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Unformatted text preview: A proof that e is transcendental Jonathan L.F. King University of Florida, Gainesville FL 32611-2082, USA [email protected] Webpage http://www.math.ufl.edu/ ∼ squash/ 13 January, 2012 (at 10:25 ) Abstract: Uses calculus and divisibility to show that e is not algebraic. The file has two proofs, a short one due to Hilbert and a long one, probably of Hermite. Notation. For posint N , let ≡ N means “mod- N congruent”. Let :: N :: mean ≡ N ! ie., congruence mod N-factorial. 1 : Lemma. For k a natnum, integral J k := R ∞ x k e x d x equals J k = k ! . ♦ Proof. IByParts yields J k = k · J k- 1 . And J = 1. 2 : Corollary. For each natnum M and intpoly f : Z ∞ f ( x ) · x M e x d x :: M +1 :: f (0) · M ! . ♦ Hilbert’s proof that e is transcendental The Set-up. FTSOC, suppose e is algebraic of de- gree D ∈ Z + . We thus have an intpoly h ( x ) := B D x D + ··· + B 1 x + B , with B D 6 = 0 , such that h ( e ) = 0. And B 6 = 0, since h () has mini- mal degree. Proof of transcendentality. For a posint exponent r to be chosen later, define Φ( x ) := [ x- 1][ x- 2][ x- 3] ··· [ x- D ] , and I u ‘ := Z u ‘ x r Φ( x ) r +1 e x d x. Thus 0 = 0 · I ∞ = h ( e ) I ∞ = U ( r ) + L ( r ) where we have split each integral into an Upper part and a Lower part: U ( r ) := B I ∞ + X D K =1 B K e K I ∞ K , L ( r ) := X D K =1 B K e K I K . We have that U ( r ) r ! + L ( r ) r ! = 0 . The contradiction will come by showing that U ( r ) r ! is always a non-zero integer; then showing that r can be chosen large enough that L ( r ) r ! is less than 1. Upperbounding L ( r ) . Over all x in the com- pact interval [ .. D ] , let A be an upperbound for the abs-value of Φ( x ) and of x · Φ( x ). For each K ∈ [ 1 .. D ] , then, | I K | is bounded by A r +1 ....
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