Mathematics - 2014-12-09 (page 7 of 7)

@evinda maybe wolframalpha can help?

@anon can you explain where the 1+3+6+10+15 part came from, then?

anon

@MathyPerson it counts the triangles with all three vertices in the array, which are "upright" ...

evinda

@MathyPerson I don't think so.. That is the exercise I meant: math.stackexchange.com/questions/1059418/…

Is there someone to which I could ask a question about algebraic geometry?

anon

you can ask Ted

That's indeed his 'zone'

teadawg1337

11:04 PM

Ahh... Just woke up from a nap

Hello!! I need some help in measure theory... Is there someone that can help me??

@anon Well, so far i counted 27 triangles with 3 vertices on the triangulat lattice

*triangular

@MaryStar Not sure if I know enough, but I can try

@Studentmath Let $1 \leq p_1 \leq p_2 \leq +\infty$. Show that in a Lebesgue measurable $E\subset R^d$ with $0<m(E)<+\infty$ we have that $L^{p_2} \subsetneq L^{p_1}$.

Using Hölder's inequality I got that $||f||_{p_1} \leq ||f||_{p_2} \mu (E)^{1/p_1 \cdot q}$.

Is this correct so far?? How could I continue??

Or is there an other way to show this??

Ach so, a bit too advanced for me, sorry

11:16 PM

@anon 27 is the most I can find so far

teadawg1337

Hello @Pedro

Heya.

@PedroTamaroff hey there...

@anon What I do now is that the number of elements in at least 2 sets is #(A intersect B) + #(A intersect C) + #(B intersect C) - 2*#(A intersect B intersect C)

@anon However, I'm not sure what the A, B, and C would be.

@robjohn Hello Rob.

11:19 PM

@MaryStar what more do you think you need to show?

@robjohn I have to show that it is strictly $<$. Now I have shown that it is $\leq $, right??

evinda

Hey @PedroTamaroff Do you maybe know if we find that f and h have a common tangent from f(x,0) and h(x,0) at this exercise?

http://math.stackexchange.com/questions/1052007/intersection-multiplicity-of-the-curves

@MaryStar why do you need to show that it is strictly less than?

@robjohn Shouln't I show that to prove that $L^{p_2} \subsetneq L^{p_1}$ ??

@anon Are you still there..??

11:23 PM

@MaryStar what does that inclusion mean?

@robjohn That an element of $L^{p_2}$ doesn't belong to $L^{p_1}$, right??

@MaryStar wait... isn't that backwards?

@MaryStar where in there is strict inequality needed?

@robjohn How can I show it then??

what do you mean? you just said what you needed to show.

@MaryStar I think you are going in the wrong direction in places.

@robjohn Do I have to take an element of $L^{p_2}$ that doesn't belong to $L^{p_1}$?? Is such an element the function $f(x)=\frac{1}{x^{p_2}}$, if $0<x \leq 1$ and $f(x)=0$, if $x>1$ ??

11:31 PM

To show that $L^{p_2}\subsetneq L^{p_1}$ you need to show that any element of $L^{p_2}$ is also in $L^{p_1}$ and that there is some element of $L^{p_1}$ that is not in $L^{p_2}$.

@MaryStar You won't find an element of $L^{p_2}$ not in $L^{p_1}$ since $L^{p_2}\subsetneq L^{p_1}$

@robjohn I got stuck right now...

@MaryStar You need to show that there is an element of $L^{p_1}$ that is not in $L^{p_2}$

evinda

@PedroTamaroff Have you taken a look at the exercise? :)

@evinda Not really, no.

Quick, probably silly question - $D_n$ is a subgroup of $D_{2n}$, right..?

11:38 PM

@Studentmath Yes, you can draw a picture of the $n$-gons.

Phew, alright

Right

@robjohn To show this could we use the function $f(x)=\frac{1}{x^{p_2}}$, if $0<x \leq 1$ and $f(x)=0$, if $x>1$ ??

$f(x) \in L^{p_1} : \int |f(x)|^{p_1}=\int_0^1\frac{1}{x^{p_2 p_1}}=-p_2p_1+1<\infty$

$f(x) \notin L^{p_2} : \int |f(x)|^{p_2}=\int_0^1\frac{1}{x}=[\ln x ]_0^1=-\infty$

Is this correct??

Thanks!

@MaryStar How do you know what $E$ is?

Can anyone help me with this problem? Let d_n be the number of ordered sequences of die rolls (i.e., sequences of integers from 1 to 6) that add up to n. For example, d_4=8, because a total of 4 can be rolled in 8 ways

and d_0=1, since 0 can be rolled in one way (roll no dice).

Let D(x) be the generating function D(x) = d_0 + d_1x + d_2*x^2 + d_3*x^3 +.... Then 1/{D(x)} is a polynomial. What polynomial is it?

11:41 PM

@MathyPerson What have you tried?

@robjohn What do you mean??

@evinda Martin has already answered your question, it seems.

@Behaviour What is your avatar?

@PedroTamaroff Actually, I'm not even sure where to start!

Behaviour

@PedroTamaroff Source credited at the bottom of "about me" box.

@MaryStar aren't you working in a measurable $E$ in $\mathbb{R}^n$? You want to show this no matter what $E$ is.

11:43 PM

@robjohn Ok...How can I do that?? How can I use the information that $0<m(E)<\infty$ ??

Chris's sis

@Behaviour I saw you have some Shog9 quotes. I've never heard before of such quotes. Where are they coming from?

Mike Miller

Ah, I'm sorry to hear you're a Pet Shop Boys fan, @Behaviour.

@MathyPerson Let me see if I understood.

You can roll any amount of times, yes?

Behaviour

@MikeMiller Heh. BTW quid picked up on it on his/her own.

@MathyPerson I assume you consider the roll $1+2$ different from $2+1$, too, say.

11:50 PM

@PedroTamaroff Yes. a total of 4 can be rolled in 8 ways:

\begin{array}{*4c} 4 & 3+1 & 2+2 & 1+3 <br /> <br /> ~2+1+1~ & ~1+2+1~ & ~1+1+2~ & ~1+1+1+1~ \end{array}

Mike Miller

@Behaviour You referred to the detriment of moderating another forum pre-moderating stack exchange. Could you expand a bit on what you mean by that? (What could go wrong, I mean - I'm not asking about the people involved.)

I suspect I know to whom you refer, but I never got the whole story from old meta posts - details were usually sparse.

Hrm, but $D_n$ and $<\alpha>$ share $\alpha$ in $D_{2n}$, right?

Behaviour

@MikeMiller Basically, some people with experience in Internet forums of other kinds may be treating Math.SE as the same thing, plus MathJax and some reputation nonsense. (Others are willing to adapt; I don't want to paint everyone with the same brush). I don't believe we need a moderator with such an approach.

Here's a little sample (the end of comment):

@fbueckert Alas, explaining that to a non-mathematical audience would require a 30 page paper, since to appreciate the points requires a nontrivial understanding of mathematics and the various ideas on how it is effectively taught (online). Certainly that will not fit in an SE comment. My views on such matters are informed by a few decades of heavy involvement in such (going back to the early days of usenet newsgroups). — Bill Dubuque Aug 3 at 17:11

Where $\alpha$ is the.. reflection, that's it.

@PedroTamaroff are you still there?

11:58 PM

@MathyPerson Yes. I'm a bit busy now, but I'll get back to you. It's a nice problem.