e is transcendental Robert Hines November 6, 2015 e is irrational and transcendental numbers exist The irrationality of e is straightforward to prove, and has been known since at least Euler (who first called e,\e"). Theorem. e is irrational. n X xk Proof. Let H (x) = ;H = H (1). Then n k! n n k=0 1 X 1 e − H = n k! k=n+1 1 1 1 = 1 + + + ::: (n + 1)! n + 2 (n + 2)(n + 3) 1 1 1 < 1 + + + ::: (n + 1)! n + 1 (n + 1)2 1 1 1 n + 1 1 1 = 1 = = : (n + 1)! 1 − n+1 (n + 1)! n n n! p If e = q , then q!(e − Hq) 2 Z, but 1 0 < q!(e − H ) < ; q q a contradiction. [Fun fact: Hn(x) 2 Q[x] is irreducible for all n]. The moral of the proof is that Hn (a rational number) approximates e too well. Consider the following Lemma (Liouville, 1844). If ξ is a real algebraic number of degree n > 1, then there is a constant A > 0 (depending on ξ) such that h A ξ − > : k kn 1 Proof. Suppose p(x) 2 Z[x] is irreducible of degree n with p(ξ) = 0. Then p(ξ) − p(h=k) = (ξ − h=k)p0(α) for some α between ξ and h=k by the mean value theorem. The left hand side is a non- zero rational number (p(ξ) = 0 and p is irreducible so h=k is not a root) with denominator less than kn so that we get 1 h 0 ≤ ξ − supfp (x): x 2 (ξ − 1; ξ + 1)g: kn k The above result can be improved to Theorem (Thue-Siegel-Roth). For all > 0, there are only finitely many rational solu- tions to h 1 ξ − < k k2+ if ξ is algebraic and irrational. The 2 + exponent is the best possible since we have the following Proposition. If ξ 2 R is irrational then there are infinitely many rationals p=q such that jξ − p=qj < 1=q2: Proof. This is an application of the pigeonhole principle. Two of the n + 1 numbers 1; fkξg (the fractional part of kξ) for 1 ≤ k ≤ n must lie in one of the n subintervals (i=n; (i + 1)=n]; 0 ≤ i ≤ n − 1 of (0; 1]. Hence there is a p and 1 ≤ q ≤ n such that jqξ − pj < 1=n; i.e jξ − p=qj < 1=nq ≤ 1=q2: Infinitely many of these p=q must be distinct, else jξ − p=qj takes on a minimum value, say larger than 1=n for some n, and the above construction gives a contradiction. Transcendental numbers exist (by cardinality arguments - thanks Cantor!), but let's exhibit one explicitly (as Liouville did). P1 −n! Proposition. ξ = n=0 10 is transcendental. j! j! Pj −n! Proof. Let kj = 10 ; hj = 10 n=0 10 . Then (hj; kj) = 1 (as hj ≡ 1 mod 10) and 1 1 hj X −n! X −n ξ − = 10 < 10 kj n=j+1 n=(j+1)! 1 10 = 10−(j+1)! = (10j!)−j 1 − 1=10 9 · 10j! −j < Ajkj ; where A(j) ! 0 as j ! 1 so that ξ is transcendental by the lemma above. e is transcendental We now begin the proof that e is transcendental (Hermite, 1873). We have to be able to simultaneously approximate ex at different values to obtain a contradiction similar to that given above for the irrationality of e. P1 (i) For a polynomial f(x), let F (x) = i=0 f (x). Integrating by parts a bajillion times, we get Z x ex f(t)e−tdt = F (0)ex − F (x) (the \Hermite identity"): 0 Consider the specific polynomial xp−1(x − 1)p ····· (x − n)p f(x) = : (p − 1)! (n will be the degree of the fictitious minimal polynomial for e over Q and p will be a large prime). We have the following estimate for 0 ≤ k ≤ n: Z k k k −t je F (0) − F (k)j = e f(t)e dt 0 ≤ nen sup ff(t)g t2[0;n] np−1(np)n = (p − 1)! which goes to zero as p ! 1 for a fixed n. [Mildly interesting: this proof requires the existence of infinitely many primes.] We now show that such an estimate is impossible if Pn k e is algebraic by showing that k=1 F (0)e − F (k) is an integer between 0 and 1. 1. F (0) 2 ZnpZ for p > n: We have f(x) = a(x)b(x) where xp−1 a(x) = ; b(x) = (x − 1) ····· (x − n)p; (p − 1)! so that N X N f (N)(x) = a(i)(x)b(N−i)(x) : i i=0 Note that a(i)(0) = 0 unless i = p − 1 in which case a(p−1)(0) = 1. Hence 1 1 N X X X N F (0) = f (N)(0) = a(i)(0)b(N−i)(0) i N=0 N=0 i=0 1 X N = b(N−(p−1))(0) p − 1 N=p−1 p2 p = b(0) + p(::: ) = (−1) n! + p(::: ) 2 ZnpZ (remember that p > n): 2. F (k) 2 pZ for 1 ≤ k ≤ n: We have f(x) = c(x)d(x) where (x − k)p xp−1(x − 1)p ····· (x − n)p c(x) = ; d(x) = : (p − 1)! (x − k)p Note that c(i)(k) = 0 unless i = p in which case c(p)(k) = p. Hence 1 N X X N F (k) = c(i)(k)d(N−i)(k) i N=0 i=0 1 X N = p d(N−p)(k) 2 p : p Z N=p Pn k Now, if e were algebraic, say k=0 cke = 0; ck 2 Z; c0 6= 0, and p > jc0j, then n X 1 ≤ ckF (k) (because c0;F (0) 2 ZnpZ;F (k) 2 pZ) k=0 n n ! X X k = ckF (k) − cke F (0) k=0 k=0 n X k = ck F (k) − e F (0) k=0 n X k ≤ M jF (k) − e F (0)j (where M = maxfjckjg) k k=0 which is less than 1 for p large as shown above. Hence e is transcendental. What about π? Here is a proof that π is irrational in the spirit of Hermite. P1 n (2n) For a polynomial f(x), let F (x) = n=0(−1) f (x) (this mimics sin x in the way we mimicked ex before). We have d Z π (F 0(x) sin x − F (x) cos x) = f(x) sin x; f(x) sin xdx = F (0) + F (π): dx 0 If π = a=b were rational, consider the polynomial bn 1 f(x) = xn(π − x)n = xn(a − bx)n 2 [x]: n! n! Q We have bounds Z π bnπ2n Z π 2(π2b)n 0 < f(x) sin xdx ≤ sin xdx = ! 0 as n ! 1: 0 n! 0 n! Note that f (k)(0) = 0 for 0 ≤ k < n as f has a zero of order n at zero. We also have f (k)(0) 2 Z for k ≥ n by the following (easy) lemma. Lemma. For p(x) 2 Z[x], k! divides all the coefficients of p(k)(x). dk n n n−k Proof. dxk x = k! k x for k ≤ n and higher derivatives are zero. Hence f (k)(0) 2 Z for all k. Finally note that f(x) = f(π − x); f (k)(x) = (−1)kf (k)(π − x) so that f (k)(π) = (−1)kf (k)(0) 2 Z for all k. Therefore F (0) + F (π) 2 Z, a contradiction, and π is irrational. We can prove that π is transcendental using the methods we used for e, although the details are slightly more tedious. We start with the identity eπi + 1 = 0. If πi were algebraic (degree n), we would have n m Y γ X P γ X α 0 = (1 + e i ) = e i i i = a + e i i=1 i2f0;1g i=1 n where the γi are the galois conjugates of πi, a = 2 −m are the number of zero exponents in the first sum (note that a ≥ 1), and the αi are the non-zero exponents in the first sum. Thinking about symmetric functions for a while (details omitted), we see that n ! Y X φ(x) = x − iγi 2 Q[x]: i2f0;1g i=1 Divide by xa and clear denominators to get a polynomial m X i (x) = bix 2 Z[x]; bm > 0; b0 6= 0 i=0 whose roots are exactly the αi. Furthermore, assume bmαi is an algebraic integer for all i. Once again we apply the \Hermite identity," this time to the polynomial (m−1)p mp m bm b Y f(x) = xp−1 p(x) = m xp−1 (x − α )p: (p − 1)! (p − 1)! i i=1 Plug in x = αi and sum over i to get m m Z αi X X αi −t −aF (0) − F (αi) = e f(t)e dt: i=1 i=1 0 Our goal, as before, is to show that the LHS is a non-zero integer but that the RHS can be made arbitrarily small. We have !p mp mp Y F (0) = (−1) bm αi 2 ZnpZ i for large p. We also have m X mp X p−1 Y p F (αi) = pbm αi (αi − αj) 2 pZ i=1 i j6=i mp for large p because it is symmetric in αi and the denominator is cleared by bm .