Math: Fleeting Thoughts

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Re: Math: Fleeting Thoughts

Postby Patashu » Mon Jul 12, 2010 3:39 am UTC

Maybe it doesn't hold both ways. It's clear that you can go from right to left, but left to right loses information.

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Re: Math: Fleeting Thoughts

Postby jestingrabbit » Tue Jul 13, 2010 9:48 pm UTC

Patashu wrote:I suspect something like this holds: a + bw + cw^2 === (a + 0.5b - 0.5c) + (sqrt(3)/2)bcj


Having thought a little more, I think that the way to nudge you is to ask you to find elements of the split complex numbers, the complex numbers or the dual numbers such that their cube is 1. Hint:
Spoiler:
there is a reason we usually only hear about the complex numbers.


Patashu wrote:Maybe it doesn't hold both ways. It's clear that you can go from right to left, but left to right loses information.


Well, commutativity doesn't really have a direction. Either wz = zw for all z and w, or not. But talking about losing information is on the right track it seems.

Anyway, new set of assumptions. We are interested in [imath]K = \{a + bk : a,b\in \mathbf{R} \}[/imath] under the operations + and *. We assume that it is a commutative additive group, with 0 + 0k the identity element, that multiplication distributes over addition and also that multiplication is associative ie (ab)c = a(bc). I actually needed that before as well, and it was rather foolish of me not to see it.

Another technicality is that I implicitly assume that [imath]rk = r\cdot k[/imath]. This seems obvious, but it does seem to be an underlying assumption so I should state it.

With those assumptions its possible to identify at least one other algebra. There may well be others too, but it comes down to a question whose answer I don't know.

Spoiler:
Addition is as before, as you'd expect. So we're left with the problem of defining multiplication. In general, a product will look like

(a+bk)*(c+dk) = ac + bkc + adk + bkdk

To get further, let [imath]k\cdot r = A(r) + B(r) k[/imath]. We need to start wrapping our head around what the functions A and B can look like.

[math]A(rs)+B(rs)k = k\cdot (rs) = (A(r) + B(r)k)\cdot s = A(r)s + B(r)A(s) + B(r)B(s)k[/math]

and

[math]A(r+s) + B(r+s)k = k\cdot(r+s) = k\cdot r + k\cdot s = A(r) + A(s) + (B(r) +B(s))k.[/math]

Equating coefficients, which we have enough to guarantee is okay, we have that

  1. A(r+s) = A(r)+A(s),
  2. A(rs) = A(r)s + B(r)A(s),
  3. B(r+s) = B(r)+B(s) and
  4. B(rs)=B(r)B(s).

3 and 4 mean that B is such that either B(r) = r for all r, or B(r) = 0 for all r. We can completely determine the outcome of assuming that B(r)=0 for all r, so we'll do that first.

Using, 2, we have [imath]A(r) = A(r\cdot 1) = r A(1).[/imath] So all of A is determined by A(1). Now consider
[math]\begin{align*}(-A(1) + k)(a+bk) &= -aA(1) - A(1)bk + ka + kbk \\
&= -aA(1) - A(1)bk + aA(1) + bA(1) k \\
&= 0\end{align*}[/math]
Regardless of what A(1) is, we always have some element k'= -A(1) +k such that [imath]k'z = 0[/imath] for all [imath]z\in K[/imath] and all these algebras are identical, and we might as well pick that instance where A(1) = 0 as the representative of this class. So, if we have K and state that [imath]k\cdot z = 0[/imath], then that defines a particular algebra. Its not very interesting though, imo anyway.

The interesting and unresolved case arrives when B(r) = r for all r. Using 1 we can prove that A(qr)=qA(r) if q is rational. Applying 2 to the case where r=s=1 we have A(1) = 2A(1), so A(1) = 0. From this we have that A(q)=0 for rational q. We could then assume continuity of A and get back to the commutative case, but that seems lazy to me.

We can simplify 2 and end up with A(rs) = rA(s) + sA(r). Now, this is more than a little weird, because it brings in the derivative, and its in fact pretty straight forward to prove that if p is a polynomial with rational coefficients, A(p(r)) = Dp(r) A(r), where Dp is the derivative of p. From this we can conclude that if a is algebraic, then A(a)=0. But from there I'm pretty stuck.


So, apart from the case where [imath]k\cdot 1 = 0[/imath] and things isomorphic to that, to resolve the issue you need to know what functions satisfy A(r+s) = A(r)+A(s) and A(rs) = rA(s)+sA(r).

Edit: So, A is a derivation apparently. More research...
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Re: Math: Fleeting Thoughts

Postby Eastwinn » Wed Jul 14, 2010 3:40 am UTC

Two things that are interesting:

The geometric mean of the natural numbers up to [imath]n[/imath], i.e. [imath]\sqrt[n]{n!}[/imath]

And then this beast, which is defined for [imath]x\; \epsilon \; C[/imath]...
[math]\lambda (x) = \sum_{k=1}^{\infty }\frac{x^{k}}{k^{k}}[/math]
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Re: Math: Fleeting Thoughts

Postby squareroot » Wed Jul 14, 2010 4:43 am UTC

Eastwinn wrote:Two things that are interesting:

The geometric mean of the natural numbers up to [imath]n[/imath], i.e. [imath]\sqrt[n]{n!}[/imath]

And then this beast, which is defined for [imath]x\; \epsilon \; C[/imath]...
[math]\lambda (x) = \sum_{k=1}^{\infty }\frac{x^{k}}{k^{k}}[/math]


Try taking the ratio of n and the geometric mean of the numbers up to n for large n... it's identities like that make me love math (and please, nobody explain exactly why to me - It'll just lose some of the magic that way. EDIT: Nvm, I learned why... )-: )

I don't understand what the second function is supposed to do, should it just grow very fast?
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Re: Math: Fleeting Thoughts

Postby Patashu » Wed Jul 14, 2010 7:48 am UTC

Hmmm, I'm finding it hard to prove what the cube roots of 1 in complex numbers are, outside of converting to modulus-argument form and using the obvious geometric interpretation - except I don't know what the geometric interpretation of a split-complex number is. Probably something hyperbolic, given what I remember of conics.
Dual numbers only have 1 as a cube root of 1, of course.

*looks it up*

Aah: exp(j*theta) = cosh(theta) + j * sinh(theta), according to wikipedia.

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Re: Math: Fleeting Thoughts

Postby jestingrabbit » Wed Jul 14, 2010 2:51 pm UTC

try equating real and imaginary parts in

1 = (a + bi)3 = a3 - 3ab2 + (b3 + 3a2b)i

and similarly for the split complex'. You should be able to work out b2 in terms of a2 using the imaginary part (and this has to be the same for both algebras). Sub that into the real part and see how simple (or not) what you get is.
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Re: Math: Fleeting Thoughts

Postby the tree » Wed Jul 14, 2010 5:52 pm UTC

If you're trying to prove rather than find, then just go through them, cube them all - and quote FTA to say that there aren't any more.

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Re: Math: Fleeting Thoughts

Postby Eastwinn » Wed Jul 14, 2010 8:43 pm UTC

squareroot wrote:I don't understand what the second function is supposed to do, should it just grow very fast?


It does have an interesting curve on the positive side, but all the action is on the negative side where it takes a big dive, come up a bit, and then slowly levels out. I don't know whether that leveling out is asymptotic or if it's more like a log, where it grows incredibly slowly.

Edit: Also Mathematica doesn't appear to know how it works.
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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Fri Jul 16, 2010 9:13 pm UTC

The Boy or Girl paradox revisited

Please don't hurt me! :)
You overhear a parent: "Well, I've got two kids, but John's gone to camp this week". What's the probability that the other kid left at home is a girl?

When I first saw this problem a day or so ago, I thought "Ah, the old Boy or Girl paradox that I read about in Martin Gardner's books when I was in high school. Easy!"

So I reasoned that the answer must be 2/3, i.e. the "Smith" case from the Wikipedia page, but now I'm not so sure. Firstly, there's the issue that both kids are named John. I'd like to reject this possibility, partly because of the parent's phrasing, but also on the grounds that it's a bad idea to import red herrings into simple probability / combinatorics puzzles. Secondly, there's the possibility that John is a girl, which I'd also like to reject, for the much the same reason.

But then there's the "gone to camp" v "left at home" distinction. I'm beginning to think that this is equivalent to the older v younger distinction, which would make the scenario equivalent to the "Jones" case from the Wikipedia, making the answer 1/2. But I'm not quite convinced.

So, is it really the "Jones" case in disguise, or is this yet another red herring? And am I justified in rejecting the other red herrings?

Help!

I didn't want to start a thread on this, considering the other thread I found on the classic version of the paradox was locked (for good reason, IMHO), but I'd be happy for it to move into its own thread if people think it merits one.

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Re: Math: Fleeting Thoughts

Postby Xanthir » Fri Jul 16, 2010 9:52 pm UTC

With the assumption that the other child is *not* at camp, then it falls into the Jones case, as one child holds a distinguished position. You just need to impose an ordering on the children, such that you know the ordinal of the kid who's gender was provided, to make it a Jones case.

If both children could plausibly be at camp, but the parent is just for some reason only mentioning the camp-ness of one of them, then you lose that ordering and return to the Smith case.
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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Sat Jul 17, 2010 10:25 pm UTC

Thanks, Xanthir! Much appreciated.

Edit:

In case you're wondering, this puzzle arose on another forum I frequent. The debate still rages, with some people totally not getting the Smith scenario, although most of the discussion now centres on the wording of the original question (which I quoted verbatim above). Some say it's ambiguous, or that the at camp / at home distinction isn't the same as the older / younger distinction.

I'm still (mostly) convinced that it is equivalent to the Jones scenario, and that comparing it to the case of tossing a gold coin & a sliver coin seems to be valid.

But I just had a troubling thought while thinking about the at camp / at home thing.

How do we tackle the following question?

Q: I have two children, Chris and Pat. One of them is at camp, one is at home.
What's the probability that Chris is the one at camp?

A1: There's not enough information to give a valid probability.
A2: We don't know how the decision was made to send one of the children to camp, so the best probability we can assign is 1/2.

As you might guess, I'm not particularly happy with either of these answers, or this state of affairs. But is my new question even relevant to the original puzzle?
:(

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Re: Math: Fleeting Thoughts

Postby Tirian » Sun Jul 18, 2010 12:34 am UTC

To me, the trouble with these questions is that when people volunteer information, you have to analyze why they offered that piece of information and not a different true statement.

Let me put it this way. Let's say that Susan took a math test and an English test today. Based on past data, you expect a 50% probability that she did well on either of them, and those are independent events. So you see Susan and have the following discussion:

You: How did you do on the tests?
Susan: I did well on the math test.

What is the probability that she did well on the English test? As a matter of psychology, I think you'd be naive to assume that it was still 50%, or else she'd have said that she did well on both. You have no formal evidence that the result of the English test is dependent on her statement, but the circumstantial evidence is heavy. But we'd be arguing that until the cows came home if we were of a mind to.

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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Sun Jul 18, 2010 2:38 am UTC

PM 2ring, the problem you are having is that you have not defined "probability." Presumably here you are asking which is more likely to be the case: That a person named A is going to camp, or that a person named B is going to camp. In this case, it is obvious that the two are equal. Assuming exactly one of the two scenarios is true, then clearly the probability of each must be 0.5.

However, if you are asking which of those two specific people is more likely to go to camp, then you do not know, because you do not know anything about them. However, given no more information than their name, the best guess you can make is assigning an equal probability to each outcome. This is the equivalent of the question, "Which is more likely, event A or event B?" Since you know nothing about the event, the best answer is, "I don't know, but supposing one is more likely than the other, it could just as easily be the one labeled A as the one labeled B, so they are equally likely given the information at hand."

In either case, given the information you have, the probabilities are 50/50. Probabilities are always calculated given a limited knowledge of the situation.

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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Sun Jul 18, 2010 3:07 am UTC

Tirian wrote:To me, the trouble with these questions is that when people volunteer information, you have to analyze why they offered that piece of information and not a different true statement.

Good point, Tirian, and thanks for you comments & example. This issue has been raised on the other forum in reference to the original question. My stance is that it's unknown why they offered that particular piece of information, so we can't let it bias our probability calculation.


Eebster the Great wrote:PM 2ring, the problem you are having is that you have not defined "probability."

Gosh! I didn't realize I was allowed to do that. :) * ponders the implications *

Eebster the Great wrote:Presumably here you are asking which is more likely to be the case: That a person named A is going to camp, or that a person named B is going to camp. In this case, it is obvious that the two are equal. Assuming exactly one of the two scenarios is true, then clearly the probability of each must be 0.5.

Ok, that sounds reasonable.

Eebster the Great wrote:However, if you are asking which of those two specific people is more likely to go to camp, then you do not know, because you do not know anything about them. However, given no more information than their name, the best guess you can make is assigning an equal probability to each outcome. This is the equivalent of the question, "Which is more likely, event A or event B?" Since you know nothing about the event, the best answer is, "I don't know, but supposing one is more likely than the other, it could just as easily be the one labeled A as the one labeled B, so they are equally likely given the information at hand."

In either case, given the information you have, the probabilities are 50/50.

I think I've covered those two viewpoints in my A1 and A2, haven't I?

Eebster the Great wrote:Probabilities are always calculated given a limited knowledge of the situation.

Thanks for reminding me of that, Eebster. I guess it's no different to a standard coin toss: with sufficiently accurate knowledge of the forces involved in the toss, we could use the laws of physics to predict the outcome before the coin comes to rest, but without that information we can't do better than assigning the usual 50:50 probability.

Of course, if we were looking at a fundamental quantum situation, eg a single radioactive decay event, then we can't even do the prediction using the laws of physics, but I don't want to open that can of worms at this stage. :)

Please keep the comments coming, folks. I feel that thinking about this stuff is giving me a deeper appreciation for how probability really works, and what it means. Or maybe I'm just confusing myself by over-thinking. And under-sleeping. :)

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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Mon Jul 19, 2010 10:39 pm UTC

Well quantum mechanics do not use the same statistics as as classical probability, so it isn't exactly equivalent. But you are correct that quantum events exhibit true randomness.

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Re: Math: Fleeting Thoughts

Postby vookaloop » Fri Jul 23, 2010 12:19 am UTC

afarnen wrote:
Bassoon wrote:It hasn't been said yet, but I hate it when people say "timesing" or "minusing" instead of "multiplying" or "subtracting". But it's overly ridiculous when people say "plussing" instead of "adding". It makes me want to scream.

Although I have heard people use "minus" as a verb, I don't think I've heard it used in gerund form. I also don't believe I've heard "times" or "plus" as verbs at all. Interesting...



Here the *ing is being used as a verb which makes it the present participle. A gerund would be the noun form of the verb with *ing.

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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Fri Jul 23, 2010 1:57 am UTC

vookaloop wrote:
afarnen wrote:
Bassoon wrote:It hasn't been said yet, but I hate it when people say "timesing" or "minusing" instead of "multiplying" or "subtracting". But it's overly ridiculous when people say "plussing" instead of "adding". It makes me want to scream.

Although I have heard people use "minus" as a verb, I don't think I've heard it used in gerund form. I also don't believe I've heard "times" or "plus" as verbs at all. Interesting...



Here the -ing form is being used as an adjective which makes it the present participle. A gerund would be the noun form of the verb with -ing.


Fix'd

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Re: Math: Fleeting Thoughts

Postby vookaloop » Fri Jul 23, 2010 12:21 pm UTC

Eebster the Great wrote: an adjective


If you are saying "I am timesing these two numbers together" it would be a present participle verb. If used in the sense of an adjective, i.e. "I have to do these timesing problems" then it would be a gerund (noun) acting in an adjective function as a premodifier of a noun. Considering the analog was "multiplying" I would assume verb but that is just my interpretation. Anyway I'll stop because this is math thoughts, just had to respond cause of the altered text.

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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Sat Jul 24, 2010 3:40 am UTC

vookaloop wrote:
Eebster the Great wrote: an adjective


If you are saying "I am timesing these two numbers together" it would be a present participle verb. If used in the sense of an adjective, i.e. "I have to do these timesing problems" then it would be a gerund (noun) acting in an adjective function as a premodifier of a noun. Considering the analog was "multiplying" I would assume verb but that is just my interpretation. Anyway I'll stop because this is math thoughts, just had to respond cause of the altered text.


False. "I am verbing" is the present continuous, which does use the present participle in its construction, but as a whole it is a finite verb (specifically, one in the indicative mood, active voice). Notice that this construction is not inconsistent with the participle being an adjective, since the subject is being described as "verbing." That is to say, if I say "I am working," then I am describing myself as "working," which can alternatively be seen as a predicate nominative.

If on the other hand I say, "I have to do these timesing problems," you are correct that it is a gerund, since it is describing the problems as "involving multiplication," but if instead I say, "The running man was next to the timesing woman," I mean that a man is running next to a woman who is performing multiplication. Both are adjectives, but the former is a gerund and the latter is a participle. An example of the gerund used as a noun might be, "I hate timesing," indicating my dislike of multiplication itself. This is similar to the infinitive, "I hate to times."

At least, that's the way I see it. It gets more complicated if we want to consider gerundives and periphrasis.

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Re: Math: Fleeting Thoughts

Postby letterX » Sat Jul 24, 2010 4:42 am UTC

Make it stop... please never say that word again! It makes my ears bleed.

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Re: Math: Fleeting Thoughts

Postby vookaloop » Mon Jul 26, 2010 3:21 am UTC

Eebster the Great wrote:False. "I am verbing" is the present continuous, which does use the present participle in its construction, but as a whole it is a finite verb (specifically, one in the indicative mood, active voice). Notice that this construction is not inconsistent with the participle being an adjective, since the subject is being described as "verbing." That is to say, if I say "I am working," then I am describing myself as "working," which can alternatively be seen as a predicate nominative.


Interesting. Thanks for the explanation.

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Re: Math: Fleeting Thoughts

Postby Macbi » Sun Aug 08, 2010 7:13 pm UTC

MFT: In science, "model" has exactly the opposite meaning to "model" in mathematics.
One of them is a set of axioms based on a phenomenon they describe, the other is a construct obeying a set of axioms.
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Re: Math: Fleeting Thoughts

Postby skeptical scientist » Wed Aug 18, 2010 8:48 am UTC

Why is the adjective form of "arithmetic" spelled "arithmetic" but pronounced "air-ith-maa-tic"?

Meta-fleeting-thought: is this a mathematics fleeting thought, or a linguistics fleeting thought?

Macbi wrote:MFT: In science, "model" has exactly the opposite meaning to "model" in mathematics.
One of them is a set of axioms based on a phenomenon they describe, the other is a construct obeying a set of axioms.

That's true; I never thought of it that way. Funny.
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Re: Math: Fleeting Thoughts

Postby Xanthir » Thu Aug 19, 2010 7:19 am UTC

skeptical scientist wrote:Why is the adjective form of "arithmetic" spelled "arithmetic" but pronounced "air-ith-maa-tic"?

Clearly your spelling is broken. Spell it with an A. Problem solved.
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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Fri Aug 20, 2010 4:04 am UTC

Xanthir wrote:
skeptical scientist wrote:Why is the adjective form of "arithmetic" spelled "arithmetic" but pronounced "air-ith-maa-tic"?

Clearly your spelling is broken. Spell it with an A. Problem solved.

Well spelling is far more standardized than pronunciation, so that seems to be pretty backward.

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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Fri Aug 20, 2010 3:14 pm UTC

skeptical scientist wrote:Why is the adjective form of "arithmetic" spelled "arithmetic" but pronounced "air-ith-maa-tic"?

Wah? In my dialect, the noun is əRITHmətic, the adjective is ArithMETic.

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Re: Math: Fleeting Thoughts

Postby Macbi » Mon Aug 23, 2010 2:24 pm UTC

What's the smallest molecule which is topologically non-planar (where it's atoms are nodes and its bonds are edges)?
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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Mon Aug 23, 2010 2:50 pm UTC

Macbi wrote:What's the smallest molecule which is topologically non-planar (where it's atoms are nodes and its bonds are edges)?
Ammonia? (Shouldn't this be in Science rather than Mathematics?)

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Re: Math: Fleeting Thoughts

Postby letterX » Mon Aug 23, 2010 3:05 pm UTC

Because there's no Science: Fleeting Thoughts thread?

Also, Ammonia is non-planar when you look at it in its physical configuration, but he was asking about non-planarity as a graph property. And, after thinking for several minutes, I don't actually know any non-planar molecules. Even something like a fullerene is planar when you blow up one of the faces to be the exterior face, and arrange everything else inside it. But then, I know very little about chemistry.

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Re: Math: Fleeting Thoughts

Postby PM 2Ring » Mon Aug 23, 2010 4:32 pm UTC

but he was asking about non-planarity as a graph property.
Ah. Good point.

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Re: Math: Fleeting Thoughts

Postby letterX » Mon Aug 23, 2010 4:51 pm UTC

Indeed. Even your avatar is planar (if indeed it even is a molecule). If anyone had any examples, I would really like to see them. A molecule with K5 or K3,3 as a minor would be really interesting.

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Re: Math: Fleeting Thoughts

Postby Macbi » Mon Aug 23, 2010 5:07 pm UTC

I found lots in this paper (pdf), but the smallest one (the Simmons-Paquette molecule, right at the bottom) still uses 35 atoms.
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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Tue Aug 24, 2010 5:36 am UTC

Macbi wrote:I found lots in this paper (pdf), but the smallest one (the Simmons-Paquette molecule, right at the bottom) still uses 35 atoms.

There is also a three-rung molecular Mobius ladder (at least 60 atoms, but not all such molecules are intrinsically chiral according to this link) and a molecular trefoil knot (much larger). Some even occur naturally, such as certain proteins or circular DNA strands (one book lists the result of the action of enzyme TN3 Resolvase on circular DNA).

Here is a book on the subject.

This article suggests that Au34 nanoclusters have intrinsic chirality, one less atom than the Simmons-Paquette molecule (if you can call nanoclusters "molecules").

Unfortunately, I can't find any source on the smallest synthesized intrinsically chiral molecule.

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Re: Math: Fleeting Thoughts

Postby Blatm » Sat Aug 28, 2010 12:45 pm UTC

skeptical scientist wrote:Why is the adjective form of "arithmetic" spelled "arithmetic" but pronounced "air-ith-maa-tic"?


This has always bothered me.

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Re: Math: Fleeting Thoughts

Postby The Scyphozoa » Mon Aug 30, 2010 2:35 am UTC

Same as "address", noun and verb, I guess. (AD-dress vs. ad-DRESS)
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Eebster the Great
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Re: Math: Fleeting Thoughts

Postby Eebster the Great » Mon Aug 30, 2010 5:09 pm UTC

The Scyphozoa wrote:Same as "address", noun and verb, I guess. (AD-dress vs. ad-DRESS)

I think he's objecting to /ɛ/ being realized as [ɛɪ], but I don't think that's the only time that happens in English. I can't think of another example off the top of my head.

I also think the pronunciation varies with the region.

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Eastwinn
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Re: Math: Fleeting Thoughts

Postby Eastwinn » Mon Aug 30, 2010 6:44 pm UTC

I was trying to find a way to expand (2n)! without using [imath]\pi[/imath] but was unsuccessful :| .
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Re: Math: Fleeting Thoughts

Postby squareroot » Tue Aug 31, 2010 3:57 am UTC

Eastwinn wrote:I was trying to find a way to expand (2n)! without using [imath]\pi[/imath] but was unsuccessful :| .

Funny, I've been toying around with (2n choose n) as (n goes to infinity) for a bit... at one point, I thought it went to c*4^n, and I thought I had c, but then I noticed a flaw... I'm thinking it's something on the order of (4^n)/n now, but again I'm not sure.

EDIT: Wolfram-Alpha, with a bit of guess and check, told me it's sqrt(pi)*(4^n)/sqrt(n). Hmm.
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Re: Math: Fleeting Thoughts

Postby letterX » Tue Aug 31, 2010 5:45 am UTC


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Talith
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Re: Math: Fleeting Thoughts

Postby Talith » Tue Aug 31, 2010 6:26 pm UTC

Just need to get these thoughts written down cos it's bugging me:

Suppose we have an m x n rectangle where m and n are integers and the rectangle is partitioned itself into integer edge-length rectangles (can be any size and position as long as they partition the large rectangle into at least two smaller rectangles). Call this partition of the rectangle [imath]R_0[/imath] and call [imath]R_{i+1}[/imath] a refinement of [imath]R_i[/imath] if it can be formed by replacing a proper subset of the 'tiles' with their union in the obvious way (this union must itself be a rectangle).

Intuitively we're starting off with a set of rectangles and then rubbing out lines on the partition so that we still have a partition (and never rub out every line).

Call [imath]R_k[/imath] a total refinement if [imath]R_k[/imath] is the partition of the m x n rectangle in to exactly two other rectangles (and so cannot be further refined). Call [imath]R_k[/imath] a non-trivial refinement if [imath]R_k[/imath] is a partition of the m x n rectangle in to more than two rectangles and no further refinement exists (this is why the proper subset property was needed for refinements).

Conjecture; Given [imath]R_0[/imath] and it is known that [imath]R_k[/imath] is a non-trivial refinement, no other distinct non-trivial refinement exists. That is, [imath]R_{k'}=R_k[/imath].

Ok, I knew writing it down would help. A counter example exists: a 2x2 square with a 'border' of 1x1 squares has at least 2 distinct non-trivial refinements (mirror images of each other) and a few more that aren't mirror images.

Ok, new conjecture; Any [imath]R_0[/imath] for which more than one non-trivial refinement exists can also be totally refined.

I'll have a think about this one as a counter-example seems a little bit more out of reach for this one. Yay for fleeting thoughts. Feel free to contribute guys, I thought it was a pretty nice construction of a problem.


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