Mathematics - 2017-10-22 (page 4 of 5)

MatheiBoulomenos

2:01 PM

hi chat

hi

hey

Yo

Hello !

what are the units in R[x] ?

let me state that in a better way

MatheiBoulomenos

do you know anything about R?

Kasmir Khaan

2:06 PM

can it a function that is not constant

be a unit?

Like R =Z/2

MatheiBoulomenos

generally for R commutative, a polynomial is a unit iff the coefficients of the non-constant terms are nilpotent and the constant term coefficient is a unit in R

Kasmir Khaan

do we only define constant functions as potential units or it can be a poly

so from what I understand it is a yes :D

if we have R = Z say

then the units are polys of the form

-x+x^2 ect

MatheiBoulomenos

you should not use the term "function" when you're talking about polynomials

no, the only units of Z[x] are 1 and -1

Kasmir Khaan

what is the right word?

ÍgjøgnumMeg

elements

Leaky Nun

2:08 PM

@KasmirKhaan polynomials.

MatheiBoulomenos

just use polynomial, duh

Kasmir Khaan

okay:D

so hmm

I feel like I got something wrong here

Leaky Nun

@KasmirKhaan what makes you think this is a unit?

Kasmir Khaan

1 and -1 ?

they are invertible in Z

Leaky Nun

@KasmirKhaan what

yes they are

Kasmir Khaan

2:10 PM

oh stupid ><

they cant be units

if we multiply polys , powers add up

MatheiBoulomenos

that's right

Kasmir Khaan

and then we only get bigger power never 0

:DDD

MatheiBoulomenos

but does this hold in any ring?

Kasmir Khaan

well been up for quite a while , my brain is on low enegry

MatheiBoulomenos

consider Z/(4)[x]: what's (2x+1)^2?

Kasmir Khaan

2:11 PM

hmm let me think

Leaky Nun

@MatheiBoulomenos strange choice of brackets

Kasmir Khaan

oh

this can also be a unit

4x^2 +4x +1

which is 1 mod 4

ÍgjøgnumMeg

(Y)

Leaky Nun

we don't say mod 4 here @KasmirKhaan

in Z/4, 2+2 is 0

Kasmir Khaan

@MatheiBoulomenos Btw mathei, i finished yesterday sylow theorem =P

MatheiBoulomenos

2:12 PM

great!

Leaky Nun

in (Z/4)[X], (2x) + (2x) is 0

Kasmir Khaan

okay @LeakyNun

Ill have to repeat all the book properly after exam

MatheiBoulomenos

So what property does Z have that Z/(4) doesn't, such that this phenomenon doesn't occur in Z[x]?

Kasmir Khaan

well Z is integral domain

MatheiBoulomenos

exactly!

Kasmir Khaan

2:13 PM

ab =0 iff a = 0 or b =0

ÍgjøgnumMeg

(Y)

Kasmir Khaan

:D

Leaky Nun

no

that isn't it

Kasmir Khaan

Okay kasmir will come shortly with more questions =p but now he gotta do more stuff

MatheiBoulomenos

Can you write down a formal proof that the units of R[x] are the units of R if R is an integral domain?

Leaky Nun

2:15 PM

@ÍgjøgnumMeg read Mathei's answer above

@KasmirKhaan that isn't it

ÍgjøgnumMeg

@LeakyNun what are you talking about

Leaky Nun

@MatheiBoulomenos no

Kasmir Khaan

ehm am comfused now

Leaky what is it?

Leaky Nun

8 mins ago, by MatheiBoulomenos

generally for R commutative, a polynomial is a unit iff the coefficients of the non-constant terms are nilpotent and the constant term coefficient is a unit in R

MatheiBoulomenos

@Leaky integral domain implies no non-zero nilpotents, so this is fine

Leaky Nun

2:15 PM

you don't see that behaviour in (Z/6)[X] for example

Kasmir Khaan

@MatheiBoulomenos We are not gonna have proofs , so i read it and skipped it =p

well Z/6 is not integral domain

because 2*3 =0

this also happens in Z/4

2*2 =0

MatheiBoulomenos

Leaky is right, this phenomenon does not occur in Z/6[x]

but if you didn't have nilpotents in your class that may be too confusing for you

Kasmir Khaan

ah of nilpotent?

I know nilpotent matrices

But i understand what you mean

something if we apply it twice is like applying it 1000 times:D

projection is one of those

MatheiBoulomenos

that's idempotent, not nilpotent

Kasmir Khaan

><

Leaky Nun

2:18 PM

@KasmirKhaan what's a nilpotent matrix?

Kasmir Khaan

powers of it gives identity

Leaky Nun

@KasmirKhaan yes, but you can't have a non-constant unit in (Z/6)[X]

@KasmirKhaan no, that's... what's the word for that

do we even have a word for that

Kasmir Khaan

Leaky !

MatheiBoulomenos

As I said, if you didn't have nilpotents elements of a ring in your class, then this is probably overkill right now

Kasmir Khaan

This is not the good time for this :D

ÍgjøgnumMeg

2:19 PM

@KasmirKhaan I think leaky needs to work a little on pedagogy ;)

Leaky Nun

@KasmirKhaan $a$ is nilpotent if $a^n=0$ for some $n \in \Bbb N$

@ÍgjøgnumMeg stares

Kasmir Khaan

Yepp exactly , mathei gets me ! =p ill keep reading :)

@ÍgjøgnumMeg He expects allways much from others thats y he push to limits ;D but he is a good teacher

ÍgjøgnumMeg

@LeakyNun With all due respect, it'd be helpful for you to gauge the level of the person you reply to before replying hahaha

Balarka Sen

2:39 PM

Waitwaitwait something happened to CringeGum

Kasmir Khaan

@MatheiBoulomenos Can you give me an example of element inR[x] when I have to reduce it in that ring or soemthing of that nature? :)

Balarka Sen

awww shit jake paul posted a roast video on ricegum ahahaha

ÍgjøgnumMeg

When talking about sequences $0 \to N \to M \to M/N \to 0$ are all of the maps meant to be considered the "canonical" maps in a sense? I.e. (in the context of modules) $N$ is a submodule of $M$ and so the map $N \to M$ should just be the inclusion of $N$ in $M$ and the map $M \to M/N$ is just the natural projection of $M$ down to $M/N$?

Balarka Sen

I would think so

Alessandro Codenotti

most likely, unless they explicitely specify different maps

ÍgjøgnumMeg

2:47 PM

Alright cheers

Alessandro Codenotti

Also because generally if $0\longrightarrow L\stackrel{f}{\longrightarrow}M\stackrel{g}{\longrightarrow}N\longrightarrow 0$ is exact then $L\subseteq M$ and $N=M/L$ with the natural identifications induced by $f$ and $g$

ÍgjøgnumMeg

Yis that makes sense

Thanks

Lozansky

How can I see that $y=y_1-y_2$ isn't a solution to $$\ddot{y}(t)+e^{t^2}\dot{y}(t)+e^ty(t) = \cos(t)$$ if $y_1$ and $y_2$ are solutions?

I suspect it's because of the $\cos$-term?

Akiva Weinberger

Hm yeah probably

Balarka Sen

Yes, plug it in.

Lozansky

2:59 PM

Maybe I should just derive

Yeah

Akiva Weinberger

I'm guessing $y=y_1-y_2$ would be a solution to ${\rm blah}=0$

rather than ${\rm blah}=\cos(t)$

Lozansky

Yes

Balarka Sen

Only solutions to linear ODE's with zero constant coefficient forms a vector subspace of the function space.

Akiva Weinberger

"Forms a vector space" means "You can add them together and multiply them by constants and they still work"

Alessandro Codenotti

Affine subspace if you have nonzero constant term, but the important part is that the ODE is linear

Akiva Weinberger

3:02 PM

So you can only do that if the thingy on the right is zero (and it's linear)

Lozansky

Well plugging in $y=y_1-y_2$ should yield $0$

Akiva Weinberger

I guess the relevant word here is "nullspace" (which is a Linear Algebra thing, don't know if you've studied it yet)

Lozansky

So obviously it's not a solution

MatheiBoulomenos

@KasmirKhaan can you be more specific? I don't really understand your question

Kasmir Khaan

well =p

its like when have to prove that some polynomials are irreducible in some Ring

or to factor them

Meow Mix

3:05 PM

Chocolate rain

some stay dry but others feel the pain

Alessandro Codenotti

Is $2X-8$ irreducible in $\Bbb Z[X]$? And in $\Bbb Q[X]$? @Kasmir

Kasmir Khaan

hmm let me think :D

well in Z[x] it is

ÍgjøgnumMeg

@KasmirKhaan Are you sure?

Alessandro Codenotti

What's the definition of irreducible element of a ring?

Kasmir Khaan

it cant be factored into smaler elements right?

like prime numbers

Alessandro Codenotti

3:10 PM

That's good intuition but not a proper definition

ÍgjøgnumMeg

@KasmirKhaan Give something precise!

Kasmir Khaan

2x-8 has deg =1

ÍgjøgnumMeg

@KasmirKhaan Try and factor it in the "naive" way and see why you might want to rethink your answer ;)

Kasmir Khaan

well we can do it 2(x-4)

but that is not allowed right?

ÍgjøgnumMeg

@KasmirKhaan Who said that wasn't allowed?

Alessandro Codenotti

3:12 PM

Check the exact definition of irreducible element and then decide whether it is allowed or not

Kasmir Khaan

do we allow constant polys?

Okay !

I think it is better to read more than ask ><

ÍgjøgnumMeg

@KasmirKhaan It depends whether or not the constant polynomials are units in your ring

Akiva Weinberger

We don't care about units, though, so like $-(-2X+8)$ doesn't matter

Kasmir Khaan

Exam is tomorrow this is why doing thsi things more sloppy

ÍgjøgnumMeg

That's fine, I'm sure everybody here has crammed before

Akiva Weinberger

3:13 PM

Kinda like how $5$ is prime in $\Bbb Z$ even though it's $(-1)(-5)$

We don't care about units (invertible elements)

In $\Bbb Q[X]$, all constant polynomials are units. So that's why we don't care about them there.

In $\Bbb Z[X]$, only $1$ and $-1$ are units.

Kasmir Khaan

hmm this is new =P

Alessandro Codenotti

A nonzero, non unit $a$ is called irreducible iff whenever $a=bc$ either $b$ is a unit or $c$ is a unit

Akiva Weinberger

@KasmirKhaan Right, so take something like $X^2+1$ in $\Bbb Q[X]$

ÍgjøgnumMeg

@Akiva Almost all constant polynomials!

Akiva Weinberger

@ÍgjøgnumMeg Meh, 0 doesn't exist /s

2

Kasmir Khaan

3:14 PM

okay that is a good defintion :)

ÍgjøgnumMeg

hahahah

Akiva Weinberger

@KasmirKhaan If we cared about *units, we could write $\frac12(2X^2+2)$

and $\frac43(\frac34X^2+\frac34)$

and nothing would be irreducible, and it would be a stupid definition

Kasmir Khaan

Ehm so in our case 2 (x-4)

I see how that would be problematic =p

but in since Q is a field

all elements are units exept zero

Akiva Weinberger

Yeah. So, in $\Bbb Z[X]$, $2X-8$ is not irreducible ('cause it's $2(X-4)$), and in $\Bbb Q[X]$ it...

Kasmir Khaan

it is irreducible

in Q =P

Akiva Weinberger

3:17 PM

yeah

Kasmir Khaan

:D thanks guys :)

I think is it time for me to sleep for a whiel >< been up since yesterady

Akiva Weinberger

Another thing, in $\Bbb C[X]$, the only irreducible elements are linear ones

Kasmir Khaan

Ill wake up early and do some old exams =p

hmm because of i ?

Lozansky

A critical point $\mathbf{x}^{(0)}$ to the system $\mathbf{x}' = \mathbf{f}(\mathbf{x})$ is stable iff for every $\epsilon,\delta >0$ $$||\mathbf{x}(0)-\mathbf{x}^{(0)}||<\delta \Rightarrow ||\mathbf{x}(t)-\mathbf{x}^{(0)}||<2\epsilon$$ for all $t\geq 0$

Akiva Weinberger

(ex: $X^2+1$, which is irreducible in $\Bbb R[X]$, can be written as $(X+i)(X-i)$ in $\Bbb C[X]$)

Lozansky

3:17 PM

I want to say that's false

Alessandro Codenotti

There is a theorem called Gauss lemma saying that polynomials irreducible in $\Bbb Z[X]$ are irreducible in $\Bbb Q[X]$ (well the full version is much more general, but this is an important case)

Kasmir Khaan

@AkivaWeinberger ah good example thanks, i think I got the hang of it :)

Lozansky

Because of the $2 \epsilon$

Argon

Is it the case that $j\cdot\left<i\right>=\{j,-j,k,-k\}$ where $\left<i\right>$ is a subgroup of the quaternions?

ÍgjøgnumMeg

@KasmirKhaan $\Bbb C$ is algebraically closed

Lozansky

3:18 PM

But is that reason enough to say it's false?

Kasmir Khaan

@ÍgjøgnumMeg that sounds cool but I dont know what it is ><

Akiva Weinberger

@KasmirKhaan The (hard) theorem is that every polynomial has a root in $\Bbb C$. That's the Fundamental Theorem of Algebra.

Because of it, every nonlinear polynomial can be split into linear ones.

Alessandro Codenotti

(the only exceptions in the other direction, that is polynomials irreducible in $\Bbb Q[X]$ that are reducible in $\Bbb Z[X]$, are of this kind, polynomials with integer coefficents that all share a common divisor)

ÍgjøgnumMeg

@KasmirKhaan It just means that every polynomial with coefficients in $\Bbb C$ has roots in $\Bbb C$

Kasmir Khaan

but what about linear ones?

x+i eg

Alessandro Codenotti

3:19 PM

linear polynomials are always irreducible over a field

Kasmir Khaan

that is its lowest form

Akiva Weinberger

Sure, you can split it into linear polynomials

$(x+i)$

You can just split it into one linear polynomial

:P

Kasmir Khaan

:D

well guys ! thanks again for all your help !

Ill come back in 7 hours or so

Akiva Weinberger

Gnight

Kasmir Khaan

Good night! :)

Argon

3:22 PM

If $H = 9 \mathbb{Z}_{36}$, $g = 16$, am I correct in asserting $gH = \{0\}$?

ÍgjøgnumMeg

@Argon Yep

Argon

Thanks.

ÍgjøgnumMeg

Just by observing that $H = \langle 9 \rangle$ and since $36 = 4\times 9$ and $4\mid 16$, you'll just get multiples of $36$

Argon

One more: If $H=\left<i\right>$ and $H$ is a subgroup of the quaternions, am I correct to say $j\cdot\left<i\right>=\{j,-j,k,-k\}$?

ÍgjøgnumMeg

yes you're correct

Argon

3:30 PM

Hm I wonder why I am being marked incorrectly by this software

ÍgjøgnumMeg

Well it's just by multiplication right? You have $\langle i \rangle = \left\lbrace 1, -1, i, -i \right\rbrace$ and $ji = -k$.

Argon

That's what I was thinking

ÍgjøgnumMeg

What software is it?

Argon

WebWorK

I could literally check the source code, maybe I'll do that :P

ÍgjøgnumMeg

Hahaha do it

Leaky Nun

3:34 PM

Conjecture: for $a,b \in \Bbb R^3$ with $\langle a,b \rangle = 0$, we have $(a \times b) \times b = -|b|^2 a$

Argon

@ÍgjøgnumMeg github.com/openwebwork/webwork-open-problem-library/blob/… Here we go

Leaky Nun

@Argon lol

cheating level > 9000

4

Argon

Open source problem libraries are never a good idea :P

Well the source code says the answer should be $\{16,25,34,7\}$.

Leaky Nun

can anyone help me prove my conjecture?

Argon

@LeakyNun Would something like (18) do the trick: mathworld.wolfram.com/CrossProduct.html

Leaky Nun

3:47 PM

@Argon thanks!!!

I wonder how you prove that

Argon

@LeakyNun For your case, you want the brackets more like this: wikimedia.org/api/rest_v1/media/math/render/svg/…

Leaky Nun

@Argon sure

Argon

Triple product

In vector algebra, a branch of mathematics, the triple product is a product of three 3-dimensional vectors, usually Euclidean vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and, less often, the vector-valued vector triple product. == Scalar triple product == The scalar triple product (also called the mixed product, box product, or triple scalar product) is defined as the dot product of one of the vectors with the cross product of the other two. === Geometric interpretation === Geometrically, the scalar triple product ...

Leaky Nun

my guess was right: you have to decompose the vector

Lozansky

@LeakyNun Tensor notation?

Leaky Nun

3:51 PM

@Lozansky I don't know what tensor is

Lagrange's formula: $$A \times (B \times C) = B \langle A , C \rangle - C \langle A, B \rangle$$

Lozansky

$[\mathbf{A} \times (\mathbf{B} \times \mathbf{C})]_i = \epsilon_{ijk}A_j \epsilon_{klm}B_lC_m = [\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}]A_jB_lC_m = A_mB_iC_m - A_lB_lC_i =...= \mathbf{B}(\mathbf{A} \cdot \mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B})$

Leaky Nun

@Lozansky interesting, though I can't understand it, lol

Jenna Sloan

@Pseudohuman What's a tensor?

Pseudohuman

1 | noun | a generalization of the concept of a vector
2 | noun | any of several muscles that cause an attached structure to become tense or firm

Jenna Sloan

@Pseudohuman What's the mathematical definition of a tensor?

Pseudohuman

3:59 PM

An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and m^n components and obeys certain transformation rules. Each index of a tensor ranges over the number of dimensions of space. However, the dimension of the space is largely irrelevant in most tensor equations (with the notable exception of the contracted Kronecker delta). Tensors are generalizations of scalars (that have no indices), vectors (that have exactly one index), and matrices (that have exactly two indices) to an arbitrary number of indices.

Jenna Sloan

@Pseudohuman thanks

Lozansky

@LeakyNun You can also prove it by projecting the triple product on a unit vector and then carry out the cross products in that direction

Pseudohuman

@JennaSloan You're welcome!

MatheiBoulomenos

a tensor is just an element of a tensor product of some spaces

Lozansky

Is that a recursive definition?

MatheiBoulomenos

4:03 PM

it's not

the tensor product is defined by a universal property (you still have to show that it exists, but that's not too difficult)

Lozansky

Oh

MatheiBoulomenos

Tensor product

In mathematics, the tensor product V ⊗ W of two vector spaces V and W (over the same field) is itself a vector space, together with an operation of bilinear composition, denoted by ⊗, from ordered pairs in the Cartesian product V × W into V ⊗ W, in a way that generalizes the outer product. The tensor product of V and W is the vector space generated by the symbols v ⊗ w, with v ∈ V and w ∈ W, in which the relations of bilinearity are imposed for the product operation ⊗, and no other relations are assumed to hold. The tensor product space is thus the "freest" (or most general) such vector space,...

Lozansky

I think we only did outer product and contraction

MatheiBoulomenos

You can say that the tensor product basically reduces multilinear algebra back to linear algebra

Daminark

Hey

MatheiBoulomenos

4:18 PM

hi

Lozansky

When finding solutions to an eq. system on the form $\mathbf{x}' = A \mathbf{x}$, should we favor the larger eigenvalue of $A$ or the lower?

anon

4:34 PM

what do you mean by "favor"? the general solution will involve all of them.

Lozansky

Oh I was thinking if the eigenvalues are complex

anon

I don't see how that's relevant to what I asked/said

Lozansky

If they are complex, then a solution is given by $\mathbf{v}e^{t\lambda}$ where the real and imaginary part form two linearly independent solutions?

Oh wait

I think that's correct

anon

I'm guessing you're assuming $A$ is real

Lozansky

Yes

anon

4:39 PM

if $\lambda$ is a complex eigenvalue, the real/imaginary parts of $e^{\lambda t}{\bf v}$ will not be solutions

Lozansky

What?

anon

oops, no, they will be

the reason is that the real/imaginary parts are linear combinations of $\exp(\lambda t)$ and $\exp(\overline{\lambda}t)$

Lozansky

Yes

And if $Im(\lambda)=0$ we have to include both eigenvalues

anon

so what's your question?

@Lozansky if ${\rm Im}(\lambda)=0$ then $\overline{\lambda}$ and $\lambda$ are the same eigenvalue no?

Lozansky

Say that eigenvalues of $A$ are complex. Should we favor $\lambda_{+}$ or $\lambda_{-}$?

Yes?

anon

4:42 PM

what do you mean by "favor"?

Lozansky

When finding a fundamental matrix

anon

again, what do you mean by "favor"? are you under the impression the fundamental matrix will involve one complex eigenvalue but not its conjugate?

Lozansky

Yes?

One solution ($\lambda_{-}$) will be decaying and one ($\lambda_{+}$) will be growing exponentially

anon

no

Lozansky

Hm

anon

4:45 PM

the two complex conjugates have the same real part

which means they're either both decaying or both growing (or both spiraling without amplitude change)

Lozansky

But if $X=\mathbf{v}e^{\lambda t}$, then $X$ will be decaying for $\lambda_{-}$ and the other growing, no?

anon

if you're cool with your fundamental matrix taking complex values, then you use both eigenvalues. if you want only real-valued column entries, then take the real and imaginary parts of $\exp(\lambda t)$ (which will be the same as those of $\exp(\overline{\lambda}t)$, only with a sign change)

@Lozansky No. For example, $\exp((2+3i)t)$ and $\exp((2-3i)t)$ are both growing.

Lozansky

Oh right

But that sign change, is not important?

Regarding stability

anon

the stability of which solution?

the sign change in the fundamental matrix won't affect a specific solution, only how you express the specific solution in terms of the fundamental matrix

Lozansky

Okay, so there is no preference?

Vrouvrou

4:56 PM

please i have this integral $\int_{\Omega\cap B_r(y_n)} f(u_n(x))dx>a$ by a change of variable $v_n(x)=u_n(x+y_n)$ why we obtain that $\int_{\Omega\cap B_r(0)-y_n} f(v_n(x))dx>a$

???

Akiva Weinberger

5:36 PM

What does $\Omega\cap B_r(0)-y_n$ mean? Are you translating the set $\Omega\cap B_r(0)$ by the amount $y_n$?

Hm actually I think I see it (if we ignore the ${}-y_n$). First substitute $x+y_n$ into $x$, and then substitute $v_n(x)$ for $u_n(x+y_n)$.

@Vrouvrou

You know how $\int_a^bf(x)dx=\int_{a-y}^{b-y}f(x+y)dx$? This is like that

Sha Vuklia

okay, I had to visit the chat, because how the FUQ didn't you know what tunak tunak was @Akiva :P

also @Balarka I am still looking for the ultimate Russian/soviet/communist meme. Something that will just blow my mind forever. The Soviet Army hard bass one gets close, very close, but I am looking for something ever better:P and Serbia Strong is also really really close, like 99.9% perfection. But one day, I will find the meme I'm looking for, I know that

Vrouvrou

@AkivaWeinberger $B_r(0)\cap \Omega-\{y_n\}$

Akiva Weinberger

Wait doesn't deleting one point from integrals not change the value?

Vrouvrou

5:51 PM

i don't understand you

Sha Vuklia

also @Balarka I disagree with you that tunak tunak is the best indian/bollywood meme. How about the Bollywood Thriller? I think that one beats tunak tunak by far:P anyhow, I'll stop talking about old memes now lol

Akiva Weinberger

@Vrouvrou Does $-$ mean relative complement here? Like is $A-B$ the set of elements of $A$ not in $B$?

Vrouvrou

6:24 PM

it is not saying what is - here

@AkivaWeinberger

Akiva Weinberger

I think it makes sense if you get rid of the $-y_n$ bit

Balarka Sen

6:41 PM

@ShaVuklia You're right. Bollywood on the whole is itself a massive ass meme

Hi @Daminark

Daminark

How's it go?

Balarka Sen

It go's well

Daminark

Oof, our complex prof sent out midterm solutions and in one of the problems he was like "It is amazing that only 3 students got this right"

Like goddamn this guy is spitting some fire

Balarka Sen

lol

yikes

Daminark

To be fair he's right, it was a 2-part problem. First was about z^2 being injective on the right half plane (exclude imaginary axis), and find the complement of the image

Balarka Sen

6:48 PM

Right, z^2 wraps C around itself twice

A quarter of the plane maps injectively

Daminark

And the complement of the image is negative reals by an argument argument

Balarka Sen

Yup