International Journal of Algebra, Vol. x, 200x, no. xx, xxx - xxx
Fundamental Theorem of Algebra Constructively Alvaro H. Salas Universidad de Caldas, Manizales, COLOMBIA Universidad Nacional de Colombia email :
[email protected] FIZMAKO Research Group Abstract In this paper we prove the Fundamental Theorem of Algebra (FTA) by using techniques that allow to develop an algorithm to …nd zeros of polynomials with complex coe¢ cients. This algorithm is based on d’Alembert’s proof of the FTA. We called it d’Alembert’s Descent Method.
1
Introduction.
The Fundamental Theorem of Algebra is one of the most fundamental results in mathematics. Though several attempts at proofs had been given, the proof by Frederick Carl Gauss in 1799 is generally considered the “…rst complete proof” of this substantial theorem. Although Gauss himself considered his proof ‡awless, he still assumed a key idea that was not proven until the mid 1800s with the works of Bolzano and Weierstrass. Prior to the proof by Gauss was a proof by d’Alembert in 1746. Although this proof is now known to also be correct, at the time it had far too many gaps to be considered by the mathematical world as accurate. Interestingly enough, we now know that the holes in d’Alembert’s proof are easier to …ll than the assumption in Gauss’s proof. Using Descartes’s discoveries about polynomial factors, d’Alembert was able to come up with an intuitive proposition to prove the FTA. The key to d’Alembert’s proof was this proposition now respectfully called d’Alembert’s Lemma. The Lemma states that if p(z) is a non-constant complex polynomial and p(z0 ) 6= 0, then any neighborhood of z0 contains a point z1 such that jp(z1 )j < jp(z0 )j. The proof of this lemma was given by Puiseux in 1850. Since d’Alembert could not prove his proposition, his proof was considered too weak to be an actual proof to the FTA.
2
A. H. Salas
2
The Fundamental Theorem of Algebra and its proof.
Let p(z) = a0 + a1 z + a2 z 2 + be a polynomial of degree n
+ an z n , an 6= 0
(1)
1 with real or complex coe¢ cients.
Lemma 1. For any complex number z0 and for every " > 0 there exists = ("; z0 ) > 0 such that j jp(z0 +h)j jp(z0 )j j
jp(z0 +h) p(z0 )j < " for any h satisfying jhj < : (2)
Proof. We have p(z0 + h) = c0 + c1 h + c2 h2 +
+ cn hn ,
(3)
where ck = ck (z0 ) =
1 (k) p (z0 ), k = 0; 1; 2; :::; n: k!
Observe that c0 = p(z0 ). Let M = maxfjc1 j; :::; jcn jg = M (z0 ; h):
(4)
It follows from (3) and (4) that jp(z0 + h)
p(z0 )j
jc1 hj + jc2 hj2 +
+ jcn hjn
M jhj(1 + jhj +
+ jhjn 1 )
= M jhj If we de…ne
1 jhjn 1 jhj
M jhj : 1 jhj
1 " ; 2 2M + 1 it follows from (5) that
(6)
= min then for any h satisfying jhj < jp(z0 + h)
p(z0 )j
M jhj = 2M jhj 1 1=2j
(5)
2M
" < ": 2M + 1
(7)
Lemma 1 says us that both jp(z)j and p(z) are continuous functions on the whole complex plane. In particular, If zj ! z0 (j ! 1) then p(zj ) ! p(z0 ) and jp(zj )j ! jp(z0 )j (j ! 1) If h ! 0 then p(z0 + h) ! p(z0 ) and jp(z0 + h)j ! jp(z0 )j for all z0 :
(8)
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FTA Constructively
Lemma 2. For any R > 0 there exists r > 0 such that jzj > r implies jp(z)j > R:
(9)
lim p(z) = 1:
(10)
C = maxfja0 j; :::; jan 1 jg:
(11)
In other words, z!1
Proof. Let We have an z n j
jp(z)
+ jan 1 jjzjn
ja1 jjzj + ja2 jjzj2 + C(1 + jzj + = C
jzjn jzj
1
+ jzjn 1 )
(12)
1 1
Let jzj > 1. It follows from (12) that jp(z)j = jan z n + p(z) jan jjzjn
C
= jan jjzjn 1 jan jjzjn 1
an z n j
jzjn jzj
jan z n j
an z n j
jp(z)
1 1 (13)
C 1 jzj n jan j jzj 1 C 1 jan j jzj 1
:
We have proved that jp(z)j
jan jjzjn 1
Let r = max
C 1 jan j jzj 1 s n
for any jzj > 1:
! 2R 2C ; +1 jan j jan j
(14)
(15)
It is clear from (14) and (15) that if jzj > r then jp(z)j
jan jjzjn 1
C jzj 1
> jan j
2R jan j
1
1 2
= R:
(16)
4
A. H. Salas
Lemma 3 (d’Alembert’s Lemma) Let + an z n , an 6= 0
p(z) = a0 + a1 z + a2 z 2 +
be a polynomial of degree n 1 with complex coe¢ cients. If is a complex number such that p( ) 6= 0 then there exists a complex number 0 such that jp( 0 )j < jp( )j. Proof. In view of (15) with z0 = and h = z we may write p(z) in the form p(z) = b0 + b1 (z ) + b2 (z )2 + + bn (z )n ; (17) where 1 (k) p ( ), k = 1; 2; :::; n k!
bn = an 6= 0, b0 = p( ) 6= 0 and bk = Let 1
m
(18)
1:
n be the least integer for which bm 6= 0. Consider the polynomial (19)
q(z) = p(z + ): We have q(z) = b0 + bm z m + bm+1 z m+1 +
+ bn z n :
(20)
We may write q(z) in the form q(z) = b0 + bm z m + z m+1 g(z); where g(z) =
+ bn z n
bm+1 + bm+2 z + 0 if m = n:
Consider some number by
such that 0
0
(26)
5
FTA Constructively
and choose the number
= ( ; ) subject to 0
N: (32)
By Lemma 2, for R =
+ 1 there exists r > 0 such that
jp(z)j >
+ 1 for any jzj > r:
(33)
r for any j > N:
(34)
It is clear from (32) and (33) that jzj j
We see that the sequence (zj )j is bounded. By the Bolzano-Weierstrass theorem there exists a convergent subsequence of this sequence. Without loss of generality we may assume that zj ! z0 (j ! 1). By Lemma 1 and from (31), jp(zj )j ! jp(z0 )j =
(j ! 1):
(35)
We claim that p(z0 ) = 0: Indeed, if p(z0 ) 6= 0 we could apply d’Alembert’s Lemma to = z0 to obtain a complex number 0 such that jp( 0 )j < jp(z0 )j = ; which is a contradiction.
6
3
A. H. Salas
Solving polynomial equations
Suppose that at every point z of the complex plane a perpendicular is erected whose length (for the given scale unit) is equal to the modulus of the value of the polynomial p(z) at this point, that is, is equal to jp(z)j. The end points of the perpendiculars will, in view of the above proved continuity of the modulus of a polynomial, constitute some continuous curved surface situated above the complex plane. See Figure 1. The Lemma 2 shows that as jzj increases this surface recedes from the complex plane, though quite naturally the recession is not in the least monotonic.
Figure 1. The surface corresponding to jp(z)j Using this geometric point of view, we can describe d’Alembert’s lemma in the following fashion [1]. Given that jp( )j > 0. This means that the length of the perpendicular erected to the complex plane at point is nonzero. Then, by d’ Alembert’s lemma, there is a point 0 such that jp( 0 )j < jp( )j; that is, the perpendicular at the point 0 will be shorter than at the point z and, consequently, the surface formed by the endpoints of the perpendiculars will at this new point 0 be somewhat closer to the complex plane. Obviously, the roots of the polynomial p(z) will be those complex numbers (or those points of the complex plane) at which the surface formed by the
7
FTA Constructively
endpoints of the perpendiculars touches this plane. It is impossible to prove the existence of such points by relying on d’Alembert’s lemma alone. Indeed, using this lemma it is possible to …nd an in…nite sequence of points z0 = , z1 = 0 , z2 ,...,zn ,...such that jp(z0 )j > jp(z1 )j > jp(z2 )j >
(36)
However, it does not follow from this that there exists a point z such that p(z) = 0, all the more so that the decreasing sequence of positive real numbers (36) does not necessarily have to tend to zero. Despite of this, we sometimes may obtain a good approximation to a zero of a polynomial by using D’Alembert’s idea. We will call this D’Alembert’s descent method. Observe that, in view of (28), m jp( 0 )j 1 + m+1 D jp( )j < jp( )j (37) and then jp( 0 )j
'( ) jp( )j ,
(38)
where
1 : (39) D The quantity '( ) measures the descent in the sense that jp( 0 )j is approximately the 100'( ) % of jp( )j. The less the number '( ) the deepest the descent. The quantity '( ) attains its absolute minimum for '( ) = 1
= If
m (m+1)D
m
m+1
D < 1, 0
r. This means that if jp(z)j 1 then jzj r. In particular, all roots of the equation p(z) = 0 lie on the disk jzj r:
8
A. H. Salas
We …rst choose some complex number jp( )j > 0. Let m = minf1
j
satisfying conditions j j
(j)
n j p ( ) 6= 0g and
r
=
m
a0 (m) p (
)
r and
(41)
:
If m < n we de…ne a number D by D = D( ; ) =
bm
(jp(m+1) ( )j + jp(m+2) ( ) j +
+ jp(n) ( )
n m 1
j) > 0: (42)
Finally, let
= ( ; )=
8 > > > > < > > > > :
m (m+1)D
if m < n and
m (m+1)(D+1)
m (m+1)D
< 1:
m (m+1)D
if m < n and
1:
(43)
if m = n:
Then jp( 0 )j < jp( )j, where 0 = + . If p( 0 ) = 0 we have found a root. Otherwise, we apply D’Alembert’s lemma to 0 to get a complex number 00 satisfying jp( 00 )j < jp( 0 )j and so on. De…nition 1. Given a positive number "; we will say that the complex number r is an "-root of the polynomial p(z) if jp(r)j < ". We may apply D’Alembert’s Descent Method until we get a reasonable good approximation to a root of polynomial (29) in the sense of De…nition 1. Following example illustrates the method. Example 1. Let us …nd an "-root of the following tenth degree polynomial for " = 10 3 : p(z) = 6z 10
3z 9 + 2z 8
5z 7 + 9z 6
7z 5 + 15z 4
2z 2 + 17z
8:
(44)
Let z0 = = 0:5. Then jp( )j = jp(0:5)j = 0:828125 > 0. From (40),(41),(42) and (43) we …nd that r =
20 6:7, m = 1, 3 0:0185474:
0:03403 and
0:0383572, D
Then all zeros of p(z) lie on the disk jzj 20 and z1 = 3 Observe that jp( 0 )j 0:4337 < jp( )j = 0:828125:
0
= +
(45)
0:48145259.
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FTA Constructively
Calculations are shown in Table 1. This table says us that the number r = z10 = 0:46046219 is an "-root of the polynomial (44) . j
zj
0 1 2 3 4 5 6 7 8 9 10
0:50000000 0:48145259 0:47127374 0:46593407 0:46319866 0:46181419 0:46111772 0:46076841 0:46059350 0:46050597 0:46046219
D 0:0383572 0:03403 0:0206995 0:01679 0:0107679 0:008291 0:00549332 0:004112 0:00277461 0:002047 0:00139436 0:001021 0:000698957 0:0005097 0:000349923 0:0002547 0:000175073 0:0001273 0:0000875643 0:00006365 0:0000437891 0:00003182
100'( )%
jp(zj )j
52:44 51:23 50:62 50:31 50:15 50:08 50:04 50:02 50:01 50:00 50:00
0:828125 0:4337 0:2221 0:1124 0:05655 0:02836 0:01420 0:007106 0:003554 0:001778 0:0008888
Table 1.
4
Conclusions.
We gave an indirect proof of the Fundamental Theorem of Algebra. This proof is non constructive. However, it is based on d’Alembert’s Lemma which helps us to seek a root of a given polynomial. In this sense we have proved the FTA constructively.
5
References
[1] A. Kurosh, Higher Algebra, Mir Publishers, 1980. [2] Fine and Rosenburg, The Fundamental Theorem of Algebra, SpringerVerlage, New York (1993). [3] J. Fraleigh, A First Course in Abstract Algebra Seventh Edition, AddisonWesley, Boston (2003). [4] J. Stillwell. Mathematics and Its History, Second Edition, Springer Undergraduate Texts in Mathematics, New York (2000).
