Mathematics - 2014-05-31 (page 1 of 3)

1:40 AM

Indeed, rather helpful here - though I do want to try your way, need to practice up my integration skills.

Prof. @Ted, my day and night are upside down now, had to drive my parents to the airport.

Ted Shifrin

Oh sure, blame it on them.

Easiest way out, blame your parents on everything.

@seaturtles

Can you help me with something?

It's about Hilbert's theorem for $A[[X]]$.

Ted Shifrin

Is mr @turtles here?

I don't know.

Do you think you can help Ted?

Ted Shifrin

1:47 AM

Um, probably not. 35 years ago I could have ...

sea turtles

@PedroTamaroff eh?

@seaturtles That if $A$ is noetherian so is $A[[X]]$.

@seaturtles There's a part of the argument that I am not buying.

2:08 AM

Why have you stopped?

Hi

@JasperLoy Maybe pedro ended up purchasing the part after all.

@KarlKronenfeld Karl could you help? =)

@PedroTamaroff Sure

@KarlKronenfeld OK. Do you happen to have a copy of Lang's Algebra?

2:20 AM

yep

@PedroTamaroff where is the proof?

My copy has it around page 175 or so.

section?

"Noetherian Rings and Modules."

OK. What I don't "buy" is that he's proving that for each $d>r$ we can subtract linear combinations of $f_{ri}$ to $g$ so that $g-\sum c_i^{(d)}X^{d-r}f_{ri}\in (X^{d+1})$, and this process goes on and on, but somehow one needs to justify why $g=g_1(X)f_{r1}+\cdots+g_{n_r}(X)f_{rn_r}$ where $g_i(X)=\sum_{j\geqslant d}c_i^{(j)}X^{j-r}$

I can see it is related to $\bigcap (X^m)=0$.

people.brandeis.edu/~igusa/Math205bS10/Math205b_S10_71.pdf

I did find that ^

It has a very similar proof.

But it ends in the same deal "Continuing in this way..."

@PedroTamaroff Expand the $g_i$ in $g_i(X)f_{ri}$

@KarlKronenfeld Come again?

2:34 AM

Expand all of the $g_i$ in their power series forms and collect terms

@KarlKronenfeld And?

You get $g$

If you didn't, you could determine the lowest degree term of the difference

@KarlKronenfeld Right, but my point is this.

The $c_i^{(d+1)}$ depend on the $c_i^{(d)}$. And so on.

Why shouldn't we allow that?

No, wait. I'm not saying that's incorrect.

My point is that there's kind of an infinite sum thing going on here, which we then disguise as a finite sum of the $g_i$.

2:39 AM

They are power series...

Yes, of course. But the $g-{\rm stuff}\in (X^j)$ are linear finite combinations.

Ok

Victor

Hi, Any body can help me on this? math.stackexchange.com/questions/815565/…

I'm not seeing the problem @PedroTamaroff

Perhaps there is something to not like in the infinite sum of the various $f_{ri}$ rather than of the $X^r$.

But, it is still easily checked that at every degree the sum of the coefficients has finitely many terms.

@KarlKronenfeld OK, let's see if I can fix thing in my head. =P

For each $R$ there are $c_i^{(v)}$ with $v=d,\ldots,d+R$ and $i=1,\ldots,n_r$ such that $$g-\sum_{v=d}^{d+R}\sum_{i=1}^{n_r}c_i^{(v)}X^{v-r}f_{ri}\in (X^{d+R})$$

2:48 AM

Is anyone else having trouble uploading images on the main site?

It failed here, too.

Now let $$g_i=\sum_{v\geqslant d}c_i^{(v)}X^{v-r}$$

The claim is that $$g=g_1f_{r1}+\ldots+g_r f_{rn_r}$$

We know that $f_{ri}=a_{ri}X^r+\text{higher order terms}$.

3:01 AM

@KarlKronenfeld I guess my point is just that it is not at all evident that $$g=g_1f_{r1}+\ldots+g_r f_{rn_r}$$

3:32 AM

@PedroTamaroff ?

Hi @TedShifrin

or bye

3:47 AM

Hey @MikeMiller

@seaturtles Are you there?

hello eyebrows

4:03 AM

@MikeMiller I'm stuck.

i'm mike

Suppose that $M$ is an $A$-module, and $a\in \bigcap {\rm supp}\, M$. How can I show that for each $x\in M$ there is $n$ such that $a^nx=0$? That is, how to show that $a\in\bigcap_{x\in M} \sqrt{(0:x)}$?

What Mr. Lang is showing is that $$\bigcap {\rm supp}M=\bigcap\sqrt{(0:x)}$$

The body bounded by $x+y=\frac{\pi}{2};z=0;y=0;y=\sqrt{x}$ is the same as the boundries $0\le z \le \frac{\pi}{2}-x$, $0\le y \le \sqrt{x}$, $0\le x \le \frac{\pi}{2}$, right?

Got it.

@Studentmath What do you think?

4:19 AM

@Pedro pretty sure so. $z$,$y$ is obvious, I think $x$ should follow from them - thus we have to bound it from 0 (for $y$) and to $\frac{\pi}{2}$ (for $z$). Then again I am noobish and don't trust my logic :P

Is it right in this case (and generally that's the logic to follow)?

I am confused. You didn't give upper bounds for $z$ in the first part.

Darn me and darn my sleepiness. Sorry. It's $x+z=\frac{pi}{2}$

I really hate $z$ (or like $y$)

4:37 AM

In that case it is correct, right?

4:57 AM

Nah. It's wrong.. now to find out why.

5:11 AM

@FernandoMartin Derivador mundial.

Fernando Martin

Gracias, gracias.

5:26 AM

Or it just gets me a really low value, ach so, will roll with the low value. Seems right otherwise.

dppt

doot

hoot

6:59 AM

I hate my uncertainty..

9:26 AM

This chat is dead.

Hippalectryon

I guess so

9:42 AM

The French are here, lol.

You're Indian @Jasper ?

@G.T.R Nope, Singapore.

Ohhh I didn't know

I want to be born in France or Germany my next life. =)

10:12 AM

@DanielFischer Are you there? I have another Futile Attempts quickie if I may...

@MattN. Last chance, from tomorrow on it's June ;)

@DanielFischer No! Are you leaving the site then?

@MattN. Calendar. May/June. Bad pun.

Hahaha, sorry now I get it : D

Regarding Futile Attempts: I was reading this answer here and I think it's a whole lot of rubbish. It claims an invertible operator is automatically such that its inverse is also continuous. That's just not true, or is it?

Between Banach spaces and some others, it is true. Let me read.

10:16 AM

I also don't believe the "it depends on the context".

It does. A linear map may be linear-algebra-invertible without being invertible in the category of topological vector spaces, for example. Given the context (Ha!), with the quote "Observe that every invertible linear map is bounded below, as is every isometric linear map" from the book, I'd agree with the interpretation that it is used in the sense "isomorphism in the category $\mathbf{EVT}$".

@DanielFischer Ok, if you say so I will believe it. But what is the E in EVT?

Espaces, "Espaces vectoriels topologiques".

Heh. When suddenly wild French appeared. >o<

And I thought it meant Electronic Vector Spaces.

LOL

10:24 AM

Ok. Thank you for your help! @DanielFischer

De nada, @Matt.

(Da nich für)

I understand, I speak Spanish.

I also briefly thought about escalating this and replying in Japanese but then decided not to because I made a similar joke once before and tb accused me of showing off.

: )

No problem with showing off

Me neither.

I'll be back.

@MattN. Schwarzenegger style?

11:16 AM

Hi all, what is wrong with my wolfram alpha query? n=n+Which[n%6==5, 0, n%6==0, 1, n%6==1, 0, n%6==2, 3, n%6==3, 2, n%6==4, 1]

reference.wolfram.com/mathematica/ref/Which.html

I'm trying to find out an equation for this. I don't really know what to call this type of sequence or what to google so I can learn how (what steps to take) to solve my problem.

11:34 AM

Greetings

@r9m Starting from the fact

$$f(x^2)+f(y^2)\le2 f(\sqrt{x y})$$
one may easily prove by induction that
$$\frac{1}{n}\sum_{k=1}^nf(x_k^2)\le f\left(\sqrt[n]{\prod_{k=1}^{n} x_k}\right)$$
Hence, by setting $\displaystyle x_k=\frac{k}{n}$ above , and letting $n\to\infty$, we get that
$$\int_0^1 f(x^2) \ dx \le f\left(\frac{1}{e}\right)$$
Now, letting $x^2\mapsto x$ combined with the integration by parts, we conclude that
$$f(1)-f(1/e)\le \int_0^1 \sqrt{x} f'(x) dx$$

Q.E.D.

@r9m don't post my solution.

Balarka Sen

Ahoy @skullpatrol

@BalarkaSen Ahoy my friend :-)

I've just created some new stuff ...

11:49 AM

... how creative of you :D

Could anyone please give me a push in the right direction? For example, name what I'm doing (I'm not a native english speaker)?

askaway

I want to learn/understand how I can translate repeating sequence of values to an equation (or find out/recognise an equation is not possible and I need a function).

after 2 day's of searching I came up with: wolfram alpha query: n=n+Which[n%6==5, 0, n%6==0, 1, n%6==1, 0, n%6==2, 3, n%6==3, 2, n%6==4, 1]

But that to didn't work.

this is a sequence of 6 by the way.

I do not know what to google that can help me further.

I'm going to propose it for a contest $$\lim_{n\to\infty} \left(\frac{1}{\displaystyle \binom{2014}{2014}}+\frac{1}{\displaystyle \binom{2014}{2014+1}}+\cdots+\frac{1}{\displaystyle \binom{2014}{2014+n}}\right)$$

12:26 PM

@skullpatrol :D

12:43 PM

:D

12:53 PM

@skullpatrol Let me show you something nice ...

$$\int_0^{\infty} \frac{1-\cos(x)}{x^2} \int_0^{ x}\frac{\sin(t)}{t} \ dt \ dx=\frac{\pi^2}{8}+\log(2)$$

(it's newly created)

►

@DanielFischer Exactly : )

1:08 PM

lol, cool vid

ROFL tex.stackexchange.com/a/94910

No-one that could point me into the right direction?

@GitaarLAB did you try mathematica.stackexchange.com ?

1:42 PM

@G.T.R: sorry, just noticed your response. I'm in doubt between math.SE and mathematica.SE. Further-more, whilst usually having no problems with English, It turns out that my math-related English is not adequate enough to even describe my problem in a way that I can google it. Thus, I at least hoped someone could point me in the right direction: name what I'm doing so I have a starting-point to google.

@GitaarLAB If I understood well, you have a periodic sequence of integers, and you want to find a closed form for it ?

I usually have no problem thinking out of my box, but this rare occasion has me 'punching holes in the dark' so to say.

I'm hoping to find an equation (if this is the correct terminology) instead of a function (like something that has if/else or loops etc). Using the latter, in code would be no problem for me. BUT I'm guessing this one can be done with an equation.

Essentially I'm trying to 'synchronise' any positive integer to: oeis.org/A007310

I already found out I can generate that sequence with the simple equation: n = n+((n+3)mod 6)

That obviously (at least in my environment) is a lot more efficient than some kind of sub-loop and extra variables.

@GitaarLAB ok, so you're trying to generalise this to any sequence ?

yes: wolfram alpha query: n=n+Which[n%6==5, 0, n%6==0, 1, n%6==1, 0, n%6==2, 3, n%6==3, 2, n%6==4, 1]

but that gave an error in wolfram alpha.

Whilst I'm happy with just the answer (that I can decompose and surely understand how that works), I'm more interested in learning how to come to such an answer in the first place.

by answer, I meant equation.

But, are you willing to do this for a finite or an infinite sequence ?

1:50 PM

in the 'formula' from wolfram alpha, n could theoretically be infinite (but I'm limiting to the IEEE floating point max 2^53 = 9.007.199.254.740.992

or, some values under it.

(to keep capacity for the calculation in memory)

I lost you actually. I understood n = n+((n+3)mod 6) is a formula for oeis.org/A007310 . What do you want to do next ?

I found another sequence that (if I know the equation to it) could be plugged into another (dumber) equation (so at least I have something to put into wolfram alpha, hoping it can help me from there on, at least name what I'm doing): 1,0,5,5,5,5

@G.T.R sorry. Should have explained that. That equation works perfectly as long as n is a valid point on the line.

Now I want an equation that can put any value for n on that line. From there on I can use the n = n+((n+3)mod 6)

Put in other words, given any n, compute the next or previous point on the line (exept when n is already on the line, because of the logic from - to, from should include n, if it's a valid point on the line). Hope that makes it more clear, however this is now more specific to what I'm currently trying to do. In general I'd like to know how to (try) to find an equation for a given repeatable sequence.

explained (more specific) in another way: n=n+(if n mod 6 == 5 then 0 else if n mod 6 == 0 then 1 else if n mod 6 == 1 then 0 else if n mod 6==2 then 3 else if n mod 6==3 then 2 else if n mod 6==4 then 1)

as one can see, the sequence is finite: a repetition of 6 values.

2:09 PM

I wonder what is the asymptotic behaviour of $$\int_0^1 \int_0^1 \cdots \int_0^1 \frac{1}{\sqrt{x_1^2+x_2^2+\cdots +x_n^2}} \ dx_1 \ dx_2\cdots dx_n$$

@DanielFischer is the question above familiar to you? Just let me know if you have some ideas on it.

@Chris'ssis For $n=1$ the integral blows up.

@PedroTamaroff Indeed.

@Chris'ssis $1$

@Chris'ssis Wait, all the terms are infinite past the first... sorry $\infty$

I answered a question =)

@robjohn What do you mean by $\infty$?

2:20 PM

@Chris'ssis the first term is $1$, then the denominator of each other term is $0$

@Chris'ssis Did you mean the top term to increase instead of the bottom term of the binomial?

@robjohn I think my notations above failed ...

@robjohn Sorry, I have to rewrite it ... :-(

@robjohn $$\lim_{n\to\infty} \left(\frac{1}{\displaystyle \binom{2014}{2014}}+\frac{1}{\displaystyle \binom{2014+1}{2014}}+\cdots+\frac{1}{\displaystyle \binom{2014+n}{2014}}\right)$$

@Chris'ssis That's what I was suggesting above :-)

@robjohn Yeah, I saw now. :-)

Partially answered math.stackexchange.com/questions/815650/… ^^

@Chris'ssis I have to go walk the dog... bbl

2:33 PM

@robjohn OK

@Chris'ssis Could you check numerically the first conjuctured closed form for a few values Chris?

I feel it is wrong, but do not trust maple

@robjohn By the way, yesterday I also posted a very nice limit involving Fibonacci numbers plugged into a nested radical :-)

Maybe I'll repost it a bit later since I have to leave now.

Nevermind it looks correct, sigh

@seaturtles

3:45 PM

@DanielFischer "on est pas venu pour beurrer les sandwiches !"

@G.T.R A propos de quoi?

@DanielFischer Hello!

@TedShifrin Hello, too!

Hi

@DanielFischer la réplique dans Les Tontons Flingueurs :P

Hi @Pedro, @Ted, and @MrEyeglasses.

@G.T.R J'comprends pas.

3:48 PM

@DanielFischer I was trying to show $\mathfrak r(\mathfrak a)=\mathfrak a$ implies that, if $\mathfrak a$ is decomposable, all associated primes are minimal, but then I can't. =)

@Daniel je suis sûr que tu connais le film. A un moment ils vont mettre du beurre sur des sandwiches et boire de l'alcool fort

@Chris'ssis I'm back... I think it should be $\frac{2015}{2014}$

@PedroTamaroff Well, you could ask an algebraist for help ;)

@N3buchadnezzar LEL.

@DanielFischer Heh, I tried with anon. No lucks. I was also trying to show things like $(x)\cap(y,z)=(zx,zy)$.

3:52 PM

@robjohn I got a different result though ... $$\displaystyle \frac{2014}{2013}$$

@G.T.R I think I don't know it.

@Chris'ssis Actually, I was off by one in my head... I was using $$\dfrac1{\binom{n+1}{k+1}}=\dfrac{k+1}{k}\left(\dfrac1{\binom{n}{k}}- \dfrac1{\binom{n+1}{k}}\right)$$

@PedroTamaroff What is $\mathfrak{r}(\mathfrak{a})$, what does "decomposable" mean, and what is the general setting we're in?

@Daniel not possible, not that one. It stars Lino Ventura, Bernard Blier and Jean Lefevre

@robjohn OK

3:55 PM

mmm

@DanielFischer Primary decomposition of ideals. By $\mathfrak r(-)$ I mean the radical of the ideal.

@Chris'ssis so yes, $k=2013$ and the sum is $\frac{2014}{2013}$

@G.T.R Well, I wasn't a regular cinema-goer when it came out, and I'm not sure it got much airing in the German television.

Why do we say that $\{ a \}$ is open if it lies in the topology, is that just how we define open ?

@N3buchadnezzar Yes, by definition a set is open if and only if it is in the topology.

3:56 PM

@robjohn Indeed.

@robjohn Is there any way to block someone to read me on chat?

@Chris'ssis Nope. You can block yourself from seeing someone else, but not the other way around.

@DanielFischer I am having some problems connecting this with the open definition of intervals

@robjohn OK ...

MickLH

would be impossible to enforce anyways, someone could just sign out and view public logs

@N3buchadnezzar Why? They belong to the (standard) topology, so they're open.