[math-fun] Systems of continued fractions
Given a finite continued fraction, you can start at the last term and build up the cf by | b d c ab + dc | - -> a + --- = ------- | c b b where 'b/c' is the current fraction, 'a' is the current term, and 'd' is usually 1. This operation is nicely represented with a 2x2 matrix: | [ a d ] [ b ] = [ ab + dc ] | [ 1 0 ] [ c ] [ b ] I'm looking at a situation where I have multiple fractions: | b0 b1 b2 | --, --, -- | c0 c1 c2 and an update rule that takes these to | d00 c0 d01 c1 d02 c2 | a0 + ------ + ------ + ------, | b0 b1 b2 | | d10 c0 d11 c1 d12 c2 | a1 + ------ + ------ + ------, | b0 b1 b2 | | d20 c0 d21 c1 d22 c2 | a2 + ------ + ------ + ------, | b0 b1 b2 Is there a way to do this case with matrices? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
Reposting, since I never saw it on the list and got no replies... On Fri, Aug 28, 2009 at 9:28 AM, Mike Stay<metaweta@gmail.com> wrote:
Given a finite continued fraction, you can start at the last term and build up the cf by
| b d c ab + dc | - -> a + --- = ------- | c b b
where 'b/c' is the current fraction, 'a' is the current term, and 'd' is usually 1. This operation is nicely represented with a 2x2 matrix:
| [ a d ] [ b ] = [ ab + dc ] | [ 1 0 ] [ c ] [ b ]
I'm looking at a situation where I have multiple fractions:
| b0 b1 b2 | --, --, -- | c0 c1 c2
and an update rule that takes these to
| d00 c0 d01 c1 d02 c2 | a0 + ------ + ------ + ------, | b0 b1 b2 | | d10 c0 d11 c1 d12 c2 | a1 + ------ + ------ + ------, | b0 b1 b2 | | d20 c0 d21 c1 d22 c2 | a2 + ------ + ------ + ------, | b0 b1 b2
Is there a way to do this case with matrices? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
Looks like we're the only two left, Mike --- did we miss Armageddon? WFL On 8/31/09, Mike Stay <metaweta@gmail.com> wrote:
Reposting, since I never saw it on the list and got no replies...
On Fri, Aug 28, 2009 at 9:28 AM, Mike Stay<metaweta@gmail.com> wrote:
Given a finite continued fraction, you can start at the last term and build up the cf by
| b d c ab + dc | - -> a + --- = ------- | c b b
where 'b/c' is the current fraction, 'a' is the current term, and 'd' is usually 1. This operation is nicely represented with a 2x2 matrix:
| [ a d ] [ b ] = [ ab + dc ] | [ 1 0 ] [ c ] [ b ]
I'm looking at a situation where I have multiple fractions:
| b0 b1 b2 | --, --, -- | c0 c1 c2
and an update rule that takes these to
| d00 c0 d01 c1 d02 c2 | a0 + ------ + ------ + ------, | b0 b1 b2 | | d10 c0 d11 c1 d12 c2 | a1 + ------ + ------ + ------, | b0 b1 b2 | | d20 c0 d21 c1 d22 c2 | a2 + ------ + ------ + ------, | b0 b1 b2
Is there a way to do this case with matrices? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
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I expected Gosper to step in, since it's his area. Doesn't the obvious generalization to 6x6 matrices work? Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Monday, August 31, 2009 4:25 PM To: math-fun Subject: Re: [math-fun] Systems of continued fractions Looks like we're the only two left, Mike --- did we miss Armageddon? WFL On 8/31/09, Mike Stay <metaweta@gmail.com> wrote:
Reposting, since I never saw it on the list and got no replies...
On Fri, Aug 28, 2009 at 9:28 AM, Mike Stay<metaweta@gmail.com> wrote:
Given a finite continued fraction, you can start at the last term and > build up the cf by > | b d c ab + dc | - -> a + --- = ------- | c b b
where 'b/c' is the current fraction, 'a' is the current term, and 'd' is usually 1. This operation is nicely represented with a 2x2 matrix:
| [ a d ] [ b ] = [ ab + dc ] | [ 1 0 ] [ c ] [ b ]
I'm looking at a situation where I have multiple fractions:
| b0 b1 b2 | --, --, -- | c0 c1 c2
and an update rule that takes these to > | d00 c0 d01 c1 d02 c2 | a0 + ------ + ------ + ------, | b0 b1 b2 | | d10 c0 d11 c1 d12 c2 | a1 + ------ + ------ + ------, | b0 b1 b2 | | d20 c0 d21 c1 d22 c2 | a2 + ------ + ------ + ------, | b0 b1 b2
Is there a way to do this case with matrices? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
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On Tue, Sep 1, 2009 at 8:18 AM, Schroeppel, Richard<rschroe@sandia.gov> wrote:
I expected Gosper to step in, since it's his area. Doesn't the obvious generalization to 6x6 matrices work?
Rich
Um. It's not obvious to me--how would that work? In ab + dc, it's a sum of terms with two variables, so it's a product of two matrices. In the example system of fractions, you get a0 b0 b1 b2 + d00 c0 b1 b2 + d01 c1 b0 b2 + d02 c2 b0 b1 as a numerator, so I thought it may be the case that you have to multiply four matrices, and I'm not sure how that would work either.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Monday, August 31, 2009 4:25 PM To: math-fun Subject: Re: [math-fun] Systems of continued fractions
Looks like we're the only two left, Mike --- did we miss Armageddon? WFL
On 8/31/09, Mike Stay <metaweta@gmail.com> wrote:
Reposting, since I never saw it on the list and got no replies...
On Fri, Aug 28, 2009 at 9:28 AM, Mike Stay<metaweta@gmail.com> wrote: > Given a finite continued fraction, you can start at the last term and > build up the cf by > > | b d c ab + dc > | - -> a + --- = ------- > | c b b > > where 'b/c' is the current fraction, 'a' is the current term, and 'd' > is usually 1. This operation is nicely represented with a 2x2 matrix: > > | [ a d ] [ b ] = [ ab + dc ] > | [ 1 0 ] [ c ] [ b ] > > I'm looking at a situation where I have multiple fractions: > > | b0 b1 b2 > | --, --, -- > | c0 c1 c2 > > and an update rule that takes these to > > | d00 c0 d01 c1 d02 c2 > | a0 + ------ + ------ + ------, > | b0 b1 b2 > | > | d10 c0 d11 c1 d12 c2 > | a1 + ------ + ------ + ------, > | b0 b1 b2 > | > | d20 c0 d21 c1 d22 c2 > | a2 + ------ + ------ + ------, > | b0 b1 b2 > > Is there a way to do this case with matrices? > -- > Mike Stay - metaweta@gmail.com > http://math.ucr.edu/~mike > http://reperiendi.wordpress.com >
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
On Aug 31, 2009, at 6:25 PM, Fred lunnon wrote:
Looks like we're the only two left, Mike --- did we miss Armageddon? WFL
Yes, and it's called "start-of-the-semester". But here's something for torus fans: Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus? If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set z(s,t) = x(s) + y(t), (s,t) in [0,1]^2. Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ?
Oh, man, I cannot rigorize this at all, but I have a *strong* intuition that the answer is "no". It really feels like there is a simple topological fixed-point argument that would make it clear that this is impossible. Maybe start by finding a vector around which the winding number of x(s) is 1; then look at the vector y'(t) dotted with that vector. Surely this must change sign at least once. Here I trail off. On Tue, Sep 1, 2009 at 12:23 PM, Veit Elser <ve10@cornell.edu> wrote:
On Aug 31, 2009, at 6:25 PM, Fred lunnon wrote:
Looks like we're the only two left, Mike --- did we miss Armageddon? WFL
Yes, and it's called "start-of-the-semester".
But here's something for torus fans:
Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus?
If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set
z(s,t) = x(s) + y(t), (s,t) in [0,1]^2.
Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Wed, Sep 2, 2009 at 05:45, Allan Wechsler<acwacw@gmail.com> wrote:
Oh, man, I cannot rigorize this at all, but I have a *strong* intuition that the answer is "no". It really feels like there is a simple topological fixed-point argument that would make it clear that this is impossible.
Do degenerate solutions count? If the x(s) and y(t) surfaces are flat and coplanar, the Minkowski sum is also flat. For instance if x(s) and y(t) are unit circles in the (x,y) plane, then z(s,t) is the disk with radius 2. Cheers, Seb
On Tue, Sep 1, 2009 at 12:23 PM, Veit Elser <ve10@cornell.edu> wrote:
Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus?
If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set
z(s,t) = x(s) + y(t), (s,t) in [0,1]^2.
Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ?
On Wed, Sep 2, 2009 at 6:05 AM, Seb Perez-D <sbprzd+mathfun@gmail.com<sbprzd%2Bmathfun@gmail.com>
wrote:
On Wed, Sep 2, 2009 at 05:45, Allan Wechsler<acwacw@gmail.com> wrote:
Oh, man, I cannot rigorize this at all, but I have a *strong* intuition that the answer is "no". It really feels like there is a simple topological fixed-point argument that would make it clear that this is impossible.
Do degenerate solutions count?
If the x(s) and y(t) surfaces are flat and coplanar, the Minkowski sum is also flat. For instance if x(s) and y(t) are unit circles in the (x,y) plane, then z(s,t) is the disk with radius 2.
The question was whether the sum could be a smooth torus, not a smooth manifold with boundary. So this doesn't count. My intuition is no, but I haven't found a proof yet.
Cheers,
Seb
On Tue, Sep 1, 2009 at 12:23 PM, Veit Elser <ve10@cornell.edu> wrote:
Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus?
If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set
z(s,t) = x(s) + y(t), (s,t) in [0,1]^2.
Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ?
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-- Andy.Latto@pobox.com
On Sep 2, 2009, at 7:53 AM, Andy Latto wrote:
The question was whether the sum could be a smooth torus, not a smooth manifold with boundary. So this doesn't count.
My intuition is no, but I haven't found a proof yet.
Maybe this way of stating the problem helps: Consider two smooth maps from S^1 to R^3 given by s --> x(s) t --> y(t) Question: can the map S^1 x S^1 to R^3 given by (s,t) --> x(s) + y(t) be smooth? Veit
On Wed, Sep 2, 2009 at 8:40 AM, Veit Elser <ve10@cornell.edu> wrote:
On Sep 2, 2009, at 7:53 AM, Andy Latto wrote:
The question was whether the sum could be a smooth torus, not a smooth
manifold with boundary. So this doesn't count.
My intuition is no, but I haven't found a proof yet.
Maybe this way of stating the problem helps:
Consider two smooth maps from S^1 to R^3 given by
s --> x(s) t --> y(t)
Question: can the map S^1 x S^1 to R^3 given by
(s,t) --> x(s) + y(t)
be smooth?
I believe that this is actually a different question, and that the answer to this question is "yes", while the answer to the original question is "no"! That is, the two curves in R^3 give a map from the torus to R^3. I believe that this map can be an immersion (that is, smooth, but not 1-1, so the torus in R3 is smooth, but self-intersects) but not an imbedding (a 1-1 smooth map, so that the image 'looks like' a torus). How to choose x and y to get an immersion: A smooth curve x give us a smooth map s-> x'(s)/|x'(s)| from S^1 to S^2. This map must intersect every closed hemisphere in S^2; otherwise, the original space curve is always going "up" and never "down", so it can't close up. Further, unless it is a great circle, it touches every open hemisphere of S^2, for the same reason. I believe that any smooth map from S1 to S2 that satisfies these constraints comes from a closed space curve (hand-waving argument: if it doesn't close up, see what direction it fails to close up in, and increase the magnitude of x' when it has a component in that direction). The map from the torus to R3 fails to be smooth exactly when x'(s) is a multiple of y'(s). So to immerse the torus smoothly, we need two smooth closed curves on the sphere, each of which touches every open hemisphere. Take a sin wave around the equator, and move it slightly up and slightly down (but by an amount less than the amplitude) to get two such curves. I don't have a proof yet that the immersion has to be an imbedding. But you can think of the process of making the torus as taking the curve x(t), and "moving it around" along the path y(t) to get a surface, which is a closed surface when you finish because y(t) is closed. When you move a circle around in a circle to make a torus in the conventional way, you "turn the circle around" as you go, and this is necessary to avoid self-intersection. The definition of "turn around as you go" involves some induced map to O(3), and the loop involved has to be of the non-identity homotopy class. And the minkowski sum construction will give you something in the identity class. But I don't have the details worked out yet. Andy
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participants (7)
-
Allan Wechsler -
Andy Latto -
Fred lunnon -
Mike Stay -
Schroeppel, Richard -
Seb Perez-D -
Veit Elser