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2 数論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行当時のものです.

3 Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published in Japan by arrangement with DUNOD Editeur through le Bureau des Copyrights Français, Tokyo FAX

4 i ( ( ,2 Z 7 12 (A. D

5 ii d e π p Σ e ax + by = c

6 iii Z F p ϕ(n lcm(1, 2,,n r 2 (n (1 x α

7 iv x 5 + y 5 = z (p y 2 = x 3 + k

8 1 1 1 d e π ( 1.4 ( 1.5 ( d 1.1 d d d d = a/b a, b 2 b 2 d = a 2 d p k d = p 2k+1 δ p δ p 2k+1 a 2 p k+1 a a = p k+1 α b 2 δ = pα 2 p b p a p b a b d 1 d Q p, q Q p + q d =0 p = q =0 q 0 d = p/q Q 1.1 q =0 p = e 1.2 e n

9 2 1 e = n k=0 1 k! + R n, R n = k=n+1 R n > 0 1 R n = (n +1! + 1 (n +2! + 1 (n +3! + 1 ( = 1+ 1 (n +1! n (n +2(n (n +2(n +3(n ( < 1+ 1 ( 1 2 ( 1 3 (n +1! = 2 2 (n +1!. (1.1 0 <n! e n! n k=0 1 k! (1.1 1 k! < 2 n +1. (1.2 e = a/b a b (1.2 0 <n! a bα n < 2b n +1. n 1 α n = n! k=0 k! β n = n! a bα n n e 1.3 π 1.3 π P (x 2n F (x =P (x P (x+p (4 (x +( 1 n P (2n (x P (xsinx =(F (xsinx F (xcosx π 0 P (xsinxdx= F (0 + F (π. (1.3 π = a/b (a, b N (1.3 P (x = 1 n! xn (a bx n I n = π P (xsinxdx (0,π P (xsinx 0 I n > 0 (0,π x(a bx a2 4b I n 1 ( a 2 n n! π 4b

10 1.4 3 lim I n =0 n n F (0 F (π Z ( 1.1 I n Z (n 0 I n π 1.4 q C, q > 1 T q (x = n=0 x n q n(n+1 2 (x C (1.4 T q (qx =1+xT q (x ( q Z, q 2 x Q T q (x K C f(x = a n x n (a n K R p, q, r p<r p + q +1 K P, Q Q 0, deg P p, deg Q q g(x = b n x n ( x <R n=0 n=0 Q(xf(x+P (x =x r g(x. ( ( 1.4 x = α/β (α, β Z T q (x = µ (µ, ν Z ν ( x (1.5 T q q n = A n να n, A n Z (n 0 ρ α/q ρ < K = Q, f(x = T q (x, p = q =2ρ, r =3ρ P Q

11 4 1 (1.6 P Q Q(xT q (x+p (x =x 3ρ g(x. (1.7 P, Q deg P 2ρ, deg Q 2ρ, Q 0 g (1.7 T q C (1.4 x =0 g 1 0 σ 0 h Q(xT q (x+p (x =x 3ρ+σ h(x, h(0 0. (1.8 x = α/β x/q n = α/(βq n να n β 2ρ q 2ρn B n ( B n = να3ρ+σ α nh ( α β ρ+σ q ρ+σ βq n ( α ( α ( α = β 2ρ q 2ρn Q βq n να n T q βq n + να n β 2ρ q 2ρn P βq n. (1.9 ( α ( α ( α β 2ρ q 2ρn Q βq n να n T q βq n, β 2ρ q 2ρn P βq n Z B n Z n ( B n να3ρ+σ α nh(0 β ρ+σ q ρ+σ ( a n b n lim a n/b n =1 ρ n α/q ρ < 1 lim B n =0 α 0 h(0 0 n n B n 0 0 B n α f α P n f(qn 0 Q n (n, (1.10 P n /Q n e T q (α/β e (1.2

12 <e P n 2 < 2. (1.11 Q n (n +1Q n Q n n 1 P n = n! k=0 k!,q n = n! T q (α/β (1.5 T q ( α βq n = k n α n T q( α β + l n α n (k n,l n Z (1.9 ( α T q P n α n 1 c β Q n q ρ+σ Q n c. (1.12 Q n Q n = k n β 2ρ q 2ρn Q(α/βq n, P n = α n β 2ρ q 2ρn P (α/βq n +l n Q n /k n,c= max n N α 3ρ+σ β ρ σ h(αβ 1 q n e T q (α/β (1.11 P n Q n n (1.12 P n Q n P Q 1.1 P Q 2 Q n Q n f(q n α R 0 < α P n ε(n Q n Q n (n 0, lim ε(n =0 n P n /Q n α 1.3 ( α P n /Q n (n 0 0 < α P n 1 < Q n Q 2. n

13 α Q>1 p/q 1 q<q, 0 < qα p 1/Q Q +1 0, 1,α [α], 2α [2α],, (Q 1α [(Q 1α] ([x] x 0 x [x] < 1 [0, 1] aα + b a b 0 a Q 1 [0, 1] Q [0, 1/Q], [1/Q, 2/Q],, [(Q 1/Q, 1] Q +1 2 ( ξ 1 = a 1 α + b 1,ξ 2 = a 2 α + b 2 0 a 1 Q 1, 0 a 2 Q 1, a 1 a 2 ( ξ 1 =0, ξ 2 =1 a 1 >a 2 ξ 1 ξ 2 1/Q ξ 1 ξ 2 = (a 1 a 2 α + b 1 b 2 1/Q 0 <a 1 a 2 Q ( 1.6 Q>1 1.2 P 1 /Q 1 0 < α P 1 /Q 1 < 1/(QQ 1 1 Q 1 <Q (α 0 < α P 1 /Q 1 < 1/Q Q 1/Q < α P 1 /Q 1 P 2 /Q 2 0 <Q 2 <Q 0 < α P 2 /Q 2 < 1/(QQ 2 < 1/Q 2 2 P 1 /Q 1 P 2 /Q 2 P 1 /Q 1 P 2 /Q 2 P 1 /Q 1 α P 2 /Q 2 α 0 P 1 /Q 1 α > 1/Q, P 2 /Q 2 α < 1/(QQ 2 1/Q P n /Q n (n 0 α P n /Q n < 1/Q 2 n d d = a/b d = a /b,a <a, b <b ( a/b b a a>a >a > 2 e, π T q (α/β 0

14 n +1 n 1 2 {u n } n N u m = u n m, n (m n P (0 = P (0 = = P (n 1 (0 = 0 F (0 Z F (π Z α = β =log α P (x =x d + a d 1 x d a 1 x + a 0, d 1 α P m Z,m 2 (2 n=0 m n2 1.8 α 2 a, b, c ac 0 aα 2 + bα + c =0 1 Φ =(1+ 5/2 2 2 e F n F n =2 2n +1(n 0 χ = P x < 1 X x X 2n x 2n f(x =, g(x = 1 x 2n 1+x 2n n=0 1 f(x g(x =2 f(x x 1 x n=0 n=0 1 F n

15 (1 x α,α Z ( 8.3 ( f(x = a k x k ( x <R 0 (m, n N 2 k=0 a b c (Q m,p n,r m,n f [m/n] a Q m,p n deg Q m m, deg P n n b R m,n (x = b k x k ( x <R 0 k=0 c Q m (xf(x+p n (x =x m+n+1 R m,n (x ( x <R m = n (Q n,p n,r n,n 1.1 f n S n (x = 1 f(x S n (x =x n+1 a n+k+1 x k. k=0 n k=0 a k x k [0/n] (Q m,p n,r m,n f [m/n] λ C (λq m,λp n,λr m,n

16 8.2 2 (1 x α (Q n,p n,r n,n f Q n (0 0,R n (0 0(n 0 n 0 Q n (x P n (x Q n+1 (x P n+1 (x = c nx 2n+1, c n 0 Q n (xf(x+p n (x =x 2n+1 R n (x Q n (x P n (x Q n+1 (x P n+1 (x = Q n (x P n (x+q n (xf(x Q n+1 (x P n+1 (x+q n+1 (xf(x Q n (x x 2n+1 R n (x = Q n+1 (x x 2n+3 R n+1 (x Q n (x Q n+1 (x = x 2n+1 ( Q n+1 (xr n (x+x 2 Q n (xr n+1 (x P n (x P n+1 (x = x2n+1( Q n+1 (0R n (0 + b k x k. (8.1 2n +1 (8.1 b k =0(k 1 c n = Q n+1 (0R n (0 k= (1 x α a (a 0 =1 (a n = a(a +1 (a + n 1, n 1 (8.2 a, b, c, x C x < 1 2 F 1 ( a, b (a n (b n 2F 1 x = x n. (8.3 c (c n n! n=0 8.1 a = b = c =1 ( 1, 1 2F 1 x = x n = x. n=0

17 a = α, b = c 2 ( α, b ( α( α +1 ( α + n 1 2F 1 x = x n =(1 x α. b n! n=0 8.1 (8.3 c 0, 1, 2, a 2 F a, b 1 c x a c<a c (a n 0 (c n 0 a c c<a 2 F a, b 1 c x a F 1 ( a, b c x x(1 xy +(c (a + b +1xy aby =0 (8.4 c Z (8.4 (0, 1 ( a, b ( a c +1,b c +1 y = A 2 F 1 x + Bx 1 c 2F 1 x c 2 c (A, B C ( ( a, b ( c a, c b 2F 1 x (1 x a+b c = 2 F 1 x (8.6 c c ( 8.2 ( (1 x α 8.2 α Z, m,n N x < 1 ( m, n + α ( n, m α 2F 1 x (1 x α 2 F 1 x m n m n = x m+n+1 ( 1m ( m α m+n+1 m+nc m (m + n +1! 2F 1 ( n +1 α, m +1 m + n +2 x. ε [ 1/2, 1/2] (8.6 a = m, b = (n + ε +α, c = m (n + ε ( m, (n + ε+α ( (n + ε, m α 2F 1 x (1 x α = 2 F 1 x. (8.7 m (n + ε m (n + ε ( (8.7 ε 0 m 2 F m, n + α 1 m n x

18 8.2 2 (1 x α 103 (8.7 3 ( (n + ε, m α n ( (n + ε k ( m α k 2F 1 x = x k m (n + ε ( m (n + ε k k! k=0 m+n ( (n + ε k ( m α k + x k ( (n + ε k ( m α k + x k. (8.8 ( m (n + ε k k! ( m (n + ε k k! k=n+1 k=m+n+1 ( ε 0 ( F n, m α 1 m n x 2 0 n +1 k m + n ( n ε k =( n ε( n ε +1 ( n ε + k 1 ε 3 g(x ( n ε k =[( n ε ( ε] [( ε +1 ( ε + m] [( ε + m +1 ( ε + m +1+(k m n 1 1], ( m n ε k =[( m n ε ( n 1 ε] [( n ε ( ε] [( ε +1 ( ε +1+(k m n 1 1], ( m α k =[( m α (n α] [(n +1 α (n +1 α +(k m n 1 1], k! =(m + n +1! [(m + n +2 (m + n +2+(k m n 1 1]. g(x ( n ε k /( m n ε k ( ε( ε 1 ( ε n g(x l = k m n 1 g(x = [( ε +1 ( ε + m] [( m α (n α] [( m n ε ( n 1 ε] (m + n +1! xm+n+1 l=0 ( ε + m +1 l (n +1 α l x l ( ε +1 l (m + n +2 l. x < 1 x ε [ 1/2, 1/2] ε 0 3 m!( m α (n α (m +1 l (n +1 α l ( m n ( n 1 (m + n +1! xm+n+1 x l l!(m + n +2 l (8.8 ε l=0

19 b>0, x <b ( a, b 2F 1 x (a n = (1+ 1 ( 1+ 2 ( 1+ n 1 x n c b (c n=0 n b b b n!. x b [2 x, ( a, b lim b 2 F 1 x (a n x n = c b (c n n!. x C ( a (a n x n 1F 1 x = c (c n n!. (8.9 n=0 n=0 8.2 a1,a 2,,a p pf q x b 1,b 2,,b q q p 1 p, q N a1,a 2,,a p X pf q x = b 1,b 2,,b q n=0 2 F 1, 1 F 1 (a 1 n (a 2 n (a p n x n (b 1 n (b 2 n (b q n n! (8.10 1F x x/b ( F 1 ( ac x xy +(c xy ay = 0 (8.11 c Z (8.11 (0, ( a ( a c +1 y = A 1 F 1 x + Bx 1 c 1F 1 x c 2 c (A, B C (8.12 ( a 1 F ac 1 x 8.4 n, m N

20 ( m ( n 1F 1 x e x 1 F 1 x m n m n ( 1 m x m+n+1 ( m +1 = ( m + n (m + n +1! 1 F 1 x n + m +2 m (x C. 8.2 x x/α lim (1 + α x/α α = e x ( m, n + α lim α 2 F 1 x m n α ( m k = lim (1 n ( 1 n k +1 ( x k α ( m n k α α k! k=0 ( m = 1 F 1 x. m n ( n, m α lim α 2 F 1 x ( n = 1 F 1 x, m n α m n ( n +1 α, m +1 lim α 2 F 1 x ( m +1 = 1 F 1 x. m + n +2 α m + n K = Q( d 2 α K e α K α Q e α ( 3.14 ( k N π k π k Q k = m d d α =(π ki k Q( d e α = e ikπ = ( 1 k Q( d 8.5 e x 8.4 m = n

21

22 e (1873 ( π e π - ( ( ( K A K α A P K[x], P 0 P (α =0 α K A K K Q C f Ω f f C(x P 0,P 1,,P d C[x], P d 0 x Ω P d (x(f(x d + P d 1 (x(f(x d P 0 (x = 0 (12.1 f f (C(z

23 f(x =(1+x 1/2 2 f(x (f(x 2 (1 + x = e x e x P d (xe dx + P d 1 (xe (d 1x + + P 0 (x =0, (12.2 P 0,,P d C[x] d P d 0 x C P d (x = P d 1 (xe x P 0 (xe dx (12.3 x R, x ( P d = Ω = {x C x < 1} f(x = x 2n (12.4 n=0 f f(x 2 =f(x x. (12.5 f x Ω (f(x d + Q d 1 (x(f(x d Q 0 (x = 0 (12.6 Q i C(x d 1 (12.6 x x 2 (12.5 (f(x x d + Q d 1 (x 2 (f(x x d Q 0 (x 2 =0 (f(x d +(Q d 1 (x 2 dx(f(x d 1 + =0 (12.6 d Q d 1 (x =Q d 1 (x 2 dx Q d 1 (x =A(x/B(x A, B C[x]

24 A(xB(x 2 =A(x 2 B(x dxb(xb(x 2. (12.7 B(x 2 A(x 2 B(x B(x 2 A(x 2 B(x 2 B(x B(x =b C (12.7 A(x =A(x 2 bdx deg A 1 deg A =0 A(x C bdx =0 bd (1.6 α Z α 1 (12.8 Z K A K ( 8.3 K 2 ( α C d α 1 = α, α 2,,α d P α α α =max α i (12.9 i 12.1 α K α σ 1,σ 2,,σ m K α, β α =max σ i (α. (12.10 i α + β α + β, (12.11 αβ α β. ( α 0 α

25 (12.8 (12.13 α 0 α den α ( α d α α d+1 (den α d. ( (1929 (12.5 (12.4 f(x 12.5 α, 0 < α < 1 f(α = α 2n n=0 2 2n 1 f(x, (f(x 2, (f(x 3, ( k f(x = b kn x n, n=0 n=0 b kn Z m N 1 m +1 m P i (x = a in x n (i =0, 1,,m g m (x n=0 P m (x(f(x m + P m 1 (x(f(x m P 0 (x =x m2 g m (x (12.14 P i (x x n (n = 0, 1,,m m i=0 min(m,n k=0 a ik b i,n k =0 (n =0, 1,,m 2 1. (12.15

26 (m +1 2 a ik Z b i,n k 1 m 2 Q Z g m g m =0 P i 1 f 12.2 g m (x =x σ h m (x, σ 0, h m (0 0 ( f(α K = Q(α, f(α, d=[k : Q] a =denα, α = β/a, β A K, b =denf(α, f(α =γ/b, γ A K (12.5 n 1 f(x 2n =f(x x 2k (n 1 (12.17 f(α 2n = k=0 A n ba 2n 1, A n A K (n 1 (12.14 (12.16 x α 2n m ( β 2 n ( An i P i = α (m 2 +σ2 n h m (α 2n (12.18 a 2n ba 2n 1 i=0 (12.18 B n B n K P i deg P i m den B n a m2n b m a m2n 1 b m a m2n+1 ( (12.18 h m (0 0 n (12.20 B n h m (0α (m2 +σ2 n (n. (12.20 B n 0 (n (12.21 B n c 1 α m2 2 n (n 1 (12.22 c 1 n 4 B n 12.2

27 B n m ( m a ij α j2n ( f(α 2 n i. i=0 j=0 c 2 =max i,j a ij, c 3 =max(2, α, f(α ( 12.5 B n c 2 (m +1 2 c m2n+1 3 (n 1. (12.23 B n (12.13 (12.21 B n 0 B n K deg B n d (12.19 (12.22 (12.23 c 1 α m2 2 n (c 2 (m +1 2 c m2n+1 3 d+1 (b m a m2n+1 d md log b +( d +1log(c 2 (m n ( m 2 log α +2m(( d +1logc 3 d log a log c 1 (12.24 α < 1 m m 2 log α +2m(( d +1logc 3 d log a > 0 m n ( [19] (12.14 x m2 g m (x 2 β B n 3 β, den β, β 4 β 0 (12.20

28 (12.13 β 1 K A K ( 12.5 Z 1 a ik Z ( {A ij } 1 i m,1 j n n>m A N max A ij A i,j (x 1,,x n Z n 0 < max x i (na m n m, (12.25 i A 11 x 1 + A 12 x A 1n x n =0 A 21 x 1 + A 22 x A 2n x n =0. ( A m1 x 1 + A m2 x A mn x n =0. i =1, 2,,m V i ( W i {A i1,,a in } ( V i + W i na X N (x 1,,x n Z n (0 x i X n y i = A ij x j (i =1, 2,,m j=1 (y 1,,y m V i X y i W i X (i = 1,,m (x 1,,x n E (X+1 n (y 1,,y m F (nax +1 m X = [ ] (na m n m (X +1 n m > (na m (X +1 n > (X +1 m (na m (nax +1 m. card E>card F 2 (x 1,,x n, (x 1,,x n (y 1,,y m n A ij (x i x i =0 (i =1,,m j=1

29 , , , , 31, 40, ( , , , 54, , , ( , 96

30 , , , , , , , , , , 131, , 138 p 11

31 260 p 11 6, , x 2 + y 2 = z 2 52 x 3 + y 3 = z 3 58 x 4 + y 4 = z 4 53 x 5 + y 5 = z , 60, , 43, ( , , 80 34, , ζ e ( n

32 2003 JCLS Printed in Japan ISBN

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます.   このサンプルページの内容は, 初版 1 刷発行時のものです. 微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)

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