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e_transcendental

# e_transcendental - A proof that e is transcendental...

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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 0 x k e x d x equals J k = k ! . Proof. IByParts yields J k = k · J k - 1 . And J 0 = 1. 2 : Corollary. For each natnum M and intpoly f : Z 0 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 0 , with B D 6 = 0 , such that h ( e ) = 0. And B 0 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 0 = h ( e ) I 0 = U ( r ) + L ( r ) where we have split each integral into an Upper part and a Lower part: U ( r ) := B 0 I 0 + X D K =1 B K e K I K , L ( r ) := X D K =1 B K e K I K 0 . 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 [ 0 .. 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 0 | is bounded by A r +1 . Let B := Max K [ 1 .. D ] | B K | · e K . It follows that L ( r ) 6 D · BA r +1 . When divided by r !, this quantity goes to zero.

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