[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )

Similar documents
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


4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx


1

- II

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

, 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

x i [, b], (i 0, 1, 2,, n),, [, b], [, b] [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ] ( 2 ). x 0 x 1 x 2 x 3 x n 1 x n b 2: [, b].,, (1) x 0, x 1, x 2,, x n

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 =

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

2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p

DVIOUT

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) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)

2009 IA I 22, 23, 24, 25, 26, a h f(x) x x a h

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

高等学校学習指導要領

高等学校学習指導要領

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

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

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

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.

body.dvi

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

A S hara/lectures/lectures-j.html ϵ-n 1 ϵ-n lim n a n = α n a n α 2 lim a n = 0 1 n a k n n k= ϵ

Chap11.dvi

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0


J1-a.dvi

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

2011de.dvi

() 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.

f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) B p.1/14

(1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9

1 yousuke.itoh/lecture-notes.html [0, π) f(x) = x π 2. [0, π) f(x) = x 2π 3. [0, π) f(x) = x 2π 1.2. Euler α

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K

No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

no35.dvi

f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f

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

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

II K116 : January 14, ,. A = (a ij ) ij m n. ( ). B m n, C n l. A = max{ a ij }. ij A + B A + B, AC n A C (1) 1. m n (A k ) k=1,... m n A, A k k

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (

[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

(iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y = 0., y x, y = x. (v) 1x = x. (vii) (α + β)x = αx + βx. (viii) (αβ)x = α(βx)., V, C.,,., (1)

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (

Microsoft Word - 信号処理3.doc

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

°ÌÁê¿ô³ØII

v er.1/ c /(21)

t = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z

i

main.dvi

(2000 )

2014 S hara/lectures/lectures-j.html r 1 S phone: ,

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

φ s i = m j=1 f x j ξ j s i (1)? φ i = φ s i f j = f x j x ji = ξ j s i (1) φ 1 φ 2. φ n = m j=1 f jx j1 m j=1 f jx j2. m


Note.tex 2008/09/19( )

A

B [ 0.1 ] x > 0 x 6= 1 f(x) µ 1 1 xn 1 + sin sin x 1 x 1 f(x) := lim. n x n (1) lim inf f(x) (2) lim sup f(x) x 1 0 x 1 0 (

CALCULUS II (Hiroshi SUZUKI ) f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b)

熊本県数学問題正解

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s



( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

III ϵ-n ϵ-n lim n a n = α n a n α 1 lim a n = 0 1 n a k n n k= ϵ-n 1.1

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

II 2 II

応力とひずみ.ppt

p03.dvi

B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:

4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X

6. Euler x

基礎数学I


1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

F S S S S S S S 32 S S S 32: S S rot F ds = F d l (63) S S S 0 F rot F ds = 0 S (63) S rot F S S S S S rot F F (63)

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

DE-resume


FX ) 2

Transcription:

1 1.1 [] f(x) f(x + T ) = f(x) (1.1), f(x), T f(x) x T 1 ) f(x) = sin x, T = 2 sin (x + 2) = sin x, sin x 2 [] n f(x + nt ) = f(x) (1.2) T [] 2 f(x) g(x) T, h 1 (x) = af(x)+ bg(x) 2 h 2 (x) = f(x)g(x) T, 2 f(x) g(x), f(x) T 1, g(x) T 2, p 1 (x) = af(x) + bg(x) T 2 T 1 T 2 2 f(x) g(x) p 2 (x) = f(x)g(x),?

[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t ), f(x) f(x) ()

1.2 [] x T f(x) g(x) (f, g) = T f(x)g(x) dx (1.4) (f, g) =, f(x) g(x) x T a b, a, b θ( θ ) a b = a b cos θ a, b a = (a 1, a 2,, a n ), b = (b 1, b 2,, b n ), n a b = a i b i i=1, a, b,, a b = [] {1, cos x, sin x, cos 2x, sin 2x,, (1) 1 cos mx dx = (m = 1, 2, ) (1.5) 1 sin mx dx = (m = 1, 2, ) (1.6) cos mx sin nx dx = (m = 1, 2, n = 1, 2, ) (1.7) cos mx cos nx dx = (m n m = 1, 2, n = 1, 2, ) (1.8) sin mx sin nx dx = (m n m = 1, 2, n = 1, 2, ) (1.9) (2) 1 2 dx = 2 (1.1) cos 2 mx dx = (m = 1, 2, ) (1.11) sin 2 mx dx = (m = 1, 2, ) (1.12)

(1.5) (1.12), [ sin mx 1 cos mx dx = m 1 sin mx dx = ] 2 = 1 m (sin 2m sin ) = 1 ( ) = (1.5) m [ ] cos mx 2 = 1 m m (cos 2m cos ) = 1 (1 1) = (1.6) m cos mx sin nx dx = 1 {sin (m + n)x sin (m n)x dx 2 = 1 { sin (m + n)x dx sin (m n)x dx 2 = 1 [ ] 2 [ ] 2 cos (m + n)x cos (m n)x 2 m + n m n = (m n) (1.7) cos mx sin nx dx = = 1 2 cos mx sin mx dx {sin (m + m)x sin (m m)x dx = 1 sin 2mx dx 2 = 1 [ ] cos 2mx 2 2 2m = (m = n) (1.7) cos mx cos nx dx = 1 {cos (m + n)x + cos (m n)x dx 2 = 1 { cos (m + n)x dx + cos (m n)x dx 2 = 1 [ ] 2 [ ] 2 sin (m + n)x sin (m n)x + 2 m + n m n = (m n) (1.8) sin mx sin nx dx = 1 {cos (m n)x cos (m + n)x dx 2 = 1 { cos (m n)x dx cos (m + n)x dx 2 = 1 [ ] 2 [ ] 2 sin (m n)x sin (m + n)x 2 m n m + n = (m n) (1.9)

1 1 dx = [x] 2 = 2 (1.1) cos mx cos mx dx = = 1 2 = 1 2 { = 1 2 cos 2 mx dx (1 + cos 2mx) dx 1 dx + cos 2mx dx [ ] sin 2mx 2 [x] 2 + 2m { = 1 {2 + 2 = (1.11) sin mx sin mx dx = (1.12)

1.3 [] f(x) 2 f(x), f(x) a 2 + a 1 cos x + b 1 sin x + a 2 cos 2x + b 2 sin 2x + a 3 cos 3x + b 3 sin 3x + = a (a n cos nx + b n sin nx) (1.13), f(x), cos nx, sin nx(n = 1, 2, ) 2, 2 n,, 2 { (2L) f(x) = a (a n cos nx + b n sin nx) {{ () e x = 1 + x 1! + x2 2! + + xn n! + sin x = x x3 3! + x5 5! + ( 1)n x 2n+1 (2n + 1)! + cos x = 1 x2 2! + x4 x2n + ( 1)n 4! (2n)! + f(x) = f() + f () 1! f (n) () = x n n! x + f () 2! x 2 + f () x 3 + 3! { x, x, ()

[] 2 f(x), a n b n f(x) = a (a n cos nx + b n sin nx) (1.14) (1) a n (1.14) cos mx(m = 1, 2, ), 2 f(x) cos mx dx = { 2 a (a n cos nx + b n sin nx) cos mx dx (1.15) = a cos mx dx + 2 a n cos nx cos mx dx + b n sin nx cos mx dx {{ = a [ ] sin mx 2 { n m + a n 2 m n = m = + a m a = a m (m = 1, 2, ) (1.16) f(x) dx = { 2 a = a 2 (a n cos nx + b n sin nx) dx dx + a n cos nx dx +b n sin nx dx {{{{ = a 2 [x]2 + = a (1.17), (1.16) (1.17) a n = 1 f(x) cos mx dx = a m (m =, 1, 2, ) (1.18) f(x) cos nx dx (n =, 1, 2, ) (1.19) (2) b n (1.14) sin mx(m = 1, 2, ), 2 f(x) sin mx dx = { 2 a (a n cos nx + b n sin nx) sin mx dx (1.2)

= a sin mx dx + 2 a n cos nx sin mx dx +b n sin nx sin mx dx {{ = a [ ] { cos mx 2 n m + b n 2 m n = m = + b m = b m (m = 1, 2, ) (1.21) b n = 1 f(x) sin mx dx = b m (m = 1, 2, ) (1.22) f(x) sin nx dx (n = 1, 2, ) (1.23)

1.4 ( 2) [] : f( x) = f(x), y x x cos x x f( x) = x = x = f(x) f( x) = cos ( x) = cos x = f(x) : f( x) = f(x), x sin x x x f( x) = x = f(x) f( x) = sin ( x) = sin x = f(x)

[] (P1) f(x) f(x) = f e (x) + f o (x) f e = f(x) + f( x), f o = 2 f(x) f( x) 2 f e (x) f o (x) f(x) (P2),, h(x) = f e (x)g e (x) : h( x) = f e ( x)g e ( x) = f e (x)g e (x) = h(x) h(x) = f o (x)g o (x) : h( x) = f o ( x)g o ( x) = { f o (x){ g o (x) = f o (x)g o (x) = h(x) h(x) = f e (x)g o (x) : h( x) = f e ( x)g o ( x) = f e (x){ g o (x) = f e (x)g o (x) = h(x) (P3) f e (x), f o (x), f e (x) dx = 2 f o (x) dx = f e (x) dx

[] ( 1) f e (x), b n = b n = 1 {{ f e (x) sin {{{{ nx dx = a n = 1 {{ f e (x) cos {{{{ nx dx = 2 f e (x) cos nx dx ( 2) f o (x), a n = a n = 1 {{ f o (x) cos {{{{ nx dx = b n = 1 {{ f o (x) sin {{{{ nx dx = 2 f o (x) sin nx dx ( 3) f(x) f e (x), f o (x), f(x) = f e (x)+ f o (x) a n = 1 = 1 = 1 = 2 b n = 1 = 1 = 1 = 2 f(x) cos nx dx {f e (x) + f o (x) cos nx dx f e (x) cos nx dx f e (x) cos nx dx f(x) sin nx dx {f e (x) + f o (x) sin nx dx f o (x) sin nx dx f o (x) sin nx dx ( 4) f(x) g(x), a n (f), b n (f) a n (g), b n (g), c d cf(x)+dg(x), ca n (f)+da n (g), cb n (f) + db n (g)

2 f(x) ( P.16[ 2](1)) { 1 ( x < ) T = 2 f(x) = ( x < 2) (1) a n a n = 1 f(x) cos nx dx = 1 { f(x) cos nx dx + = 1 { 1 cos nx dx + = 1 cos nx dx = 1 [ ] sin nx (n ) n = 1 (sin n sin ) = n n = a = 1 f(x) dx = 1 { 1 dx + (2) b n f(x) cos nx dx cos nx dx dx = 1 dx = 1 [x] = 1 [ ] = 1 b n = 1 f(x) sin nx dx = 1 { f(x) sin nx dx + = 1 { 1 sin nx dx + = 1 sin nx dx = 1 [ ] cos nx n = 1 (cos n cos ) n = 1 n {( 1)n 1 = { 2 n f(x) sin nx dx sin nx dx (n ) (n ) ( b 2n 1 = ) 2 (2n 1) f(x) = a (a n cos nx + b n sin nx) = 1 [ ] 1 sin nx {1 ( 1)n n = 1 2 + 2 (sin x + 1 3 sin 3x + 1 sin 5x + ) 5 [ = 1 2 + 2 ] 1 sin (2n 1)x 2n 1

2 f(x) ( P.18[ 2](2)) T = 2 f(x) = x ( x < ) f(x), ( 1), b n = (n = 1, 2, 3, ), a n (n =, 1, 2, ) a n = 1 = 2 f(x) cos nx dx f(x) cos nx dx = 2 x cos nx dx = 2 ( ) sin nx x dx (n ) n = 2 { [ ] sin nx x 1 sin nx dx n n = 2 [ ] cos nx n n = 2 n 2 {( 1)n 1 = { (n ) (n ) 4 n 2 ( ) 4 a 2n 1 = (2n 1) 2 n = a = 1 f(x) dx = 2 x dx = 2 [ ] x 2 2 = 2 ( 2 2 ) = f(x) = a (a n cos nx + b n sin nx) = [ ] 2 cos nx {( 1)n 1 [ = 2 4 = 2 4 ( cos x + n 2 ) cos 3x cos 5x + + 3 2 5 ] 2 cos (2n 1)x (2n 1) 2

2 f(x) ( P.18[ 2](3)) T = 2 f(x) = x( x < ) f(x), ( 2), a n = (n =, 1, 2, ), b n (n = 1, 2, 3, ) b n = 1 = 2 f(x) sin nx dx f(x) sin nx dx = 2 x sin nx dx = 2 ( ) cos nx x dx n = 2 {[ ( )] cos nx x + 1 n n = 2 { [ ] sin nx ( 1) n + n n = 2 { 2 (n ) n ( 1)n+1 = n (n ) 2 n cos nx dx f(x) = a (a n cos nx + b n sin nx) = = 2 n+1 sin nx 2( 1) n ( sin x sin 2x 2 + sin 3x 3 )

1 [ 2 ] f(x) = a (a n cos nx + b n sin nx) a n = 1 f(x) cos nx dx (n =, 1, ) a n = 1 f(x) cos nx dx (n =, 1, ) b n = 1 f(x) sin nx dx (n = 1, 2, ) b n = 1 f(x) sin nx dx (n = 1, 2, ) f(x) = x ( x < 2) f(x) 2 2 x, f(x) = x ( x < 2) b n = 1 f(x) sin nx dx = 1 x sin nx dx b n = 1 f(x) sin nx dx = 1 x sin nx dx, f(x) = x ( x < ), f(x) = x ( x < 2) f(x) = x ( x < ), b n = 1 f(x) sin nx dx = 1 + 2) sin nx dx + (x 1 x sin nx dx f(x) 2 x

1.5 [] f(x) x, x { f( x) ( x ) f e (x) = (1.24) f(x) ( x ), x 2 f e (x + 2n) = f e (x) x <, n =, 1,, 1, (1.25) f(x) = a 2 + a n cos nx, a n = 2 f(x) cos nx dx (n =, 1, 2, ) (1.26) [] f(x) x, x { f( x) ( x ) f o (x) = (1.27) f(x) ( x ), x 2 f o (x + 2n) = f o (x) x <, n =, 1,, 1, (1.28) f(x) = b n sin nx, b n = 2 f(x) sin nx dx (n = 1, 2, ) (1.29) f(x) x < () f e (x) < x < < x < f e (x) f o (x) < x < < x < f o (x) ( 1) ( 2) ()

1.6 ( 2L) [ ( 2L) ] x = L t h(t) = a (a n cos nt + b n sin nt) t : 2 (1.3) t x 2 2L, 2 t 2L x ( ), f(x) 2L(L > ), t = L x f(x) = h( L x) = a (a n cos n L x + b n sin n L x) x : 2L (1.31) a n dt = (dt = dx) dx L L a n = 1 = 1 = 1 L L L L L h(t) cos nt dt h( L x) cos n L x L dx f(x) cos n x dx (1.32) L b n b n = 1 L L L f(x) sin n L x dx (1.33) [ ( 2L) ] f(x) x L f(x) = a 2 + a n = 2 L L a n cos n L x (b n = ) (1.34) f(x) cos n x dx (1.35) L f(x) = b n sin n L x (a n = ) (1.36) b n = 2 L f(x) sin n x dx (1.37) L L

2L f(x) { 1 ( x < L) T = 2L f(x) = (L x < 2L) (1) a n a n = 1 2L f(x) cos n L L x dx = 1 { L f(x) cos n 2L L L x dx + f(x) cos n L L x dx = 1 { L 1 cos n 2L L L x dx + cos n L L x dx = 1 L cos n L L x dx = 1 [ L L n sin n ] L L x (n ) = 1 (sin n sin ) = n n = a = 1 2L f(x) dx = 1 { L 2L 1 dx + dx = 1 L dx = 1 L L L L L [x]l = 1 [L ] = 1 L (2) b n 2L b n = 1 f(x) sin n L L x dx = 1 { L f(x) sin n 2L L L x dx + f(x) sin n L L x dx = 1 { L 1 sin n 2L L L x dx + sin n L L x dx L = 1 sin n L L x dx = 1 [ L L n cos n ] L L x = 1 (cos n cos ) n = 1 n {( 1)n 1 = { 2 n (n ) (n ) f(x) = a (a n cos n L x + b n sin n L x) = 1 [ ] 1 n {1 ( 1)n sin n L x (sin L x + 1 3 sin 3 L x + 1 5 sin 5 ) L x + [ = 1 2 + 2 = 1 2 + 2 1 2n 1 (2n 1) sin x L ]

L = ( 2L = 2) f(x) = 1 2 + 2 (sin x + 1 3 sin 3x + 1 sin 5x + ) 5, 2 P.17[ 2](1) f(x) L = 1( 2L = 2), T = 2 f(x) = f(x) = 1 2 + 2 (sin x + 1 3 sin 3x + 1 sin 5x + ) 5 { 1 ( x < 1) (1 x < 2)

2 [ 2 f(x) ] f(x) = a (a n cos nx + b n sin nx) a n = 1 f(x) cos nx dx (n =, 1, 2, ) b n = 1 f(x) sin nx dx (n = 1, 2, ) x f(x) f(x) = a 2 + a n cos nx, a n = 2 f(x) cos nx dx (n =, 1, 2, ) x f(x) f(x) = b n sin nx, b n = 2 f(x) sin nx dx (n = 1, 2, ) x x, L L [ 2L f(x) ] L f(x) = a (a n cos nx L a n = 1 L L L b n = 1 L L L f(x) cos nx L f(x) sin nx L + b n sin nx L ) dx (n =, 1, 2, ) dx (n = 1, 2, ) L x L f(x) f(x) = a 2 + a n cos nx L, a n = 2 L L f(x) cos nx L dx (n =, 1, 2, ) L x L f(x) f(x) = b n sin nx L, b n = 2 L L f(x) sin nx L dx (n = 1, 2, )

1.7 [] a x b a = x < x 1 < x 2 < < x n = b x 1, x 2,, x n 1,,, lim f(x i + e) = f(x i + ) (e >, i =, 1, 2,, n) e lim f(x i e) = f(x i ) (e >, i =, 1, 2,, n) e, f(x) a x b 1, 2 () x i [] x i, x i+1 f(x) xi+1 e 2 xi+1 lim f(x) dx = f e 1,e 2 i (x) dx x i +e 1 x i, f i (x), f(x) [x i, x i+1 ], f(x i + ) fi (x) = f(x) f(x i+1 ) (x = x i ) (x i < x < x i+1 ) (x = x i+1 ), f(x) b a f(x) dx = x1 x f (x) dx + x2 x 1 f 1 (x) dx + + xn x n 1 f n 1(x) dx,,,

[] f(x) f (x) a x b, f(x) [] f(x) 2, f(x) [] f(x) x, f(x) x f(x), x f(x + ) + f(x ) 2

[ ( )] P.17[ 2](2), f(x) = x ( x < ) f(x) = a (a n cos nx + b n sin nx) = 2 [ ] 2 cos nx {( 1)n 1 = 2 4 [ = 2 4 ( cos x + cos 3x + 3 2 cos (2n 1)x ] (2n 1) 2 n 2 cos 5x 5 2 +, f(x) x <, x = 2 4 ( ) cos 3x cos 5x cos x + + + 3 2 5 2, x =, f() = =,, cos = 1, = 2 4 (1 + 1 3 2 + 1 5 2 + ) 2, 4 1 + 1 3 + 1 2 5 + + 1 2 (2n 1) + = 2 2 8, ) P.29 1-6 2