Are there higher dimensional analogues of periodic functions

instead of period = 2p, it's 3p?

@Obliv What?

I don't understand...

Idk either but the fundamental period of fourier series got me thinkin..

modular forms

@Obliv Well, I am not sure that I can answer your question without knowing what you mean.

for a differential equation the ansatz $f(t,x)=X(x)Y(t)$ (separation of variables) supposes a particular type of solution. What if the solution cannot be separated like this? How would you find the solution?

@geocalc33 You would use some other technique.

hmm @XanderHenderson Actually what if A solution CAN be expressed as a separation of variables but those aren't ALL the solutions

I guess you use some other technique to find the rest of the soutions

@geocalc33 Do you have an example? Uniqueness theorems are going to play a role, here...

@XanderHenderson Yes $tu_{tt}=-xu_{x}.$ Using seperation of variables you can reduce this to ODE's and find solutions

but this separation technique doesn't give the distributional solution, or fundamental solution

Why are Jordan measures only defined for bounded sets? Why not define them for unbounded sets?

@QuitMSE How is the semi-direct product defined?

@TedShifrin do you now the answer to my Q sir?

Don’t you need a finite cover by rectangles?

@TedShifrin G be a group and H, K<G with H normal in G. Let $\phi:K\to Aut(H) $ be a homomorphism. Then the set $\{(h, k) :h\in H, k\in K\}$ is a group under the operation $ (h_1, h_2) ×_{\phi}(h_2, k_2) =(h_1\phi_{k_1}(h_2), k_1k_2) $ is a group, called semi direct product of H and K induced by $\phi$.

temperature here is 31 deg C (87.8 F) with 79% humidity so it feels like 40 deg C (online sources)

K acts on H via conjugation

So if the action is trivial, doesn’t that formula say abelian?

@TedShifrin Yes. Then semi direct product is nothing but direct product.

Oh wait. We’re trying to do the converse? If the result is abelian, then what does that tell us?

@TedShifrin but we can just define an infinite cover

Does the definition allow that?

@TedShifrin semi direct product abelian implies (h_1, k_1) (h_2, k_2) commute. Then h_1\phi_{k_2}(h_2) =h_2\phi_{k_2}(h_1)

Different $k$, actually?

Sorry. Need to leave for doctor’s appointment.

Well. $\phi_k(h) =khk^{-1} $ then abelianness of H, K forces $\phi$ to be a trivial one.

@TedShifrin Yes. Left side should conation $k_1$

@TedShifrin OK.

@TedShifrin why can’t we extend the def, since it matches better worth our intuition of area

@Shinrin-Yoku How? How are you going to extend the definition?

@QuitMSE is your question answered now?

@QuitMSE here, you can substitute $k_1=$ identity.

then use the fact that H is Abelian.

@XanderHenderson by defining infinite covers by rectangles

allowing infinite unions of rectangles @XanderHenderson

and then following the usual def

surely intuitively unbounded area makes sense

After all that’s what improper integrals are

@DLeftAdjointtoU: Hi :-). Where can I study Zeta function (of the hypersurface defined by a polynomial)?

Ireland Rosen has a chapter on this but I'm having difficulty understanding it.

let $f(x_0,x_1,x_2)\in F[x_0, x_1, x_2] $ be a nonzero homogeneous polynomial that is nonsingular over every algebraic extension of F.

What is the meaning of 'non singular over ...'??

@Shinrin-Yoku What is the area of such a cover?

@Koro It's a wrong room.

this is an extract from chapter 11 of Ireland and Rosen's. I don't recall 'non singular over every algebraic extension...' being defined earlier in the book.

@Koro Why not you post it as a question in the main site?

If I posted this question, it could get closed due to lack of context so I ask it here.

@Koro You will get some comments at least.

← 6 messages moved from CURED

@XanderHenderson you just define the area to be the sum of infinite disjoint rectangles

@Shinrin-Yoku Which will always be infinite?

No it could be finite @XanderHenderson after all there are multiple improper integrals which are finite

@Shinrin-Yoku Sorry I was unclear---I am thinking of the outer Jordan measure.

Meh... not always infinite. There are going to be counter-examples.

But you are going to have real problems with getting control over the bounds on the covers of infinite sets.

I would expect that the "right" extension is to consider the measure of $B(0,r) \cap E$ as $r \to \infty$.

But the lesbegue outer measure allows infinite covers no? (Is not that the only difference between the two measures?) @XanderHenderson

@Shinrin-Yoku No... not quite.

Though I suppose it depends on precisely how you define the Lebesgue measure in higher dimensions.

In any event, if you allow infinite covers, then you have defined the Lebesgue measure.

The Jordan outer "measure" of an unbounded set is always going to be infinite, because you covering the set with a finite number of rectangles. So there is going to be at least one unbounded rectangle of infinite area no matter how you cover the set. This is bad.

IF you want to consider countable covers of sets, you are just defining the Lebesgue measure.

This (and other similar issues, like the content of a bounded set of rationals) is probably why Lebesgue measure exists.

Ok, that makes sense so the lebesgue outer measure is the correct measure to talk of areas of unbounded functions. @XanderHenderson . Another question I have probably because I don't know measure theory but why is measure defined using the weird caratheodory criteria when one can just define the outer and inner measure and say a set is measurable if they are equal like in the jordan case?

You want your measure to be closed with respect to your usual set operations, e.g. taking unions and intersections. The Caratheodory condition does that for you.

There is a pretty good MO discussion of this: mathoverflow.net/questions/34007/…

Ah! Here is the answer I was looking for: mathoverflow.net/a/308902/113150

@QuitMSE Is that a good book for measure thwory?

@Shinrin-Yoku I don't know but I am really enjoying.

i like bruckner^2's Elementary Real Analysis

For $f(t) = \begin{cases} t, & 0 \leq t < 1\\ 2 - t, & t \geq 1 \end{cases}$ we have $f(t) = t + (2-2t)\mathscr{U}(t-1)$ when I take $\mathcal{L}\{f(t)\}$ I get $$\mathcal{L}\{t\} + \mathcal{L}\{(2-2t)\mathscr{U}(t-1)\} = \frac{1}{s^2} + (\frac{2}{s} - \frac{2}{s^2})e^{-s}$$ But that's wrong :(

I'm not supposed to have the $\frac{2}{s}$ term for some reason

@Shinrin-Yoku It's good for measure theory but I think it's not good enough for Lebesgue integral ( atleast not good as Stein's book)

@XanderHenderson the second answer says "need not be jordan measurable" , but isnt it not jordan measurable by definition?

What are the odds that my textbook is wrong

I have looked at it up down sideways can't seem to see the flaw in my reasoning

I am studying problem 3d) in Spivak's Calculus, chapter 22. The problem statement is as follows:

Consider the sequence $$\frac12,\frac13,\frac23,\frac14,\frac24,\frac34,\frac15,\frac25,\frac35,\frac45,\frac16,\ldots.$$

For which numbers $\alpha$ is there a subsequence converging to $\alpha$?

Is there a formula to the sequence or something? I can not get my head around it.

bets on the table

for any real number between 0 and 1

inclusive

how do you see this?

you can find 1/100, 1/1000, 1/10000, etc.

for convergence to 0

I see now.

for convergence to 1, you can find 99/100, 999/1000, etc.

yes, thanks

np

Is there a common distribution (not the Dirac Delta distribution of course) that takes the form of the red line?

I asked a question on mathoverflow and everyone laughed at it

If a solution exists for all time does that mean $t>0$?

The answer is yes

what's the definition of inequality i want to prove $k>0 \iff \forall j \in \mathbb{N}: k \geq \frac{1}{j}$

@geocalc33 I don't know what you mean.

@shintuku I don't know what you mean, either.

I figured it out

shin: what is k here (i.e., real numbers - and maybe, how constructed/realized, or what do you know about them)? should that be a "there exists"?

i think it's a real number

it's a magical step in a measure theory proof

$\mu(\{|u|>0\}) = \mu\left(\bigcup_{j \in \mathbb N}\{|u| \geq \frac{1}{j}\}\right)$

in the proof of Markov's inequality

i mean i could just take it for granted

like a barbarian

nvm it's not in the proof of markov's inequality, but it's sensible enough i will believe it

ok, ignoring the application of a measure for the moment, you have a set equality, {x : f(x) > 0} = union_{j in N} {x: f(x) > 1/j}, which is certainly true. you do mean 'exists' there

not the most exciting fact in the world, but often you see sets rewritten as countable unions or intersections of other sets in measure theory

whoops, i do mean exists

thanks leslie

But what if we tried extending coprimality to the reals :thinking:?

And what is the meaning of the values satisfying Wilson's theorem but for the Gamma function? :thinking: I digress

@TedShifrin Hello Ted! are you busy ?