Bernoulli polynomials

Transcription

1 Bernoulli polynomials Jordan Bell Department of Mathematics, University of Toronto February 2, 26 Bernoulli polynomials For k, the Bernoulli polynomial B k (x) is defined by ze xz e z = B k (x) zk, z < 2π. () The Bernoulli numbers are B k = B k (), the constant terms of the Bernoulli polynomials. For any x, using L Hospital s rule the left-hand side of () tends to as z, and the right-hand side tends to B (x), hence B (x) =. Differentiating () with respect to x, B k(x) zk = z2 e xz e z = B k (x) zk+ = z k B k (x) (k )!, k= so B (x) = and for k we have B k (x) = B k (x) (k )!, i.e. B k (x) = kb k (x). Furthermore, for k, integrating () with respect to x on [, ] produces = ( hence B (x)dx = and for k, ) z k B k (x)dx, z < 2π, The first few Bernoulli polynomials are B k (x)dx =. B (x) =, B (x) = x 2, B 2(x) = x 2 x + 6, B 3(x) = x x2 + 2 x.

2 The Bernoulli polynomials satisfy the following: B k (x + ) zk = ze(x+)z e z = zexz (e z + ) e z = ze xz + zexz e z x k z k+ = + x k z k = (k )! + k= B k (x) zk B k (x) zk, hence for k it holds that B k (x + ) = kx k + B k (x). In particular, for k 2, B k () = B k (). Using (), hence for k, Finally, it is a fact that for k 2, B k ( x) zk = ze( x)z e z = zez e xz e z = ze xz e z = ze xz e z = B k (x) ( z)k, B k ( x) = ( ) k B k (x). 2 Periodic Bernoulli functions sup B k (x) 2ζ(k) x (2π) k. (2) For x R, let [x] be the greatest integer x, and let R(x) = x [x], called the fractional part of x. Write T = R/Z and define the periodic Bernoulli functions P k : T R by P k (t) = B k (R(t)), t T. 2

3 For k 2, because B k () = B k (), the function P k is continuous. For f : T C define its Fourier transform f : Z C by f(n) = f(t)e 2πint dt, n Z. T For k, one calculates P k () = and using integration by parts, P k (n) = (2πin) k for n. Thus for k, the Fourier series of P k is P k (t) P k (n)e 2πint = (2πi) k n k e 2πint. n Z For k 2, n Z P k (n) <, from which it follows that n N P k (n)e 2πint converges to P k (t) uniformly for t T. Furthermore, for t Z, 2 P (t) = π n= n sin 2πnt. n For f, g L (T) and n Z, ( ) f g(n) = f(x y)g(y)dy e 2πinx dx T T ( ) = g(y) f(x y)e 2πinx dx dy T T ( ) = g(y) f(x)e 2πinx e 2πiny dx dy T T For k, l and for n, = f(n)ĝ(n). P k P l (n) = P k (n) P l (n) = (2πin) k (2πin) l = (2πin) k l = P k+l (n), and P k P l () = = P k+l (), so P k P l = P k+l. cf. 2 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 499, Theorem B.2. 3

4 3 Euler-Maclaurin summation formula The Euler-Maclaurin summation formula is the following. 3 If a < b are real numbers, K is a positive integer, and f is a C K function on an open set that contains [a, b], then a<m b f(m) = b a ( )K K! f(x)dx + b a K ( ) k (P k (b)f (k ) (b) P k (a)f (k ) (a)) k= P K (x)f (K) (x)dx. Applying the Euler-Maclaurin summation formula with a =, b = n, K = 2, f(x) = log x yields 4 m n Since e +O(n ) = + O(n ), log n = n log n n + 2 log n + 2 log 2π + O(n ). n! = n n e n 2πn( + O(n )), Stirling s approximation. Write a n = log n + m n m. Because log( x) is concave, a n a n = ( n + log ) + n n =, which means that the sequence a n is nonincreasing. For f(x) = x, because f is positive and nonincreasing, m n f(m) n+ f(x)dx = log(n + ) > log n, hence a n >. Because a n is positive and nonincreasing, there exists some nonnegative limit, γ, called Euler s constant. Using the Euler-Maclaurin summation formula with a =, b = n, K =, f(x) = x, as P (x) = [x] 2, <m n m = log n + 2n n x 2 dx n R(x) x 2 dx, which is <m n m = log n R(x) x 2 dx + n R(x) x 2 dx; 3 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 5, Theorem B.5. 4 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 53, Eq. B.25. 4

5 as R(x)x 2 x 2, the function x R(x)x 2 is integrable on [, ). Since n R(x)x 2 dx n x 2 dx = n, m n m = log n + C + O(n ) for C = R(x)x 2. But log n + m n m γ as n, from which it follows that C = γ, and thus m n 4 Hurwitz zeta function m = log n + γ + O(n ). For < α and Re s >, define the Hurwitz zeta function by ζ(s, α) = n (n + α) s. For Re s >, Γ(s) = t s e t dt, and for n do the change of variable t = (n + α)u, For real s >, Then where n N Γ(s) = (n + α) s u s e (n+α)u (n + α)du = (n + α) s u s e nu e αu du. (n + α) s Γ(s) = (n + α) s Γ(s) = f N (s, u) = n N u s e nu e αu du. u s e nu e αu du = {u s αu e (N+)u e e u > u u =. f N (s, u)du, f N (s, u) and the sequence f N (s, u) is pointwise nondecreasing, and { lim f u s e αu N (s, u) = f(s, u) = e u > u N u =. 5

6 By the monotone convergence theorem, which means that, for real s >, Write f N (s, u)du ζ(s, α)γ(s) = f(s, u)du = Now, by (), for < u < 2π, f(s, u)du + f(s, u)du, f(s, u)du. f(s, u) = u s e αu e u = u s 2 ue αu e u = u s 2 B k (α) ( u)k = ( ) k B k (α) uk+s 2. f(s, u)du. For k 2, real s >, and < u < 2π, by (2), B k(α) uk+s 2 2ζ(k) (2π) k uk+s 2 ( u ) k = 2ζ(k) u s 2, 2π which is summable, and thus by the dominated convergence theorem, f(s, u)du = = ( ) k B k (α) uk+s 2 du ( ) k B k (α) k + s. Check that s ( )k B k (α) k+s is meromorphic on C, with poles of order or at s = k+, k (the order of the pole is if B k (α) = ), at which the residue is ( ) k B k (α).5 On the other hand, check that s f(s, u)du is entire. Therefore ζ(s, α)γ(s) is meromorphic on C, with poles of order or at s = k +, k and the residue of ζ(s, α)γ(s) at s = k + is ( ) k B k (α). But it is a fact that Γ(s) has poles of order at s = n, n, with residue ( )n n!. Hence the only pole of ζ(s, α) is at s =, at which the residue is. 5 Kazuya Kato, Nobushige Kurokawa, and Takeshi Saito, Number Theory : Fermat s Dream, p

7 Theorem. For n and for < α, ζ( n, α) = B n(α) n. Proof. For n, because ζ(s, α) does not have a pole at s = n and because Γ(s) has a pole of order at s = n with residue ( )n (n )!, lim s n (s ( n))γ(s)ζ(s, α) = ζ( n, α) lim s n = ζ( n, α) Res s= n Γ(s) = ζ( n, α) ( )n (n )!. (s ( n))γ(s) On the other hand, ζ(s, α)γ(s) has a pole of order at s = n with residue ( ) n B n (α) n!. Therefore i.e. for n and < α, ζ( n, α) ( )n (n )! = ( )n B n (α) n!, ζ( n, α) = B n(α) n. 5 Sobolev spaces For real s, we define the Sobolev space H s (T) as the set of those f L 2 (T) such that f() 2 + f(n) 2 n 2s <. n Z\{} For f, g H s (T), define f, g H s (T) = f()ĝ() + f(n)ĝ(n) n 2s. n Z\{} This is an inner product, with which H s (T) is a Hilbert space. 6 6 See 7

8 For c C Z, if s > r + 2, c n e 2πinx n N Cr (T) = sup sup c n (2πin) j e 2πinx j r x T n N c 2 + sup sup c n (2πin) j e 2πinx j r x T n N c 2 + (2π) r c n n r = c 2 + (2π) r c 2 + (2π) r n N n N n N c n n s n (r s) c n 2 n 2s /2 c 2 + (2π) r (2 ζ(2s 2r)) /2 n N n N n (2s 2r) c n 2 n 2s For f H s (T), the partial sums n N f(n)e 2πinx are a Cauchy sequence in H s (T) and by the above are a Cauchy sequence in the Banach space C r (T) and so converge to some g C r (T). Then ĝ = f, which implies that g = f almost everywhere. For k, P k () = and P k (n) = (2πin) k for n. For k, l > s + 2, P k, P l Hs (T) = n Z\{} = n Z\{} (2πin) k (2πin) l i k+l (2πn) k l = i k+l (2π) k l 2 ζ(k + l). Thus if k > s + 2 then P k H s (T), and in particular P k H k (T) for k. For s > r + 2, if f Hs (T) then there is some g C r (T) such that g = f almost everywhere. Thus if r + 2 < s < k 2, i.e. k > r +, then there is some g C r (T) such that g = P k almost everywhere. But for k, P k is continuous, so in fact g = P k. In particular, P k C k 2 (T) for k 2. /2. /2 8

9 6 Reproducing kernel Hilbert spaces For x T and f : T C, define (τ x f)(y) = f(y x). We calculate τ x f(n) = f(y x)e 2πiny dy T = e 2πinx f(y)e 2πiny dy T = e 2πinx f(n). Let r. For x T, define F x : T R by For n Z, F x () =, and for n, For f H r (T), F x = + ( ) r (2π) 2r τ x P 2r. F x (n) = δ (n) + ( ) r (2π) 2r e 2πinx P2r (n). F x (n) = ( ) r (2π) 2r e 2πinx (2πin) 2r = n 2r e 2πinx. f, F x Hr (T) = f() F x () + = f() + n Z\{} = f() + = f(x). n Z\{} n Z\{} f(n) F x (n) n 2r f(n) n 2r e 2πinx n 2r f(n)e 2πinx This shows that H r (T) is a reproducing kernel Hilbert space. 7 Define F : T T R by F (x, y) = F x, F y Hr (T) = F x (y) = + ( ) r (2π) 2r P 2r (y x). Thus the reproducing kernel of H r (T) is 8 F (x, y) = + ( ) r (2π) 2r P 2r (y x) cf. Alain Berlinet and Christine Thomas-Agnan, Reproducing Kernel Hilbert Spaces in Probability and Statistics, p. 38, who use a different inner product on H r (T) and consequently have a different expression for the reproducing kernel. 9