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. [30] 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)[25] Use problem E1 to construct an explicit transcendental number. (b)[22] 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
 Peters

Click to edit the document details