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Author Topic: Mathematics Help Thread  (Read 229105 times)

Simmura McCrea

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Re: Mathematics Help Thread
« Reply #840 on: October 08, 2011, 03:00:42 pm »

Oh. Hurr fucking durr.

*Simmura goes to find something to hit his head against.*
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ed boy

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Re: Mathematics Help Thread
« Reply #841 on: October 14, 2011, 05:57:16 am »

I was discussing a potential problem with a physicist, and there is a bit of a question that I have, about modular number systems.

Consider a modular system of integers. That is we can consider the integers mod n to be {0,1,2,...,n-1}. Addition and multiplication can be defined just fine, as can the additive inverse, subtraction. But what about multiplicative inverse (division)?

It is certainly possible to define fractions for some modular integer systems. For example, 1/2 is defined as 'the number such that, when multiplied by 2, equals 1'. In integers mod 7, we have that 1/2=4, as 4 multiplied by two equals 8=1. However, in the integers mod 8, 1/2 cannot exist, as it would requide double an integer to be equal to nine.

When then brings us onto the main question. What about a modular system of real numbers? I've only ever seen modular systems of integers before, never seen people work with modular real numbers. There do exist systems that people work with that are modular, and can accept real numbers (such as eix), but working with modular real numbers (that is, a half-open interval [0,k)) I have not seen done.

Certainly, over the real numbers, it would be possible to define addition, multiplication, and subtraction, but what about division?

The real question which I want to answer, it is it possible to define a vector field over such a space? A vector field requies modular reals to be a field, which requires multiplicative inverses to exist.

EDIT: I have had a last minute brainwave about this. It turns out that division can only exist on the interval (k-1,k). However, what is to stop us modifying our definition of addition so the new addition has an additive identity of k-1? We would then have a field that is analogous to [0,k-k-1), or have I gone wrong somewhere?
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Another

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Re: Mathematics Help Thread
« Reply #842 on: October 14, 2011, 09:56:58 am »

I am not a specialist in this and it is a wild guess but what about modifying the definition of "mod k" operation so that for e>=0:
k+e=k-1+e mod* k,
k-1-e=k-e mod* k?
(0-e=k-1-(k-1+e)=k-(k-1+e) mod*k,
0+e=k-1-(k-1-e)=k-(k-1-e) mod*k)
Applied iteratively until the number lands into (k-1,k) or 0  it would be kind of an asymmetric "mod* k" operation. There could be a complication with 0=k-k-1 mod*k that would require a set of additional rules for some additive operations and it may cause this construction to fail being something like a field. Some nice properties of the usual "mod k" would probably be lost. Probably for large enough k some properties of such construction would become quite close to [0, +inf) for reals and some properties will be really unusual for k close to 1.

What may be related to physics is that for any fixed k some integrals will probably not diverge (built-in regularization). Or maybe not.
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Christes

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Re: Mathematics Help Thread
« Reply #843 on: October 14, 2011, 03:47:27 pm »

Short answer: You certainly can't get a field, and probably can't get a ring.

I'm not sure how much modern algebra everyone has had, but the way we usually construct the integers mod n is by defining an equivalence relation on the integers where a~b precisely when a=b+kn from some integer k.  Taking equivalence classes gives the integers mod n.

Try the same construction on the reals:

Fix a real number r.  Set a~b when a=b+kr for some integer k (you want an integer here - if you let k be any real number, everything is equal trivially).  Now take equivalence classes.  You basically get a set that looks like [0,r).  So far, so good.

Now we want to define addition and multiplication the same way as for the integers mod n.  (I'd be happy if we get any ring at all, let alone a field)  Here's where we start having problems.  You can define addition easily enough.  (As an abelian group, it's just a copy of the unit circle in the complex plane)

But how to define multiplication?  Well, if you want to do it the same way as we did it for integers mod n, you're going to have trouble.  I'm pretty sure there is no way to define multiplication here that is consistent with multiplication on the reals.

e.g. Let r = 3/2.  We have 2~1/2,   2*2=4~1.  But clearly 1/2*1/2=1/4 and 1/4 is not equivalent to 1.

What we're really trying to do here is mod out the ring R by the subset {kr| k an integer}.  This is not an ideal so you don't get a ring.  (It *is* an abelian subgroup, so you do get an abelian group)

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EDIT: I have had a last minute brainwave about this. It turns out that division can only exist on the interval (k-1,k). However, what is to stop us modifying our definition of addition so the new addition has an additive identity of k-1? We would then have a field that is analogous to [0,k-k-1), or have I gone wrong somewhere?

I'm not entirely certain what you mean, but you're probably going to really break the distributive property doing that.  Also, it's a theorem that for any ring, the additive identity times anything is the additive identity.
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k+e=k-1+e mod* k,
k-1-e=k-e mod* k?
(0-e=k-1-(k-1+e)=k-(k-1+e) mod*k,
0+e=k-1-(k-1-e)=k-(k-1-e) mod*k)

I'm very confused by what's going on here.  Are you just adding more rules to the typical "mod"?
« Last Edit: October 14, 2011, 06:06:43 pm by Christes »
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Another

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Re: Mathematics Help Thread
« Reply #844 on: October 14, 2011, 04:58:49 pm »

Quote
k+e=k-1+e mod* k,
k-1-e=k-e mod* k?
(0-e=k-1-(k-1+e)=k-(k-1+e) mod*k,
0+e=k-1-(k-1-e)=k-(k-1-e) mod*k)

I'm very confused by what's going on here.  Are you just adding more rules to the typical "mod"?
Exactly. And I am not sure that either this modification of "mod" could be of any practical use as common "mod" is or that even a division ring could be constructed with it and some additional rules.
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Christes

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Re: Mathematics Help Thread
« Reply #845 on: October 14, 2011, 06:02:54 pm »

Well, is this occurring in the reals (or any field for that matter)?  If I'm understanding you correctly, I think you'll have trouble even making a ring for reasons I stated above.

But even if you did, it would almost certainly have zero divisors since for any nonzero real a, you can always find some b so that ab=k.
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PsyberianHusky

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Re: Mathematics Help Thread
« Reply #846 on: October 28, 2011, 05:48:53 pm »

Anyone comfortable with stats, right now I more or less understand what I am doing with a "t-test" but I am unsure of what to do with a "z-test" according to the problem a mean, a standard deviation and a population size.

Mostly I am looking for an explanation that is more in English then mathematics notation.
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Thank you based dwarf.

Dutchling

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Re: Mathematics Help Thread
« Reply #847 on: November 02, 2011, 02:00:46 pm »

So.
I have a math test tomorrow about proving stuff and I think I understand most of it. 'Most' does not include this one:
Spoiler (click to show/hide)
Don't  mind my paint art :P the pic I made wasn't very visible..
« Last Edit: November 02, 2011, 05:52:49 pm by Dutchling »
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Tellemurius

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Re: Mathematics Help Thread
« Reply #848 on: November 02, 2011, 02:07:19 pm »

Well if P is 90 degrees you should be about to use SOH by using AP

ILikePie

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Re: Mathematics Help Thread
« Reply #849 on: November 02, 2011, 02:13:54 pm »

cos 60 = 5/AS
1/2 = 5/AS
AS = 10
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Dutchling

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Re: Mathematics Help Thread
« Reply #850 on: November 02, 2011, 02:14:43 pm »

I don't think that qualifies as exact.

fakedit: but that does! Thanks.

edit: The angle 30 degrees though, the total is 60 degrees. Although it should still work correctly.
« Last Edit: November 02, 2011, 02:16:50 pm by Dutchling »
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Urist Imiknorris

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Re: Mathematics Help Thread
« Reply #851 on: November 02, 2011, 02:19:06 pm »

Then it's cos 30 = AS/5 = √3/2.
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ed boy

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Re: Mathematics Help Thread
« Reply #852 on: November 04, 2011, 11:32:29 am »

Here's a question. I feel that it should have a relatively straightforward answer, but I can't seem to find it by anything other than trial and error. Is there a more efficient solution?

The problem is as such:

We are working in a two-dimensional plane. There is a target circle of radius r>0, with a centre at (0,0). You can place as many circles as you want in the plane, subject to the restriction that no two circles (including the target circle) can overlap. The circles you place can be of any size, and they do not have to be the same size. You want to place circles so that any straight line with an endpoint in the target circle passes through at least one other circle. You want to do this with the minimum number of circles.
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Another

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Re: Mathematics Help Thread
« Reply #853 on: November 04, 2011, 12:47:24 pm »

I can think of a simple asymmetrical pattern of 3 giant (relative to r) almost touching circles that would satisfy the requirements and a trivial proof that it can't be done with 2 circles. Do you want me to elaborate?
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Christes

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Re: Mathematics Help Thread
« Reply #854 on: November 04, 2011, 02:04:49 pm »

Here's a question. I feel that it should have a relatively straightforward answer, but I can't seem to find it by anything other than trial and error. Is there a more efficient solution?

The problem is as such:

We are working in a two-dimensional plane. There is a target circle of radius r>0, with a centre at (0,0). You can place as many circles as you want in the plane, subject to the restriction that no two circles (including the target circle) can overlap. The circles you place can be of any size, and they do not have to be the same size. You want to place circles so that any straight line with an endpoint in the target circle passes through at least one other circle. You want to do this with the minimum number of circles.

Some pedantic questions, because they really do make a difference:

When you say circle, do you mean the interior, boundary, or both?  ("Circle" usually means just the boundary, but the phrase "in the target circle" makes me wonder)

Also, do tangent circles count as overlapping?

If the answer to 1 is either boundary or both and the answer to 2 is no, then something like what the previous poster said should work, I think.
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