Mathematics - 2017-03-31 (page 2 of 5)

Semiclassical

16:00

Right.

On the other hand, the natural numbers with zero are a monoid under addition which people like.

My default convention is that 0 isn't a natural number, though.

alan2here

this (0 and successor) process can be illustrated visually with a type of simple graph called a path graph

Semiclassical

(I'd say it's best to just do "positive integers" and "non-negative integers", but I know that in some languages people take "non-negative" to mean positive!)

alan2here

I can see the confusion, 0 could be called positive, negative, both, neither, etc...

I don't mind if 0 is included for now

by recursively repeating this sucessor/increment, the other hyper-operations can be used

and by taking the inverses of these functions, the possible values expand beyond the natural numbers, all the way into the complex numbers

Semiclassical

Well, it matters in that you either have an operation (adding 0) which leaves the element the same, or not.

And that 'all the way' disguises quite a bit of legwork.

First, you go from natural numbers to the integers by introducing additive inverses. Then you go to the rationals by introducing multiplicative inverses.

alan2here

+0 is like "repeating increment (1-1) times", I cannot imagine it being missing, but ok, it could be either way for various benefits and losses

Semiclassical

16:11

But to get to the reals you have to consider infinite sequences of rationals.

I don't think you can get to them just through 'inverses.'

alan2here

does it matter though what route you take, all the hyperoperations (all infinite of them) could be discovered then the inverses could be discovered afterwards and it'd be complex numbers either way

oh, well, rational complex numbers then perhaps

MATH ASKER

@skullpetrol ohh ok i thought it was that but i didn't know how i could convert degrees into minutes thanks

Semiclassical

Well, if you're thinking of hyperoperations, you can think of taking an infinite number of smaller and smaller additions.

Hence you get sequences and therefore real numbers, I guess.

alan2here

:)

Semiclassical

But it definitely has to be -hyper-operations.

alan2here

16:16

So approximately, it's the steps:

1. A value (call it 0)
2. successor on step 1 to yield more values eventually expanding into the natural numbers (or there about)
3. Can visually illustrate this with a path graph.
4. Hyper-operations recursively using the function from step 2.
5. Inverses
6. All the complex numbers.

Now, if by step 3 we had a different type of simple graph, like a cyclic graph, or a fully connected graph, or something more general and less symmetric like a planar graph,

because, for example the function in step 2 was different,

and the rest of the steps continued as before.

What would the result be, presumably not the complex numbers, but in simple terms, is this done, can this be useful?

Aetos

Is anyone here good with algebraic geometry?

alan2here

brb

back

Vrouvrou

hello, please how we know if the space is complete or not

i know that a metric space i say to be complete if every cauchy sequence is convergent

alan2here

@Semiclassical Any thoughts on this semi-classical?

Vrouvrou

but for example if i have the space $(R,d)$ where $d(x,y)=|x^3-y^3|$

Semiclassical

16:28

Not really.

Vrouvrou

is it a complete space

Semiclassical

This is past the realm where I have anything interestint to say.

alan2here

What do you mean?

I apreceate your input with the various stages though, I'm just not confident enough to formulate a SE question or understand the answer :P

Zophikel

mathb.in/135784

$$\frac{f^{(n-1)}(z+h)-f^{(n-1)}(z)}{h}=\frac{(n-1)!}{2 \pi i}\int_{C}f(\zeta)\frac{1}{h}\Bigg[\frac{1}{\zeta -z-h^{n}} - \frac{1}{(\zeta-z)^{n}}\Bigg]d \zeta$$

I'm having trouble justifying the RHS side of the formulated difference quotient anyone wanna take a stab :)

BAYMAX

Hey any good reference for problem based approach on Integral equations ?

Vrouvrou

16:34

can someone help me on metric space?

Semiclassical

16:47

@Zophikel Should the first fraction in the integrand be $\frac{1}{(\zeta-z-h)^n}$?

If so, then this is just an instance of the Cauchy integral formula

Zophikel

@Semiclassical interesting I didn't consider that could you elaborate further

Semiclassical

Suppose you've got a function $f(z)$ which is analytic in a complex nbhd of $z=a$.

Zophikel

all right

what do you mean by nbhd do you mean neighborhood

Semiclassical

Yeah.

Then it'll have a Taylor series expansion $f(z)=\sum_{n=0}^\infty f^{(n)}(a)\frac{(z-a)^n}{n!}$.

Zophikel

all right

Semiclassical

16:53

Now suppose we integrate over a contour $C$ which encloses $z=a$.

Zophikel

all right

Vrouvrou

@AlessandroCodenotti hello

Daminark

Hey everyone!

Zophikel

@Semiclassical after we integrate then we have $\frac{1}{(\zeta-z-h)^n}

?

Semiclassical

More precisely, consider $\int_C f(z)\frac{dz}{z-a}$. The above series expansion gives this as $$\int_C\left[\frac{f(0)}{z-a}+f'(a)+f''(a)(z-a)+\cdots\right]dz$$

Vrouvrou

16:56

hello, can someone help me on complete metric space

Zophikel

@Semiclassical all right I see where you are going with this

Semiclassical

Yeah.

The only term in there that'll survive this integration is the first one, so you'll get $2\pi i\cdot f(0)$.

On the other hand, you could just as well have divided by $(z-a)^{n+1}$.

In that case, every integral except the $n$th would vanish and you'd get $2\pi i\cdot f^{(n)}(a)/n!$.

Zophikel

By performing a series expansion you able to make the assertion that $\frac{1}{(\zeta-z-h)^n} was just an instance of the Cauchy integral formula

Semiclassical

(There's probably a more rigorous way to present that. But for these purposes it's fine.)

Right.

Wikipedia refers to it as Cauchy's differentiation formula.

Zophikel

ahh all right @Semiclassical could you latex this up for me plz

Semiclassical

16:58

No.

Zophikel

i'll have to write down to make sure I really get it

Semiclassical

Do it yourself.

Zophikel

all right

Daminark

@vrouvrou I kinda know what's going on with complete metric spaces

Also hey @s.harp

Semiclassical

Probably the rigorous thing to do is write $$\int_C\left[\frac{f(z)}{(z-a)^{n}}-\frac{f^{(n)}(a)}{n!}\right]\frac{dz}{z-a}$‌$ and somehow argue that this integral is zero.

Bah, why isn't that rendering.

Oh well.

Daminark

17:02

@Semi OK it's not rendering for you either lmao

I was like, wait is my thing off?

Semiclassical

yeah :/

Zophikel

@Semicalssical how do you get latex to render for mircosfot edge

Semiclassical

use the latex in chat link in the room desc.

Vrouvrou

@Daminark please is this metric space is complete $(R,d)$ where $d(x,y)=|x^3-y^3|$

s.harp

@Daminark hello, this time you've got me when I'm here :)

Daminark

17:03

Mark your calendars, we're here at the same time!

Semiclassical

$$\int_C\left[\frac{f(z)}{(z-a)^n}-\frac{f^{(n)}(a)}{n!}\right]\frac{dz}{z-a}$$

Huh. Now it's working.

Zophikel

@Semiclassical thank you

Semiclassical

Okay, I have no idea why that wasn't working the first time.

Zophikel

lol

Daminark

Note that $|x^3 - y^3| = |x-y||x^2 + xy + y^2|$

So let's say you have a Cauchy sequence $\{x_n\}$

Semiclassical

17:06

Huzzah for geometric sums :)

Daminark

First, a disclaimer, never dismiss the possibility that I do something completely stupid, so this comes with no warranty

But, choose the point $x$ that it would converge to under the classic norm

SoumyoB

can someone explain to me what is a homogeneous change of variables? I'm talking in the context of quadratic forms

Vrouvrou

have you an idea @Daminark ?

Aetos

Is $\Delta^n$ the polydisc?

Daminark

I think that should work

Since $|x_n - x|$ tends to $0$, and then $|x_n^2 + xx_n + x^2|$ can be bounded

So it should be complete?

I'll roll with that for now, but double check my argument thoroughly to be sure

Semiclassical

17:11

@SoumyoB Looking at a separate source, it would seem that such a change of variables amounts to $$(x_1,x_2,\ldots)\mapsto (x_1',x_2',\ldots)=(\lambda x_1,\lambda x_2,\ldots)$$

SoumyoB

so just multiplying them by a common constant?

not even using combinations of more than one variable inside one coordinate of the image under the map that constitutes the change of variables?

Semiclassical

Right. In particular, one often works in a context where the relevant variables are $(x_0,x_1,\ldots)$ and one takes $\lambda=1/x_0$. Then $(x_0,x_1,x_2,\ldots)\mapsto (1,x_1/x_0,x_2/x_0,\ldots)$.

Hmm.

SoumyoB

that is for example, $(x,y) \mapsto (x+y,x-y)$?

Semiclassical

I'm not sure I'd rule that outo, no.

Vrouvrou

@Daminark but to prove that a pace is complete we must consider a Cauchy sequence and then prove that is convergent

Semiclassical

17:13

Hmmmm

SoumyoB

@Semiclassical thanks

I've got to do a presentation on classical groups tomorrow morning and hardly any time is left for preparing and I do not know crap about this subject

Semiclassical

The utility of a transformation like the above is that you set one of the coordinates to 1.

SoumyoB

I've studied too much probability and almost nothing on groups, analysis or any other area of math

I see

Semiclassical

In the projective plane, for instance, one uses so-called homogeneous coordinates $[x_0,x_1,x_2]$ where there's the equivalence relation $[x_0,x_1,x_2]\sim [\lambda x_0,\lambda x_1,\lambda x_2]$.

Taking $\lambda=1/x_0$ then amounts to 'projection' into the $x_1x_2$-plane.

Daminark

@Vrou Look what I did, I said that given $\{x_n\}$ Cauchy to choose the point that it would converge to under the typical norm, which I think ought exist because of the identity (to be fair, I'm a bit lazy to check, so perhaps I'm wrong there, which kills the argument).

Call that $x$

Semiclassical

17:16

(The logic of the equivalence being that all such points in R^3 would lie on the same line through the origin.)

SoumyoB

@Semiclassical you mean projecting a quotient space of $\mathbb{R^3}$ onto the plane $z=1$?

Semiclassical

Sounds right.

I don't know the formal details that well.

SoumyoB

I see

Daminark

And then just say that the other term in the identity is bounded, so since $|x_n - x|$ gets tanked, so should $|x_n^3 - x^3|$, so that's it

SoumyoB

thanks for the help

Semiclassical

17:18

But, yeah. Taking the intersection of the line through the origin with $x_0=1$ gives a specific point in the plane.

But you could equally well have taken the intersection with the plane $x_1=1$, and there's a different projection for that.

They'd all be intersections with the same line, just different planes.

SoumyoB

but wait isn't $\lambda$ supposed to be a constant?

Daminark

Actually wait @Vrou you should look at cube roots

Like, $\{x_n\}$ is Cauchy in this metric, so $\{x_n^3\}$ is Cauchy in $\mathbb{R}$, so it converges to some $c$

Semiclassical

Well, in the projective context it's taken to be equivalence with respect to any $\lambda\neq 0$

Daminark

Then let $x = \sqrt[3]{c}$

So that $d(x_n,x) = |x_n^3 - c| \to 0$

Boom

Semiclassical

So you'd be able to make it work there.

For quadratic forms, though, I don't know.

Vrouvrou

17:23

i f like this every Cauchy sequence is convergent

Daminark

Yes

Semiclassical

@SoumyoB The post here seems to be using it in quite a different way than what I was saying, though: artofproblemsolving.com/community/…

Daminark

So it is complete

This is something I'm more sure of. I'm willing to put a warranty on this proof which confers no benefits if I fucked up, but it's still a warranty nonetheless

Semiclassical

In that version one starts with $(x_0,x_1,x_2)$ and substitutes $(x_1,x_2)=(\lambda_1 x_0, \lambda_2 x_0)$.

SoumyoB

@Semiclassical I suppose then it's probably a loosely used term

Semiclassical

17:25

hi @ted

Daminark

Hey @Ted!

SoumyoB

I couldn't find any formal definitions after running a google search

Semiclassical

well, I searched the entire phrase "homogeneous change of variables"

Ted Shifrin

Hi Semiclassic, Demonark, Soumy.

Semiclassical

And the only elementary one I saw was that.

Daminark

17:26

How's everything going?

Ted Shifrin

What's this homogeneity you guys are talking about?

Semiclassical

Something something quadratic forms / classical groups, apparently.

Vrouvrou

so if i chage the distance $d(x,y)=\ln(1+(x+y))$ i say let $(x_n)$ be Cauchy sequence , then it converge to some c @Daminark

Ted Shifrin

Is there a specific sentence/question?

Semiclassical

@SoumyoB Starting here, @ted

SoumyoB

17:28

@TedShifrin I had started it

I'm reading Larry C. Grove's Classical Groups and Geometric Algebra

Ted Shifrin

@Vrouvrou: That isn't a distance.

On my iPad I can't do these links.

SoumyoB

and it's explaining something about quadratic forms, in relation to which I had read that you can do some homogeneous change of variables to reduce any quadratic form to a diagonal one

Vrouvrou

ipad@Daminark ?

i do a mistake $d(x,y)=\ln(1+|x+y|)$

Ted Shifrin

@SoumyoB: My guess is that they're talking about projective quadrics. It's just the spectral theorem you use to diagonalize.

Rescaling won't diagonalize.

@Vrouvrou: What is $d(x,x)$?

Vrouvrou

sorry $d(x,y)=\ln(1+|x-y|)$

Ted Shifrin

17:34

Better.

Aetos

How is the absolute value of a point in $\mathbb C^2$ defined?

Ted Shifrin

As if it were in $\Bbb R^4$, but you can use the hermitian inner product if you want.

Aetos

I know all norms on finite-dimensional vector spaces are equivalent. So I guess I can say $|(z_1,z_2)|=|z_1|+|z_2|$ on $\mathbb C^2$.

Semiclassical

Uh, no.

Ted Shifrin

No, it's standardly $\sqrt{|z_1|^2+|z_2|^2}$.

Semiclassical

17:39

If you take $z_1=z_2=1$ you get $|(1,1)|=|1|+|1|=2$ when it should be $\sqrt{2}$.

(It'd work if you took $|z|=|x|+|y|$ but that's not standard)

Vrouvrou

how i can prove that the spce is complete @Daminark

s.harp

@Vrouvrou the important information is that $\ln(1+x)\approx x$ for small $x$

and $\ln(1+x)≤x$ for $x≥0$

SoumyoB

@TedShifrin could you give me an example? I'm not familiar with projective quadrics

Ted Shifrin

What is the precise thing you're trying to understand? Precise.

Semiclassical

@SoumyoB This might help: math.stackexchange.com/questions/1830202/…

s.harp

17:48

Is it called quadric because there are four real variables?

ie a quintic is something like $x_1^2+x_2^2+x_3^2+x_4^2+x_5^2-1=0$

Daminark

@Vrou I already did that, took a Cauchy sequence and showed that it converged to something

Ted Shifrin

No, quadric like quadratic ...

Any number of variables.

s.harp

oh

Ted Shifrin

:)

s.harp

so an example of a quintic would be the solutions of $x^5-1=0$

Ted Shifrin

17:50

Any poly of degree 5 is a quintic.

Vrouvrou

@Daminark i can't see that $|x_n-x|\rightarrow0$

can we prove that ?

Daminark

So we said that $\{x_n^3\}$ is Cauchy in $\mathbb{R}$, so it converges to some $c$

Meaning $|x_n^3 - c|\to 0$

Now let $x = \sqrt[3]{c}$

$d(x_n,x) = |x_n^3 - x^3| = |x_n^3 - c| \to 0$

Vrouvrou

but this because we have |.|

here we have $\ln(1+|.|)$

Daminark

Oh this is a new norm?

Sorry I didn't realize we changed it

Ted Shifrin

Pay attention to what @s.harp said above. Use basic facts about $\log(1+x)$ for $x$ small.

Vrouvrou

17:55

yes

Daminark

^^

Ted Shifrin

By the way, we assume you know how to show that $\log(1+|x-y|)$ defines a metric (i.e., that triangle inequality holds).

user193319

Hello. I am trying to prove that if G is an abelian group with elements a and b of order m and n, respectively, then the order of ab is the least common multiple of m and n. I am having a little trouble with this, however. Denote the LCM by L. Then clearly (ab)^L = e, since L is a multiple of both m and n, which means that |ab| \le L. I can't figure out how to show that |ab| = L. I could use some hints.

Topologicalife

Hi.

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