This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Modern Analysis, Homework 5 Possible solutions The following problems are optional , and are worth 5 pts each: E1.  Let P ( x ) be a degree n polynomial with integral coefficients and no rational root. Suppose that z is a root of P . Prove that there exists A > 0 such that for all rationals p/q ∈ Q we have z- p q > 1 A | q | n . Suggestion: First try for n = 2. Hint: Suppose not, then consider the integers q n P ( p q ). Solution: Let’s start with the following observation which will be useful later. Given a polynomial P ( x ) (with real or complex coefficients) of degree n , and given a fixed real (or complex) z , then we can write P ( z + x ) = P ( z ) + x f P z ( x ) for some degree n- 1 polynomial f P z ( x ), whose coefficients depend on z . This is a direct consequence of the binomial theorem: write P ( x ) = ∑ n i =0 a i x i . Then P ( z + x ) = n X i =0 a i ( z + x ) i (1) = n X i =0 a i z i + n X i =0 a i i X j =1 i j z j x i- j (2) Also, given any polynomial P ( x ) (again with real or complex coefficients), we may form the real number S ( P ), which is the sum of the absolute values of the coefficients of P . It follows from the triangle inequality that if | x | ≤ 1 then | P ( x ) | ≤ S ( P ). Back to the problem at hand. Suppose that the statement is false . This would mean that for all A > 0, there exists a rational p/q ∈ Q such that z- p q ≤ 1 A | q | n . Given our polynomial P and root z , set A = S ( f P z ) + 1, using the notations from above. Then there exists a rational p/q such that z- p q ≤ 1 S ( f P z ) + 1 · 1 | q | n (3) (note that this is less than or equal to 1). Notice 1. q n P ( p/q ) is an integer, since P has integral coefficients and has degree n . 2. q n P ( p/q ) 6 = 0 since P has no rational roots. 1 By previous discussion, we may write | q n P ( p/q ) | = | q | n | P ( z- ( z- p/q )) | = | q | n P ( z ) + ( z- p/q ) f P z ( z- p/q ) ≤| q | n 1 S ( f P z ) + 1 · 1 | q | n S ( f P z ) < 1 But this implies that the nonzero integer | q n P ( p/q ) | is less than 1, which is impossible! E2. (a) Use problem E1 to construct an explicit transcendental number. (b) Use problem E1 to construct uncountably many explicit 1 transcendental numbers....
View Full Document
This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
- Spring '11