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Swedish Mathematician Lennart Carleson Wins Abel

William Robinson writes "Sci Tech is reporting that Swedish mathematician Lennart Carleson has won the Abel Prize on Thursday for proving a 19th century theorem on harmonic analysis. His theorems have been helpful in creating iPod. Prof Carleson's major contributions have come in two fields - the first has subsequently been used in the components of sound systems and the second helps to predict how markets and weather systems respond to change. One of Carleson's many triumphs was settling a conjecture that had remained unsolved for over 150 years. He showed that every continuous function (one with a connected graph) is equal to the sum of its Fourier series except perhaps at some negligible points."

22 of 144 comments (clear)

  1. Except at some negible points? by Opportunist · · Score: 2, Insightful

    My math prof used to shoot chalk pieces onto me for saying something like that!

    That's when I decided that statistics is more my kinda speed.

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    1. Re:Except at some negible points? by gowen · · Score: 5, Informative

      Well they mean "almost everywhere", which has a very precise meaning. i.e. except at a set of measure zero (finite or countably infinite set of points.) Of course, that countable set could theoretically be the rationals, so I don't know whether I'd call it negligible.

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    2. Re:Except at some negible points? by Anonymous Coward · · Score: 2, Informative

      Well they mean "almost everywhere", which has a very precise meaning.

      The original article on Sci-Tech Today does use the phrase "almost everywhere" instead of the incorrect "negligible". I suppose it is excusable for someone who is not familiar with the area not to realise that it has a precise meaning, so maybe it would have been better to use the more common "presque partout", and let people reach for their French dictionaries.

    3. Re:Except at some negible points? by Capt'n+Hector · · Score: 3, Informative

      Incorrect. A set of measure zero can be uncountable. (cf the Cantor set)

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    4. Re:Except at some negible points? by F�an�ro · · Score: 2, Interesting

      Isn't the set of all computable numbers also countable infinite?
      So you could have a continuous function which diverges from the sum of its Fourier series in all computable points?

  2. Young people today by gowen · · Score: 5, Funny

    Young people today. You tell them about a deep result in real analysis, and the only thing they're interested in is how it relates to their iPod. And get off my lawn.

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    Athletic Scholarships to universities make as much sense as academic scholarships to sports teams.
    1. Re:Young people today by LarsWestergren · · Score: 4, Interesting

      I'm 31 and have recently started doing a lot of maths in my spare time so that I can get a real computer science and engineering degree one day (I have a degree, but it is CS light... now that I work as a programmer I know how much I'm really lacking), so it is nice to see that at least for some people the old saying by Hardy, "mathematics is a young man's game" isn't true. Carleson is 78 today, and around 40 back when he did the main breakthroughs he is honored for today.

      Hardy's saying is a bit of slight against all female mathematicians too, come to think of it...

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  3. Wiki Article by zaguar · · Score: 5, Informative
    For those of you who want more than a press release, heres a start :

    Wiki Article on the Breakthrough

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  4. Re:negligible? by Anonymous Coward · · Score: 3, Informative

    It would have been better if they had said "almost everywhere". That is, the set of points at which a Fourier series diverges has Lebesgue measure zero. There is a quasi-converse, due to the Israeli analyst Katznelson, which, given any set of Lebesgue measure zero -- let's call it Bill -- constructs a continuous function whose Fourier series diverges everywhere in Bill. For more info, see Tom Korner's excellent "Fourier Analysis".

  5. nice work, but no iPod by penguin-collective · · Score: 5, Informative

    The result he proved is nice mathematics, but you don't need it for iPods or audio coding. First of all, for many engineering purposes, it only matters that it works, not that you can prove that it works theoretically. Secondly, audio coding is done over discretely sampled signals, and most of those theorems become simple linear algebra in that case.

  6. Obligatory CmdrTaco quote on the subject by grand_it · · Score: 4, Funny
    His theorems have been helpful in creating iPod.

    No wireless. Less decimals than pi. Lame.

  7. iPod? by liangzai · · Score: 3, Insightful

    Yeah, it might be that the MPEG-4/AAC/H.264 algorithms are based in part on Fourier analysis, but I fail to understand how Carleson's theorems have been used in making the iPod. Cupertino is hardly knowledgable in the more esoteric realms of theoretical mathematics, and there is simply no need to incorporate such stuff in an mp4 player.

    This is bad journalism, written by bad reporters who lack the most basic understanding of mathematics and engineering. He just thought it might be cool to clam in an iPod in the mess.

  8. iPod Reference Misleading by Chrononium · · Score: 4, Informative

    The iPod reference is completely misleading, as simple harmonic analysis is way bigger than just an iPod. It's merely talking about this guy proving that Fourier was basically right, validating harmonic analysis and expanding the horizons for signal processing. That's the biggie: signal processing, not the bloody iPod. The stupid article probably includes iPod just for the sake of hits.

  9. WTF does this have to do with iPods?! by pslam · · Score: 5, Informative
    His theorems have been helpful in creating iPod.

    Oh really? Search Wikipedia entries, the articles, all links - no mention of iPod except in those annoying side adverts. Why? Because it has nothing to do with it

    Credit where credit is due, and none is due here.

    If you want credit, how about: Shannon, Fourier and Huffman. Then there's all the folks involved in working out noise masking and all the oddities of human hearing that I don't have the names of.

    I seriously need a "No iPod mentions whatsoever" checkbox for my slashdot profile to pull some more signal out of the slashdot article noise.

  10. a continuous function is NOT by ratta · · Score: 2, Informative

    one with a connected graph. For instance the function equal to sin(1/x) for x != 0 and 0 for x = 0 does have a connected graph but is NOT continuous.

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  11. His Fourier paper... by Bromskloss · · Score: 2, Informative

    ...was published in the Swedish maths journal Acta Mathematica and is calles On the convergence and growth of partial sums of Fourier series.

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  12. Re:Indeed by The+Cow+of+Pain · · Score: 3, Interesting

    This is sometimes mis-stated as 'you can draw the graph without taking your pen off the paper'.

    That's not a mis-statement in the case of a real function of a real variable. It's not that informative, but definitely correct in the sense that a function (real etc.) is continuous iff the graph is path-connected (i.e. every two points on the graph can be connected by a continuous path (and by saying 'continuous path' I have of course made the definition self-referential and thus silly, but it is still true)).

  13. Re:Indeed by Bromskloss · · Score: 2, Interesting
    I recall correctly: a continuous function from one space to another is a function such that the inverse images of open sets are open.>

    Yes, you do recall correctly! ;-) Your version is a more general one. In a metric space, where your parent's version is applicable, the two are equivalent. Sorry, I probably shouldn't tell you this, you certainly knew it already. However, someone else might be interested.

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  14. Re:iPod? by glesga_kiss · · Score: 3, Insightful
    I fail to understand how Carleson's theorems have been used in making the iPod.

    The iPod reference got this story greenlighted on slashdot. Otherwise it might not have made it. If you want to guarantee acceptance, mention something bad about M$, something good about Linux, or anything about Apple (preferably good, but there is the odd flame article).

    This advice was brought to you by someone with a 100% submission record. (ok, one of one ;-)

  15. Re:Indeed by The+Cow+of+Pain · · Score: 2, Informative

    Your version is a more general one. In a metric space, where your parent's version is applicable, the two are equivalent.

    Math geek warning!

    Actually the equivalence goes much further than for metric spaces. In all topological spaces you have a sense of convergence of a sequence, and so it makes sense to ask the question "Does x_n->x imply f(x_n)->f(x)?". If f is continuous, the answer is always yes, but the converse need not be true in general - it is however true if the topological space is a so-called 1st-countable space, e.g. if the space is metric (as you said).

    For even more general topological spaces you need the concept of a net, but in that setting the equivalence is actually total; a function f is continuous if and only if x_i->x implies f(x_i)->f(x) for every net (x_i).

  16. What is convergence in a non metric space? by Bromskloss · · Score: 2, Interesting

    This is interesting, please give us more!

    In all topological spaces you have a sense of convergence of a sequence

    I must admit I didn't know of any way to speak of convergence without the notion of a metric. How is that possible?

    For even more general topological spaces you need the concept of a net

    More general than what? And do you mean we need the "net" to replace the sequence? If you say so I'll believe you. However, one must still be able to define a sequence (a function from "the set of all natural numbers" to "the topological space in question"), since it doesn't really require much of the space, right?

    (We need more maths on Slashdot!)

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    1. Re:What is convergence in a non metric space? by The+Cow+of+Pain · · Score: 2, Informative

      I must admit I didn't know of any way to speak of convergence without the notion of a metric. How is that possible?

      In a topological space you have a notion of a neighbourhood of a point, i.e. a set containing the point in it's interior. You then say that the sequence (x_n) converges to the point x if for every neighbourhood U of x, there is a number N, such that x_n is in U whenever n>N. Basically this is a translation of the epsilon-N-formalism for convergence in metric spaces (since in a metric space U is a neighbourhood of x exactly when there is an epsilon such that the epsilon-ball around x is contained in U).

      More general than what?

      More general than 1st-countable spaces. That is general topological spaces with no structure but the existence of a family of open sets.

      And do you mean we need the "net" to replace the sequence?

      Yes. The concept of a convergent sequence is not in general strong enough to capture topological properties such as continuity (i.e. the preimage of an open set is open), closedness (the complement is open), and compactness (every open cover has a finite subcover), but replacing sequences by nets, you actually get the classical characterisations of these properties known from metric spaces; a function is continuous iff x_i->x implies f(x_i)->f(x), A is closed iff any net (x_i) in A, which converges in your larger space has the limit contained in A, and C is compact iff any net has a convergent subnet with limit in C. Nets are really nice objects for doing point set topology.

      However, one must still be able to define a sequence (a function from "the set of all natural numbers" to "the topological space in question"), since it doesn't really require much of the space, right?

      Yes, no problems there. They just don't provide enough information once we go to the more obscure topological spaces.