Slashdot Mirror
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."
http://babelfish.altavista.com/babelfish/trurl_pag econtent?url=http%3A%2F%2Fwww.fontys.nl%2Fnieuws%2 Fnieuws_artikel.asp%3Fdocid%3D3487&lp=nl_en
- Leon Mergen
http://www.solatis.com
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'...
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.
TI-89 + solver/roots function = roots of polynomial
Without RTFA I can categorically state that it's all Dutch to me...
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)
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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"
I have discovered a truly remarkable formula to solve any polynomial, but my site has too little bandwidth for me to post it here.
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.
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.
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.
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Due to lack of disk space this user has been discontinued
That's dutch, not german.
quidquid latine dictum sit altum videtur.
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.
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.
Avantslash: low-bandwidth mobile slashdot.
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
(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.
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
I mean, come on. A Dutch student?
postmodernsideshow.com
Yeah, I proved that 11 years ago. Unfortunately for the rest of humankind, the margin was too small for me to write everything down.
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.
... + 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.
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) +
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.)
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.
--- Sigmentation Fault - Comments Dumped
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.
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.
(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
Not so. Exact solutions like the ones provided by mathematical formulas are still useful for a number of reasons:
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.
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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 |
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.
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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
There are even computer implementations of this for limited cases (called "generate and test" algorithms). But I wouldn't advocate running big headlines claiming -- MarkusQmore 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 |
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
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 discoveryWhile 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 PuzzlesEver 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.'
PolynomialsGeert-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!'
PublicationGeert-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.'
Less Drang.
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
What I'm surprised at, though, is that nobody's pointed out the most obvious problems with this scheme:
- 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?
- 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!Shitsurei shimashita *cuts off finger*
"A witty saying proves nothing." ~Voltaire
"d'Oh!" ~Homer
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: