(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (
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1 B / 9/ 3/ y = x 2 x x =
2 (, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) ( ( ( = ) 2 ( ) 2 2 ( ) 3 2 ( ) ) = 4 ( ) k = 4 k= 6 = k= 4 6 k 2 = 5 32 =.46 (4 + )(2 4 + ) n ( n 2 ) = k 2 = n (n + )(2n + ) 6 k= ) 2 2
3 8 4 S S [, ] 8 S S 8 n n 3. [, ] n S n S n 3
4 n n S n lim S n n lim S n = lim n n = lim n (n + )(2n + ) 6n2 ( + ) ( n n = 6 2 = 3 ) lim n S n 3 [, ] 3 y = x2 x x = 3 S n lim S n n 3 [, ] ( ) k 2 k =, 2, 3 4 s 4 s 4 = 4 3 k= ( ) k 2 = k= k 2 =
5 4. [, ] 8 s s 8 5. [, ] n s n s n n s n lim n s n lim s n = lim n n = lim n (n )(2n ) 6n2 ( ) ( 2 6 n n = 6 2 = 3 [, ] 3 n s n < S < S n lim s n = lim n S n = 3 y = x 2 x x = S 3 S x 2 dx ) 2 y = x 2 f(x) y = f(x) x x = x = b < b 5
6 f(x) =x x x 2 x 3... x n =b [, b] n = x, x, x 2,..., x n, x n = b x k x k (k =, 2,..., n) f(x k ) (k =, 2,..., n) n n S n =f(x )(x x ) + f(x 2 )(x 2 x ) + + f(x n )(x n x n ) n = f(x k )(x k x k ) k= S = lim n k= S b n f(x k )(x k x k ) b f(x) dx f(x) dx f b integrl n lim f(x k ) (x k x k ) n k= b f(x) S sum dx S 6
7 f(x) f(x) dx n ( ) f(x) = x 2 S = x 2 k 2 dx lim n n n k= n lim f(x k )(x k x k ) b n k= b b f(x) dx = f(x) dx = c f(x) dx = f(x) dx+ b f(x) dx b c f(x) dx 2 b b f(x) f(x) f(x) x x f(x) dx f(x) f(x) dx f(x) f(x) F (x) f(x) F (x) = f(x) b f(x) dx = F (b) F () 2. F (x) = x f(x) dx f(x) 2. f(x) F (x) b f(x) dx = F (b) F ().2 f(x) F (x) = x F (x) = f(x) 7 f(x) dx
8 6. f(x) = x [, 5] F (x) =. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x 7. f(x) = 2x [, 5] F (x) = x. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x x dx x dx = x 2x dx 2x dx = 8. f(x) = 2 x [, 5] F (x) = x. F (), F (), F (2), F (3), F (4), F (5) 2. x 5 F (x) = 3. F (x) F (x) = x 2 2 x dx = ( x 5) x x ( x 5) 2x x x dx x dx ( x 5) 2 x 9 (). [, b] f x b F (x) = x f(x) dx F F (x) = f(x) f(x) F (x) x f(t) dt F (x) 2x dx x 2 x dx 8
9 Proof. F (x) = x f(x) dx F (x) x f(x) f(x) S = F (x) S x x+h x f(x) dx x+h S f(x) F (x) S = F (x + h) F (x) S F (x) h f(x) h f(x + h) f(x) h < S < f(x + h) h h h > f(x) < S h < f(x + h) S = F (x + h) F (x) h f(x) < F (x + h) F (x) h < f(x + h) F (x + h) F (x) lim f(x) < lim < lim f(x + h) h h h h F (x + h) F (x) lim f(x) = f(x) lim h h h F (x) = f(x) = F (x) lim h f(x + h) = f(x) Remrk. x f(x) dx x x f(x)dx x x 2 x t x f(t) dt x x b t t x 9
10 .3 f(x). f(x) f(x) F (x) = f(x) F (x) f(x) 2 ( 2). F (x) f(x) b f(x) dx = F (b) F () Proof. G(x) = x f(x) dx 9G (x) = f(x) F (x) f(x) F (x) = f(x) G(x) F (x) (G(x) F (x)) = G (x) F (x) = f(x) f(x) = C x G() = x b G(x) F (x) = C G() F () = C f(x) dx = C = F () b G(x) F (x) = F () f(x) dx = G(b) = F (b) F () F (b) F () [ ] b F (x) F (x) b b [ ] b f(t) dt = F (x)
11 x 2 dx x 2 F (x) x 2 [ ] F (x) F () F () 2 x 2 dx f(x).4 ( ) 3 x3 = x 2 ( ) 3 x3 x 2 3 x3 + 3 = x 2 ( 3 x3 + 3 x 2 3 x3 + ) 2 = x 2 3 x3 + 2 x 2 x 2 4. f(x) f(x) f(x)dx f(x) dx f(x) f(x) x 2 C 3 x3 + C x 2 x 2 dx = 3 x3 + C 5. f(x) F (x) f(x) C F (x) + C C = F (x) f(x)dx = F (x) + C C f f
12 f(x) f(x) dx x dx =? x x x dx x x dx? (?)? x (?) = x? 2 x2 x x x x ( ) 2 x2 + C x dx 2 x2 + C x dx = 2 x2 + C f(x).4. x n x dx x 2 dx x 2 x 2 dx (?)? x n dx x 2 dx =? (?) = x 2? = 3 x3 + C x 2 dx = 3 x3 + C x n dx =? (?) = x n? = n + xn+ + C 2
13 6. x n dx = n + xn+ + C (n =,, 2,... ) dx = x + C dx dx 7. C. x 4 dx 3. x 6 dx 5. x dx 2. x 5 dx 4. x 99 dx kf(x) dx = k f(x) dx k ( f(x) + g(x)) dx = ( f(x) g(x)) dx = f(x) dx + f(x) dx g(x) dx g(x) dx Proof. 2. ( f(x) + g(x) (f(x) + g(x)) dx (?)? f(x) dx + ( g(x) dx) = ( f(x) dx) + (f(x) + g(x)) dx =? g(x) dx) = f(x) + g(x) (?) = f(x) + g(x)? = f(x) dx + g(x) dx (f(x) + g(x)) dx = f(x) dx + g(x) dx 9. 3
14 2. (8x 3 5x + 3) dx = 8 = 8 x 3 dx 5 4 x4 5 x dx + 3 = 2x x2 + 3x + C dx 2 x2 + 3x + C 2.. (3x 2 4x 2) dx 2. (2x + x 3 ) dx 3. (x )(2x 3) dx 4. t 3 (t 6) dt 5. (2y + 5) 2 dy 6. 5 dx 7. 7 dy (2x 3 3x 2 ) dx (x 3)(2x + ) dx (3x + 2) dx (x 2 6x + 5) dx (8x 3 + 2x 2 + 4x) dx (x + 2)(x 2) dx (3x + 7)(x ) dx, 2, F (x). F (x) = x 2 4x, F (3) = 4 () F (x) x 2 4x F (x) = (x 2 4x) dx = (b) F (3) = 4 C C = (c) F (x) = 2. F (x) = 6x 2 4x 5, F (2) = 2 3. F (x) = 3 x, F () = F (x) = 5x 2 2x + 6, F ( ) = 9 5. F (x) = (3x )( x), F () = 3 4,5 f(x) F (x) f(x) dx F (x) 4
15 23.. y = f(x) (, ) (x, f(x)) 3(x 2 ) f(x) 2. y = f(x) (, 2) (2, ) (x, f(x)) x 3 x f(x).4.3 x α 24. α x α dx = xα+ α + + C f(x) dx dx f(x) Remrk 25. α = α = x dx dx x 3 dx x 2 dx x x 6 dx 3 x dx x x dx x 4 dx dx x n n n ( x 2 x 3 ) dx 3. (x x) 2 dx 2. ( x + x ) 2 dx 4. x 3 2x 2 dx x 5 5
16 f(x) b F (x) b f(x) dx = 29. f(x) dx f(x) [ ] b F (x) = F (b) F (). 4 x 3 dx x 3 dx = F (x) F (4) F () (y 2 2y 3)dy (x 3) 2 dx < 6
17 dx x. 3 2 x 2 dx 5. 3 (t 3) 2 dt 2. 3 x 2 dx dx 6. 3 (y 3 4y)dy 3. 3 x 3 dx 3. (2x 2 + 3x + 5)dx 3. β α (x α)(x β) dx = 6 (β α) f(x) f( x) = f(x) f( x) = f(x) 34. f(x) f(x) f(x)dx = f(x)dx = 2 f(x)dx y 7
18 x 2 dx x 4 dx (x 3 + 2x 2 + ) dx (4x 3 + 6x 2 9x ) dx (x 3) 2 dx [, b] f(x) y = f(x) x 2 x =, x = b S f(x) S = b f(x) S = f(x) dx b f(x) dx f(x) S = b f(x) dx f(x) y = x 3 3x 2 + 2x x S Proof.. x y = x 3 3x 2 +2x = x(x 2)(x ) y = x =,, 2 2. S = (x 3 3x 2 + 2x) dx 2 (x 3 3x 2 + 2x) dx = x S 8
19 . y = 3x 2 x 3 2. y = x 3 + x 2 2x 3. y = x 2 5x y = x 2 x x 2 x =, x = 4. y = 4x x 2 2. y = x 3 9
20 4. [, b] f(x) g(x) 2 y = f(x), y = g(x) 2 x =, x = b S S = b ( ) f(x) g(x) dx. g(x) b f(x) g(x) S = = b b f(x) dx b g(x) dx ( ) f(x) g(x) dx 2. f(x), g(x) y m f(x) + m g(x) + m f(x) g(x) S = = = b b b (f(x) + m) dx b (g(x) + m) dx ( ) f(x) + m (g(x) + m) dx ( ) f(x) g(x) dx b 4. y = x 2 + 2x + 4 y = x x 2 + 2x + 4 = x 2 2x 2 + 2x + 4 = 2(x + )(x 2) = 2. 3 S = 2 2(x + )(x 2) dx = (x + )(x 2) dx = 2 6 (2 + )3 = 9. y = x 2 4 y = x 2 2. y = 4x x 2 y = x 4 3. y = x 2 3x 4 y = 2 + x x 2 4. y = 2x 2 7x + 9 y = 5x x 2 2
21 43.. y = x 2 3x + 5 y = 3 3. y = 2x 2 +3x 5 y = x 2 +4x 3 2. y = 2x 2 + 3x 5 y = x 4. y = x 3 x 2 y = x 44.. y = x 2 x 2. y = x 3 x y 4. y = x 2 + 2x + 3 x 5. y = x 2 y = x 2 + 3x + 5 x = x = 2 3. y = x 2 x 3 x 45. x x =, x = 2. y = x 2 2x y = x 2 + 5x f(x) = x + b f(x) dx =, xf(x) dx = f(x) = x 2 + bx +. f () = 4, 2. f(2) = 3, f(x) dx = xf(x) dx = 3 n lim f(x k ) (x k x k ) n k= b f(x) dx 2 2
22 (km/ ).5 5 ( ) km/ 5.5km/.5km km/ 5 5.5km/ km/ km = km 5 5 f() 2 f(2) 3 f(3) 4 f(4) 5 f(5) = 7.5 (km/ ) ( ) 5 f(x) dx km/h h = km kg/h h = kg /kg kg = / = / = / = 22
23 48. 2 f(x) = x 2 3x x kg kg y y = 5 4x + 6x 2 3 kg 7 kg 4 kg 5. 3 t y = 4 3 x 4 9 x kg 3 x kg 3x 2. x kg 2..4 kg kg 4. x kg 52. x kg kg /kg y = x 2 6x kg 23
24 [],, [2] 4, 5, IIIC+α [3] I II III,,, [4], [5], [6], S.,, [7], [8] ei =-,, ; (2/) [9], [], [], [2], [3], [4] mth stories,, (29//) [5]!!, [6] :,,, ; (23/3/9) [7],,,,, [8], A.C., K. [9] mth stories, [2], [2],, 24
25 [22], [23] 96, [24], [25],, (23/5) [26]!,, (2/7/6) [27],,, [28], [29],, [], [2], [3] [4], A.C., K. [5], [6], [7], [8], [9] 4 5 6, [], [] 3 3, [2], [3] :, 25
26 2.3 (f(g(x))) = f (g(x)) g (x) f(x + b) dx =? (?) = f(x + b)? f(x + b) f(x + b) dx (?)? F (t) f(t) F (t) = f(t) ( ) F (x + b) = F (x + b) = f(x + b)? = F (x + b) + C 53. F (t) f(t) f(x + b) dx = F (x + b) + C f(t) t x+b (x+b) α sin(x + b) 54. (2x ) dx Proof. t = 2x f(t) = t F (t) = (2x ) dx = 2 (2x ) + C = 22 (2x ) + C t dt = t + C f(x + b) 2 x + b t 26
27 55.. (2x + ) 2 2. (5 4x) 3 3. (x + 3) 2 4. (3x 2) 4 5. (4 3x) (2x + 3) 2 3x + 8. (3x 2) 2 9. ( 4x) 3. ( ) x +. (x + 2) 2. ( 5x) ( 2x) x Proof. (2x ) dx (2x ) dx = [ 22 (2x ) 2x + 3 ( x ) 3 8x + 7 ] = (x 3) 2 dx (4 x) 3 dx (2x ) 3 dx (3 2x) dx (2x ) 3 dx 5 2x dx + 3x dx (2x 3) 4 dx dx 3x 5 5 2x dx + 3x dx (2x 3) 4 dx dx 3x 5 27
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B
1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a
3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (
y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) (velocity) p(t) =(x(t),y(t),z(t)) ( dp dx dt = dt, dy dt, dz ) dt f () > f x
I 5 2 6 3 8 4 Riemnn 9 5 Tylor 8 6 26 7 3 8 34 f(x) x = A = h f( + h) f() h A (differentil coefficient) f f () y = f(x) y = f( + h) f(), x = h dy dx f () f (derivtive) (differentition) (velocity) p(t)
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)
Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =
1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A
2010 II / y = e x y = log x = log e x 2. ( e x ) = e x 3. ( ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0
2010 II 6 10.11.15/ 10.11.11 1 1 5.6 1.1 1. y = e x y = log x = log e x 2. e x ) = e x 3. ) log x = 1 x 1.2 Warming Up 1 u = log a M a u = M a 0 log a 1 a 1 log a a a r+s log a M + log a N 1 0 a 1 a r
() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.
![() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.](/thumbs/89/98384840.jpg "() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.")
() 6 f(x) [, b] 6. Riemnn [, b] f(x) S f(x) [, b] (Riemnn) = x 0 < x < x < < x n = b. I = [, b] = {x,, x n } mx(x i x i ) =. i [x i, x i ] ξ i n (f) = f(ξ i )(x i x i ) i=. (ξ i ) (f) 0( ), ξ i, S, ε >
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i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
ii 3.,. 4. F. (), ,,. 8.,. 1. (75%) (25%) =7 20, =7 21 (. ). 1.,, (). 3.,. 1. ().,.,.,.,.,. () (12 )., (), 0. 2., 1., 0,.
24(2012) (1 C106) 4 11 (2 C206) 4 12 http://www.math.is.tohoku.ac.jp/~obata,.,,,.. 1. 2. 3. 4. 5. 6. 7.,,. 1., 2007 (). 2. P. G. Hoel, 1995. 3... 1... 2.,,. ii 3.,. 4. F. (),.. 5... 6.. 7.,,. 8.,. 1. (75%)
1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1
sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p
2012 IA 8 I 1 10 10 29 1. [0, 1] n x = 1 (n = 1, 2, 3,...) 2 f(x) = n 0 [0, 1] 2. 1 x = 1 (n = 1, 2, 3,...) 2 f(x) = n 0 [0, 1] 1 0 f(x)dx 3. < b < c [, c] b [, c] 4. [, b] f(x) 1 f(x) 1 f(x) [, b] 5.
- II
- II- - -.................................................................................................... 3.3.............................................. 4 6...........................................
, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x
![, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x](/thumbs/93/113428934.jpg ", 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x")
1 1.1 4n 2 x, x 1 2n f n (x) = 4n 2 ( 1 x), 1 x 1 n 2n n, 1 x n n 1 1 f n (x)dx = 1, n = 1, 2,.. 1 lim 1 lim 1 f n (x)dx = 1 lim f n(x) = ( lim f n (x))dx = f n (x)dx 1 ( lim f n (x))dx d dx ( lim f d
body.dvi
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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. http://www.morikita.co.jp/books/mid/008142 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. Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published
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