it under, at least.

@meow-mix dot product

oh people arent just typing the normal $\cdot$

ok

So you mean the fact that the Tchebychev polynomials' roots are the best for interpolating functions ?

Yeah. See the paragraph starting about here at the DLMF for some details

OK,so there is no value satisfying that relation so both equals 1?

@Ramanujan something like that

Aaaanyways

The thing I'm more familiar with myself is the fact that they naturally arise when doing trigonometric polynomials

47th question how to do?

namely, $T_n(\cos \theta)=\cos(n\theta)$ and $U_n(\cos\theta)=\dfrac{\sin(n\theta+\theta)}{\sin \theta}$

This stills does not quite answer my question though

There's also a rather interesting determinantlal representation of the second-kind polynomials $U_n(x)$

I know most of their properties, but I was wondering what motivated Tchebychev into studying them for polynomial interpolation

namely, equation 19 here: mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html

And that representation points to their link to the study of orthogonal polynomials in general

namely, one has Favard's theorem: Any three-term recurrence relation (which amounts to the determinant of a tridiagonal matrix with $x$ on the diagonal) gives rise to a sequence of polynomials which are orthogonal w/r/t an appropriate weight function.

But now I'm rambling on orthogonal polynomials :)

I think that stuff does link with the minimax stuff, but it doesn't get to the history.

Yes I can (barley) see why

@Ramanujan I have no idea lol

The biography for Chebyshev here includes the following quote of another source

"Chebyshev was probably the first mathematician to recognise the general concept of orthogonal polynomials. A few particular orthogonal polynomials were known before his work. Legendre and Laplace had encountered the Legendre polynomials in their work on celestial mechanics in the late eighteenth century."

"Laplace had found and studied the Hermite polynomials in the course of his discoveries in probability theory during the early nineteenth century. Other isolated instances of orthogonal polynomials occurring in the work of various mathematicians is mentioned later."

It was Chebyshev who saw the possibility of a general theory and its applications. His work arose out of the theory of least squares approximation and probability; he applied his results to interpolation, approximate quadrature and other areas. He discovered the discrete analogue of the Jacobi polynomials but their importance was not recognized until this century."

"They were rediscovered by Hahn and named after him upon their rediscovery. Geronimus has pointed out that in his first paper on orthogonal polynomials, Chebyshev already had the Christoffel-Darboux formula.

Nice

The linkage of orthogonal polynomials to his work on approximation theory seems relevant.

@DHMO answer is 4th option

@Ramanujan I'm writing an answer now

We are however, dealing with a semigroup analogue of permutations: en.wikipedia.org/wiki/Transformation_semigroup

@Astyx Also, if you google the source of that quote---"The work of Chebyshev on orthogonal polynomials"---the first hit is from Google books, and previews the article.

I'll look that up

Thanks a lot !

Hint:

You could probably read Chebyshev directly, though. It looks like he published in French

e.g. "Sur les fractions continues" for his work relating orthogonal polynomials to continued fractions

Really ? Was he not Russian ?

He was

Then I will if I find time to

That might eb in a while though

Looking at a citation, it seems that particular paper was originally published in Russian (1855) but fairly quickly translated to French (1858)

Of course, it may be that Chebyshev's name being attached to those polynomials specifically is a bit of an anachronism, as often happens in math

He definitely did work on orthogonal polynomials re: approximation theory, but his work on Chebyshev polynomials need not have been the highlight of his research in that area.

Anyways, lots of interesting history

@Astyx For something shorter, check out this pdf: math.montana.edu/jobo/writing/documents/cookson.pdf

@DHMO nice!my math teacher left this problem by saying much confusing problem

@Ramanujan your teacher didn't know how to approach the problem?

@DHMO can I know what's your age?

section 2.3 includes a mention of when he first used Chebyshev polynomials

@DHMO may be

Will do

@Ramanujan no, (dv/dx)(dx/dy) = (dv/dx)(v) = v(dv/dx)

Got

@DHMO but du/dx is not equal to d^2x/dy^2

@Ramanujan du/dx = d^2y/dx^2

Ok,great!bye

bye

Hello!

morning

I'm returned from LaTeX workshops.

Does anyone know what the subgroup of integers containing only the odd integers is called when you endow it with a + and * operator?

@MickLH it isn't a group

and a group only has one operator

I believe it's a Field? Or a Ring? I'm thinking of

no, it isn't even closed under the operator

You can define + as $F(x, y) = x + y - 1$

it is closed under the usual multiplication though

(2n-1)(2m - 1)=4nm-2(n+m)+1=odd

Therefore closed under *

And define * as $F(x, y) = (2x+1)(2y+1)/2 - 1$

Oops

I screwed up the unmapping there but I'm on my phone :(

You don't get inverses in the integers for multiplication, though

no additive identity either

Can I treat space $\mathbb{R}^2$ as identical with $\mathbb{C}$?

@java-devel yes

well, if $F(x,y)=x+y-1$ then $1$ is the identity

right, 1 is the additive identity, and 3 is the multiplicative identity

@java-devel You can identify them, just keep in mind that it'll be an R^2 with a specific kind of multiplication

So I can also treat dual numbers space as identical with $\mathbb(R}^2$.

i.e. $(x+i y)(u+i v)=(xu-yv)+(yu+x v)i$ in C translates to $(x,y)*(u,v)=(xu-yv,yu+xv)$ in R^2

@MickLH is your structure isomorphic to (Z,+,*)?

Probably under $n\mapsto 2n+1$ mapping all integers to odd integers.

@DHMO I think so? Yes I believe that mapping does it

yeah, $(2n+1)+(2m+1)-1=2(n+m)-1$.

then it is a ring, as (Z,+,*) is a ring.

@Semiclassical +

woops, yeah. should've been 2(n+m)+1.

How can I understand the concept of a (open) sphere about center a and radius r, where r > 0 with respect to a certain set $\mathbb{R}^p$, where p is a dimension number?

Hello again,

@java-devel Think the interior of a p-sphere. An open sphere is one that does not include its p-1 dimensional surface

This is a specific example of an open set

$\sum_{k=1}^p x_k^2<r^2$.

@Secret Can you explain how do you understand that open sphere is one that does not include its p-1 dimensional surface?

Take p=3, the equation of this open sphere is given by $\{x \in \mathbb{R}^3 : d(x,a)<r\}$

the set of all points at distance from the point $a$ such that the distance is less than r

If you are using the eucliedian metric, then $d(x,a)=\sqrt{x^2-a^2}$

and the equation of this sphere is $(x-a)^2+(y-a)^2+(z-a)^2=r^2$

How's equation of a closed sphere?

$\{x \in \mathbb{R}^3 : d(x,a)<=r\}$?

Yup

yes

Is open sphere does not cover the plane (area, sphere) of the set?

@Semiclassical OH! Thank you! Yes I forgot to mention an utterly critical detail... The set is finite also

Specifically, all operations are modulo a power of two

oh, Z_2?

Well, in that case, you've only got either 0 or 1.

Ah.

I dunno about finite fields.

is two groups have coprime orders, then the only homomorphism between them is the one mapping all elements to the identity, correct?

I think so yes

the inverse of an (invertible) linear map is a linear map, right?

What do you think ?

obviously

If you doubt it, try and prove it

Otherwise some day you'll ask a question, someone will answer wrong and you'll let yourself beleive them

It's almost a field, it has closure, associativity, commutativity, distributivity, identities, and additive inverses, but multiplicative inverses are given by $x^{-1} = \frac{x+3}{x-1}$ which is only defined for half of the elements

Is there a structure that captures this constraint effectively?

a "Mick Algebra" ;)

For a simple way to answer, figure out how to write $\binom{x}{1}\mapsto \binom{y(x)}{1}$ with $y(x)$ linear

(in all seriousness, no idea)

(also when the inverse doesn't exist, it can still "almost exist" if there's a way to formalize that?)

maybe look at Wikipedia's page on algebraic structures

@MickLH what do you mean almost exist?

nilpotence?

I mean that some fraction of the elements with no inverse, have some small number of "candidate" inverse elements

still not sure what you mean

whats the property of these "candidate" inverses?

might be a semiring?

a semiring is a ring without the requirement of additive inverses

I guess you want it to not have multiplicative inverses, though.

They do mention the Boolean semiring, though.

...my explanation became unwieldy so I'd like to acknowledge both of these links and say thanks @Semiclassical, while I try again to explain in less than 2 paragraphs lol

lol, np

i have good google-fu

(there's a comment that literally says that, hah)

@meow-mix can you prove this?

@Alessandro prove what?

Click on the arrow next to my message

@Alessandro well subgroups map to subgroups

If some $H \leq G$ is non-Hopfian, does that imply that $G$ is also non-Hopfian? :0

@Alessandro wait, this property only works for finite groups, so i CAN say this

hmmm ill think about it

How can they have coprime orders if they're not finite?

^

Hello, i have this definition $W^{1,\Phi}(\Omega})=\ovrline{C_0^{\infty}(\Omega)}$

can we deduce from this

that

$u=0$ on $\partial \Omega$ ?

^ \overline