Kenneth Kuttler

8.12. THE FUNDAMENTAL THEOREM OF ALGEBRA 181

20. Let a,b be two positive numbers and let p > 1. Choose q such that

1p+

1q= 1.

Now verify the important inequality

ab≤ ap

Hint: You might try considering f (a) = ap

p + bq

q − ab for fixed b > 0 and examineits graph using the derivative.

21. Using Problem 20, show that if α > 0, p > 1, it follows that for all x > 0(p−1

px+

px1−p

≥ α.

22. Using Problem 21, define for p > 1 and α > 0 the following sequence

xn+1 ≡p−1

pxn +

px1−p

n , x1 > 0.

Show limn→∞ xn = x where x=α1/p. In fact show that after x1 the sequence decreasesto α1/p.

23. Consider the sequence{(

1+ xn

)n}∞

n=1 where x is a positive number. Using the bino-mial theorem show this sequence is increasing. Next show the sequence converges.

24. Consider the sequence{(

1+ xn

)n+1}∞

n=1where x is a positive number. Show this

sequence decreases when x > 2. Hint: You might consider showing (1+ y)(x/y)+1 isincreasing in y provided x > 2. To do this, you might use the following observationrepeatedly. If f (0) = 0 and f ′ (y)> 0, then f (y)≥ 0. There may also be other waysto do this.

25. Let zez−1 = ∑

∞n=0

bnn! zn. The bn are called the Bernoulli numbers. Show that

0!n!+

1!(n−1)!+

2!(n−2)!+ · · ·+ bn−1

(n−1)!1!=

{1 if n = 10 id n > 1

Hint: You might use the Cauchy product after multiplying both sides by ez−1.

8.12 The Fundamental Theorem of AlgebraThe fundamental theorem of algebra states that every non constant polynomial having co-efficients inC has a zero inC. IfC is replaced byR, this is not true because of the example,x2 + 1 = 0. This theorem is a very remarkable result and notwithstanding its title, all thebest proofs of it depend on either analysis or topology. It was proved by Gauss in 1797.The proof given here follows Rudin [24]. See also Hardy [14] for a similar proof, more dis-cussion and references. You can also see the interesting article on Wikipedia. You googlefundamental theorem of algebra and go to this site. There are many ways to prove it. This

8.12.20.21.22.23.24.25.THE FUNDAMENTAL THEOREM OF ALGEBRA 181Let a,b be two positive numbers and let p > 1. Choose g such that1 1—-+-=1.PqNow verify the important inequalityPpaab< ~ 4.pP gqHint: You might try considering f (a) = £ + a —ab for fixed b > 0 and examineits graph using the derivative.Using Problem 20, show that if a > 0, p > 1, it follows that for all x > 0—1 a p(Potx+ Sav) >a.Pp PUsing Problem 21, define for p > 1 and @ > 0 the following sequencea _Xn = int om Pix, >0.Show limy-0o.X, =x where x = c'/? Tn fact show that after x, the sequence decreases1/ptoa’.Consider the sequence { (1 + xy where x is a positive number. Using the bino-mial theorem show this sequence is increasing. Next show the sequence converges.co. 1 . ws .Consider the sequence {(1 + xynt \ where x is a positive number. Show thisn=1sequence decreases when x > 2. Hint: You might consider showing (1+ y)@/ + igincreasing in y provided x > 2. To do this, you might use the following observationrepeatedly. If f (0) =0 and f’ (y) > 0, then f(y) > 0. There may also be other waysto do this.Let 2 =Yn-o Pa 2M The b, are called the Bernoulli numbers. Show thatT ter Oidn>1bo br bo bot ff Lifn=1Oln! | 1(n—1)! | 2!(n—2)! (n—1)!!Hint: You might use the Cauchy product after multiplying both sides by e* — 1.8.12 The Fundamental Theorem of AlgebraThe fundamental theorem of algebra states that every non constant polynomial having co-efficients in C has a zero in C. If C is replaced by R, this is not true because of the example,x* +1 =0. This theorem is a very remarkable result and notwithstanding its title, all thebest proofs of it depend on either analysis or topology. It was proved by Gauss in 1797.The proof given here follows Rudin [24]. See also Hardy [14] for a similar proof, more dis-cussion and references. You can also see the interesting article on Wikipedia. You googlefundamental theorem of algebra and go to this site. There are many ways to prove it. This