Mathematics - 2017-07-05 (page 8 of 10)

Typhon

20:00

@EricStucky huh?

Semiclassical

sort of moot here, since 1 is most definitely the multiplicative identity of Z[sqrt(3)]

Typhon

@Faust7 a ring is a set of numbers closed under addition, negation, and multplication.

Faust7

a ring is an abelian group that is well defined with a second binary operation

Typhon

@Faust7 you're getting overcomplicated

Eric Stucky

Finish your writeup, Faust ;)

Faust7

20:02

its important that your do have multiplicative inverses

do not*

dammit

lol

Typhon

@Faust7 no it isn't. fields are rings as well.

Semiclassical

Just to cite from Wikipedia for clarity: "A ring is an abelian group with a second binary operation that is associative, is distributive over the abelian group operation, and has an identity element." (the last one is optional here)

Typhon

the abelian group operation is just addition, is it not?

at least, from context it appears to be

Semiclassical

Yeah, in this case it definitely is.

Typhon

it always is, is it not?

at least when we are dealing with numbers

Semiclassical

20:04

Eh, I imagine one could think of an example where it's not called addition for some reason

Typhon

this isn't set theory or boolean algebra

@Semiclassical oh

Semiclassical

That said, even the Wiki page basically just says "yeah, one's addition and the other is multiplication"

Typhon

figures

still

a generator for Z[root 3] cannot exist

as there is no positive smallest element

Semiclassical

not by itself, at any rate

Typhon

yeah it would be a generating set

which I argue is 1 and root three

as all extended integers are defined to be a linear combination of 1 and root 3

Semiclassical

20:07

That does seem the most obvious statemen.t

Typhon

faust says it cannot be

idk why

it seems like valid candidate

Semiclassical

well, wait for him to come back

Typhon

yeah

@Semiclassical I think he is looking for a set of one element

meanwhile, I'm solving the modified goldbach conjecture

Semiclassical

about the only way I can see to have one generator is if you count sqrt(3)^0=1 as being generated by sqrt(3). but that seems highly dubious.

Typhon

he said it was things of the form

g+g+g+g+g+g+g+g...

which, while I may be wrong seeks out the smallest element

Semiclassical

20:11

yeah, that'd generate some subset of Z[sqrt(3)]

Typhon

$(2+\sqrt{3})^\-infty$

Semiclassical

\infty

Typhon

smallest unit

unless its conjugate is smaller

XD

wait no

conjugate to infinity

Semiclassical

Smallest as least magnitude, or as most negative?

Typhon

least magnitude

the extended integers are a dense set

like the rationals are dense

Semiclassical

20:13

in that case, I'd just say that there is no smallest unit.

Typhon

@Semiclassical that was my point

Semiclassical

ah.

Typhon

there is no smallest element in magnitude

so no singleton generator

Semiclassical

So the units are of the form $(2\pm \sqrt{3})^n$ for integer $n$. Neat.

Faust7

$ R= \{ i+j \sqrt 3 | i,j \in \mathbb {Z} \}$
let $ b \in R $ define $ b= i+j \sqrt 3 $ and define $b^{'} = i-j \sqrt 3 $
then $bb^{'} = i^2 - 3j^2 $ define this to be N(b) where N is the norm of b.

Since $i^2 - 3j^2 \in \mathbb {Z} $ it implies that $ N(b) \in \mathbb {Z} $
now the only units of $\mathbb {Z} = \pm 1$ so N(b) is a unit for $\mathbb {Z}$ iff $N(b) = \pm 1$ this implies $i^2 - 3j^2$ is unit for $\mathbb {Z}$ iff $i^2 - 3j^2= \pm 1$

Using the fact that $N(bb')=N(b)N(b')$.

So if $b$ is a unit, there exists $b'$ such that $bb'=1$ we deduce that $N(b)N(b')=1$, since $N(b), N(b…

Semiclassical

20:19

Well, that's consistent with $i=2,j=1$.

Typhon

@Semiclassical yup. I proved it by showing that units are closed under multiplication and then showing that if a unit is greater as a real number than either coefficient is greater. I also showed that they increase as real numbers as a function of n. Then I assumed that there was one "in between" two powers of the unit and divided to show that such a unit would be in-between 1 and 2+\sqrt{3}. Contradiction.

therefore all units are of that form

without loss of generality, that is true for negatives as well

note

1/(2+\sqrt{3}) is 2 - \sqrt{3}

so without loss of generality we can just say all integral 2 + sqrt{3}

Semiclassical

sure.

Typhon

professor gave me one bad mark on it

something about real analysis and lower upper bounds or something

(when proving the units were monotonically increasing)

Semiclassical

eh. I can understand that on the grounds of "don't describe something twice"

In any case, I'm not sure where this is going.

Typhon

@Semiclassical no. It was more like: instead of saying that "supposing not implies a maximum" you say "supposing not implies a minimum"

Semiclassical

20:23

the units are of the form $\pm (2+\sqrt{3})^n$ with $n$ being any integer (including zero).

Typhon

yup

:p

Faust7

how do u get -1 then?

Semiclassical

Ah. So it was in the context of your argument.

Fixed.

Faust7

think thats all of them then

Typhon

@Faust7 positive units

Semiclassical

20:25

Again, though, not seeing where this is going.

Faust7

its a circle semi

Typhon

@Semiclassical just making small talk about how i proved those were the units.

I was told the method was "very clever"

Semiclassical

No, I mean.

Faust stepped out to do something, and what he came back with seems to be a result which both of you were already aware fo.

Typhon

yeah

Semiclassical

So...where is this going?

Typhon

20:27

@Faust7 dude. We studied units for a week in the class. I thought you were looking for the generators?

@Faust7 ellipse. it is an ellipse. So what?

Faust7

i was trying to explain you could prove what you did easier using the fact that the only generators of Z were $ \pm 1 $

Typhon

uuuh

Faust7

which is what my gibberish above does

Typhon

that if a norm could be factored into a norm and an integer than the integer is a norm?

Faust7

its using the conjugate of an element

but yeah the norm is an acceptable definition

Typhon

20:30

huh?

that doesn't prove what I was proving at all.

3

Q: Proof for elements of $\textbf{Z}[\sqrt{3}]$ regarding the existence of the norm.

So for some context, I was in a proof writing class a couple months back. I really liked it and did quite well, but midway through the course we were doing things regarding the norm of these other kinds of integers (elements of $\textbf{Z}[\sqrt{3}]$). Basically things like the fact that there is...

@Faust7 I'm proving that for some integer b that if $N(z) = N(x)b$ then there is some element with a norm of b. All you did was prove it for the case that z and x are both units...

which is beyond being merely trivial

This statement directly implies that if any integer is not a norm of an extended integer, than any odd powers of that integer is also not a norm

because all squares are norms of integers

Liad

i need to find the surface of the set $\{(x,y,z) : f(x) \ ^ 2 = z \ ^ 2 + y \ ^ 2\} $ where $f>0$ is a continuously diferentiable function. someone can help?

$f:(a,b) \to \Bbb R$

Typhon

um

there's a problem

"f(x)"

f(a,b) -> R

Liad

i dont see the problem

Typhon

those both contradict each other

SteamyRoot

What do you mean with "find the surface" ?

Typhon

20:37

f takes in two arguments

yet you passed it one

Eric Stucky

haha I had this problem too Typhon: (a,b) is an interval

Typhon

oooh

lol

@Liad partial differential equations are just beyond my level of knowledge

SteamyRoot

You can notice that it will have rotational symmetry along the $x$-axis

Semiclassical

well, there's no derivatives in there

Typhon

it is still a differential equation... sorta.

Semiclassical

20:39

uh, where?

Liad

@SteamyRoot to find a param. to this regular surface and calculate the integral $\int f(p(u,v) ) \dfrac{\partial p}{\partial u}\times \dfrac{\partial p}{\partial v}dudv $

Eric Stucky

zeroth order ;P

SteamyRoot

By moving to cylindrical coordinates where $y = r \cos \theta, z = r \sin \theta$.

Oh, you want the surface area

Typhon

@Semiclassical f being continuously differentiable implies that we need derivatives to find the solutions such that f is differentiable.

SteamyRoot

Yeah, I would suggest you do the polar transformation, since then your surface will only depend on the radius $r$ and $x$

Semiclassical

20:40

uh, no. he's given that $f$.

Liad

i got it!

Semiclassical

The point is to figure out the surface area for an arbitrary such $f$.

Liad

$p(t , \theta ) = (t,f(t)cos(\theta),f(t)sin(\theta))$

Typhon

@Semiclassical no he just wants to find the surface...

the parameterization

Liad

$t \in (a,b) \theta \in [0,2\pi] $

Typhon

20:41

"i need to find the surface of the set"

no mention of area

SteamyRoot

Err, yeah

it only depends on $x$ and $\theta$, not $x$ and $r$ :P

Semiclassical

take a look at the integral he's written down.

that's for surface area.

Typhon

ugh

i hate surface area

the integrals are.... hairy

Semiclassical

it's not the best day, no.

Liad

@SteamyRoot does my param. is what you meant?

Semiclassical

20:44

though in this case I think it helps to relabel the coordinates so that it's $x^2+y^2=f(z)^2$

in which case in cylindrical coordinates this is just $r=f(z)$.

Typhon

^^

Liad

i think this is a bit confusing :P

at least for me

Typhon

wait that's it!

Semiclassical

One thing to keep in mind is that $x,y,z$ are just labels of things.

SteamyRoot

Your parametrisation seems fine, but...

Semiclassical

20:45

The surface area can't depend on those names.

Typhon

this problem is trivial!

SteamyRoot

ah, nevermind, you have $f > 0$

Liad

@SteamyRoot hey, no but

SteamyRoot

in that case it's fine

Liad

great :)

SteamyRoot

20:45

the one thing that does bother me, though, is that in your integral you wrote $f(p(u,v))$

Liad

yea

Semiclassical

So if it being labelled with the axis of symmetry as being $x$ and you'd rather it were $z$, you're free to change those labels.

SteamyRoot

but $p$ seems to be a $\mathbb{R}^2 \to \mathbb{R}^3$ function?

Liad

that's how surface integrals are defined, doesn't it?

SteamyRoot

and $f: (a,b) \to \mathbb{R}$ ?

Liad

20:46

yea right

Semiclassical

should presumably be $p(u,v)_x$ or some such.

Typhon

@Semiclassical i feel like there is a formula for things with rotational symmetry... something called cylinders of revolution? You solved it.

Liad

hm i did not meant $f$ from the exercise

its the integral of $1$ doesnt it?

Semiclassical

I'm sure I've done it, yeah.

SteamyRoot

You should just have that cross product

Akiva Weinberger

20:47

Surfaces of revolution?

SteamyRoot

but it should be inside a norm

Liad

yea

Typhon

@AkivaWeinberger cylinder was what my calculus class called it

Liad

it is surface integral of type 1 or something like that, right?

SteamyRoot

No idea :P

Liad

20:48

well maybe i mixed some terms in the translation to English :)

Typhon

@Liad you know how you did integrals of solids of revolution formed by rotating y = f(x) about the y axis? Do that with f(x) and you have your answer.

it will be much more trivial

and you'll have the surface area

Liad

@SteamyRoot its the integral of $\int_a ^ b \int_0 ^ \{2\pi} ||\partial p / \partial t \times... || $ right ?this is what you said earlier ?

Typhon

@Liad you're overcomplicating it big time

Liad

Huh?

You have a simpler param. ?

Typhon

Surface of revolution

A surface of revolution is a surface in Euclidean space created by rotating a curve (the generatrix) around an axis of rotation. Examples of surfaces of revolution generated by a straight line are cylindrical and conical surfaces depending on whether or not the line is parallel to the axis. A circle that is rotated about any diameter generates a sphere of which it is then a great circle, and if the circle is rotated about an axis that does not intersect the interior of a circle, then it generates a torus which does not intersect itself (a ring torus). == Properties == The sections of the surface...

r = f(z)

Semiclassical

20:55

Think geometrically.

Typhon

the surface is just the surface of revolution of f(x) when rotated about the y axis

swap x and z in your original equation

SteamyRoot

Well, the mathjax doesn't check out but it looks okay

Though, since it's a surface of revolution, you'll be able to simplify it and get rid of the rotation part

Typhon

wikimedia.org/api/rest_v1/media/math/render/svg/…

Semiclassical

If you take the square root (and you can do that because $f>0$) then your equation is $\sqrt{y^2+z^2}=f(x)$.

Typhon

^^^

Semiclassical

20:56

The LHS is the distance from the x-axis, and the RHS is some distance $f(x)$.

Typhon

mathworld.wolfram.com/SurfaceofRevolution.html

Liad

so far im getting $2\pi \int_a^b \sqrt{f'(t) \ ^ 2 f(t) \ ^ 2 + f(t) \ ^2 }dt$

Semiclassical

So if you pick some value of $x$, that's just a circle of radius $f(x)$ around the $x$-axis. As you move along the axis, the radius of this circle can change.

Typhon

^^your area from x = a to x = b

according to wolfram alpha

Semiclassical

And that's exactly what's going on in the picture for the Wiki link

Typhon

20:58

^^^

Liad

i can use $u = f(t) $

SteamyRoot

@Liad Seems good!

You can pull the $f^2$ out of the root, though

since $f > 0$

Typhon

^^

Semiclassical

Beyond that there's not all that much to say.

Typhon

@SteamyRoot not a necessary condition.

SteamyRoot

20:59

@Liad That won't get you anywhere, I think

Liad

right.

Typhon

@Liad what does the problem ask you to find or prove?

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