证明π是一个超越数~

The Transcendence of pi Steve Mayer November 2006 Abstract The proof that pi is transcendental is not well-known despite the fact that it isn’t too difficult for a university mathematics student to follow. The purpose of this paper is to make the proof more widely available. A bonus is that the proof also shows that e is transcendental as well. The material in these notes are not mine; it is taken from a supplement issued by Ian Stewart as an adjunct to a Rings and Fields course in 1970 at the University of Warwick. Definition. A complex number is algebraic over Q if it is a root of a polynomial equation with rational coefficients. Thus a is algebraic if there are rational numbers α0, α1, . . . , αn not all 0, such that α0a n + α1a n−1 + . . .+ αn−1a+ αn = 0. Definition. A complex number is transcendental if it is not algebraic, so it is not the root of any polynomial equation with rational coefficients. In proving that it is impossible to ’square the circle’ by a ruler-and-compass construction we have to appeal to the theorem: The real number pi is transcendental over Q The purpose of this supplement is to indicate, for those who may be interested, how this theorem may be proved. It is possible to prove that there exist transcendental real numbers by using infinite cardinals, as was first done by Cantor in 1874. Earlier Liouville (1844) had actually constructed transcendentals, for example ∞∑ n=1 10−n! is transcendental.1 1A proof can be found at http://rutherglen.ics.mq.edu.au/math334s106/m2334.Dioph.Liouville.pdf 1 However, no naturally occurring real number (such as e or pi) was proved tran- scendental until Hermite (1873) disposed of e. pi held out until 1882 when Linde- mann, using methods related to those of Hermite, disposed of that. In 1900 David Hilbert proposed the problem: If a, b are real numbers algebraic over Q, if a 6= 0 or 1 and b is irrational, prove ab is transcendental. This was solved independently in 1934 by the Russian, Gelfond, and a German, Schneider. Before proving transcendence of pi we shall prove a number of similar theorems, using simpler versions of the final method, as an aid to comprehension. The tools needed are first-year analysis.2 Theorem 1. pi is irrational Proof. Let In(x) = ∫ +1 −1 (1− x2) n cos(αx) dx Integrating by parts we have α2In = 2n(2n− 1)In−1 − 4n(n− 1)In−2 (n ≥ 2) which implies that α2n+1In = n! (Pn sin(α) +Qn cos(α)) (*) where Pn, Qn are polynomials of degree < 2n+ 1 in α with integer coefficients. Remark. degPn = n, degQn = n− 1 Put α = pi 2 , and assume pi is rational, so that pi = b a , a, b ∈ Z From (*) we deduce that Jn = b2n+1In n! is an integer. On the other hand Jn → 0 as n→∞ since b is fixed and In is bounded by∫ +1 −1 cos (pix 2 ) dx Jn is an integer, → 0. Thus Jn = 0 for some n. But this integrand is continuous, and is > 0 in most of the range (−1,+1), so Jn 6= 0. Contradiction. � 2This was true in 1970. Is it still true today? 2 Theorem 2. pi2 is irrational (so pi does not lie in any quadratic extension of Q) Proof. Assume pi2 = a b , a, b ∈ Z. Define f(x) = xn (1− x)n n! , G(x) = bn [ pi2nf(x)− pi2n−2f ′′(x) + . . .+ (−1)npi0f (2n)(x)] (superscripts indicating differentiations). We see that the value of any derivative of f at 0 or 1 is either 0 or an integer. Also G(0) and G(1) are integers. Now d dx [G′(x) sin(pix)− piG(x) cos(pix)] = [G′′(x) + pi2G(x)] sin(pix) = bnpi2n+2f(x) sin(pix) since f (2n+2)(x) = 0 = pi2an sin(pix)f(x) so that pi ∫ 1 0 an sin(pix)f(x) dx = [ G′(x) sin(pix) pi −G(x) cos(pix) ]1 0 = 0 +G(0) +G(1) = integer. But again the integral is non-zero and → 0 as n → ∞. Thus again we have a contradiction. � Getting more involved, now: Theorem 3 (Hermite). e is transcendental over Q Proof. Suppose ame m + . . .+ a1e+ a0 = 0 (ai ∈ Z). WLOG a0 6= 0 Define f(x) = xp−1(x− 1)p(x− 2)p . . . (x−m)p (p− 1)! where for the moment p is arbitrary and prime. Define F (x) = f(x) + f ′(x) + . . .+ f (mp+p−1)(x). Now if 0 < x < m, |f(x)| ≤ m p−1mmp (p− 1)! = mmp+p−1 (p− 1)! 3 Also d dx (e−xF (x)) = e−x [F ′(x)− F (x)] = −e−xf(x) so that aj ∫ j 0 e−xf(x) dx = aj [−e−xF (x)]j 0 = ajF (0)− aje−jF (j). Multiplying by ej and summing over j = 0, 1, . . .m we get m∑ j=0 aje j ∫ j 0 e−xf(x) dx = F (0).0− m∑ j=0 ajF (j) = − m∑ j=0 mp+p−1∑ i=0 ajf (i)(j). We claim that each f (i)(j) is an integer, divisible by p except when j = 0 and i = p − 1. For only non-zero terms arise from terms where the factor (x − j)p has been differentiated p times, and then p! cancels (p− 1)! and leaves p, except in the exceptional case. We show that in the exceptional case the term is NOT divisible by p. Clearly f (p−1)(0) = (−1)p . . . (−m)p. We CHOOSE p larger than m, when this product cannot have a prime factor p. Hence the right-hand side of the above equation is an integer 6= 0. But as p→∞ the left-hand side tends to 0, using the above estimate for |f(x)|. This is a contradiction. � Theorem 4 (Lindemann). pi is transcendental over Q Proof. If pi satisfies an algebraic equation with coefficents in Q, so does ipi (i =√−1). Let this equation be θ1(x) = 0, with roots ipi = α1, . . . , αn. Now eipi + 1 = 0 so (eα1 + 1) . . . (eαn + 1) = 0 We now construct an algebraic equation with integer coefficients whose roots are the exponents of e in the expansion of the above product. For example, the exponents in pairs are α1 +α2, α1 +α3, . . . , αn−1 +αn. The α s satisfy a polynomial equation over Q so their elementary symmetric functions are rational. Hence the elementary symmetric functions of the sums of pairs are symmetric functions of the α s and are also rational. Thus the pairs are roots of the equation θ2(x) = 0 with rational coefficients. Similarly sums of 3 α s are roots of θ3(x) = 0, etc. Then the equation θ1(x)θ2(x) . . . θn(x) = 0 4 is a polynomial equation over Q whose roots are all sums of α s. Deleting zero roots from this, if any, we get θ(x) = 0 θ(x) = cxr + c1x r−1 + . . . cr and cr 6= 0 since we have deleted zero roots. The roots of this equation are the non-zero exponents of e in the product when expanded. Call these β1, . . . βr. The original equation becomes eβ1 + . . . eβr + e0 + . . . e0 = 0 ie ∑ eβi + k = 0 where k is an integer > 0 (6= 0 since the term 1 . . . 1 exists) Now define f(x) = csxp−1 [θ(x)]p (p− 1)! where s = rp− 1 and p will be determined later. Define F (x) = f(x) + f ′(x) + . . .+ f (s+p)(x). d dx [ e−xF (x) ] = −e−xf(x) as before. Hence we have e−xF (x)− F (0) = − ∫ x 0 e−yf(y) dy Putting y = λx we get F (x)− exF (0) = −x ∫ 1 0 e(1−λ)xf(λx) dλ. Let x range over the βi and sum. Since ∑ eβi + k = 0 we get r∑ j=1 F (βj) + kF (0) = − r∑ j=1 βj ∫ 1 0 e(1−λ)βjf (λβj) dλ. CLAIM. For large enough p the LHS is a non-zero integer. r∑ j=1 f (t) (βj) = 0 (0 < t < p) by definition of f . Each derivative of order p or more has a factor p and a factor cs, since we must differentiate [θ(x)]p enough times to get 6= 0. And f (t) (βj) is a polynomial in βj of degree at most s. The sum is 5 symmetric, and so is an integer provided each coefficient is divisible by cs, which it is. (symmetric functions are polynomials in coefficients = polynomials in ci c of degree ≤ s). Thus we have r∑ j=1 f (t) (βj) = pkt t = p, . . . , p+ s. Thus LHS = (integer) + kF (0). What is F (0)? f (t)(0) = 0 t = 0, . . . , p− 2. f (p−1)(0) = cscpr (cr 6= 0) f (t)(0) = p(some integer) t = p, p+ 1, . . . . So the LHS is an integer multiple of p+cscprk. This is not divisible by p if p > k, c, cr. So it is a non-zero integer. But the RHS → 0 as p → ∞ and we get the usual contradiction. � 6