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General Solution for Polynomial Equations?

An anonymous reader writes "On september 9, several media reported that a young Dutch student found a formula to determine the roots of any polynomial equation. Does this conflict with Abel's proof that such a formula cannot exist? Here is the news item (in Dutch) on his school's homepage." Another reader writes "A Dutch student at the Fontys school of physics has solved a math problem of several centuries old: finding the roots of any polynomial equation. Arxciv copy here. Although an exact solution has been proven impossible for higher orders, this is not the case for numeric solutions."

124 of 482 comments (clear)

  1. The fish by leonmergen · · Score: 5, Informative
    --
    - Leon Mergen
    http://www.solatis.com
    1. Re:The fish by Ctrl-Z · · Score: 4, Funny

      Is there a math-to-English translator for those of the Slashdot community that can't understand the PDF? Theoretically, I should be able to read it -- I have a degree in mathematics -- but we aren't all so lucky.

      --
      www.timcoleman.com is a total waste of your time. Never go there.
    2. Re:The fish by Anonymous Coward · · Score: 4, Funny

      Why did I keep expecting the words "bork bork bork" to come while reaing that?

    3. Re:The fish by 31415926535897 · · Score: 4, Funny

      I should be able to read it -- I have a degree in mathematics

      I'm guessing your degree was for English Mathematics. This paper is clearly in Dutch Mathematics (so, just head over to Babelfish).

    4. Re:The fish by Dashing+Leech · · Score: 5, Funny
      This paper is clearly in Dutch Mathematics

      At least it isn't in Polish Mathematics. Not only would it be difficult to decipher, you'd also have to read it backwards.

    5. Re:The fish by Josh+Booth · · Score: 4, Funny

      Does that mean you can read Reverse Polish math forewards?

    6. Re:The fish by quantaman · · Score: 2, Funny
      Is there a math-to-English translator for those of the Slashdot community that can't understand the PDF? Theoretically, I should be able to read it -- I have a degree in mathematics

      Hey, thanks for volunteering!!

      :)

      --
      I stole this Sig
    7. Re:The fish by jlcooke · · Score: 2, Informative

      bork bork bork? That's sweedish.

      That's fun is the Muppetizer that was on the www.muppets.com website before the evil Disney took it over. I have a copy of it here.

      Even funnier are the swear words they replaced!!!

    8. Re:The fish by apankrat · · Score: 4, Funny

      Sure, if your stack is big enough.

      --
      3.243F6A8885A308D313
    9. Re:The fish by thelenm · · Score: 2, Funny

      I have no stack. What? Where was I...

      --
      Use Ctrl-C instead of ESC in Vim!
  2. Right in the middle of my Calc class too... by JoshieCK · · Score: 5, Funny

    Last quarter's PreCalc class said this was impossible? Now it's possible?
    Dang it, that means I'll have to buy a new math book for this quarter's Calc class, won't I?

    Ah, the world, she is a changin'...

    1. Re:Right in the middle of my Calc class too... by BlueCup · · Score: 2, Funny

      Man... I was thinking this was really cool. But now I'll probably have to buy a new book as well =( I think this is just a ploy by "the man" to sell more calc books... bastard.

      --
      WANNAWIKI Wannawiki WannaWiki WANNAWIKI!
    2. Re:Right in the middle of my Calc class too... by Xyrus · · Score: 5, Insightful

      It's a NUMERIC solution, not an ALGEBRAIC solution.

      Abel's proof showed that polynomials with a degree higher than 4 could not be solved algebraically (i.e through a finite number of additions, subtractions, multiplications, etc.). Abel's proof did no say it was impossible to solve the equations (indeed, numerical solutions to these equations are solved regularly).

      This is similar to how some integral equation solutions cannot be expressed in simple terms. However numerical answers are rather easy to obtain (even easier with a computer) :).

      The method presented is a simpler way to find the roots of polynomial equations numerically by treating it like a power series (x, x^1, x^2,...,x^n) and applying standard differential techniques.

      Pretty cool if you ask me. :)

      ~X~

      --
      ~X~
    3. Re:Right in the middle of my Calc class too... by pjt33 · · Score: 4, Interesting

      It's a shame the paper doesn't say why this is better than using Sturm sequences, which work perfectly well.

    4. Re:Right in the middle of my Calc class too... by Xyrus · · Score: 2, Informative

      I agree the paper is lacking somewhat in describing why this algorithm is better. I looked into Sturm sequences. It appears that part of the algorithm is relying on a similar principal. The difference seems to come in when trying to converge on the root, which the student is using differential methods.

      But as you pointed out, I'm not sure what the advantage is in doing so. I wouldn't really classify that document as a paper, more like an empirical proof. It is not by any stretch rigorous (for instance you don't use the word "should" to describe a result from a step in a proof.

      ~X~

      --
      ~X~
    5. Re:Right in the middle of my Calc class too... by slashrogue · · Score: 4, Funny

      I would agree with it being cool if I understood any of what you just said. :\

    6. Re:Right in the middle of my Calc class too... by MemoryAid · · Score: 2, Funny
      Last quarter's PreCalc class said this was impossible?

      Anything is possible at Zombo.com.

      --
      Language students: Don't try to learn English here. This ain't it.
    7. Re:Right in the middle of my Calc class too... by Austerity+Empowers · · Score: 3, Informative

      In the event you weren't joking (I'm sure lots of people don't know what he means), I believe it means that there will never be a magic equation to 4th order polynomials and above like one learned in algebra to the 2nd order polynomial:
      ax^2+bx+c =0
      which is:
      x= (-b +- sqrt(b^2-4ac))/2a
      (aka the quadratic equation)

      Instead you can get VALUES for the roots by using algorithmic approaches, but you cannot come up with a generalized equation like the quadratic equation above.

      If it still doesn't make sense, then it probably doesn't matter, but some lines of engineering require finding roots of high order polynomials. I haven't done it since school, so my whole post may be wrong, but there's your explanation =)

    8. Re:Right in the middle of my Calc class too... by aardvarkjoe · · Score: 4, Informative
      I believe it means that there will never be a magic equation to 4th order polynomials and above like one learned in algebra to the 2nd order polynomial.
      Fifth order, I believe; there actually is a quartic equation to solving a 4th-order polynomial algebraically. (Although admittedly, even with the equation I'd hate to try to do it by hand.)
      --

      How can we continue to believe in a just universe and freedom to eat crackers if we have no ale?
    9. Re:Right in the middle of my Calc class too... by logicnazi · · Score: 2, Interesting

      I already made a long post explaining the article (if you are really curious look at my recent posts) but I needed to point out here that you forgot radicals in your descriptiong. Even the quadratic isn't solveable in terms of finite additions subtractions, multiplications. Abel's theorem says that you can't solve the general polynomial of degree five or higher using addition, multiplication and taking nth-roots.

      Personally, I am entierly unimpressed with this paper. The numerical method involved looks horridly inefficent (newton's method is probably pragmatically more interesting) and the math itself seems pretty uninspired.

      --

      If you liked this thought maybe you would find my blog nice too:

    10. Re:Right in the middle of my Calc class too... by logicnazi · · Score: 2, Interesting

      Hmm...I don't have a source but I'm pretty sure you are incorrect. Sure *fixed width* floating point solutions won't work but this is hardly enough to show there is no general numeric solution.

      I'm pretty sure there is a general numerical solution doing something fairly simple (a little better than newton's method). Of course it is entierly inefficent and practically useless.

      --

      If you liked this thought maybe you would find my blog nice too:

  3. There's no conflict... by Anonymous Coward · · Score: 5, Interesting

    a formula to determine the roots of any polynomial equation. Does this conflict with Abel's proof that such a formula cannot exist?

    Although an exact solution has been proven impossible for higher orders, this is not the case for numeric solutions.


    No conflict here. Saying that an exact solution does not exist is consistent with saying that numeric solutions do exist.

    A numberic solution is a solution that is "close enough", but not exact. Sort of like saying 2.0000000000000001 = 2. They aren't equal, but for many purposes, they are equivalent.

    1. Re:There's no conflict... by Captain+Salty+Pete · · Score: 2, Interesting

      Why should an algebraic solution necessarily be an exact solution?

    2. Re:There's no conflict... by kieran · · Score: 2, Funny

      A numberic solution is a solution that is "close enough", but not exact. Sort of like saying 2.0000000000000001 = 2. They aren't equal, but for many purposes, they are equivalent.

      Absolutely - just ask Intel.

    3. Re:There's no conflict... by bonniot · · Score: 3, Informative
      Why should an algebraic solution necessarily be an exact solution?

      Because otherwise it wouldn't be called a solution.

      "Numeric solution" is used here sloppily, it should be "numeric approximation": a number that is close enough to the real solution, or better, a process to progressively come with numbers closer and closer to the solution. In the second case, you can stop the process when you got close enough to the solution, depending on the precision you need.

  4. Better formula by StevenHenderson · · Score: 4, Funny

    TI-89 + solver/roots function = roots of polynomial

    1. Re:Better formula by dasmegabyte · · Score: 4, Funny

      Don't mock the TI-89. That's what they're using to serve this article.

      --
      Hey freaks: now you're ju
  5. Maths == Dutch by Anonymous Coward · · Score: 5, Funny

    Without RTFA I can categorically state that it's all Dutch to me...

    1. Re:Maths == Dutch by penguinoid · · Score: 4, Funny

      RTFA? FTFA, as if anyone could understand it anyways.

      --
      Don't waste your vote! Vote for whoever you want, unless you live in a swing state it won't matter anyways
    2. Re:Maths == Dutch by AllUsernamesAreGone · · Score: 4, Funny

      Hmm.. if we can assume that Maths := Dutch then, given that this is Dutch Maths, we can substitute Dutch for Maths to prove that ti is completely Double Dutch...

    3. Re:Maths == Dutch by OneOver137 · · Score: 2, Funny

      Begone with your MathCad symbology! Repent and pronounce allegiance to Mathematica!

  6. Man does the impossible by gowen · · Score: 5, Insightful

    Popular media today reports that someone has done what is well established to be impossible. Now, which one is more likely:
    i) Abel's proof contains a flaw that generations of extremely talented mathematicians have failed to spot in their years and years of teaching it.
    ii) Student mistaken; popular media talking out of arse.

    (Can't read PDF; slashdotted)

    --
    Athletic Scholarships to universities make as much sense as academic scholarships to sports teams.
    1. Re:Man does the impossible by bs_testability · · Score: 2, Funny

      I'd say it's about 50/50

    2. Re:Man does the impossible by Chris_Jefferson · · Score: 5, Insightful

      Actually, I'm fairly certain it is:
      iii) Student comes up with interesting (and possibly new, I don't know) result of generating infinate series which converges to root of polynomial. Someone (popular media?) believes that this violates Abel's proof (it doesn't, his proof is for finite representations of the roots).

      This has NOTHING to do with Abel, or Godel, or anyone other related theories, as they do not consider the case of infiniate series.

      --
      Combination - fun iPhone puzzling
    3. Re:Man does the impossible by gr8_phk · · Score: 4, Interesting
      iii) Media doesn't understand the difference between an exact solution in radicals and a numeric algorithm. Neither does the public.

      Now I'll RTFA to determine which it really is....

    4. Re:Man does the impossible by WolfWithoutAClause · · Score: 5, Insightful
      Wow. That's news. At least to anyone completely unaware of Newton's method.

      Yeah, but Newton's method isn't guaranteed to converge. This method claims to converge; although you don't get the exact answer in a finite number of steps.

      Whether this method is useful or not, probably depends on how fast it converges and how long it takes to do each step.

      --

      -WolfWithoutAClause

      "Gravity is only a theory, not a fact!"
    5. Re:Man does the impossible by orangesquid · · Score: 4, Funny

      Actually, here's a really easy exact formula for Pi (written in base Pi, just convert the answer to base 10): 10

      --
      --TheOrangeSquid Is it any wonder things seem so awry? We swim in a sea of confusion and don't have to think to survive
    6. Re:Man does the impossible by pete-classic · · Score: 2, Funny
      (Can't read PDF; slashdotted)


      I have discovered a truly remarkable proof. which this web server is too small to contain.

      -Peter
    7. Re:Man does the impossible by graf0z · · Score: 2, Informative
      i) Abel's proof contains a flaw that generations of extremely talented mathematicians have failed to spot in their years and years of teaching it.

      ii) Student mistaken; popular media talking out of arse.

      iii) Abel's theorem holds ("you cannot solve all polynomial equations by radicals"); student solves all polynomial equations not using radicals but using differential equations and power series; popular media like /. do not understand that this method is known for more than hundred years and that there is no inconsistence.

      /graf0z.

      ps: a link provided by the author himself: solving the quintic

    8. Re:Man does the impossible by grape+jelly · · Score: 2, Insightful

      This is completely untrue. If you look at any number base system, "10" indicates the number of the base. Take base 10 for example:

      1,...,9,10

      or base 8:

      1,...,7,10

      Your argument that 1base(pi) should be pi is ridiculous because 1 in any base (short of base 1, which is equally ridiculous) is really just plain 1. Things only get interesting once numbers exceed the base value.

      Posted at +2 just for fun

    9. Re:Man does the impossible by Anonymous Coward · · Score: 3, Funny

      All your base are belong to us!

    10. Re:Man does the impossible by koali · · Score: 2, Insightful

      That's all fine and dandy, but he said 10(base pi

      10(base 2 =2(base 10
      10(base 8 =8(base 10

      etc.

  7. Re:/.ed after 4 comments by Anonymous Coward · · Score: 2, Informative
  8. Re:Damn, 11 Years too late by Alien54 · · Score: 4, Informative
    Unfortunately, the solution requires a knowledge of calculus:

    To do so, we express x as a powerseries of s, and calculate the first n-1 coefficients. We turn the polynomial equation into a differential equation that has the roots as solutions. Then we express the powerseries' coefficients in the first n-1 coefficients. Then the variable s is set to a0. A free parameter is added to make the series convergent.

    The short paper has more details.

    --
    "It is a greater offense to steal men's labor, than their clothes"
  9. Re:/.ed after 4 comments by cunniff · · Score: 5, Funny

    I have discovered a truly remarkable formula to solve any polynomial, but my site has too little bandwidth for me to post it here.

  10. Mirror by avalys · · Score: 5, Informative

    Apparently some people can't get to the site, which is funny because I'm having no problem, but here is a mirror.

    The Roots of any Polynomial Equation

    --
    This space intentionally left blank.
    1. Re:Mirror by SydShamino · · Score: 2, Interesting

      Off-topic, but if you are at MIT, then perhaps your requests are routing through Internet 2? Just a thought. Try tracerouting to the site and see what you get.

      --
      It doesn't hurt to be nice.
  11. Its no big deal by moss1956 · · Score: 5, Informative

    The theorem of Abel (or Galois) that is being referred to merely claims that you can't find a general formula built from just the arithmetic operations plus taking nth roots. It has been known for a long time that there is a general formula using elliptic functions.

    The student just used the method of formal power series to solve the equation. This approach dates back at least to Cauchy ~1850 and probably can be found in the works of Euler.

    1. Re:Its no big deal by the_2nd_coming · · Score: 2, Funny

      it is basically like this:

      student gets ahead of teacher's lessons plan...news at 11.

      --



      I am the Alpha and the Omega-3
    2. Re:Its no big deal by NonSequor · · Score: 5, Funny

      Heh, just about everything can be found in the works of Euler. It's like they say, "In Mathematics, it is customary to name things after the first person after Euler to discover them."

      --
      My only political goal is to see to it that no political party achieves its goals.
    3. Re:Its no big deal by proverbialcow · · Score: 5, Funny

      student gets ahead of teacher's lessons plan...news at 11.

      In the U.S., that is a big deal. ;)

      --
      The only surefire protection against Microsoft infections is abstinence. - The Onion
    4. Re:Its no big deal by stromthurman · · Score: 3, Funny

      Naturally, Gauss will have claimed to have discovered this first, but felt it was too trivial to publish.

      --
      I have discovered a truly remarkable sig which this margin is too small to contain.
  12. No sample code or algorithm? by mikael · · Score: 4, Insightful

    I'm surprised he didn't include some sample Matlab, Java applet or C code in his paper. It would be useful to have a demonstration that this really works.

    --
    Vintage computer adverts: http://www.vintageadbrowser.com/computers-and-software-ads
    1. Re:No sample code or algorithm? by Darth_Burrito · · Score: 2, Interesting

      Wrong. A mathematical proof demonstrates to (some) other mathematicians that the approach works for all cases. Meanwhile, this translates to the rest of the world as "the math guys say they think this works." To many of us, this is roughly equivalent to hearing someone say, "Jim thinks he has fixed the mail server."

      To lend validity to either of these statements, you need to add understandable, accessible evidence. Jim thinks he fixed the mail server. It's been up and running for 30 days without any problems. Here is a proof showing how to numerically solve any polynomial equation and here's a program that runs the algorithm and shows several examples that work.

  13. I am old by TheVidiot · · Score: 2, Interesting


    This is yet another reminder of how long it has been since I was in university....

    I can recognized the names of the equations involved, but that's about it...

    In many ways, that illustrates how useful the knowledge has been over the years!

  14. I guess this is obsolete now... by Paster+Of+Muppets · · Score: 4, Funny

    LISTER: Yeah, the Skutters managed to smuggles something out of the medi-lab for us, y'know that stuff that helps impotent guys put the zest back in their love lives?

    KRYTEN: 'Boing!', the virility enhancement drug!?

    LISTER: That's the stuff, and we've Mickey Finn'd their drinks.

    RIMMER: Within seconds, you're harder than a quadratic equation, and, it doesn't wear off for seven hours.

    KRYTEN: For seven hours those guys are going to be like catapults!

    Red Dwarf, Series 8, Episode 6

    --
    Due to lack of disk space this user has been discontinued
  15. Re:run away! by slavemowgli · · Score: 4, Informative

    That's dutch, not german.

    --
    quidquid latine dictum sit altum videtur.
  16. Rule of equations in school by Kohath · · Score: 5, Funny

    The rule of equations (at least in school) is:

    The more complicated the equations for the math problem looks, the more likely the answer is 1.

    1. Re:Rule of equations in school by PureCreditor · · Score: 2, Funny

      we as /.-ters all know that as complexity of the problem reaches infinity, the answer approaches 42. =)

    2. Re:Rule of equations in school by Mignon · · Score: 4, Funny

      When I was teaching calculus I made an exam where every answer was 2. I was sort of amused watching the students faces as they at first started doubting themselves, then slowly got the "joke."

    3. Re:Rule of equations in school by revscat · · Score: 4, Insightful

      Time for some mega nerdiness: I was captain of the math team when I was in high school.

      Feh, screw that "nerdiness" crap. Good for you. Math is a powerful tool, worthy of dedication. I wish I were better at it, and respect those who are. I think being captain of the math team is far and away a better thing than being the captain of the freakin football team.

    4. Re:Rule of equations in school by ralmeida · · Score: 2, Funny

      Shame on you. It should've been 42.

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    5. Re:Rule of equations in school by HikeFanatic · · Score: 2, Funny

      I had one math professor who decided to include "Prove or disprove: Fermat's Last Theorem" as a bonus question on an exam.

      The class knew he was joking, but some decided to try it anyway...

  17. Isn't it an approximation method? by hankwang · · Score: 5, Interesting

    I am a phycisist, not a professional mathematician, and I didn't understand all steps in the whole paper. However, the author mentions a series expansion with an infinite number of terms in equation (6), although only the first n terms are ever used in defining the solution. That sounds a bit strange to me. In any case, the exact solution for a third-order equation (n=3) involves lots of cube roots and I don't see those anywhere, which also suggests that it's all about an approximation method.

    1. Re:Isn't it an approximation method? by Leadhyena · · Score: 5, Insightful

      You got it... instead of a solution by radicals (which Abel's proof shows does not exist for general polynomials with degree 5 and higher) he takes it into differential equations and creates a powerseries, which essentially gives an approach to the real number root, which doesn't necessarily have a radical decomposition. Plus, the proof looks like a lot of handwaving at a cursory glance. I'm more inclined to believe that this is a wash.

    2. Re:Isn't it an approximation method? by famebait · · Score: 2, Informative

      I thought the issue with older methods was that although they converge easily on any root, it becomes non-trivial to find out where to start in order to hit some of the roots when you get to the higher orders. If this method reliably finds them efficiently and is suitable as an automated algorithm, that is certainly a useful result, though it might be not exactly revoloutionary to the field of mathematics.

      --
      sudo ergo sum
  18. /.'ed, try this Coraled URL using another source by arnoroefs2000 · · Score: 2, Informative
  19. My brain seems to have shut down. by D_Nice · · Score: 3, Funny

    While I was in HS and College, this would have made so much sense to me. Looking at all the work behind it just makes my head hurt now. I think I replaced my math knowledge with coding ability.

    --
    Technology's a battle between companies producing more idiot-proof systems and nature producing bigger and better idiots
  20. Easy! by Bluesman · · Score: 3, Funny

    (1) Let Sa be the set of all possible roots of polynomial equations.

    (2) From [1], we have determined that the correct roots, a1...an, exist in Sa.

    (3) Let the set Sb be the set that contains only a1...an.

    (4) The intersection of sets Sa and Sb will thus be the roots of the polynomial equation.

    Therfore, we derive the formula:

    Sa ^ Sb = roots

    --
    If moderation could change anything, it would be illegal.
    1. Re:Easy! by daveashcroft · · Score: 5, Funny

      You seem to have forgotten the final step:

      5) Profit!

      (awaits an ass-whooping by the mods)

  21. europe by nnnneedles · · Score: 5, Funny

    The present:
    european academic finds solution to very hard problem.

    2 years later:
    a) americans find way of turning said solution into entertainment technology and make billions of dollars.b) European academic still unemployed and eating pasta all week.

    We need more GREED in europe.. :/

    --
    Will code a sig generator for food
    1. Re:europe by vidnet · · Score: 2, Funny

      Phh! Spammers are making billions on rooting already!

  22. Obvious Hoax by wondafucka · · Score: 4, Funny

    I mean, come on. A Dutch student?

    1. Re:Obvious Hoax by troon · · Score: 2, Insightful

      Hey, the Dutch are dudes.

      • Huygens
      • I'm sure there are others, too
      --
      Ydco co ,df C erb-y go. a Ekrpat t.fxrapev
  23. what no TeX? by Carbon+Unit+549 · · Score: 2, Insightful

    From the looks of the ugly type, he must of used word. Oh the horror. The horror.

    --

    nohup rm -rf ~/. >& zen &

  24. 11 Years too late by blibloblu · · Score: 4, Funny

    Yeah, I proved that 11 years ago. Unfortunately for the rest of humankind, the margin was too small for me to write everything down.

  25. Re:Very Skeptical by gtoomey · · Score: 2, Funny

    So this is just another numerical method to approximate roots of polynomials? Newton did it 400 years ago. Talk about media beatup.

  26. This doesn't seem likely by 808140 · · Score: 5, Interesting

    The article in question is slashdotted, but my guess is either that this is media sensationalism, or the writeup is claiming something different from the student -- it seems like perhaps a new way to numerically approximate polynomial roots has been discovered.

    However, from what I remember, Abel's theorem was proven using Galois Groups and Field extensions. This implies that what it actually proves is that analytical solutions using a particular set of functions -- in particular, the field operations (addition, subtraction, multiplication, division by non-zero) extended to include radicals (square, cubic, etc roots), composed in any way possible (as in a ruler and compass construction proof) cannot possibly generate an analytical formula depicting the solution for polynomials of order greater than 4.

    Does this mean that an analytical formula using other functions is impossible? Not at all. Trivially, I will define a function called, say, omega, which, given a n-dimensional complex vector, gives a solution to one of the roots of the function a_n * x^n + a_(n-1) * x^(n-1) + ... + a_0 where a_n are the elements of said vector. Then, by repeated application of omega and polynomial long division, I have an analytical solution to any polynomial, of any order, in complex space.

    Clearly, this solution is analytical in the sense that it a) provides an exact solution and b) is algebraic in nature. However, it isn't useful, because it depends on a function (omega) which cannot itself be defined analytically in terms of other functions (or at least, not ones we know how to compute).

    The reason Abel's proof is so important is because it deals with the 4 fundamental operations that polynomials themselves use (the field ops) and adds radicals, which are inverse ops to the building blocks of polynomials themselves. So it essentially says, we cannot use the functions that we constructed the polynomial with to solve it.

    Now, my omega function may seem a little bit contrived to non-math types, but actually a large number of functions are arbitrarily defined this way. Logarithms are a good simple example. An analytical formula for the likes of log n wouldn't be possible either, and yet we study logarithms without having an express analytical means of calculating them.

    What you should ask yourself is, what does analytical mean, anyway? It really isn't useful (or correct) to say that no analytical solution exists unless you explicitly restrict what particular set of 'basic' functions/operators the analytical solution can contain. In Abel's case (and it's a beautiful proof, by the way) he uses the field operators plus radicals. But what if you added logarithms into the mix? Exponential functions?

    It's impossible to say. If you don't restrict your base, you open yourself up to the attack that I just used with the omega function (which certainly exists, after all, I just defined it.)

    1. Re:This doesn't seem likely by miskatonic+alumnus · · Score: 2, Informative

      It's not quite that simple. Your function isn't well-defined. Furthermore, I believe you would need the Axiom of Choice just to show that such functions exist.

  27. No closed formula by MoobY · · Score: 4, Insightful

    Note that the student's result is not a closed formula, and is thus not in conflict with Abel's proof. The system uses convergence (and thus, reuires an infinite number of operations) to find the correct roots.

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  28. I did something similar by Ann+Coulter · · Score: 4, Informative

    I used hypergeometric functions to solve the equation

    a x^b + c x^d + e x^f = 0

    where the exponents are integers and the coefficients can be complex. I tried to generalize it for complex exponents but I quit after a while. Google should provide some preliminary information on using hypergeometric functions to solve the quintic

    a x^5 + b x^c + e = 0

    where c is less than 5 and greater than zero.

    This is an analytic solution to the general trinomial that I found empirically (without proof). If one wants to solve to solve the quartnomial then two dimensional structures, quintnomials need 3 dimensional structures. This was computationally taxing on me and my computer so I didn't even consider the quartnomial equations.

    By the way, I have implemented a Jenkins-Traub algorithm not so long ago that gives numerical approximations to general polynomial roots. It is fast and well known.

  29. Re:Man does the well-known by Gr8Apes · · Score: 3, Interesting

    Using series to approximate the solution of differntial equations is taught in class. Heck, go a little further in mathematics and you'll conjure up polynomials functions as the solution to a set of partial differential equations, known as the Galerkin Method

    So in what way is the above news? (Hint, take a look at the link and what's stated there.)

    --
    The cesspool just got a check and balance.
  30. Go Dutchers! by Anne_Nonymous · · Score: 2, Funny

    er...

    Dutches...
    Dutchians...
    Hollandistas...
    Holla ndics...
    Netherlandites...
    Netherisks...

    Hmmmm...??? In any case, good job, people!

  31. Re:run away! by mithras+the+prophet · · Score: 4, Funny

    (Mods, parent was mistaken, but not a troll).

    My favorite word in the 503 message was geblokkeerd. That's what I'm going to use instead of "slashdotted" from now on -- "Oh no! The site is geblokkeerd!"

    --
    four nine eighteen twenty-7 thirty-nine forty-7 fiftyeight sixty-nine seventy-9 eighty-8 one-hundred-and-nine one-twenty
  32. Re: Exact solutions are useful by Alwin+Henseler · · Score: 3, Insightful
    Some may think a good approximation of a calculation problem is "good enough". Too many variables or numbers? Just throw a more powerful calculator at the problem.

    Not so. Exact solutions like the ones provided by mathematical formulas are still useful for a number of reasons:

    • Using an exact formula can provide a shortcut, that enormously reduces the amount of calculation you have to do. It also allows to do some calculations by hand.
    • More important: an exact solution provides insight in the relation between its variables. That's very important from an educational point of view. And by substituting in other formulas, this can also advance the state of the art in other areas of science. A numerical solution, or a progression towards certain values might help you believe there is a certain relation between variables. An exact solution can help you prove such a relation. That is very significant for any sort of theoretical science.
  33. Re:/.ed after 4 comments by CSG_SurferDude · · Score: 4, Funny

    Dang, I just started reading this, and you allready beat me to it! ;-)

    However, I am still typing up my GUT (I prove that there are only 17 dimensions, string theory is wrong, the Multiverse doesn't REALLY exist, and that the cat is alive or dead BEFORE you open the box), and should have it available for subscribers shortly.

  34. Here's a math prof's take on the paper by 192939495969798999 · · Score: 4, Insightful

    From a seasoned math professor's reading of it: "It looks like a mess to me.
    I don't know what his point is. He says its a "method of solving the roots"
    of a polynomial. Well, we already have very fine methods for doing that,
    interval Newton methods for instance. Using circular disk arithmetic in the
    complex plane we can find all the complex roots as well.
    There is no need whatever to make things more complicated such as going to
    differential equations. That is unneccessary. Root finding is an algebraic
    problem."

    --
    stuff |
  35. See Mathematica Poster: "Solving the Quintic" by high+na · · Score: 4, Informative

    In this poster, they discuss this topic precisely, including Abel's theorem. One of the readers was correct: although Abel proved impossibility of solution for polynomials higher than degree 5 IN TERMS OF ROOTS AND OTHER ALGEBRAIC ENTITIES, there is nothing ruling out a solution in terms of, say, hypergeometrics. This is precisely what they do, and there's a nice development of this using power series. So, although I didn't get to read the PDF, it seems from the posts here that this is what the student did. Thus, no big deal. That said, I salute the student for figuring this out on his own, and he shouldn't be discouraged by discovering something that is not new.

  36. Re:/.ed after 4 comments by mikael · · Score: 3, Funny

    You can adjust the width of the margins using the HTML command.


    <BODY TOPMARGIN=(integer) LEFTMARGIN=(integer) MARGINHEIGHT=(integer) MARGINWIDTH=(integer)>

    That way, you'll never run out of space.

    --
    Vintage computer adverts: http://www.vintageadbrowser.com/computers-and-software-ads
  37. Better than that by MarkusQ · · Score: 3, Insightful

    You can solve them if you're prepared to write the roots in terms of elliptic functions, IIRC
    You can do "better" than that. If you're prepared to write the roots in terms of logical functions, you can "solve" anything.

    Want the roots of f(x) = 0?

    They are

    {All x : f(x)=0}
    There are even computer implementations of this for limited cases (called "generate and test" algorithms). But I wouldn't advocate running big headlines claiming
    Cranky /. poster solves everything!
    -- MarkusQ
  38. Can we get a grip here by The+Subliminal+Kid · · Score: 2, Insightful

    Some Dutch kid is very bright and has found a rapidly converging power series for finding roots. It has been done before but may be this one is slightly more or less cool that the others.

    What this does have any sodding effect on is Abel's Impossibility Theorem which is a good thing because it would screw up lots of other things that take Abel as axiomatic.

    All credit to the kid I certainly never produced anything a tenth that far about my school grade and this is the slack news season.

  39. ... guaranteed solution formulae here by 192939495969798999 · · Score: 3, Informative

    more quotes from the professor: "The "range" software of Oliver Aberth (that I have on our computer) can find
    all the roots, real and complex, of any polynomial to whatever accuracy you
    specify. Of course the more you ask for, the longer it may take, but it's
    pretty fast for ten places for polynomials of degree say ten or so.
    His book "Precise Numerical Methods Using C++" describes the methods used in
    his range software."

    Those are guaranteed solutions, too, not just "i think it's pretty close, but there's no way to prove it."

    They also have guaranteed solvers for nonlinear (and/or partial) DE's... this kid is about 50 years too late.

    --
    stuff |
  40. Duh by JustNiz · · Score: 2, Insightful

    >> Dutch student found a formula to determine the roots of any polynomial equation. Does this conflict with Abel's proof that such a formula cannot exist?

    If something exists, the fact that it can exist is irrefutable.

  41. slashdot walkabout by phyruxus · · Score: 5, Funny

    How to solve a polynomial
    1) put poly in standard form and take the first n-1 derivatives.
    2) put the derivatives in terms of x(s) (for 1..n-1), or remember why you dropped calculus and goto step 9.
    3) Use the derivatives to write a differential equation with coefficients m1..mn, or remember why you dropped differential equations and goto step 9.
    4) Use the original equation to reduce the differential equation to order n, and note the use of "then" instead of "than" in the mit write-up. (sorry, mit).
    5) Substitute a formula for x(s), multiply resulting eq by it's denominator, getting another diffEq. Whee! ask a Grad student.
    6) Now substitute a power series representation. All 's' should be zero. (mutter: Aha! I knew it) Solve b_sub_i for 1..n-2 (Grad student).
    7) Substitute another power series to get an equation. (The grad students are gone, ask your hallmates, one of 'em has to be a math major.)
    8) Let b_sub_n-1 equal the determinant of a funky, unexplained matrix (here, have an aspirin).
    9) Everyone else in the class is out drinking by now, so don't worry about the next matrix, it's even funkier. Write a note on your hand to memorize it this weekend. Go drinking with peers.
    10) Wake up at 3pm tomorrow, and try to remember what the hell all those squiggles meant.
    11) Change your minor from math to polisci. Don't worry about taking Calc 1-3, DiffEq, or linear algebra. Note: many girls do not care about the roots of arbitrary polynomials, so no worries there. 8^)

    --
    "A witty saying proves nothing." ~Voltaire
    "d'Oh!" ~Homer
    1. Re:slashdot walkabout by BengalsUF · · Score: 2, Funny

      12) Profit!!!

    2. Re:slashdot walkabout by Anonymous Coward · · Score: 2, Funny

      Hrm...

      You are quick to point out the "then" vs. "than" in the writeup, yet you use "it's" when clearly "its" is called for. "It's" is never used for possession- rather, 'it's' used to mean it is (as I have so cleverly demonstrated).

  42. An applied mathematician's take by coult · · Score: 2, Insightful

    From what I can tell, it appears to be a method for transforming a polynomial into a differential equation some of whose solutions are roots of the polynomial. From that point I suppose one could use numerical methods for ODE's to find those solutions.

    People here have been commenting that Newton's method works just fine for finding roots of polynomials. But, convergence can be quite slow, especially for unidentified multiple roots though, and for highly clustered roots you can run into conditioning problems.

    The paper makes no mention of actual numerical algorithms (in particular no discussion of convergence rates or guarantees for solving the ODE numerically) so it is hard to say whether the result is actually useful or just a bunch of manipulation of symbols on paper.

    --

    All is Number -Pythagoras.

  43. Re:/.ed after 4 comments by troon · · Score: 2, Informative

    Intended humour aside, that may work in many browsers, but it's not valid HTML.

    Use CSS styling instead.

    --
    Ydco co ,df C erb-y go. a Ekrpat t.fxrapev
  44. Translation by curtvdh · · Score: 3, Informative

    Disclaimer: it has been years since I've spoken Dutch. What follows hsould be taken with a fairly large grain of salt...

    Fontys student develops important mathematical discovery

    While most students languished on the beaches this Summer, Fontys student Geert-Jan Uytdewilligen discovered the solution to one of the oldest mathematical problems. He proved an inportant step in - wait for it - the classification of the zero-points of polynomials of any order.

    This problem was already known to the ancient Egyptians. During the Renaissance, a clearer understaning of (the problem) existed, and one 19th century scientist published (a paper) on his findings that stated the problem could not be solved. But Geert-Jan Uytdewilligen, a fourth-year student at Fontys High-school of Applied Science finally shed light on the complex problem. He discovered a formula for the classification of the zero-points of any order. Mathematical proofs have thus far not come from the sixth grade.

    Difficult Puzzles

    Ever since his youth, Geert-Jan Uytdewilligen was obsessed with the solution of difficult puzzles. 'I always feel at home in abstract thought', he says, 'In elementary school, I was very good at arithmetic, and therefore in my future studies, I stuck to mathematics. At one particular point in (my) mathemetics lesson, (we) handled the parabola. From that moment, I became interested in the pure algebraic problems that flowed from that. In particular, the higher grade comparison of the zero-points intruiged me, since mathematicians had been searching for a solution to the problem for ages. This was a challenge for me, to solve the problem which is purely theoretical. I had a slight practical advantage, because one can usually fill in the numbers(?) with a computer. The problem is then solved in this manner.'

    Polynomials

    Geert-jan designed mathematical formulae that were previously regarded as not-undestandable by the layperson. Perhaps you might recognize this formula: a[n]*x^n+a[n-1]*x^(n-1)+..+a1*x+a0=0. 'This is the general form of a regular polynomial', he says. Regular polynomials are a combination of increasing powers and multiplication. If you solve for x in this formula, then you get the zero-points of the polynomial. Polynomial solutions up to the sixth order are already known. I found a formula to find the zero-points of a polynomial of any order!'

    Publication

    Geert-Jan's discovery first saw light of day in the magazine Science Guide, and generated a lot of publicity. This was the reward of two years of hard work. Geert-Jan: 'You don't expect such a vague starting-point to result in such a hit. This is strange, considering the amount of technical jargon, which make the theory hard to follow. But at the same time, the pieces of the puzzle began to come together. Yes, I had the Eureka-moment! But I remained a freely sober person, and held myself together. I didn't allow my studies to suffer because of my hobby.'

  45. I wonder if anyone will get this... by Anonymous Coward · · Score: 5, Funny
    It's a shame the paper doesn't say why this is better than using Sturm sequences

    Less Drang.

  46. Looks flawed by hanwen · · Score: 4, Informative

    Looks flawed to me. He performs a sensitivity analysis in the constant of the polynomial (which he calls "s"). It remains unclear why. After a convoluted sequence of operations, he derives a power series for x as function of s , and proves convergence by requiring |s| smaller than 1.

    Finally, he puts back a_0 into s, but conveniently forgets the case that a_0 is bigger than 1.

    Also, it is not clear whether this is in the complex plane or not. For example, for finding real roots of real polynomials, you could use Sturm Sequences. There's even sample code in graphics Gems IV (IIRC).

    In any case, the student was studying at the "hogeschool" which roughly translates to "higher professional education", a school which doesn't teach mathematics, and whose level which significantly lower than Dutch the MSc., BSc. or engineering degree.

    Han-Wen

    (yes, I am a mathematician)

    --

    Han-Wen Nienhuys -- LilyPond

    1. Re:Looks flawed by Anonymous Coward · · Score: 2, Informative

      I believe he chose the variable s due to the Laplace Transform, which I am sure you must have learned in a 1st/2nd year Calc course.....

    2. Re:Looks flawed by Anonymous Coward · · Score: 4, Informative
      Finally, he puts back a_0 into s, but conveniently forgets the case that a_0 is bigger than 1.

      Except he doesn't forget this. He tells you to divide the polynomial by a constant to make it so. Since you're a "mathematician" let me help you visualize this difficult step:

      18x^6 - 12x^3 + 3 = 0 ... divide by 6
      3x^6 - 2x^3 + 1/2 = 0

      Gee, a_0 is less than 1 now and the roots are the same. Huh, I wonder how that happened? In actuallity he tells you to divide by a larger number such that all of the coefficients are fractions, but you get the picture.

    3. Re:Looks flawed by ninja0 · · Score: 4, Insightful

      Nope, grandparent was right. You can't just divide and expect convergence to suddenly occur. I believe the leading coefficient is assumed to be 1 for the result he's using. IAA math major.

      --
      --If the world didn't suck, we'd all fall off.
    4. Re:Looks flawed by oGMo · · Score: 2, Insightful
      (yes, I am a mathematician)
      Well then,
      After a convoluted sequence of operations,
      Sounds like precise mathematical terminology and understanding of the operations.
      In any case, the student was studying at the "hogeschool" which roughly translates to "higher professional education", a school which doesn't teach mathematics, and whose level which significantly lower than Dutch the MSc., BSc. or engineering degree.
      Typical academic arrogance. Letters after your name do not make your more correct. From what I've seen, the more letters you have, the less "new stuff" you're going to come up with. You may be right in this case---but show where the mathematical errors are, don't point at credentials.
      --

      Don't think of it as a flame---it's more like an argument that does 3d6 fire damage

    5. Re:Looks flawed by hanwen · · Score: 2, Interesting
      You may be right in this case---but show where the mathematical errors are, don't point at credentials.

      The problem is that it is not "mathematics," in the sense that he doesn't write a logical sequence of formally specified lemmas, proofs or conclusions. It is not clear what the author is doing, so it is also not possible to pinpoint exactly where the errors are.

      --

      Han-Wen Nienhuys -- LilyPond

    6. Re:Looks flawed by Anonymous Coward · · Score: 3, Informative

      Read the paper. There is no such assumption that a_n=1. (Why else have a_n in the numerator of equation 5 if it's always 1?) He is quite explict about dividing by "(more than) the maximum of the absolute values of the coefficients". Dividing by any non-zero constant (even complex) does not change the roots of the polynomial at all.

      Assuming his formula are correct for the expansion coefficients, the series is guaranteed to converge once you perform the recommended normalization. The grandparent is wrong, and so are you! But good use of the slashdot technique of commenting without reading the paper, I guess that's how you get your "Interesting" score.

    7. Re:Looks flawed by eluusive · · Score: 2, Informative

      If you read the proof all the way through, you'll see that a_0 is required to be less than 1.

    8. Re:Looks flawed by Sage+Gaspar · · Score: 2, Insightful

      This kid is in what I'm guessing is the equivalent of a really good technical high school, so he deserves praise for just being at the level of messing around with this kind of stuff, let alone attempting to publish a paper. I'm sure some of his paper is lost in translation, but, commenting on the paper itself, it would never get accepted into a reputable journal for publication.

      First, he never gives any clear indication whatsoever of what he intends to prove that his method will do. What does it mean to provide a method for "solving the roots" of a general polynomial equation (determine one root, determine all roots, determine all real roots, etc)? Is the polynomial expressed over an arbitrary field, the real numbers, or the complex numbers? What is a "powerseries of s"? The power series of a function is defined as an infinite polynomial that converges over a certain interval to the value of the original function. S, as far as I can tell, is a constant, so thus a power series of s would be s itself.

      It might seem like nitpicking, but it's ambiguities like these that make it nearly impossible to read this paper and determine its correctness.

      I could pore through it, guessing at the meaning of unorthodox terminology and then proving everything that he hasn't proven to the level of mathematical rigor. But that's his job, and the job of his mentors/advisors and the referee at whatever journal he tries to submit this to.

      Furthermore, issuing a press release hyping this as the solution to a long-sought-after open problem is irresponsible. The only problem this has the potential to solve is finding a faster algorithm for computing roots (real, complex, I'm not sure, he didn't say), but the paper does not address whether this is, in fact, any better at all than current root-finding methods. Even if so, it certainly wouldn't make mathematical waves except in some very select circles, were it not that he was in high school.

      So, I apologize if this sounds harsh. I certainly couldn't have done this in high school, and it has the potential to shave some overhead off finding the roots of certain polynomials, which would be a nice little result. But I felt it necessary to explain why this paper isn't yet real mathematics.

  47. Enthusiasm without truth! by Thinkit4 · · Score: 2, Funny

    Argh, they need to make a way to skip all "funny" comments in any science article. So much enthusiasm for math and physics--so little truth.

    --
    -I am an elective eunuch.
  48. I can't believe nobody's noticed this... by sixpaw · · Score: 3, Interesting
    ...as others pointed out, this is simply the method of power series, and it's even a pretty clumsy way of getting at that power series (why generate a differential equation when you can plug power series coefficients into the original equation itself?)

    What I'm surprised at, though, is that nobody's pointed out the most obvious problems with this scheme:
    1. Your polynomial has (up to) n roots; this approach converges, maybe, to one of them. Which one do you get, and how do you get the others?
    2. For that matter, some chunk of those n roots may be complex; but all the maths in his article are real. How do you solve, say, x^6+1 = 0?
    The change in radius of convergence at the end of his post is a little dicey too, at least as I'm reading it, but I could be misreading. Still, I'm frankly stunned. Worthy of a press release? If I'd turned this in as a school assignment it wouldn't even have been worth an A!
    1. Re:I can't believe nobody's noticed this... by soroka · · Score: 2, Informative
      ...why generate a differential equation when you can plug power series coefficients into the original equation itself?...

      When you substitute power series into an algebraic equation of degree higher then 2, you get a system of true nonlinear equations for the coefficients of the series.
      Example:
      (b0 + b1 * s +... )^3 + a(b0 + ...) + s = 0
      gives
      b0^3 + a b0 = 0
      3 b0*b1^2 + a * b1 + 1 = 0
      ...
      In contrast, the paper arrives at a system of linear equations.
  49. Professional by thrill12 · · Score: 2, Interesting

    Not a mathematician here, but just a fellow dutchman who likes to add that even though the student in question isn't studying at a university, that doesn't simply mean he couldn't have come up with a nice idea.
    Besides, at the "Hogeschool" there is teaching of mathematics, just not in the same way as at an university: It is much more "practical" - ie. without going through all the 'proving-stuff' - and the level is generally lower than at a university. But the technical studies still provide an adequate level of mathematics, as necessary for any serious engineering work.

    --
    Slashdot: stuff for news, nerds that matter, matter for news, stuff that nerd
  50. damn, you're right. by phyruxus · · Score: 3, Funny
    Note to self: preview pedantic posts.

    Shitsurei shimashita *cuts off finger*

    --
    "A witty saying proves nothing." ~Voltaire
    "d'Oh!" ~Homer
  51. NUMERIC method by khrtt · · Score: 2, Interesting

    It's a nice NUMERIC method. There are other numeric methods that do the same exact thing. This one is nice. It may be better to implement, and/or easier to use than other common numeric root finders. This method is, therefore, not only valid, but perhaps very useful science. The next time I need to write a root finder, I will definitely give this a close look. BUT THIS HAS NOTHING TO DO WITH ABEL'S THEOREM!!!

    Abel's theorem proves that no exact root finder is possible. It has nothing to do with numeric root finders, except for making them a little more interesting for practical applications.

    What I mean is, even if exact root algorithm were possible, numeric root finders would still be useful. For many problems where an exact solution does exist, a numeric solution is more useful. An good example would be Gaussian elimination (for solving systems of linear equations).

    Gaussian elimination gives an exact solution. If you try to use it in practice, with coefficients given as floating numbers, you realize that for many matrices the rounding errors kill the precision of the result. A common cure for this is to run a few iterations of a numeric approximation method on the output of the Gaussian elimination, which improves the precision.

    Now, back to root finders. An exact solution is not possible for polynomials of degree higher than 5. It is possible for polynomials of degrees 1, 2, 3, and 4. However, the formulas for degree 4 get rather long. So long a numeric method would quite possibly be more accurate. It would not write the roots out in radicals, but the floating point representation (that's what you want anyways, right?) would be more precise.

    The point of this essay - when you need numeric results, numeric methods are often more useful than precise formulas, even where precise formulas exist. If Abel's theorem were wrong, you'd probably still use numeric root finders in most practical cases.

  52. I can't believe I'm about to say this, but... by whyde · · Score: 2, Funny
    If you don't restrict your base, you open yourself up to the attack that I just used with the omega function...

    If you don't restrict your base, all your base are belong to us. Take off every Omega function!! For great roots!

    I'm so ashamed of myself.

  53. Re:/.ed after 4 comments by 808140 · · Score: 2, Funny

    If Fermat had had HTML, he woulda been able to fermat his own margins...

    Do I hear crickets?

  54. Mathematical Elucidation by logicnazi · · Score: 4, Informative

    Alright, whoever wrote the article seems very confused about mathematics and abel's theorem in particular. I'm not actually an algebraist myself but I am in mathematical logician so I can comment a bit about impossibility results.

    Abel's theorem merely says you can not solve the general quintic (5th degree) or higher in terms of radicals. That is entierly in terms of multiplication, addition, and taking nth roots. If we don't put that restriction about radicals the solution is trivial. Let x be such that P(x)=0 is one obvious solution.

    Going through this again the write up is *entierly wrong*. It is completly possible to give an exact solution for the general polynomial (I just did in the paragraph above). Furthermore this distinction between exact and numerical solutions which is made so much of by our high school and college teachers is really illusionary. Writing a solution in terms of sin(3) isn't an exact value, we just have a good algorithm to approximate sin. Really what we mean when we talk about exact solutions is solvable in elementary functions, which is nothing but a certain commonly used set of functions for which we have good approximations. Unfortunatly, we still insist on students 'solving' differntial equations rather than just finding some quickly converging numerical solution even though at a deep level these are not differnt.

    Now since abel's theorem there has been considerable research on other ways to solve polynomial equations. For instance one big result was that a certain degree of polynomials could be solved in a terms of continous two place function. Possibly this result in question is another result like this one but I imagine it is much less significant. For one I'm not entierly convinced he is correct, nor novel. (Don't make the mistake of assuming if he is right he has given a continous solution of any polynomial..it isn't clear his solution is continous in the coefficents).

    --

    If you liked this thought maybe you would find my blog nice too:

  55. Can't contradict a proof by Performer+Guy · · Score: 2, Interesting

    Proofs cannot be contradicted that's the point. That's why it's called a proof.

  56. already known for 30 years by WilbertD · · Score: 2, Informative

    a formula to determine the roots of any polynomial equation. Does this conflict with Abel's proof that such a formula cannot exist? There are some misunderstandings here. Abel's result says it's not possible to solve polynomials of degree 5 or higher in terms of multiplication, addition and taking roots. If you allow fancier functions (like elliptic functions, which are similar as sin/arcsin), it's possible to solve polynomials of degree 5 [1]. The solution that the student posted is an exact solution (no numerical one). The series is exact (unless you take only a few terms of course). You can write down its coefficients explicitly in terms of the coefficients of the original polynomial. Two important remarks: 1) There exists several classes of functions (hypergeometric functions in several variables, and Siegal modular functions) which can be used to solve polynomials of any degree (this is known since 30 years or so). See bottom of [2]. 2) The method which the student used, is also known since 30 years. It can be found in the same link [3]. So, nothing new ... [1] http://mathworld.wolfram.com/QuinticEquation.html [2] http://library.wolfram.com/examples/quintic/main.h tml#higher [3]: http://library.wolfram.com/examples/quintic/main.h tml#diffeq

  57. Re:Mod up! by scapermoya · · Score: 2, Informative

    Sturm and Drang was a pre-romantic movement in Europe, somewhere around the 1780's i believe. Some of the famous Classical composers dabbled in it, I think Beethoven has a Sturm and Drang piece. It died quickly and gave rise to the Classical (then romantic) movement.
    The name of the period comes from a play by Friedrich Klinger, I think it means "Storm and Urge"

    Funny, we discussed this today in academic decathlon.

    --
    Beware the Jubjub bird, and shun the frumious Bandersnatch.
  58. Abel's theorem clarification by cletus_bojangles · · Score: 2, Insightful
    Several posters stated Abel's (or Galois'? or Ruffini's?) theorem correctly: it says that for n = 5, 6, 7, etc. there is a polynomial involving x^n whose roots cannot be expressed in terms of the four functions (+, -, *, /) and m-th roots (m = 2, 3, 4, 5...).

    But that's not the whole story. Of course, if sin(1) is a root of your polynomial, then most people would be happy and consider that a perfectly good number that we can sink our teeth into. It is in fact possible that there exist polynomials of degree 5 whose roots cannot be expressed using buttons on your calculator (assuming your calculator is somehow infinite precision). That means using exponentiation, natural logs, arcsinh, etc. A particular example is the polynomial
    2x^5 - 10x + 5

    Well, with current math we can't prove that the roots of this polynomial have this property. But if you assume the as-yet-unproven (or as-yet-disproven, take your pick) "Schanuel's Conjecture" from number theory, you can indeed prove that this polynomial's roots are in some sense "inexpressible".

    Yeah, yeah, of course you can approximate them numerically. Any australopithecine could realize that in the time it takes to gnaw an antelope femur down to the marrow.

    (Personally speaking, I find the possibility that we can't explicitly write down the roots of a quintic polynomial -- especially such a nice one -- somewhat disturbing.)

    Reference for the claims above: Timothy Chow, What is a closed-form number?, American Mathematical Monthly, May 1999, vol. 106, pages 440-448.

  59. This is news? His method seems a rehash... by dido · · Score: 2, Interesting

    The method presented in the paper looks a lot like James Cockle and Robert Harley's differential resolvent, which was new in 1862. This page gives an overview of some of the known methods for solving quintic and higher degree equations. Apparently, about twenty years ago Hiroshi Umemura found a general analytical solution for a polynomial equation of arbitrarily high degree involving Siegel modular forms, which are generalizations of the elliptic modular functions Charles Hermite used in 1858 as a solution to the quintic. Note: these don't violate Abel's Impossibility Theorem as they are not solutions in radicals.

    --
    Qu'on me donne six lignes écrites de la main du plus honnête homme, j'y trouverai de quoi le faire pendre.
  60. General Solutions to Polynomial equations? by trieck2 · · Score: 2, Interesting

    Being a "young student", he certainly deserves honorable mention for his discovery. Unfortunately though, I don't think that it is new math. I think it is the well-known "Lagrange Inversion Formula" in disguise. See http://encyclopedia.thefreedictionary.com/Lagrange %20inversion%20theorem Set "z" there to s. Set "w" there to x. Set "f(w)" = - (a_n x^n + ... + a_1 x). So "w = g(z)" is then x = g(s). I believe that the "a" there (the initial estimate for "w") would be -a_{n-1} / n a_n. Give it a try.