[] 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 )
|
|
|
- ひろき こしの
- 5 years ago
- Views:
Transcription
1 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),?
2 [] 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) ()
3 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)
4 (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)
5 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)
6 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, ()
7 [] 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)
8 = 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)
9 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)
10 [] (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
11 [] ( 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)
12 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 = (sin x sin 3x + 1 sin 5x + ) 5 [ = ] 1 sin (2n 1)x 2n 1
13 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 ] 2 cos (2n 1)x (2n 1) 2
14 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 )
15 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
16 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) ()
17 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
18 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 sin 3 L x sin 5 ) L x + [ = = n 1 (2n 1) sin x L ]
19 L = ( 2L = 2) f(x) = (sin x sin 3x + 1 sin 5x + ) 5, 2 P.17[ 2](1) f(x) L = 1( 2L = 2), T = 2 f(x) = f(x) = (sin x sin 3x + 1 sin 5x + ) 5 { 1 ( x < 1) (1 x < 2)
20 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, )
21 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,,,
22 [] 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
23 [ ( )] 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 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 , x =, f() = =,, cos = 1, = 2 4 ( ) 2, (2n 1) + = 2 2 8, ) P
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 (
http://know-star.com/ 3 1 7 1.1................................. 7 1.2................................ 8 1.3 x n.................................. 8 1.4 e x.................................. 10 1.5 sin
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
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
( ) ( ) ( ) i (i = 1, 2,, n) x( ) log(a i x + 1) a i > 0 t i (> 0) T i x i z n z = log(a i x i + 1) i=1 i t i ( ) x i t i (i = 1, 2, n) T n x i T i=1 z = n log(a i x i + 1) i=1 x i t i (i = 1, 2,, n) n
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
- II
- II- - -.................................................................................................... 3.3.............................................. 4 6...........................................
, 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
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(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
![, 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
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
![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](/thumbs/92/108343288.jpg "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, R f : R R,.,, b R < b, f(x) [, b] f(x)dx,, [, b] f(x) x ( ) ( 1 ). y y f(x) f(x)dx b x 1: f(x)dx, [, b] f(x) x ( ).,,,,,., f(x)dx,,,, f(x)dx. 1.1 Riemnn,, [, b] f(x) x., x 0 < x 1 < x 2 < < x n 1
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
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(,,
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.
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
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
< 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)
< 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) 6 y = g(x) x = 1 g( 1) = 2 ( 1) 3 = 2 ; g 0 ( 1) =
2009 IA I 22, 23, 24, 25, 26, a h f(x) x x a h
009 IA I, 3, 4, 5, 6, 7 7 7 4 5 h fx) x x h 4 5 4 5 1 3 1.1........................... 3 1........................... 4 1.3..................................... 6 1.4.............................. 8 1.4.1..............................
, 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,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 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,
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 ),
I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10
1 2007.4.13. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 0. 1. 1. 2. 3. 2. ɛ-δ 1. ɛ-n
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
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)
body.dvi
..1 f(x) n = 1 b n = 1 f f(x) cos nx dx, n =, 1,,... f(x) sin nx dx, n =1,, 3,... f(x) = + ( n cos nx + b n sin nx) n=1 1 1 5 1.1........................... 5 1.......................... 14 1.3...........................
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
![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](/thumbs/101/149159646.jpg "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")
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
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
A S1-20 http://www2.mth.kyushu-u.c.jp/ hr/lectures/lectures-j.html 1 1 1.1 ϵ-n 1 ϵ-n lim n n = α n n α 2 lim n = 0 1 n k n n k=1 0 1.1.7 ϵ-n 1.1.1 n α n n α lim n n = α ϵ N(ϵ) n > N(ϵ) n α < ϵ (1.1.1)
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
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
009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n
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
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 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
J1-a.dvi
4 [ ] 4. ( ) (x ) 3 (x +). f(x) (x ) 3 x 3 3x +3x, g(x) x + f(x) g(x) f(x) (x 3)(x +)+x+ (x 3)g(x)+x+ g(x) r(x) x+ g(x) x (x+)+ x r(x)+ g(x) x x 4x+5 r(x) g(x) x (f(x) (x 3)g(x)) g(x) x f(x) f(x)g(x) 4
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
名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
() 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 >
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
B p.1/14 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 f(x,y) (x,y) x (x,y), y (x,y) f(x,y) x y f x (x,y),f y (x,y) f(x 1,...,x n ) (x 1 x 0,...,x n 0), (x 1,...,x n ) i x i f xi
0 1-4. 1-5. (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
1-1. 1, 2, 3, 4, 5, 6, 7,, 100,, 1000, n, m m m n n 0 n, m m n 1-2. 0 m n m n 0 2 = 1.41421356 π = 3.141516 1-3. 1 0 1-4. 1-5. (1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c),
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
1 http://sasuke.hep.osaka-cu.ac.jp/ yousuke.itoh/lecture-notes.html 1.1. 1. [, π) f(x) = x π 2. [, π) f(x) = x 2π 3. [, π) f(x) = x 2π 1.2. Euler dx = 2π, cos mxdx =, sin mxdx =, cos nx cos mxdx = πδ mn,
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
II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F
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
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
no35.dvi
p.16 1 sin x, cos x, tan x a x a, a>0, a 1 log a x a III 2 II 2 III III [3, p.36] [6] 2 [3, p.16] sin x sin x lim =1 ( ) [3, p.42] x 0 x ( ) sin x e [3, p.42] III [3, p.42] 3 3.1 5 8 *1 [5, pp.48 49] sin
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
22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(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
lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d
lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3
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
: January 14, 28..,. 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, A. lim k A k = A. A k = (a (k) ij ) ij, A k = (a ij ) ij, i,
() 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)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
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 -
M3............................................................................................ 3.3................................................... 3 6........................................... 6..........................................
(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (
![(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (](/thumbs/92/107866524.jpg "(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (")
B 4 4 4 52 4/ 9/ 3/3 6 9.. y = x 2 x x = (, ) (, ) S = 2 = 2 4 4 [, ] 4 4 4 ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, 4 4 4 4 4 k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) 2 2 + ( ) 3 2 + ( 4 4 4 4 4 4 4 4 4 ( (
[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29
![[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29](/thumbs/93/114156928.jpg "[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29")
A p./29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x) + C f(x) A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x)
() 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 (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () 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 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
(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)
1. 1.1...,. 1.1.1 V, V x, y, x y x + y x + y V,, V x α, αx αx V,, (i) (viii) : x, y, z V, α, β C, (i) x + y = y + x. (ii) (x + y) + z = x + (y + z). 1 (iii) 0 V, x V, x + 0 = x. 0. (iv) x V, y V, x + y
I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
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
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 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
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 (
![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 (](/thumbs/100/143999768.jpg "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 (")
8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin
Microsoft Word - 信号処理3.doc
Junji OHTSUBO 2012 FFT FFT SN sin cos x v ψ(x,t) = f (x vt) (1.1) t=0 (1.1) ψ(x,t) = A 0 cos{k(x vt) + φ} = A 0 cos(kx ωt + φ) (1.2) A 0 v=ω/k φ ω k 1.3 (1.2) (1.2) (1.2) (1.1) 1.1 c c = a + ib, a = Re[c],
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 =
5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =
°ÌÁê¿ô³ØII
July 14, 2007 Brouwer f f(x) = x x f(z) = 0 2 f : S 2 R 2 f(x) = f( x) x S 2 3 3 2 - - - 1. X x X U(x) U(x) x U = {U(x) x X} X 1. U(x) A U(x) x 2. A U(x), A B B U(x) 3. A, B U(x) A B U(x) 4. A U(x),
v er.1/ c /(21)
12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,
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 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
i
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,
main.dvi
Chpter 5 5. [; ] f(x) 3 f(x) = X + ( n cos nx + b n nx ) (5:) n= (5.) ; ;...;b ;b ;... n; m > 3 Z Z Z mx cos mx mx nx nx cos ( : n 6= m dx = : n = m ( : n 6= m dx = : n = m nx cos dx = (5.) (5.) (5.) 3
(2000 )
(000) < > = = = (BC 67» BC 1) 3.14 10 (= ) 18 ( 00 ) ( ¼"½ '"½ &) ¼ 18 ¼ 0 ¼ =3:141596535897933846 ¼ 1 5cm ` ¼ = ` 5 = ` 10 () ` =10¼ (cm) (1) 3cm () r () () (1) r () r 1 4 (3) r, 60 ± 1 < > µ AB ` µ ±
2014 S hara/lectures/lectures-j.html r 1 S phone: ,
14 S1-1+13 http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html r 1 S1-1+13 14.4.11. 19 phone: 9-8-4441, e-mail: hara@math.kyushu-u.ac.jp Office hours: 1 4/11 web download. I. 1. ϵ-δ 1. 3.1, 3..
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
S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
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
2009 10 6 23 7.5 7.5.1 7.2.5 φ s i m j1 x j ξ j s i (1)? φ i φ s i f j x j x ji ξ j s i (1) φ 1 φ 2. φ n m j1 f jx j1 m j1 f jx j2. m j1 f jx jn x 11 x 21 x m1 x 12 x 22 x m2...... m j1 x j1f j m j1 x
4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
Note.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
A
A05-132 2010 2 11 1 1 3 1.1.......................................... 3 1.2..................................... 3 1.3..................................... 3 2 4 2.1............................... 4 2.2
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 (
![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 (](/thumbs/97/133036028.jpg "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 (")
. 28 4 14 [.1 ] x > x 6= 1 f(x) µ 1 1 xn 1 + sin + 2 + sin x 1 x 1 f(x) := lim. 1 + x n (1) lim inf f(x) (2) lim sup f(x) x 1 x 1 (3) lim inf x 1+ f(x) (4) lim sup f(x) x 1+ [.2 ] [, 1] Ω æ x (1) (2) nx(1
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)
CALCULUS II (Hiroshi SUZUKI ) 16 1 1 1.1 1.1 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) lim f(x, y) = lim f(x, y) = lim f(x, y) = c. x a, y b
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
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
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
(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
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP
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
... x, y z = x + iy x z y z x = Rez, y = Imz z = x + iy x iy z z () z + z = (z + z )() z z = (z z )(3) z z = ( z z )(4)z z = z z = x + y z = x + iy ()Rez = (z + z), Imz = (z z) i () z z z + z z + z.. z
30 I .............................................2........................................3................................................4.......................................... 2.5..........................................
23 7 28 i i 1 1 1.1................................... 2 1.2............................... 3 1.2.1.................................... 3 1.2.2............................... 4 1.2.3 SI..............................
( ) 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 +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
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
III http://www2.mth.kyushu-u.c.jp/~hr/lectures/lectures-j.html 1 1 1.1 ϵ-n ϵ-n lim n = α n n α 1 lim n = 0 1 n k n k=1 0 1.1.7 ϵ-n 1.1.1 n α n n α lim n = α ϵ Nϵ n > Nϵ n α < ϵ 1.1.1 ϵ n > Nϵ n α < ϵ 1.1.2
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 =
II 2 II
II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................
応力とひずみ.ppt
in yukawa@numse.nagoya-u.ac.jp 2 3 4 5 x 2 6 Continuum) 7 8 9 F F 10 F L L F L 1 L F L F L F 11 F L F F L F L L L 1 L 2 12 F L F! A A! S! = F S 13 F L L F F n = F " cos# F t = F " sin# S $ = S cos# S S
p03.dvi
3 : 1 ( ). (.. ), : 2 (1, 2 ),,, etc... 1, III ( ) ( ). : 3 ,., III. : 4 ,Weierstrass : Rudin, Principles of Mathematical Analysis, 3/e, McGraw-Hil, 1976.. Weierstrass (Stone-Weierstrass, ),,. : 5 2π f
B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:
![B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:](/thumbs/94/118646867.jpg "B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:")
B. 41 II: ;; 4 B [] S 1 S S 1 S.1 O S 1 S 1.13 P 3 P 5 7 P.1:.13: 4 4.14 C d A B x l l d C B 1 l.14: AB A 1 B 0 AB 0 O OP = x P l AP BP AB AP BP 1 (.4)(.5) x l x sin = p l + x x l (.4)(.5) m d A x P O
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
4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 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 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
基礎数学I
I & II ii ii........... 22................. 25 12............... 28.................. 28.................... 31............. 32.................. 34 3 1 9.................... 1....................... 1............
i 6 3 ii 3 7 8 9 3 6 iii 5 8 5 3 7 8 v...................................................... 5.3....................... 7 3........................ 3.................3.......................... 8 3 35
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 1 1 1 1. U(x, t) U(x, t) + c t x c, κ. (1). κ U(x, t) x. (1) 1, f(x).. U(x, t) U(x, t) + c κ U(x, t), t x x : U(, t) U(1, t) ( x 1), () : U(x, ) f(x). (3) U(x, t). [ U(x, t) Re u k (t) exp(πkx). (4)
1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
![1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1](/thumbs/94/121802164.jpg "1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1")
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
.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(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
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)
211 12 1 19 2.9 F 32 32: rot F d = F d l (63) F rot F d = 2.9.1 (63) rot F rot F F (63) 12 2 F F F (63) 33 33: (63) rot 2.9.2 (63) I = [, 1] [, 1] 12 3 34: = 1 2 1 2 1 1 = C 1 + C C 2 2 2 = C 2 + ( C )
#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 A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
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
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 F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ
SFGÇÃÉXÉyÉNÉgÉãå`.pdf
SFG 1 SFG SFG I SFG (ω) χ SFG (ω). SFG χ χ SFG (ω) = χ NR e iϕ +. ω ω + iγ SFG φ = ±π/, χ φ = ±π 3 χ SFG χ SFG = χ NR + χ (ω ω ) + Γ + χ NR χ (ω ω ) (ω ω ) + Γ cosϕ χ NR χ Γ (ω ω ) + Γ sinϕ. 3 (θ) 180
(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
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
DE-resume
- 2011, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 21131 : 4 1 x y(x, y (x,y (x,,y (n, (1.1 F (x, y, y,y,,y (n =0. (1.1 n. (1.1 y(x. y(x (1.1. 1 1 1 1.1... 2 1.2... 9 1.3 1... 26 2 2 34 2.1,... 35 2.2
00 3 9 ........................................................................................................................................... 4..3................................. 5.3.......................................
FX ) 2
(FX) 1 1 2009 12 12 13 2009 1 FX ) 2 1 (FX) 2 1 2 1 2 3 2010 8 FX 1998 1 FX FX 4 1 1 (FX) () () 1998 4 1 100 120 1 100 120 120 100 20 FX 100 100 100 1 100 100 100 1 100 1 100 100 1 100 101 101 100 100