Math Challenged

[caveman1917]

I think I would have called the Lebesgue integral the standard route, but no matter

I guess i'm thinking too much in terms of its physical application

In any event, in this case, we're dealing with a proper Riemann integral, provided we don't assign the endpoint the value "infinity", no?

You're right, it's an endpoint where it is undefined, not an interior point. Actually it would still be proper even if we assigned the value infinity since the distinction only matters as to the limit in an endpoint, not its value there.

ETA: it seems different authors use slightly different definitions of what constitutes an improper integral.

[Sarawak]

That was quick!

I think that is a valid answer, which entitles you to ask the next question. However, before proceeding, I would ask whether there is an answer which is different than the one you gave, which is also arguably correct?

[caveman1917]

That was quick!

I was going through Laplace transforms just a couple of days ago and used that integral for practice.

I think that is a valid answer, which entitles you to ask the next question. However, before proceeding, I would ask whether there is an answer which is different than the one you gave, which is also arguably correct?

I'm not sure, doing a series expansion and integrating term by term gives the same result, so i must be missing something here.

[Sarawak]

I'm not sure, doing a series expansion and integrating term by term gives the same result, so i must be missing something here.

Hmm, that's interesting. I am familiar with a path-integral type of approach to this one. Certainly I can derive the series, but is there an easy way to determine the limit at infinity?

The point of my question really gets at the definition of an integral. Had I been asked this question, I would have given a different answer But, it all comes down to definitions.

[caveman1917]

Hmm, that's interesting. I am familiar with a path-integral type of approach to this one. Certainly I can derive the series, but is there an easy way to determine the limit at infinity?

I just checked it numerically to see if it would approach the same value, i'm not sure if there's a simple way to determine the limit of the series analytically.

Perhaps an approach like the following might work? Since we have its derivative (ie ), we can determine its local maxima/minima, which occur at . Because it is continuous, if we can show that the sequence of maxima is strictly decreasing and the sequence of minima strictly increasing and we could also show that the sequence of maxima has a lower bound at and likewise the sequence of minima an upper bound at , we would have determined the limit at infinity at . I don't know if that approach would work though.

The point of my question really gets at the definition of an integral. Had I been asked this question, I would have given a different answer But, it all comes down to definitions.

Well, it is not absolutely convergent, so the Lebesgue integral doesn't exist. Is that what you meant?

[Sarawak]

I just checked it numerically to see if it would approach the same value, i'm not sure if there's a simple way to determine the limit of the series analytically.

Perhaps an approach like the following might work? Since we have its derivative (ie ), we can determine its local maxima/minima, which occur at . Because it is continuous, if we can show that the sequence of maxima is strictly decreasing and the sequence of minima strictly increasing and we could also show that the sequence of maxima has a lower bound at and likewise the sequence of minima an upper bound at , we would have determined the limit at infinity at . I don't know if that approach would work though.

Have to think about it.

Well, it is not absolutely convergent, so the Lebesgue integral doesn't exist. Is that what you meant?

Exactly. This is a case where



but



is undefined in the Lebesgue sense.

[grapes]

is undefined in the Lebesgue sense.

Sometimes you can take non-Lebesgue approaches.

Which is?

All standard approaches seem to give the same answer, in fact i think there's a theorem that the Lebesgue integral must equal the improper Riemann integral in this case, so i'm not sure what the added benefit is to going the Lebesgue route rather than just the standard route of dealing with improper Riemann integrals.

Non-Lebesgue approach would involve a delta function. Certainly is appropriate physically. For a sphere of radius 1, the constants that we're missing are which is also the value of the gravity in the vicinity of an infinite sheet of planar density delta--so, the value 1 at R=1 is "missing" exactly that amount to make it agree with , outside and on the sphere.

Just another interesting possibility.

And it's your turn, right?

[Sarawak]

This one would seem to exhibit similar behaviour to your example, grapes.



I find it a little easier to visualise (although your kilometrage may vary) than the original; I can picture an x-epsilon plane, with a big circular funnel right over the origin ...

[grapes]

Yes, that's the set up for gravity (or electric field) above an infinite plate, multiplied by .

From the little I've gleaned in the last few days, it's the situation that led Dirac to the delta function.

[Sarawak]

Yes, that's the set up for gravity (or electric field) above an infinite plate, multiplied by .

I got it just by tweaking your integral

[grapes]

I realize that, but physically the question is, what happens when you touch the plane? Does the field disappear, like the integral?

Or, in other words, does touching a point mean that one is coincident with that point?

[Sarawak]

I realize that, but physically the question is, what happens when you touch the plane? Does the field disappear, like the integral?

Or, in other words, does touching a point mean that one is coincident with that point?

Seems to me the whole thing is non-physical anyway - apart from the infinite place (a sphere is more palatable this way), we assume Newton's law works at any distance, including subatomic, matter is infinitely divisible, shell/plane is infinitely thin, etc.

So I would say, we conjure up such a universe, we can make our own rules for it

[grapes]

Absolutely agree.

Still, what are we allowed to do mathematically? Clearly, were not limited by Lesbegue.

The integrand in your example peaks at epsilon divided by the square root of two, with a value of .385 divided by epsilon, with a constant area of 1. Is the limit of the integrand just zero, with zero area? Point-wise, sure, but a case can be made for a delta function.

Next problem!

[caveman1917]

And it's your turn, right?

I'll let either you or Coelacanth take the turn if you want. I'm in the middle of moving to a new appartment so i'm not sure i'll be able to check up on it the next couple of days.

[caveman1917]

Is it allowed to ask questions to which you don't know the answer yourself?
If yes, see question 1, if no, see question 2.

Q1. Let .
First define a partial order on such that it is a total order.
Then define two binary operations:



How much can you make "field-like"? In other words, how many properties that would make it a totally ordered field can you preserve, given that the order must be total?
I've been wondering about this ever since the time we were trying to bijectively map to , but i can't quite solve it.

Q2. Prove that in the context of where all operations are standard and where is the neutral element of and the neutral element of
Directly from the axiomatic definition of as a totally ordered field.

[caveman1917]

How much can you make "field-like"?

Excluding the trivial solution of just defining everything in terms of the corresponding operations on .

[Sarawak]

Excluding the trivial solution of just defining everything in terms of the corresponding operations on .

Well, there goes my method of solution.

[caveman1917]

Well, there goes my method of solution.

It would make it isomorphic to the real numbers, so we wouldn't really have anything "new"

[caveman1917]

Perhaps, to get this thread going again, someone would like to tackle question 2 instead? It follows in a fairly straightforward manner from the properties of a totally ordered field.

[grapes]

I thought about it, but wasn't sure where to go with it, because if > is a total ordering, then so is <, right?

[caveman1917]

I thought about it, but wasn't sure where to go with it, because if > is a total ordering, then so is <, right?

Yes, though i don't really understand the basis of the question, since you can just define one in terms of the other.

[caveman1917]

Ah yes, i see where you're going with this, but if you take a look at the properties of a totally ordered field, you'll see that a "preferred" direction is implied.
Take a look at the first definition of "ordered field" on wiki.

[caveman1917]

Perhaps i should rephrase slightly, the direction is the same as the one you use in the definition of your ordered field.
In other words, if you use the direction you will derive and if you use you will derive .

So let's just stick to the standard first case

[grapes]

Yes, there is some ambiguity here. A field can be coupled with a total order, without the result being an "ordered field", using that definition.
http://en.wikipedia.org/wiki/Ordered_field

If you have the "totally ordered" "field" also be an "ordered field", I think we're stuck with the real numbers, even in the first problem, as Coelacanth suggested (and, of course 0 <1 ). But I'll have to think about it.

[caveman1917]

I see that the wiki page on ordered field gives the proof i'm asking for (at least an equivalent version), so don't read further than the section on the definitions

Yes, there is some ambiguity here. A field can be coupled with a total order, without the result being an "ordered field", using that definition.
http://en.wikipedia.org/wiki/Ordered_field

If you have the "totally ordered" "field" also be an "ordered field", I think we're stuck with the real numbers, even in the first problem, as Coelacanth suggested (and, of course 0 <1 ). But I'll have to think about it.

We're not stuck with the real numbers, an ordered field doesn't have to be Dedekind-complete. Every Dedekind-complete field is isomorphic to the real numbers, but there are many totally ordered fields that aren't Dedekind-complete, for example the rational numbers.

The term "totally ordered field" is the strictly correct one, since you can also have a partially ordered field. However it is not uncommon to say "ordered field" instead of "totally ordered field" and assume that if a partially ordered field is meant, it would have been specified.

and, of course 0 <1

The question is to prove that, it is not an axiom, therefor it is a theorem that should be proven from the axioms. Giving that proof is my question.

I'll list the axioms that are given:

You have a set and two binary operations and a total order such that:

(closure)

(commutative)

(associative)

(existence of identity elements)

(existence of inverse elements)

(distributive)

(then making it a totally ordered field)





Then denote










and prove

[caveman1917]

This was one of the questions actually asked by a professor of mine on our exam of Analysis I in first year. Everybody's first reaction would be "of course 0 < 1", but his point was of course that one should be rigorous and actually prove any statement one uses that is not directly embedded in the axioms one uses.

It's fairly straightforward, the hardest part is getting over the "of course it is!" reaction

[Sarawak]

I had the same problem, but the Wiki page posted two additional properties relating the ordering to the field operations.

I am thinking about a proof by contradiction.

[caveman1917]

I am thinking about a proof by contradiction.

That may indeed get you somewhere

[caveman1917]

I'll give a hint in the form of two lemmas you will need (or maybe not depending on how you solve it)

Lemma 1:
Proof: and likewise for the other

Lemma 2:
Proof:

[Sarawak]

I think I am on track then, as I had figured I needed both of these results. I had the proof of the first one, and was thinking about the proof of the second one