From: "denis-feldmann" Subject: Re: The request of an Euler devotee... Date: Thu, 29 Jun 2000 08:16:40 +0200 Newsgroups: sci.math Summary: [missing] William Hale a écrit dans le message : hale-2906000003310001@dialup106.tcs.tulane.edu... > In article <20000628212020.02197.00000279@ng-bj1.aol.com>, styble@aol.com > (Styble) wrote: > > I've heard more than once that my mathematical hero--who sadly doesn't > > get even one-seventeenth the ink that the redoubtable (but still overrated*) > > Newton generates--came up with no less than four independent proofs of the > > Fundamental Theorem of Algebra, and that the last of which was after he'd lost > > what little remained of his eyesight! > > I don't think Euler proved the Fundamental Theorem of Algebra. > Right. But he gave (with Lagrange) the essential ideas of the second Gauss proof below. > I thought Gauss was the first to prove the Fundamental Theorem of Algebra. Correct > I believe that this was his thesis. Yes (in fact, his doctoral dissertation) I believe that he gave four or so > independent proofs of this over time. Four (and seven (!) for the law of quadratic reciprocity) > > > Can any of you math mavens confirm this? And maybe, just maybe, kinda > > sorta walk me through them? (I don't expect you to set the proofs out at > > length, mind you...I'm just hoping maybe you can highlight the respective > > strategies the Swiss Master employed, if not any of his actual tactics). > > I think that the first proof was geometric. > Using "topological" properties of the lines P(x,y)=0 and Q(x,y)=0, with P and real and imaginary parts of the polynomial to solve. > I think another proof used the method of transforming the given polynomial > whose roots we want to find into another polynomial where the odd factor > of the latter polynomial is smaller than the odd factor of the former > polynomial, althought the even factor of the latter polynomial is greater > than the even factor of the former polynomial. This is the Euler-Lagrange idea. In modern terms; he reasons on polynomials on IR (as conjugation gives then the general results), and in a splitting field IK of P= \prod (z-a_i) (with a_i\in IK), he puts Q=\prod (z-(a_i+a_j+ca_ia_j)), with c \in C fixed, and i=/=j; introduce the recurrence (on the degrees , classed by d_1=2^mp eventually reduces the given polynomial to a polynomial of even degree, > which can be expressed as a product of quadratic polynomials since the > roots occur in conjugate pairs. Or, I might be confusing this with the > proof of Abel that the fifth degree polynomial is not solvable by radicals. > No, you are essentially correct. > I think the various proofs of Gauss appear in the book(s) "History of > Mathematics" by Smith. But, I am not sure about this. > The third proof essentially mimics the classical proof by Cauchy (integrating 1/P on [z|=R and using the residues theorem), but this is explained using only real integration; a 'tour de force' made necessary by the fact that complex integration was not yet made rigourous at the time. I think the fourth proof use homotopy considerations (from a modern point of view) but i am not sure. > -- > Bill Hale ============================================================================== From: hale@mailhost.tcs.tulane.edu (William Hale) Subject: Re: The request of an Euler devotee... Date: 3 Jul 2000 03:07:00 GMT Newsgroups: sci.math In article <20000628212020.02197.00000279@ng-bj1.aol.com>, styble@aol.com (Styble) wrote: > I've heard more than once that my mathematical hero > came up with no less than four independent proofs of the > Fundamental Theorem of Algebra. > > Can any of you math mavens confirm this? And maybe, just maybe, kinda > sorta walk me through them? (I don't expect you to set the proofs out at > length, mind you...I'm just hoping maybe you can highlight the respective > strategies the Swiss Master employed, if not any of his actual tactics). I may be setting out the proof at more length than you wish, but I think that it contains many interesting strategies. Pierre Samuel, in his book "Algebraic Theory of Numbers", has a two page exercise on the proof of the fundamental theorem of algebra using a method due to Lagrange. See his book for details. I will give an outline of a variation of that exercise. Samuel uses Newton's technique of employing the theory of elementary symmetric functions of the roots of a polynomial. I try to avoid this by using the theory of splitting fields that is developed just prior to proving the fundamental theorem of Galois. You might not know much about splitting fields, but most of the results seem plausible and reasonable. Strategy of proof ================= Let f(X) be a polynomial over the complex of degree d. Express d as q * 2^n where q is odd. The proof will be done by using induction on n. This is a surprise, since you would expect to be using induction on just d. But, in fact, when we have reduced the proof to the case of n-1, the q will get larger, implying that d itself will also get larger. Basic facts =========== You can easily establish the following facts which I will need later: 1) A monic polynomial p(X) over the reals of odd degree has a real root. For, since the limit of p(X) is oo as X approaches oo and the limit of p(X) is -oo as X approaches -oo, the mean value theorem of calculus guarantees that there is a real number r such that p(r) = 0. This is the only place where "analysis" is used in this proof of the fundamental theorem of algera. The rest of the proof is algebraic only. In particular, I don't need any complex analysis. 2) A polynomial p(X) over the complex of degree one has a complex root. 3) A polynomial p(X) over the complex of degree two has a complex root. See Samuel for details, although this is not hard to show. 4) Let d be an integer greater than or equal to 2. Then, the number of pairs (i, j) such that 1 <= i < j <= d is (d-1)*d / 2. Fact 4) is used for the induction step to reduce case n to case n-1. For, if d = q * 2^n where q is odd, then the number of pairs (i, j) is (d - 1) * q * 2^(n-1) where (d - 1) * q is odd. Samuel actually uses the fact that the number of pairs (i, j) such that 1 <= i <= j <= d is (d + 1)*d / 2. I think my modification will work just as well and has the advantage of making the polynomials constructed in the proof to be of a little lesser degree. Setup ===== I will just give the proof as applied to the particular case of n = 6. I think that this helps to see what is actually going on. Let C be the complex numbers. Let R be the real numbers in C. Let e(X) be a polynomial in C[X] of degree 6. I want to show that e(X) has six roots in C. By using induction on the degree of e(X), it is enough to show that e(X) has at least one root in C. Set f(X) equal to the product of e(X) and the complex conjugate of e(X). Then, f(X) is a polynomial in R[X] of degree 12. Preparation =========== I need to show that e(X) has a root r and that r is in C. I might try looking at complex numbers r in C and see if I can find one that is a root of e(X). This doesn't appear to lead anywhere unless I know some complex analysis stuff. The alternate method is to try looking at roots r of e(X) and see if I can find one that is in C. Recall that the fundamental theorem of algebra is not that e(X) has roots, but that the roots are in C. The theory of splitting fields allows me to find roots of e(X), although, at this point, I only know that these roots will be in some extension field of C. However, these roots will give me something to work with, as you will see as the proof unfolds. Let K be a splitting field of f(X) over C. Let y_1, y_2, ..., y_12 be the roots of f(X) in K. Then, K = C[y_1, y_2, ..., y_12]. Since e(X) divides f(X) and f(X) splits in K, e(X) splits in K. Let z_1, z_2,..., z_6 be the roots of e(X) in K. Let L, contained in K, be the splitting field of f(X) over R. Then, L = R[y_1, y_2, ..., y_12]. Rather than showing that there is some complex number in C that is a root of e(X), I will show that one of the roots z_i of e(X) in K actually lies in C. Polynomials for the proof ========================= The proof will analyse the following polynomials that belong to the polynomial ring K[X]. There are 148273 polynomials to consider! polynomial degree roots form of roots ---------- ------ ------------- --------------------------- e(X) 6 z_1, z_2,..., z_6 f(X) 12 y_1, y_2,..., y_12 g_1(X) 66 x_1, x_2,..., x_66 y_i + y_j + 1 * y_i * y_j g_2(X) 66 x_1, x_2,..., x_66 y_i + y_j + 2 * y_i * y_j . . . g_67(X) 66 x_1, x_2,..., x_66 y_i + y_j + 67 * y_i * y_j h_1(X) 2211 w_1, w_2,..., w_2211 x_i + x_j + 1 * x_i * x_j h_2(X) 2211 w_1, w_2,..., w_2211 x_i + x_j + 2 * x_i * x_j . . . h_2212 2211 w_1, w_2,..., w_2211 x_i + x_j + 2212 * x_i * x_j Finding a complex root of h(X) ============================== I claim that each h(X) has at least one complex root. Note that h(X) in L[X] actually is in R[X], since each automorphism of L permutes the roots of h(X) among themselves since each root is of the form x_i + x_j + c * x_i * x_j for a fixed integer c. Thus, h(X) is a monic polynomial in R[X] of odd degree 2211. Hence, h(X) has a complex root (actually, a real root). Finding a complex root of g(X) ============================== I claim that each g(X) has at least one complex root. Consider the 2212 polynomials h(X) whose roots are defined in terms of the roots of g(X). By above, each h(X) has a complex root of the form x_i + x_j + c * x_i * x_j where the c's are different. But, there are only 2211 such forms for 1 <= i < j <= 66. By the pigeon hole principle, I can find i, j, and c not equal to d such that x_i + x_j + c * x_i * x_j and x_i + x_j + d * x_i * x_j are complex numbers. By simple algebraic manipulation, I see that x_i + x_j and x_i * x_j are complex numbers. But, x_i is a root of the polynomial X^2 - (x_i + x_j) * X + x_i * x_j in C[X] of degree 2. Thus, the root x_i of g(X) is a complex number. Finding a complex root of f(X) ============================== I claim that f(X) has at least one complex root. Consider the 67 polynomials g(X) whose roots are defined in terms of the roots of f(X). By above, each g(X) has a complex root of the form y_i + y_j + c * y_i * y_j where the c's are different. But, there are only 66 such forms for 1 <= i < j <= 12. By the pigeon hole principle, I can find i, j, and c not equal to d such that y_i + y_j + c * y_i * y_j and y_i + y_j + d * y_i * y_j are complex numbers. By simple algebraic manipulation, I see that y_i + y_j and y_i * y_j are complex numbers. But, y_i is a root of the polynomial X^2 - (y_i + y_j) * X + y_i * y_j in C[X] of degree 2. Thus, the root y_i of f(X) is a complex number. Finding a complex root of e(X) ============================== I claim that e(X) has at least one complex root. By above, f(X) has a complex root y, say. Let y~ be the complex conjugate of y. Then, y~ is also a complex number since y is. Recall that f(X) = e(X) * complex conjugate of e(X). Thus, 0 = f(y) = e(y) * complex conjugate of e(y). Then, either e(y) = 0 in which case e(X) has the complex root y, or else complex conjugate e (y) = 0 in which case e(X) has the complex root y~. -- Bill Hale ============================================================================== From: Jan-Christoph Puchta Subject: Re: Fundamental Theorem of Algebra clinic, please? Date: Fri, 14 Jul 2000 11:19:51 +0200 Newsgroups: sci.math Bryan Styble wrote: > > Again, the way the Fundamental Theorem of Algebra was explained to me > by this teacher was that the reason it's "fundamental" is that it solves that > quandry for mathematicians; that is, it tells mathematicians they need never > invent any additional new number systems--that the complex numbers are all > they'll EVER need. I think the reason why this theorem is called fundamental has to be understood historically. Once upon a time, algebra meant solving of equations. And numbers meant integers, rationals, reals or complex numbers. In this context the question whether every polynomial can be solved within the complex domain is really fundamental. However, today this theorem belongs more to complex analysis then to algebra, since defining the complex domain involves topological methods. A purely algebraic statement, equivalent to the fundamental theorem is the following: If F is a field, such that every polynomial of odd degree has a root in F, and for all x in F(i) we have x^(1/2) in F(i), then F(i) is algebraically closed. Proof: Let K be a normal extension of F, G its Galoisgroup, P its 2-Sylowsubgroup. P stabilizes a subfield L of odd degree over F, take an element x in L, but not in F. The minimal polynomial of x has odd degree, hence by assumption it has a zero in F. But the minimal poynomial is irreducible, hence of degree 1, thus L=F, and G is a 2-group. But then K is a tower of extensions of degree two, and by assumption this tower has to stop at F(i). In todays language, this theorem is neither difficult nor fundamental, and the statement "C is algebraically closed" hardly algebra. But names once given tend to survive as soon as they enter a textbook. Jan-Christoph ============================================================================== From: Richard Carr Subject: Re: Fundamental Theorem of Algebra clinic, please? Date: Sat, 15 Jul 2000 02:45:56 -0400 Newsgroups: sci.math On Sat, 15 Jul 2000, Kevin Aylward wrote: :Date: Sat, 15 Jul 2000 04:15:24 GMT :From: Kevin Aylward :Newsgroups: sci.math :Subject: Re: Fundamental Theorem of Algebra clinic, please? : : :"Bryan Styble" wrote in message :news:20000713193903.17669.00000570@ng-bh1.aol.com... :> OK you brainiacs, Styble the Calculus Washout (who's always out of dough :but :> never short on "D'oh!") will pose this one again, having received only :> whimsical or fragmentary responses the first time I posted it about a :month :> ago: :> :> I'm informed that my mathematical hero Euler over his career conceived :> and executed four completely independent proofs of the Fundamental Theorem of :> Algebra, the last of which after he'd lost what little remained of his :> eyesight. :> : :So, if its really true that every equation of degree n has exactly n roots, :then what, other then the first root = -1, is the second root of x^2 + 2x + :1 = 0 Also -1. The roots are counted according to multiplicity. -1 is a root of multiplicity 2. This corresponds to the fact that x^2+2x+1=(x+1)^2. (x+1) is a repeated factor and occurs twice. The multiplicity of a factor is well-defined since C[x] is a UFD. Another way of stating the Fundamental Theorem of Algebra which doesn't involve multiplicties of roots is to say that every non-constant polynomial f(x) in C[x] can be expressed as a product of linear factors in C[x] (every non-zero polynomial if you allow the empty product to be the polynomial 1 (as would be standard) and if you allow multiplication by a unit). I don't know which proofs Euler came up with but he was blind for quite a long time and produced a huge amount of work after he became blind (perhaps more than when he could see- I'm fairly certain that at least the rate went up). Maybe he needed the money- I think he had around 20 children or so. The easiest proof of FTA is probably the one using Liouville's theorem but that's not due to Euler as Liouville was born after Euler died (unless perhaps Liouville's theorem was not due to Liouville). I don't know if you want to know this proof (or indeed if you already do). Let f(z) in C[z] be a polynomial over C and suppose f has no zeros. W.l.o.g. assume f(z) is monic. f(z)=z^n+...+a_1z+a_0 where a_n is not zero and n>=0. If n=0, let s=a_0. If n>0, let K=max{|a_0|,...,|a_{n-1}|,1/2} If |z|>=2Kn then, |f(z)| =|z^n+(a_{n-1}z^{n-1}+...+a_0)| >=|z^n|-|(a_{n-1}z^{n-1}+...+a_0)|. >=|z^n|-(|a_{n-1}z^{n-1}|+...+|a_0|) >=|z^n|-Kn|z^{n-1}| =|z^{n-1}|(|z|-Kn) >=Kn. On the other hand, if R=max{1,2Kn} then |f(z)| is continuous on the compact set {z:|z|<=R} and thus achieves a minimum, m, there. m>0 by hypothesis. Let s=min{Kn,m}>0. Thus |f(z)|>=s>0 for all z in C (in either case). i.e. f is bounded away from zero. Thus |1/f(z)|<=s for all z in C (the domain of 1/f is C, since f has no zeros). Thus 1/f is bounded in C and it is also holomorphic (being the quotient of 2 holomorphic functions the denominator not having a zero). Thus, by Liouville's theorem 1/f is constant and hence f is constant, i.e. n=0. Hence any polynomial of degree n>0 in C[x] has a zero (in C). By division we get that any polynomial in C[x] splits over C. : :You know, it took me about 10 years to realize that, when you got abort, :retry, fail in dos without having a disk in drive a: , and then pressed for :"a" for abort and it done f all, that this was an actual bug in the code, :that they could not be bothered to fix. : : :Kevin Aylward , Warden of the Kings Ale :kevin@anasoft.co.uk :http://www.anasoft.co.uk - SuperSpice "Cheap, No Shit!", a currently free :GUI xspice, unlimited component, mixed-mode Windows simulator with Schematic :Capture, waveform display, FFT's and Filter Design. :Opinions of my employer are not necessarily indicative of my own :Oscillators don't, amplifiers do" ============================================================================== From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: Fundamental Theorem of Algebra clinic, please? Date: 15 Jul 2000 15:32:17 -0500 Newsgroups: sci.math In article , Richard Carr wrote: >On Sat, 15 Jul 2000, Kevin Aylward wrote: >:Date: Sat, 15 Jul 2000 04:15:24 GMT >:From: Kevin Aylward >:Newsgroups: sci.math >:Subject: Re: Fundamental Theorem of Algebra clinic, please? ............... >The easiest proof of FTA is probably the one using Liouville's theorem but >that's not due to Euler as Liouville was born after Euler died (unless >perhaps Liouville's theorem was not due to Liouville). I find it surprising that people do not know very simple proofs. This one involves very little. Let f be a non-constant polynomial. Then |f| is continuous, and as it blows up at infinity, it has a minimum. All that has to be done is to show that the minimum is 0. Suppose it is not. We can translate the function so the location of the minimum is at 0. Then f(x) = f(0) + c*x^k +x^{k+1}*g(x), where g is a polynomial (possibly 0), and c is not 0. Let r be a k-th root of -c/f(0), and let x = b*r, b a positive real number. Then f(x) = f(0)*(1 - b^k) + b^{k+1}*h(b), where h is a polynomial. Computing bounds on the absolute value of f(x) shows that for small b the absolute value is less than that of f(0). -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558