Mathematics - 2019-04-16

user193319

12:06 AM

Problem: Let $f \in F[X]$, where $F$ is a field of characteristic $0$. Let $d(X) = gcd(f,f')$. Show that $g(X) = f(X)d(X)^{-1}$ has the same roots as $f(X)$, and these are all the simple roots of $g(X)$...Doesn't writing $d(X)^{-1}$ presuppose that $d(X)$ is an invertible element in $F[X]$? If so, wouldn't that imply that $d(X)$ is a nonzero constant in $F$? Doesn't that trivialize the problem?

loch

12:21 AM

@user193319 You're right $d(X)^{-1} \not \in F[X]$ in general, but $f(X)/d(X)$ is, since $d(X)$ divides $f(X)$

user193319

Ah, okay. Does it make sense to take the formal derivative of $f(X)/d(X)$? Does the quotient rule still hold?

Also, if $f(a)/d(a) = 0$ for some $a \in F$, may I conclude that $f(a) = 0$?

user76284

12:42 AM

@Rithaniel Near-ring with inverses?

Leaky Nun

3:07 AM

@user193319 (x-1)^2/(x-1), x=1

@user193319 you can obtain the quotient rule by taking the derivative of both sides of f=gh

Leaky Nun

4:00 AM

@Semiclassical I think I'm starting to get Green's function

BAYMAX

$f: \Bbb{Q} \rightarrow \Bbb{R}$ by $f(0)=0$ and $f(r) = \frac{p}{10^q}$ where $r = \frac{p}{q}$ with $p \in Z, q \in N$ and gcd($p,q$) =1

I was thinking whether $f$ is one-one, onto?

I thought of taking

$p=2,q=3$

giving $0.002$ same as $\frac{20}{10^4}$

Leaky Nun

gcd(20,4) is not 1

BAYMAX

but we cant take that due to the condition

yes

looke dup and seems to be 10 dyadic

Leaky Nun

however you can make it ?/10^7

20000/10^7

BAYMAX

no perhaps

as $q \in \Bbb{N}$

Leaky Nun

4:05 AM

$7 \in \Bbb N$

BAYMAX

oh ok

so its not one one

how to think of onto?

Leaky Nun

well the outputs are all rational

BAYMAX

yes

mapping of ratonals to rationals

Leaky Nun

but the codomain is $\Bbb R$

BAYMAX

cool

not onto either

Semiclassical

4:25 AM

@LeakyNun Did you manage to get the Green's function for the problem you were working on?

Leaky Nun

@Semiclassical I'm working on another Green's function problem; I'll return to the one yesterday later

Semiclassical

mmkay

Leaky Nun

is it okay if my function is discontinuous at s=1... lol

Startec

4:45 AM

Does this make sense to anyone?

I do not understand the $\nabla I ^\bot_p$

So if I calculate the gradient, how do i calculate the "orthogonal" part?

N. Maneesh

9:47 AM

@BAYMAX $f$ is not onto. since $\pi$ does not have pre-image.

But it is one-one

N. Maneesh

10:00 AM

It not one-one

since $\frac{10}{3}$ and $\frac{1}{2}$ gives the same image.

Silent

10:40 AM

I have this q/a: Let $A$ be $n\times n$ lower-triangular matrix, with all diagonal entries $0$. Prove $A^n$ is a zero matrix. The solution given is : Eigen value of A is $\lambda=0$, so $\lambda^n=0\implies A^n$ is zero matrix.

I can't understand $\lambda^n=0\implies A^n$ is zero matrix.

N. Maneesh

They used Cayley Hamilton theorem

characteristic polynomial is uniques

characteristic polynomial of $A$ is $\lamda^n.(-1)^n=0$

so, By C.H theorem

follows

Silent

Ok!! Thank you very much.

@N.Maneesh So, you mean that by plugging $A^n\cdot(-1)^n=0$ gives $A^n$ is zero matrix, right?

N. Maneesh

yes

it is obvious that characteristic polynomial is (-1)^n \lambda^n

Silent

yes, i see the char polynomial. thanks

user193319

11:36 AM

@LeakyNun Just so that we are on the same page, are you taking $f,g,h$ to be polynomials in $F[X]$, where $F$ is any field?

Leaky Nun

12:57 PM

@user193319 yes

user193319

How does one show that there is no $c > $ such that $||f||_{max} \le c ||f||_1$ for every $f \in C[a,b]$? The only way I think of proving this is using the fact that $C[a,b]$ is dense in $L^1[a,b]$. But there should be a concrete $f \in C[a,b]$ which breaks the inequality...I just can't find it...

Thanks @LeakyNun

Leaky Nun

1:12 PM

@user193319 just show explicitly that for every $c$ there is $f$ such that $\|f\|_\infty > c \|f\|_1$

user193319

Yeah, that's what I've been trying to do...

user193319

1:34 PM

@LeakyNun Any hints?

Leaky Nun

what would that function look like?

can you describe the requirement in words?

user193319

Well, since the function is continuous and Leb. Int. = Rie. Int. for continuous functions, the MVT tells me $||f|_1 = |f(z)|$ for some $z \in (0,1)$ (let's just assume $a=0$ and $b=1$). So, I need to find an $f$ such that its maximum value is at least $cf(x)$ for every $x \in [0,1]$ (I'm assuming that we can choose $f \ge 0$).

Leaky Nun

be less technical

user193319

Hmm...I

I'm not sure I follow you.

Leaky Nun

$\|f\|_1$ is the area under the graph of $f$

what is $\|f\|_\infty$?

which one do you maximize? which one do you minimize?

user193319

1:42 PM

Oh, maximize height of function, minimize its area.

So, $f$ show look like a spike?

Leaky Nun

bingo

user193319

Sweet! Thanks! Now to work out the nitty-gritty details.

Leaky Nun

good luck

yellow_flash

2:03 PM

I'm having trouble understanding this answer - math.stackexchange.com/a/1312534/257828 ; specifically, the author shows the isomorphism using the first isomorphism theorem, but how is $(I, x)$ the kernel of the map $R[x] \to R/I$ ?

Secret

2:31 PM

@Mathphile Sorry for the very late reply. I was completely bogged down with all my calculations gone wrong at the same time today and only managed to fix the load all. So to start with, besides all the points mentioned by anakhro and Semiclassical:

saying $\ln (\pi)$ is rational is the same as claming $\pi ^p = e^q$ but as we discussed earlier, whether $e,\pi$ is algebraically independent is an open question

@Semiclassical If there are uncountably many such class, then one can in theory form something that resembles the Vitali set:

Let $\mathcal{I}$ be the irrationals

Let $[r] = \{r^q : q\in \Bbb{Q}\setminus \{0\}\}$

If there are uncountably many $[r]$ and it partitions $\mathcal{I}$, then each equivalence class $[r]$ will have countably many members (in fact, rational number many)

Then one can use the axiom of choice to pick out each $r$ and this forms something like a Vitali set except with exponentiation as the equivalence relation

If it does not partition $\mathcal{I}$, then we have something more interesting for it can mean the union $R=\bigcup_{r} [r] \subsetneq \mathcal{I}$ forms some kind of exponential version of a ring ideal and the set $\mathcal{I} \setminus R$ will be mapped into $R$ by rational exponentiation (as otherwise they will satisfy the equivalence class given by $[r]$)

hmmm...

Let $s,t \in \mathcal{I}$, $q \in \Bbb{Q}$

$s^q = t$

means:

Semiclassical

2:56 PM

@Secret One thing I note in retrospect. Suppose $a^r=b$ for some rational $r$. Then $r=p/q$ for integer p,q>0, so $a^p = b^q$

Secret

yeah, that basically means $a,b$ are algebraically dependent

Semiclassical

Exactly

it's probably stronger than algebraically dependent, i.e. $a,b$ can be algebraically dependent despite $a^p\neq b^q$ for any $p,q$

Secret

Hmm, so if the structure of these equivalence class in the irrationals can be probed, it might give us a transcendental generalisation of a specific case of Gelfond's Schnider theorem

and thus pinning down $\pi^e$, $\pi^{\pi}$ etc.

Semiclassical

I do wonder if there's a name for this concept already.

right, wikipedia notes the following example: {sqrt(pi),2pi+1}

I wouldn't expect any integer power of sqrt(pi) to give an integer power of 2pi+1

But the two numbers are algebraically dependent since they're a solution of 2x^2-y+1=0

So we're looking at a specific kind of algebraic dependence, namely when the polynomial is of the form x^p-y^q

Secret

yeah

Semiclassical

3:03 PM

(i'm ignoring cases like r=-1/2 as annoying. though all that really does is force us to consider x^p y^q-1 instead)

I wonder what you'd call that, when you restrict the algebraic dependence to certain kinds of polynomials

Secret

10

Q: Is there such a thing as "quadratic independence" (and higher generalizations of linear independence)?

The notion of linear independence is very well-known and well-understood. However, is there a way to generalize the definition to other types of independence -- such as perhaps "quadratic independence", "polynomial independence", "harmonic independence", etc.? (Sorry if linear-algebra isn't a g...

linear-algebra

I don't think there is a literature name to a subset of algebraic independence

Semiclassical

judging from those answers, I'd tend to agree

Secret

25

Q: Work on independence of pi and e

It is an open problem to prove that $\pi$ and $e$ are algebraically independent over $\mathbb{Q}$. What are some of the important results leading toward proving this? What are the most promising theories and approaches for this problem?

nt.number-theory open-problems transcendental-number-theory transcendence

lol we need model theory to investigate these questions

Semiclassical

for some people, that fact will be a motivation to learn model theory

for others, it's a motivation to move on to a different problem :P

Secret

I actually don't revisit transcendental number theory often, but whenever people start writing things that makes me think of $e^e$ I cannot resist and then for a short period of time, snap back to transcendental number theory, and then fail again and return to routine

Somehow that $$e^e$$ is open really annoys me a lot more than other open questions

Btw, I think $s^e = t$ may be interesting, because it means:

Given a sequence $(s^k)$ where $k$ are the kth coefficients of the taylor series of $e$, the above being true means that the partial products of this sequence converges to $t$

The partial products also have the property of being always transcendental, since $s^q$ where $q$ is rational is always transcendental by lindlemann Weistrass

So if there is a way to confine the tail of this partial product somehow (similar to the proof of transcendence of $e$), then it might lead to somewhere

Other things about this partial product include monotonicity

Thus we always have:

$$\frac{t}{\prod_{k=1}^n s^{coeff_{e^x}(k)}} > 1$$ for any $n \in \Bbb{N}$

In particular we have:

$$\frac{t}{\prod_{k=1}^{n-1}s^{\text{coeff}(k)}} > s^{\text{coeff}(n)}$$

Thus for $s^e = t$ we have:

$$\frac{t}{\prod_{k=1}^{n} s^{\frac{1}{k!}}} > s^{\frac{1}{(n+1)!}}$$

Leaky Nun

4:54 PM

@Semiclassical do you know about Sturm-Liouville equations?

Secret

ok forget what I said earlier, that's just irrationality, not transcendentality

To be dealt with later: Investigate the irrationality of $s^e$

Semiclassical

5:16 PM

@LeakyNun strum-liouville differential equations, yeah

Leaky Nun

@Semiclassical how do I know where the boundaries are?

let's say I have $(xe^{-x} y')' = -ke^{-x} y$

Semiclassical

As in, what the domain of your solutions is?

Leaky Nun

right

why can't I pick the domain to be, say, $[-1,1]$?

why is it $[0,\infty)$?

is it where $p(x)$ is zero?

Semiclassical

Typically it’ll be motivated by context

But in the abstract, hmm

Leaky Nun

well the orthogonality relations only works for a particular domain

that's the context

Semiclassical

5:18 PM

True

I mean, you can stretch that a bit

Leaky Nun

but they won't integrate to zero anymore?

Semiclassical

Well, take the weight function in the inner product to be zero outside the region of interest

In that way you trivially extend the domain without modifying the orthogonality

But that feels like a cheat

Leaky Nun

but what if I contract the domain?

maybe I should go through a proof of the orthogonality?

do you know where I could find one?

Perturbative

How can I show that every finite group $G$ can be embedded in a group that can be generated by two elements? My guess was to take some large $D_n$ and try and embed $G$ into $D_n$

Semiclassical

I should but I’m rusty

In the realm of orthogonal polynomials, there’s Favards’s theorem:

en.m.wikipedia.org/wiki/Favard%27s_theorem

With the comparison being that knowing the recurrence relation plus initial values is enough to determine an inner product

Leaky Nun

5:28 PM

@Perturbative $S_n$ is generated by two elements

Perturbative

Damn forgot that, well then there's nothing to show

....assuming Cayleys theorem

Semiclassical

Judging from the Wikipedia page, the form of the DE determines the weight function for S-L eqs, and the endpoints are determined by the boundary conditions

I’d prefer a more definitive statement tho

Secret

[Random]

El supraHollow

$((()))$

error termination: Failure to find a mathematical object emptier than the empty set

Semiclassical

Here’s a question which I suspect should be obvious

Suppose I take the tensor product of two Hilbert spaces

I can take the tensor product of the corresponding identity operators on these spaces to get the identity on the product space

Leaky Nun

yes

Semiclassical

5:37 PM

Is this the only way to construct the identity on the product space? Ie if $A\otimes B$ is the identity on the product space, must A and B also be the identity on their respective spaces?

Feels like that should be obvious

I guess the point is that, if it acts as the identity on the product space, it also must act as the identity on the factors

(Something something universal property?)

Perturbative

6:01 PM

Quick check, a vector space $V$ of dimension $n$ over $\mathbb{Z}_p$ has cardinality $p^n$ since $V \cong (\mathbb{Z}_p) ^n$ right?

ÍgjøgnumMeg

Yep

Leaky Nun

@Semiclassical looks like $p$ needs to be zero at both ends of the domain

Perturbative

Thanks! @ÍgjøgnumMeg

Hope things are going well on your side! @ÍgjøgnumMeg

Leaky Nun

yesterday, by Leaky Nun

$\nabla^2 u = \dfrac1r (ru_r)_r + \dfrac1{r^2} u_{\theta\theta}$

why is the "axisymmetric form" much more complicated? @Semiclassical

$\nabla^2 u = \dfrac1{r^2} (r^2 u_r)_r + \dfrac1{r^2 \sin\theta} (\sin\theta ~ u_\theta)_\theta$

wait this is on a sphere

anakhro

6:17 PM

Hey hot cats.

BAYMAX

8:37 PM

@N.Maneesh cool!! thanks!

Student404Mus

9:14 PM

Hello.

Is there a way in proving such Whittaker formula

$W _ { k , m } ( z ) = \frac { \Gamma ( - 2 m ) } { \Gamma \left( \frac { 1 } { 2 } - m - k \right) } M _ { k , m } ( z ) + \frac { \Gamma ( 2 m ) } { \Gamma \left( \frac { 1 } { 2 } + m - k \right) } M _ { k , - m } ( z )$

Daminark

10:16 PM

Hey @igjo, @loch, how'd you get past the nerd guards?

Dair

11:09 PM

Goes on brilliant.org for the first time in a while. CS section got a lot harder than I remember.

Shaun

11:27 PM

I'm not good enough to be a mathematician. The chances of getting a postdoc, let alone a lectureship, are bleak. How'd you cope? I mean: I'd be happy with a menial job with enough spare time to pursue the subject as a hobby. But why not aim high?

I could have gone to St Andrews to do my PhD but literally panicked during my last exam, meaning I had to resit the next year to get my First; they couldn't defer the entry. I've settled for The University of Essex.

It's not as prestigious. Prestige helps. It shouldn't but it does.

Mathematics is all I want to do.

Okay, I could rant for ages so I'll stop now.

Daminark

There comes a point where you just have to work with the hand you've been dealt. Success may or may not be in our futures, but all we can really do is pursue it. A power outage doesn't matter when the light switch is off

2

Nicholas Roberts

Hi all, I have a basic question. Suppose I know that a linear map $T$ on $R^n$ sends integer n-tuples to other integer n-tuples. That is, $\mathbb{Z}^n$ is invariant under $T$. How can I show that the transpose of $T$ also fixes $\mathbb{Z}^n$?

Daminark

@NicholasRoberts what are some convenient integer n-tuples?

Nicholas Roberts

$e_i$ basis vectors?

Daminark

Yup, so what does our assumption say?

Nicholas Roberts

11:35 PM

We will have that $T(e_i)$ will be an n-tuple of integers

Daminark

Yup, and what do we know about the $T(e_i)$?

Nicholas Roberts

It will be the i'th collumn of the matrix representation for T?

Daminark

Yeah, pedantically the matrix representation of T in the standard basis, but exactly

Perturbative

Hey everyone!

Daminark

So T is an integer matrix. Take the transpose

Nicholas Roberts

11:38 PM

Oh, ok. So that transpose will also have all integer entries. Hence, will send integer tuples to tuples?

Perturbative

Does anyone know where I can find the definition of a bigraded vector space?

Daminark

Exactly

Nicholas Roberts

Thank you @Daminark. I appreciate your time!

loch

11:51 PM

@Daminark lmao how's it going?

Graded vector space

In mathematics, a graded vector space is a vector space that has the extra structure of a grading or a gradation, which is a decomposition of the vector space into a direct sum of vector subspaces. == ℕ-graded vector spaces == Let N {\displaystyle \mathbb {N} } be the set of non-negative integers. An N {\textstyle \mathbb {N} } -graded vector space, often called simply a graded vector space without the prefix N ...

Perturbative

@loch That doesn't have the definition of a bigraded vector space

loch

take $I=\mathbb{Z} \times \mathbb{Z}$

or $I= \mathbb{N} \times \mathbb{N}$

Perturbative

@loch Oh I see, I guess $\mathbb{N} \times \mathbb{N}$ is what's usually used

Daminark

Yo Perturbative, you're welcome Nicholas, and yeah doing alright loch, how about you?

Perturbative

Do you know of a reference other than Wikipedia or nLab? @loch

loch