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
no you couldn't.
Let me be the first to say that I welcome our new polynomial equation solving overlord. Let death come quickly to his enemies.
So any polynomial time problems can now be solved in logarithmic time or something! Run for the hills! /sarcasm
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'...
Well, I'd love to have seen the formula, but the site's gone down.
PocketGamer.org - For the gamer on the go!
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)
Athletic Scholarships to universities make as much sense as academic scholarships to sports teams.
Your "formal language statement" is like http://www.babelcode.org
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"
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.
From now on I will replace the phrase "It's Greek to me," as in, "I can't understand any of that," with the phrase "It's Dutch to me."
I will do this whenever possible in honor of this Dutch student's obviously impressive, but absolutely inscrutable (to me) breakthrough.
intellectual property law is philosophically incoherent. it is your moral duty to ignore it or sabotage it
Abel's proof, and related proofs by Galois, have stood the test of time for 200 years. It would throw most of abstract algebra and modern mathematics on its head if true.
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
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!
What are you smoking, The site is fine,
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
That's dutch, not german.
quidquid latine dictum sit altum videtur.
It's not German, it's Dutch (Nederlandic?)
What?
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.
sorry to be that crude, but it's not German - it's Dutch .. that's quite a difference ..
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.
Enough with the english jokes already. The paper itself is in english and it is only three pages. Now go read it you math people!
Actually it is dutch.
Temporarily not available!
This site is not temporarily available because of too much
data-verkeer.
If something is so important that you feel the need to post it on the internet... It probably isn't that important.
In fact that is Dutch
--
The world is divided in two categories:
those with a loaded gun and those who dig. You dig.
Sorry to scare you even more. This isn't german but dutch. Now you know why you have a saying about people talking double-dutch ;-)
thats not german, its dutch! and dutch is not even like german - i mean no(t many) germans understand dutch out of the box...
Check it out...
I think you'll find it's Dutch - didn't even RTFS(ummary)? ;)
I don't read your sig, why do you read mine?
German?
Umm, no. This is DUTCH. That's why he's from the Netherlands, where the people are called DUTCH and they speak DUTCH.
Dutch, German == Germanic Tongues
Dutch != German
Niederländisch ist nicht deutsch!
[move
WTF does german have to do with it?? Since it's in Dutch! Comparing dutch with german is a huge insult for most Dutch people, including me. Just because it might look/sound the same to foreigners, it's not. The same goes for portugesespanish.
Alright, a subject I am familiar with! As a math major, I know that polynomials of degree 5 are not solvable in terms of radicals, in general. This is Galois Theorem. It also ties in quite well in other areas of mathematics, not just algebra.
Special cases are solvable, but for the general equation x^5 + ax^4 + bx + cx + dx + e = 0, the general solution for x cannot be expressed in terms of radicals. Approximating solutions, however, is a completely different matter.
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
Just to be a complete A-hole stickler. But that is not german, it is dutch.
Tijdelijk niet beschikbaar == Temporarily Unavailable.
Sorry, I had to do it.
Disclaimer: I am half Dutch (Belgian actually).
(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.
Agreed. I can remember back in Jr. High writing a program in GW-BASIC (!!!) that would factor any polynomial equation I threw at it. No use for cheating as the teachers wanted to see your work, but was great for checking my answers.
Maybe if I could understand Dutch I'd get why this is a big deal.
IANAM, so I cannot really decipher the proof with any accuracy, but like any good Slashdot reader I will not hesitate to state my uniformed opinion.
It strikes me that there may be a problem since taking derivatives is involved. It seems like one might lose some valuable information by using derivatives to solve the equation since the zero-order term is lost.
Can a trained mathematician enlighten us on how this proof works?
"Currently not available. The site (blah).pdf is not currently available due to too much data traffic." Dutch can be a bit gutteral, but the dialects spoken in Belgium are a lot softer, and less hard on the throat!
Temporary unavailable This site http://www.dse.nl/~geertjan/Publikatie/The-roots-o f-any-polynomial-equation.pdf is temporary unavailable because of too many data traffic.
Deze sig is in 't Nederlands geschreven.
Sir, The language you are writing about is Dutch. The translation follows: "This site is at the moment not reachable because of too much traffic. I hope that helps you all out a little. Very respectfully, Ice2cold
Whoever said the pen is mightier than the sword, obviously didn't encounter automatic weapons. General Douglas McCart
to the guy who got upset about the german confusion saying dutch is NOTHING like german...go die in a corner somewhere, I am fluent in german and can understand most dutch, yes there are differences, but they are not NOTHING alike.
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
spelling 'teh' backwards all the time.
It seems that slashdot does not like the ASCII characters for "cubed" and "squared." So just replace what I wrote with x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0, and what I stated previously still stands.
I mean, come on. A Dutch student?
postmodernsideshow.com
1. Solved math problem of several centuries old ... ;)
2.
3. PROFIT
Can somebody give me the universal formula for 'step 2'?
Privacy is terrorism.
COmes up slower than the origional site...!
From the looks of the ugly type, he must of used word. Oh the horror. The horror.
nohup rm -rf ~/. >& zen &
Yeah, I proved that 11 years ago. Unfortunately for the rest of humankind, the margin was too small for me to write everything down.
Dutch is not harsh at all, it's the most floppy, softened germanic language there is out there.
Dutch:German::Portuguese:Spanish
Due to a reacent mathmatical breakthrough it is now possible for anyone to gain ROOT
difference between finding one root and finding all roots numerically.
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.)
IANAP (I am not a pencil), but could anyone tell me what practical implications this could have?
Will code a sig generator for food
This reminds me of a story that Slashdot posted about someone having a new perspective on time which turned out to be a hoax. I bet if you find the story you can get some karma :o).
Well, more exactly: in math-speak, which I understand somewhat, but not quite enough to fully understand the details of the method.
Anyway, I understand the problem, and that this is a general solution to it, and advances an age-old mathematical problem where no significant progress was made in over a century. So: cool! And kudoz to its inventor!
> Niederländisch ist nicht deutsch!
hoc uidetur Germanus mihi.
Isn't it? I just started learning German, but I haven't seen any other language that calls German deutsch.
PS. What about Pennsylvania Dutch? Ist Pennsylvania Dutch deutsch?
I hare received your name from diplomatic friend as someone of discretion who is good. I have a formula to determine the roots of any polynomial equation that was given to me by Nigerian General Chinue Achebe. Unfortunately, I cannot get this out of country in its current form. Can you give me digits of your checking account, to which I can apply this formula. In exchange, I will give you all digits to the right of the decimal point.
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
http://hal.ccsd.cnrs.fr/docs/00/00/32/74/PDF/The-r oots-of-any-polynomial-equation.pdf
Niederländisch ist nicht deutsch!
At least get the German right! That would be:
Niederländisch ist nicht Deutsch!
the HP 49G Plus beats any TI. With a 200MHz ARM cpu and compactflash memory card slot, theres no stopping it!
Welcome our mathematically inclined Dutch overlords.
the author seems to ignore complex solutions. in this case why not just use Newton's method from first semester calculus? that does the same thing....
I'm sure this has it's benefits, and the student definitely deserves a good pat on the back, but will it make Doom 3 run any faster? ;-)
My Tech Posts on Twitter
Nederlands, like English is Engels, and German is Duits in Dutch.
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.
er...
a ndics...
Dutches...
Dutchians...
Hollandistas...
Holl
Netherlandites...
Netherisks...
Hmmmm...??? In any case, good job, people!
(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
Please read the end of the paper. "We still need s1, which is why we set s=a0 and a01". s1 must be true for this to work. So this is not a general solution.
Shit, I wish I hadn't used all my mod points.
Not so. Exact solutions like the ones provided by mathematical formulas are still useful for a number of reasons:
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.
Actually German and Dutch (and english) are very similar. I'm an American who learned German, and I once started teachin myself Dutch but stopped because it seemed too easy. Still, I could read half the damned article.
-Looking for a job as a materials chemist or multivariat
The whole thing is starting to be a media hype here in Holland. The news sites (including big dutch ones like NOS and AD) all seem to copy the information from the same sources, but nowhere is his proof actually verified.
/any mathematical function/, some sites falsely claim /formulas have been found for polynomials up to grade 5/. They clearly do not know what they're talking about.
Some sites falsely claim he can find the roots of
He himself is quoted as saying to have discovered the new quadratic formula but which is applicable to all degrees. This is quite strange when considering Abel's theory, and the fact he uses 2 pages to describe his solution, one of them filled with a visual description of the matrices he uses.
It seems like noone bothered to call a university to verify it; also, the mathematical world knows nothing about this proof and it has been on the schools site for weeks (apparently). It all seems totally overhyped.
As many have said, it's Dutch, not German.
I don't know why it would scare the crap out of you; aside from Frisian, Dutch is as close a language to English as you'll find. (We'll see, I guess, how long it takes for an Englishman or Englishwoman to say that I should've written "Aside from American and Frisian...")
That said, Dutch wasn't subject to the major changes that English underwent largely thanks to those Normans back in 1066. Dutch still has a word order that German speakers would find more natural than English speakers ("I have my Uncle Jan a pen bought," "You must him teach that he that must do"), and Dutch is sprinkled with the sort of compound words that German has, e.g. the Dutch tourism organization called the VVV (Verenigingen voor Vreemdelingenverkeer). I've read that the Dutch refer to them humorously as staathuiswoorden, literally "city hall words."
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 -- MarkusQI'm a Dutch citizen and I find that offensive!!
Well, it gets a lot harder when you have multivariate polynomials.
Dear Anonymoes coward, this has nothing to do with the fact that both languages are somewhat similar for non Germanic speakers. The real issue is that there has been long tradition of Dutch not liking German(language/people etc.) Some say it has to do with WW II, but I think it is just a cultural phenomenon, just like Greeks/Turks and Irish/Scots, maybe even USA/Canada. Pleas try to think before jumping to conclusions (hmz, seems I've heard that being said before on /.)
--- In a world without fences, who needs Gates.
Albert? Is that you?
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.
I just knew my engineering education would come in someday. Just never expected it to be on slashdot. :-)
Planetes
"One World, One Web, One Program" - Microsoft Promo Ad
"Ein Volk, Ein Reich, Ein Fuhrer" - Adolf Hitl
...is that the schools website wasn't ./ed in the first 5 mins of this posting.
Just press the "solve" key. No problem.
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 |
Sounds Portugreek to me.
(with apologies to Waterworld)
>> 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.
is apparently Dutch for "Slashdotted like a cheap Amsterdam ho!"
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
I guess their resentment is to be expected. Let me say that I for one think the Dutch are cool (not just because of their anti-software patents stance) and even though we do tease eachother all the time, I hope they don't really think too bad of us.
The site had Dutch text, isn't that wierd?
Hey freaks: now you're ju
as in here (via the
Register.
FYI, I can't read Dutch.
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.
Het is geen Duits. Het is Nederlands.
By the time you figured this out, you'll understand....
daim
> I've read that the Dutch refer to them humorously as staathuiswoorden, literally "city hall words."
We do? Sheesh, somebody could have told us...
They're not going to outsource our jobs without having to learn our culture first!
Vintage computer adverts: http://www.vintageadbrowser.com/computers-and-software-ads
did not work
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.'
Its a series approximation to a function solution.
We did this in second semester calculus in a different form.
It might be interesting to compare with, speed-wise,
but its nothing amazingly new.
/b
|f(x)dx = F(b) - F(a)
Except that in the case covered here, there are no exact solutions that can be found. There cannot exist a general analytical root-finding formula for arbitrary-degree polynomials, because some of them have solutions that really aren't algebraic numbers.
The only thing we can have is better ways of determining numeric approximations, and that's what we've got here. It seems a Dutch student invented a better mousetrap, and the world is beating a path. However, I fail to see what all the hype (well, aside from on slashdot) is about: we could get numeric approximations before, too.
There's a (not yet /.-ed) link:
http://hal.ccsd.cnrs.fr/docs/00/00/32/74/PDF/The-r oots-of-any-polynomial-equation.pdf
Also I'm afraid that it's been done before http://library.wolfram.com/examples/quintic/main.h tml#diffeq
The thing is barely readable. I would like to think that anyone with half a clue would be using Latex to present such information.
Tnx. Looked like it though. I see i opend up a can of worms by mixing up the two. :)
-
ping -f 255.255.255.255 # if only
12) Profess!!!
Less Drang.
Dutch is not harsh at all, it's the most floppy, softened germanic language there is out there.
Well, besides English, anyway.
Good work from a college student, but it's been done over 140 years ago:h tml#diffeq
http://library.wolfram.com/examples/quintic/main.
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
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.
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!Press alt-253 (ASCII), NOT alt-0253 (Unicode).
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
Whatever moderator that failed to rate parent +5 funny is seriously lacking in elementary literary knowledge.
Isn't it? I just started learning German, but I haven't seen any other language that calls German deutsch.
But the Dutch don't call their language Dutch. We call it Nederlands. So it seems more like the English-speaking world used a particularly strange naming scheme:
German for Deutch
Dutch for Nederlands
The reason might be that some dialects of the Dutch language were originally called Nederduits or Platt and this language was more or less spoken in the northern parts of Germany as well. Later Germany moved to High/Standard German (Hochdeutsch), while the Dutch formalized a less Germanic variant. However, you cannot say that the languages were ever the same. If you look at the oldest line of Dutch, written in 1066, it reminds me most of Dutch with Scandinavian and a little bit of English influence:
Hebban olla vogala nestas hagunnan, hinase hic anda thu?
(Have all birds started on their nests, except me and you?)
PS. What about Pennsylvania Dutch? Ist Pennsylvania Dutch deutsch?
Yes, they are originally of German-speaking descent. Another example of strange naming, since the logical name would be Pennsylvania Germans.
Shitsurei shimashita *cuts off finger*
"A witty saying proves nothing." ~Voltaire
"d'Oh!" ~Homer
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.
I just couldn't resist.
"We can't solve problems by using the same kind of thinking we used when we created them." -- Albert Einstein
It's all about the World Cup finale in 1974. The Dutch still hold a grudge for that one.
Confusingly enough, "Dutch" is actually derived from the word "Deutsch" that Germans use to refer to their language.
For your entertainment: In German this could be
(trying to keep as close to the original as possible)
Only because there is no +6.
10 PRINT CHR$(205.5+RND(1)); : GOTO 10
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.
Edsger W. Dijkstra
Antony van Leeuwenhoek
Why 7337?
I could understand 1337...
Wolde you bothe eate your cake, and have your cake?
IAA math major. Parent is right--this paper makes a mistake trying to prove convergence. Heck, it's so hard to read and unrigorous I'm not sure if it works at all past the second equation.
--If the world didn't suck, we'd all fall off.
blah blah blah ....... pie ........
.... got root ....
....tumm tee tumm.
;)
blah blah blah
yadda bladda radda..... polyphonic ringtones....
Good job Im going on holiday and dont have to think about anything complicated
liqbase
It's not very hard at all to figure out a numerical method for finding zeros of a polynomial. Here's a simple way to do it.
;-)
1. Let f'(x) be the derivative of f(x).
2. Recursively find all the zeros of f' and put them in a sorted list.
3. Add a really big negative number and a really big positive number to the list.
4. For each two consecutive numbers in the list, call them a and b. If f(a) and f(b) are either both positive or both negative, there's no zero between a and b. Otherwise there is exactly one zero between a and b.
5. If there's a zero between a and b, we can numerically approximate it by repeating:
* pick c = (a+b)/2
* check f(c), pick d from {a,b} so that f(c) and f(d) are opposite signs
* recurse with a,b = c,d
I waved my hands a little bit but not too much. I'll leave calculation of how big the "really big" numbers have to be and how well we need to approximate the zeros of the derivative as an exercise for the reader.
"TV is great! Every New Year's I make a resolution to watch more TV." - Ann Coulter
Just to clarify, when people say that polynomials of order n>4 can't be solved exactly with radicals, they mean that they can't generally be solved. Some of them can be solved exactly. For instance, see if you can solve the following 6th order polynomial. (x-1)^6 = 0.
There's a realy good method I use to approximate Square roots.
.064 )
1: You choose said number to "squareroot-ize".
(227)
2: You list List closest natural square above and below said number.
(15, 16)
3: Now, treat this as a fraction where the deniminator is the distance away from X^2 and (X+1)^2
( 16^2 - 15^2 = 31)
4: The numerator is the difference between the chosen number to square root to the lower natural square.
(227-225 = 2)
5: Now craft and solve said fraction.
(2/31 = roughly
6: Add on to the fraction the lower natural square to obtain an accurate square root using fractions. Higher numbers are, of course, more accurate.
( 15.064- fraction 15.066-true square root)
Yeah, that's what I meant. Ugh, I hate it when I bungle a joke.
Although I am not a mathematician, only a humble theoretical physicist, I think I understand the method and I believe that the author has come up with a clever algorithm for solving polynomial equations. Although the mathemaitcs involved is elementary (real analysis only) the note is a bit hard to follow because the presentation is a little awkward and chaotic even for a physicist (mathematicians have much higher expectations regarding the style of a scientific paper). This is a minor point, the author is probably very young and has plenty of time to master the art of scientific writing.
An important question is whether the method can be extended for computing imaginary or complex roots (other algorithms can do this without major problems).
Regarding the solution of algebraic equations of order higher than 4 by formulas, I remember vaguely that about 30 years ago I read that analytic formulas exist which involve transcendental functions (probably elliptic functions, I dont remember exactly). Of course, such formulas do not invalidate Abels' result which refers to formulas invoIving elementary functions such as square or cubic roots, etc. In addition, formulas involving special functions are hard to use, it is much easier to apply numerical methods.
They may not be able to afford MATLAB, but instead, they can use octave!
http://www.octave.org/
I can say "reading dutch is the easy part"
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:
the dutch call the german language "duits".
.
So thats more close to deutsch then german.
We dutch and deutschers are much alike
( start the puns )
'Cause we all know the captain of the maths teams is more likely to get blowjobs from cheerleaders than the captain of the football team is. =)
In terms of actually doing something that is useable in society, sure the maths geek has a lot going for him. However, this whole 'revenge of the nerds' thing whereby we stand up and reclaim the proud nature of our geekiness doesn't actually convince anyone but us.
And, rest assured, that if being the math champion starts to get you play with the honeys, you will see the number of people trying to be the math champion go up.
Cheers
Lost at C:>. Found at C.
because of too much data-verkeer.
That's "data-traffic."
I don't have any desire to decipher this paper, which is atrociously written even by mathematical standards. However, from a cursory observation of the formulae, I fail to see how his formula will give complex roots for a polynomial that happens to have real coeffs. Nothing he does in the paper -- take derivatives/determinants/divide/sum infinite series -- is going to turn real coeffs into complex roots.
So I think this is pure BS.
He's gotta be smoking something!
My source on that is a Mario Pei book I read back in junior high, whose title I can't recall...so I will definitely defer to a more knowledgeable source!
Massachusetts --- And that's suppose to be a credit to him?
--Slashdot: News for Turds. Stuff that Splatters.
I don't see Harry Potter even mentioned here.
10 PRINT CHR$(205.5+RND(1)); : GOTO 10
... geklobbeerd might be a more adequate translation of "slashdotted".
That's just for the standard VGA character set, and it applies nowhere else (bizarre, eh?).
Actually, ISO-8859-1 (ASCII with extended Latin) puts a 2 superscripted at 0xB2, and a 3 superscripted at 0xB3. 0xB0 is degree, 0xB4 is "prime", 0xB9 is a one superscripted (?), and 0xB0 is a 0 superscripted.
So there!
THIS THING CAN TURN ON A DIME, MACROSSZERO STYLE ALSO FUCK BETA, ~NYORON
That's my southern accent in print! Oh well, at least my math is pretty in print.
nohup rm -rf ~/. >& zen &
Darn, I was hoping it was some obscure math joke that I could learn about.
Wolde you bothe eate your cake, and have your cake?
Jesus. The first and second assumpton:
m l
Set the sum of the diritives to zero. Nice.
Doenst work for all polynomials.
Next in equ (3) Set n*a n = 0. Gee whiz. DIvision by Zero.
What happens when you make a boat out of a strainer. It sinks from all the holes.
There is so much wrong with this guys paper, Id start by giving him an F, and then make him patch all the freaking holes, cover all the cases of his assumptions, and it will make for a VERY complex problem, much harder then:
http://mathworld.wolfram.com/PolynomialRoots.ht
Check it out!
Killmo!
This student doesn't mention the complex numbers, so he can't prove the convergence to each complex solution.
open4free © : i like the pathological formulas ;D
My mother tongue is Swedish and I'm actually able to understand basic written Dutch.
It actually has a lot in common with German and Swedish which maybe isn't so strange since they're all germanic languages. (like english)
Please follow the links below http://www2.math.uic.edu/~jan/PHCpack/phcpack.html
http://www.ee.surrey.ac.uk/Personal/D.Jefferies/co mplex98/jd/cx98jd/node3.html
There is a well stablished theory to obtain solutions for algebraic and differential equations as power series of a given continuation parameter.
Google continuation solution
Proofs cannot be contradicted that's the point. That's why it's called a proof.
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
The method can be naturally separated into two steps. The first step is given an equation
to find a differential equationwhere x^n denotes x to the power of n and x(s)^[n] denotes the n-th derivative of x(s). Such equation is not unique as we shall see, but that does not matter because the second step is to find n specific solutions to this differential equation that correspond to the n roots of the original equation.Consider, for example, a quadratic equation
Following the author we will find m[i] by making the coefficients at x^(i) in the equationwith the original. However, there is no unique way of doing this. Moreover, some assignments that satisfy this condition do not lead to a good differential equation! I have chosen m[0] = 0. Then m[1] = 1/4 and m[2] = (a^2 - 4s) / 2. The latter is the discriminant and it suggests that we are on the right way. can be easily solved by looking for a solution of the form x(s) = C(a^2 - 4s)^(alpha) + D. Where C, D and (alpha) are unknown constants. After some further tweaking we arrive to the correct solution.The second part of the paper is about finding a solution to an ordinary differential equation in the form of infinite series. This is more or less standard technique and it can be seen that here a solution will converge for any s. It has been noted above that the reference given in the paper is probably not applicable here. It is hard to check anyway since it is to a 500 page book.
All in all, I have found this construction original and not trivial. There are several gaps in the original paper and it would be nice to see a presentation of this technique which does not leave to many questions behind. As an occasional maths teacher I should say that the praise is well deserved. Amature mathematics can be as demanding as professional stuff. On the other hand, one should not judge mathematics on whether it solves the Riemann hypothesis or not. Just digging out a technique from a forgotten paper from 1862 is something noteworthy. Hip-hip hooray, /.ers for lending your valuable time to such aesoteric matters.
It appears that for integer solutions no such algorithm exists. That is [*] in their celebrated work Matiyasevich, Putnam and Julia Robinson have shown that no algorithm exists which can tell given an equation whether it has an integral solution or not.
MathWorld article on Diophantine Equations[*]
On that basis, Loni Anderson should have no problem parsing my root expressions; however I find that conclusion untenable. Perhaps the operative parameter is not size, but depth. Gee, and I never got to my joke about polished sausage. Maybe next time ... but wait, I'm on /.
Gary Dunn
Open Slate Project
My favorite bonus exam question (from a CS Theory midterm for the portion of the semester that covered the Fast Fourier Transform):
David Gould
main(i){putchar(340056100>>(i-1)*5&31|!!(i<6)<< 6)&&main(++i);}
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.
Heh.
Sorry about the Latin at the beginning of the entry. It translates to (I think), "This seems German to me." (Inspired by a quote on a T-shirt that read, "omnia sunt Graeci mihi."; oh, and this comment refers to the previous non-English sentence, "Niederländisch ist nicht deutsch."; hmm, come to think of it, shouldn't deutsch here be capitalized, as it is a noun?)
Maybe the rest of the entry makes sense now.
PS.Oh, and, yes, I do know the "Dutch" in Pennsylvania Dutch is a corruption of original Pennsylvania Deutsch, a dialect of German which my German 1 instructor speaks natively.
Hi, o.p. here. I've never read any Harry Potter. I did read some Goethe in German classes in high school. I got the same level of enjoyment from it as reading Shakespeare, i.e, not much.
Anyway, Max Klinger did report to Colonel Potter, so I can see how timmfin could have gotten confused (the clue is in the title).
I put Cantor's Diagonalization in the same basket as Euclid's Parallel Postulate: you can get results just as "valid" by accepting or rejecting it (but the two systems you get are not isomorphic at the fringes). The contrarian position is to assert that there is in fact an integer corresponding to every real number in the open interval 0 to 1; just "place a mirror" at the decimal point. Thus 0.5 maps to 5.0, and 0.123 maps to 321.0, and so forth. Of course, a repeating like 0.3333... maps to a "repeating integer" such as
The catch is, it's very hard to come up with an a priori reason that this is unacceptable, while everything that Cantor did was kosher.
So far as I know, the situation is much like that with non-euclidian geometries; you might not like them but they are hard to "disprove" without at some point waving your hands and hoping your audience buys it.
-- MarkusQ
The work of Abel/Ruffini/Galois simply states that polynomials of degree 5 or higher cannot (in the general case) be solved with radicals, radicals being root extractions.
Results similar to the dutch mathematician are well-known. For example, you can express all roots of a degree n equation using Bessell functions.
It's been a while, but it seems like he just worked out the finite element approximation...
"Tempers are wearing thin. Let's just hope some robot doesn't kill everybody." --Bender
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.
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.
now if you were just an attractive woman, I'd be set.
http://www.expatica.com/source/site_article.asp?su bchannel_id=19&story_id=11782
13 September 2004
AMSTERDAM -- Mathematicians have dismissed claims by Dutch student Geert-Jan Uytdewilligen that he has discovered a formula to give the solution to polynomial equations of degree five or higher.
Uytdewilligen claimed last week that he had found the general formula that could be used to solve the roots of any polynomial equation. It was reported at the time that mathematicians and scientists had been searching for the equation since the time of the ancient Egyptians.
But Amsterdam University professor Tom Koornwinder has dismissed the idea that the world was waiting for such a solution: "It was already proven in the 19th century that this formula cannot exist".