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Unformatted text preview: Pade Approximations and the Transcendence of Ernie Croot March 9, 2007 1 Introduction Lindemann proved the following theorem, which implies that is transcen dental: Theorem 1 Suppose that 1 , ..., k are nonzero algebraic numbers, and that 1 , ..., k are distinct algebraic numbers. Then, 1 e 1 + + k e k 6 = . The reason that this implies that is transcendental is that if were algebraic, then so is i , which would mean = e i + 1 6 = . We will not prove this general result (of Lindemann), but will instead show only that e can never equal 1 for any algebraic number , which proves is transcendental because e i = 1. The proof for the general case uses similar ideas to this special case. 2 The Proof 2.1 The Idea: Pade Approximations We begin by recalling the standard proof that e is irrational: Suppose e is rational. Then, n ! e must be an integer for all n sufficiently large; however, n ! e = I n + 1 n + 1 + 1 ( n + 1)( n + 2) + 1 ( n + 1)( n + 2)( n + 3) + , 1 where I n is the integer 2 n ! + n ! 2! + n ! 3! + + 1 . It is easy to see that for n sufficiently large, n ! e = I + , where (0 , 1), which contradicts the fact that n ! e is an integer; so, e must have been irrational. Actually, what this proof gives us is the even stronger fact that there exists an infinite sequences of integers f n and g n tending to infinity, such that f n e g n . Indeed, just take f n = n ! and g n = I n . This brings us to the following basic fact: Fact. If is some (possibly complex) number for which there exist sequences of integers f n , g n such that f n  g n , and f n  g n 6 = , then is irrational. 1 To show that e , is a nonzero rational, is irrational we will find such f n and g n . First, we begin with the case where is an integer. If we can show this, then it follows that for any rational a/b we have e a/b is irrational (on taking b th powers). Our sequence of f n s and g n s comes from what are called Pade approxi mations to e x . Basically, a Pade approximation is a pair of polynomials f ( x ) and g ( x ) such that e x f ( x ) g ( x ) for x near 0. There are methods for finding such good pairs f and g , and the simplest is to just use linear algebra. Basically, we try to find f ( x ) and g ( x ) of degree n so that the Taylor expansion of g ( x ) e x f ( x ) 1 The proof is obvious, since if = a/b were rational, then  a/b g n /f n  = 0 or is at least 1 /bf n . Multiplying through by f n gives the result. 2 about x = 0 begins c 2 n +1 x 2 n +1 + c 2 n +2 x 2 n +2 + This uniquely determines f and g up to scalar multiples. Such g and f can be found using the pade command in Maple; for example, Maple gives that in the case where g and f have degree 4, e x 1 + x 2 + 3 x 2 28 + x 3 84 + x 4 1680 1 x 2 + 3 x 2 28 x 3 84 + x 4 1680 ....
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This note was uploaded on 10/23/2011 for the course MATH 4108 taught by Professor Staff during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 Staff
 Algebra, Approximation

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