[math-fun] Chebfun: Taylor series are obsolete
If you're not aware of the "Chebfun" package developed by Trefethen, et al, at Oxford U., you will be pleasantly surprised. Basically, Chebfun approximates smooth curves over a finite interval to within machine precision. Chebfun also supports piecewise smooth curves by separately Chebbing each piece. Since different smooth curves require different numbers of terms to converge, some of these sums can have 1000 or more terms. As pointed out in the slides & video below, the whole point of this is to provide the flavor of symbolic computation with the efficiency and ease of numeric computation. For example, plotting, derivatives, integrals, root-finding are all relatively easy in this framework. The major success of Chebfun's is in ODE's, both linear & non-linear. It's a pity that Chebfun only works on Matlab; perhaps some student could be motivated to do this in Maxima. After watching the video below, I started to wonder why we bother teaching Taylor series at all. Here's a 1-hour tutorial about Chebfun (850 MBytes): http://downloads.sms.cam.ac.uk/1160831/1160835.mp4 Here are the slides for this talk (1 MByte): http://people.maths.ox.ac.uk/trefethen/chebfun.pdf Some history of Cheb approximations http://people.maths.ox.ac.uk/trefethen/cftalk.pdf
Some issues are raised by the slides, one of which lists differentiation as a built-in functional. Maybe somebody else can clarify these for me? Differentiation is numerically unstable; therefore the original function will have to be recomputed at increasing precision in order to maintain the prescribed error in higher derivatives. As far as I can see, simply increasing the number n of interpolating points is insufficient to remedy the problem, whatever approximations (Taylor, Chebychev or whatever) are employed. Furthermore allowing n to vary dynamically seems to imply that the user must express his function f as an infinite Chebychev series (or maybe combination of built-in functions). If for example he possesses only a functional equation for f, it's unclear how this can be achieved within the framework supplied. Fred Lunnon On 8/25/12, Henry Baker <hbaker1@pipeline.com> wrote:
If you're not aware of the "Chebfun" package developed by Trefethen, et al, at Oxford U., you will be pleasantly surprised.
Basically, Chebfun approximates smooth curves over a finite interval to within machine precision. Chebfun also supports piecewise smooth curves by separately Chebbing each piece.
Since different smooth curves require different numbers of terms to converge, some of these sums can have 1000 or more terms.
As pointed out in the slides & video below, the whole point of this is to provide the flavor of symbolic computation with the efficiency and ease of numeric computation. For example, plotting, derivatives, integrals, root-finding are all relatively easy in this framework.
The major success of Chebfun's is in ODE's, both linear & non-linear.
It's a pity that Chebfun only works on Matlab; perhaps some student could be motivated to do this in Maxima.
After watching the video below, I started to wonder why we bother teaching Taylor series at all.
Here's a 1-hour tutorial about Chebfun (850 MBytes):
http://downloads.sms.cam.ac.uk/1160831/1160835.mp4
Here are the slides for this talk (1 MByte):
http://people.maths.ox.ac.uk/trefethen/chebfun.pdf
Some history of Cheb approximations
http://people.maths.ox.ac.uk/trefethen/cftalk.pdf
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Let R^oo denote the real vector space that is the countable direct product of copies of the reals. I.e., all countable-tuples of reals with componentwise addition. Puzzle: What is the dimension of the real vector space R^oo ??? --Dan
Er --- why should the answer be not simply "oo" ( |N or aleph-null ) ? Admittedly, just how this might be justified in terms of a formal definition of "dimension" for a vector space is not something I have thought about. WFL On 8/26/12, Dan Asimov <dasimov@earthlink.net> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Sun, Aug 26, 2012 at 4:58 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er --- why should the answer be not simply "oo" ( |N or aleph-null ) ?
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors. Andy
Admittedly, just how this might be justified in terms of a formal definition of "dimension" for a vector space is not something I have thought about.
WFL
On 8/26/12, Dan Asimov <dasimov@earthlink.net> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Andy.Latto@pobox.com
On Sunday 26 August 2012 22:44:30 Andy Latto wrote:
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
A vector space with basis A over field F has cardinality sum {B a finite subset of A} |F|^|B| (more or less). When, as here, |F| is infinite, subject to AC we have |X| |Y| = max(|X|,|Y|), whence our vector space has dimension max(|F|,|A|). In particular, if |V| = 2^c and |F| = c then |A| = 2^c. I'm not sure how much of this works without AC. -- g
Huh? The same argument would conclude that the (vector space of) reals also has dimension c ( 2^aleph-null ) !! WFL On 8/26/12, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Sunday 26 August 2012 22:44:30 Andy Latto wrote:
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
A vector space with basis A over field F has cardinality sum {B a finite subset of A} |F|^|B| (more or less). When, as here, |F| is infinite, subject to AC we have |X| |Y| = max(|X|,|Y|), whence our vector space has dimension max(|F|,|A|).
In particular, if |V| = 2^c and |F| = c then |A| = 2^c.
I'm not sure how much of this works without AC.
-- g
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On Monday 27 August 2012 00:12:08 Fred lunnon wrote:
Huh? The same argument would conclude that the (vector space of) reals also has dimension c ( 2^aleph-null ) !! WFL
How so? Oh, I see: there's a one-word error in what I wrote; the conclusion of the first paragraph should have been "our vector space has CARDINALITY max(|F|,|A|)". Everything else is as it should be. Sorry about that. (So what this lets us conclude about the vector space of reals is that its dimension is at most c. Unless, e.g., we consider it as a vector space over Q, in which case its dimension really is c.)
On 8/26/12, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Sunday 26 August 2012 22:44:30 Andy Latto wrote:
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
A vector space with basis A over field F has cardinality sum {B a finite subset of A} |F|^|B| (more or less). When, as here, |F| is infinite, subject to AC we have |X| |Y| = max(|X|,|Y|), whence our vector space has dimension max(|F|,|A|).
In particular, if |V| = 2^c and |F| = c then |A| = 2^c.
I'm not sure how much of this works without AC.
-- g
Well, you first need to know that the vector space has a basis in the first place. I mean, it does, if you assume the Axiom of Choice -- this is probably the first Zorn's Lemma argument that most math majors ever see. (E.g. here: http://soffer801.<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> wordpress.com<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> /2011/09/24/every-vector-space-has-a-basis/<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> ) I can certainly see that there's no countable basis. Given any countable set of vectors, here's a way to construct a vector that can't be written as finite sum of them. Pick any value for the first coordinate. For each vector in your putative basis, there's at most one scalar multiple of it that matches the vector we're constructing -- which means there are at most countably many values for the second coordinate that wouldn't require a second basis vector be added to the linear combination. Keep going this way: there are always uncountably many values for the nth coordinate that force the linear combo to need at least n vectors. The vector you get at the end can't be formed by any finite linear combo at all. If you're willing to assume the Continuum Hypothesis, then we're done: there are only c many points, some subset is a basis, and no countable subset works. So c it is. But without CH, I don't know what else to say. On Sun, Aug 26, 2012 at 5:44 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Aug 26, 2012 at 4:58 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er --- why should the answer be not simply "oo" ( |N or aleph-null ) ?
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
Andy
Admittedly, just how this might be justified in terms of a formal
definition
of "dimension" for a vector space is not something I have thought about.
WFL
On 8/26/12, Dan Asimov <dasimov@earthlink.net> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Andy.Latto@pobox.com
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-- Forewarned is worth an octopus in the bush.
But what is the rationale behind the apparently arbitrary restriction to finite sums in your definition of "dimension"? A useful definition should surely be based on considerations of topology, geometry, linear operators, etc. For example, is it the case that a bounded linear operator in Hilbert space can be represented by a matrix of denumerable dimension ( |N, aleph-zero ) ? WFL On 8/27/12, Michael Kleber <michael.kleber@gmail.com> wrote:
Well, you first need to know that the vector space has a basis in the first place. I mean, it does, if you assume the Axiom of Choice -- this is probably the first Zorn's Lemma argument that most math majors ever see. (E.g. here: http://soffer801.<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> wordpress.com<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> /2011/09/24/every-vector-space-has-a-basis/<http://soffer801.wordpress.com/2011/09/24/every-vector-space-has-a-basis/> )
I can certainly see that there's no countable basis. Given any countable set of vectors, here's a way to construct a vector that can't be written as finite sum of them. Pick any value for the first coordinate. For each vector in your putative basis, there's at most one scalar multiple of it that matches the vector we're constructing -- which means there are at most countably many values for the second coordinate that wouldn't require a second basis vector be added to the linear combination. Keep going this way: there are always uncountably many values for the nth coordinate that force the linear combo to need at least n vectors. The vector you get at the end can't be formed by any finite linear combo at all.
If you're willing to assume the Continuum Hypothesis, then we're done: there are only c many points, some subset is a basis, and no countable subset works. So c it is.
But without CH, I don't know what else to say. On Sun, Aug 26, 2012 at 5:44 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Aug 26, 2012 at 4:58 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er --- why should the answer be not simply "oo" ( |N or aleph-null ) ?
Well the question is whether the answer is simply "aleph-null" or simply "2^aleph-null" (though I strongly suspect it is one or the other). The aleph-null vectors that each have a 1 in one position and zeroes everywhere else do not form a basis; for a set to be a basis, every vector has to be a *finite* linear combination of the basis vectors, and any vector with infinitely many non-zero entries is not in the span of these vectors.
Andy
Admittedly, just how this might be justified in terms of a formal
definition
of "dimension" for a vector space is not something I have thought about.
WFL
On 8/26/12, Dan Asimov <dasimov@earthlink.net> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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The rationale for the restriction to finite sums is that infinite sums are not in general meaningful in a vector space. The question I posed is purely in the category of vector spaces, so no additional structure is relevant for its solution. --Dan P.S. A basis for a vector space V can be defined in either of two equivalent ways: any minimal set of vectors that spans V, or any maximal set of vectors that is linearly independent. The dimension of V is the cardinality of any basis of V; this does not depend on the choice of basis.) On 2012-08-26, at 8:00 PM, Fred lunnon wrote:
But what is the rationale behind the apparently arbitrary restriction to finite sums in your definition of "dimension"? A useful definition should surely be based on considerations of topology, geometry, linear operators, etc.
For example, is it the case that a bounded linear operator in Hilbert space can be represented by a matrix of denumerable dimension ( |N, aleph-zero ) ?
Any finiteness restriction in the definition of "basis" for a vector space appears incompatible with that in general use elsewhere --- see for example http://en.wikipedia.org/wiki/Hilbert_space#Operators_on_Hilbert_spaces It would be interesting to know what significance attaches to the notion of "dimension" defined in this manner, which at the moment strikes me as inexplicably artificial. WFL On 8/27/12, Dan Asimov <dasimov@earthlink.net> wrote:
The rationale for the restriction to finite sums is that infinite sums are not in general meaningful in a vector space.
The question I posed is purely in the category of vector spaces, so no additional structure is relevant for its solution.
--Dan
P.S. A basis for a vector space V can be defined in either of two equivalent ways: any minimal set of vectors that spans V, or any maximal set of vectors that is linearly independent. The dimension of V is the cardinality of any basis of V; this does not depend on the choice of basis.)
On 2012-08-26, at 8:00 PM, Fred lunnon wrote:
But what is the rationale behind the apparently arbitrary restriction to finite sums in your definition of "dimension"? A useful definition should surely be based on considerations of topology, geometry, linear operators, etc.
For example, is it the case that a bounded linear operator in Hilbert space can be represented by a matrix of denumerable dimension ( |N, aleph-zero ) ?
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I admit I normally assume AC, since it is equivalent to saying that any cartesian product of (> 0) nonempty sets is nonempty, which I can't not believe. So I do assume in this problem that every vector space has a basis (which not only implies, but is equivalent to, AC). * * * Beyond AC, however, no further assumption (like CH) is necessary for solving this problem. (Ordinary cardinal arithmetic suffices.) --Dan On 2012-08-26, at 4:54 PM, Michael Kleber wrote: << Well, you first need to know that the vector space has a basis in the first place. I mean, it does, if you assume the Axiom of Choice . . .. . . . . . . If you're willing to assume the Continuum Hypothesis, then we're done: there are only c many points, some subset is a basis, and no countable subset works. So c it is. But without CH, I don't know what else to say.
participants (6)
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Andy Latto -
Dan Asimov -
Fred lunnon -
Gareth McCaughan -
Henry Baker -
Michael Kleber