p03.dvi
|
|
|
- ためひと こいたばし
- 4 years ago
- Views:
Transcription
1 3 : 1
2 ( ). (.. ), : 2
3 (1, 2 ),,, etc... 1, III ( ) ( ). : 3
4 ,., III. : 4
5 ,Weierstrass : Rudin, Principles of Mathematical Analysis, 3/e, McGraw-Hil, Weierstrass (Stone-Weierstrass, ),,. : 5
6 2π f 0. f 0 (0) = f 0 (2π). f(e jθ ) = f 0 (θ), f (0 θ < 2π). e jθ = z (f(e jθ ) = f(z)),, Stone-Weierstrass : 6
7 f p(z) = a 0 z N +a 1 z N=1 + +a N 1 z+a N p ( ),. z = e jθ, p(e jθ ) = a 0 e jnθ + a 1 e j(n 1)θ + +a N 1 e jθ +a N.. : 7
8 e jθ = cosθ +jsinθ, p(e jθ ) = N A m cos(mθ)+ m=0 N B m sin(mθ) m=1. {A m } 0 m N {B m } 1 m N : ( ). : 8
9 , (Weierstrass ) Stone-Weierstrass, ( ). A m B m,. : 9
10 t, t, f(t) N m=0 A mcos(mt) + N m=1 B msin(mt)., 2π 0 f(t)dt = c 0 2πA 0 = 2πA 0, A 0 = (1/2π) 2π 0 f(t)dt; 2π 0 f(t)cosmt = 2π 0 A m cos 2 (mt)dt = πa m, A m = (1/π) 2π 0 f(t) cos(mt)dt 2π 0 f(t)sinmt = 2π 0 A m sin 2 (mt)dt = πb m, B m = (1/π) 2π 0 f(t) sin(mt)dt. : 10
11 , 2π N f(t) A m cos(mt)+ B m sin(mt) A 0 = 1 2π A m = 1 π B m = 1 π m=0 2π 0 2π 0 2π 0 f(t)dt m=1 f(t)cos(mt)dt (1 m < ) f(t)sin(mt)dt (1 m < ) : 11
12 T, 2π T f(t) A 0 = 1 T A m = 2 T B m = 2 T m=0 T ( 2πm A m cos T t 0 T 0 T 0 ) N + m=1 ( ) 2πm B m sin T t f(t)dt ( ) 2πm f(t)cos T t dt (1 m < ) ( ) 2πm f(t)sin T t dt (1 m < ) : 12
13 e j 2πm T t f(t) m Z C m = 1 T C m e j 2πm T t T 0 f(t)e 2πm T t dt, m Z, ( ). Euler, m Z. p Ω 0, Ω 0 = 2π/T. : 13
14 f(t)? :. : Stone-Weierstrass, f,.,. III. : 14
15 ,. F(ν) = f(t)? = f(t)e j2πνt dt F(w)e j2πνt dν : 15
16 f g(ν) = f(t)e j2πνt dt g,f,f[f] (g = F[f]). g h(t) = g(ν)e j2πνt dν h, f, F 1 [g] (h = F[f]). : 16
17 f Fourier, (F 1 [F[f]] = f)., f F[f].., Ω = 2πν : 17
18 g(ω) (= g(ω/(2π))) = h(t) = 1 2π g(ω)e jωt dω f(t)e jωt dt. :
19 g g h(t) = 1 g(ω/(2π))e jωt dω, Ω = 2π 2πν, h(t) =. g(ν)e j2πνt dν, 2π, ( ),. : 19
20 (Fourier Series, FS) (Fourier Transofrm, FT) (Discrete-Time Fourier Transform, DTFT) (Discrete-Time Fourier Series, DTFS) (Discrete Fourier Transform, DFT) :
21 ( ) ( ),.,. : 21
22 . x N ( ). Ω = 2π/N. : 22
23 x : N 1 X[m] = x[k]e jωmk (0 m < N). k=0 : 23
24 x (X[0],...,X[n 1]) ( ) x[n] = 1 N 1 X[m]e jωmn (0 n < N). N m=0. : 24
25 ( p.29). X[m] = N 1 k=0 x[k]e jωmk (0 m < N) x[n] = 1 N 1 X[m]e jωmn (0 n < N) N m=0 : 25
26 x (X[0],...,X[n 1]), (X[0],...,X[n 1]) x,., (x[0],...,x[n 1]) (X[0],...,X[N 1]), x ( x )., 5 ( 3.1, 5 ),. : 26
27 X[m] 0 m < N,, m Z, X[m] = N 1 k=0 x[k]e jωmk X[m]. X. Ω = 2π/N, p Z, X[m + pn] = N 1 k=0 x[k]e jω(m+pn)k = N 1 k=0 x[k]e jωmk e j 2π N }{{ pkn } = N 1 k=0 x[k]e jωmk = X[m], X N., X (X[0],...,X[N 1]). =1 : 27
28 x F = (x[0],...,x[n 1]), X F = (X[0],...,X[N 1]),, (...,x F,x F,x F,...) (...,X F,X F,X F,...), x F X F,. x X (Analysis), x (Synthesis). : 28
29 N 1 x[n] = 1 N m=0 X[m]ejΩmn ( ), N = 1 ( ), N > 1,, N = 1. (N = 1 ). N > 1,. : 29
30 N 1 0 n < N n, z[n] = 1 N 1 X[m]e jωmn (0 n < N) N m=0. n z[n] = x[n]. X[m], z[n] = 1 N 1 N 1 x[k]e jωmk e jωmn = 1 N 1 N 1 N N. m=0 k=0 m=0 k=0 x[k]e jωm(n k) : 30
31 , k = n k n, ( z[n] = 1 N 1 ) N 1 x[n]+ x[k]e jωm(n k) N m=0 m=0 k n., 1 m, 2, ( z[n] = 1 Nx[n]+ ) N 1 x[k]e jωm(n k) N k n m=0 ( ) = x[n]+ 1 N 1 x[k] e jωm(n k) N. k n m=0 : 31
32 , 0 n,k < N, n k < N, Ω = 2π/N, e jω(n k) = e j2πn k N 1., 2, Ω = 2π/N, N 1 N 1 e j 2π N m(n k) = m=0 m=0 (e j 2π N (n k) ) m = 1 (e j 2π N (n k)) N 1 e j 2π(n k) N ( ), (e j 2π N (n k)) N = e j2π(n k) = 1,, z[n] = x[n]. : 32
33 N = 1 N = 1, x x[0]., X[0], X[m] = N 1 k=0 x[k]e jωmk m = 0 N = 1, X[0] = x[0]. N > 1 z[n], z[n] = 1 N 1 N m=0 X[m]ejΩmn, 0 n < N, N = 1 X[0] = x[0], z[0] = x[0]. : 33
34 , x, X (p )., ( ). : 34
35 : x : X[m] = 1 N 1 x[k]e jωmk (0 m < N). N k=0 x X : N 1 x[n] = X[m]e jωmn (0 n < N). m=0 1/N,. : 35
36 ( 5 ),.,, 5, x X, X = F DFT [x], x = F 1 DFT [X].,,, F 1 DFT [F DFT[x]] = x. : 36
37 ,,, F DFT [F 1 DFT [X]] = X?, F DFT F 1 DFT. e jωnk e jωnk, 1/N F DFT F 1 DFT, F DFT [F 1 DFT [X]] = X., F DFT F 1 DFT., F DFT F 1 DFT. : 37
38 F DFT [ ] F 1 DFT [ ],, F DFT F 1 DFT = I, F 1 DFT F DFT = I (I, ). X = F DFT [x] x = F 1 DFT [X], x = F 1 DFT [X] X = F DFT[X]. : 38
39 pp , x, x 1, x 2 N. x, x 1, x 2 X, X 1, X 2., X, X 1, X 2, N. : 39
40 ( ) a 1 x 1 +a 2 x 2 N 1 k=0 (a 1x 1 [k]+a 2 x 2 [k])e jωmk, F DFT [a 1 x 1 +a 2 x 2 ] = a 1 X 1 +a 2 X 2. F 1 DFT [ ], a 1 x 1 +a 2 x 2 = F 1 DFT [a 1X 1 +a 2 X 2 ]. : 40
41 , 3.2 2,. x N, Ω = 2π/N. c,k Z, X[c,k] = c+n 1 m=c x[m]e jωmk. X[c,k] = N 1 m=0 x[m]e jωmk = X[k]., p Z, c = pn +c 1, 0 c 1 < N., m 1 = m pn, m = m 1 +pn, X[c,k] = c1+n 1 x[m m1=c1 1 + pn]e jω(m1+pn)k, x[m 1 + pn] = x[m 1 ], e jω(m1+pn)k = e jωm1k e jωpnk = e jωm1k (Ω = 2π/N ), X[c,k] = c1+n 1 x[m m1=c1 1 ]e jωm1k. c 1 = 0, c 1 0., X[c,k] = N 1 x[m m1=c1 1]e jωm1k + c1+n 1 m1=n x[m 1]e jωm1k, m 2 = m 1 N, 2 c1 1 m2=0 x[m 2 + N]e jω(m2+n)k = c1 1 m2=0 x[m 2]e jωm2k ( x Ω = 2π/N ). m 2 m 1 1, X[c,k] = N 1 x[m m1=c1 1]e jωm1k + c1 1 m1=0 x[m 1]e jωm1k = N 1 m1=0 x[m 1]e jωm1k, m 1 m,. : 41
42 , c Z X[k] = c+n 1 m=c x[m]e jωmk,, N x,. : 42
43 m ( ) τ m m, Z = F DFT [τ m x], N 1 N 1 Z[n] = (τ m x[k])e jωnk = (x[k m])e jωnk k=0 k=0 N 1 = (x[k m])e jωn(k m) e jωnm = e jωnm X[n] k=0 : 43
44 m ( ) τ m m, z = F 1 DFT [τ mx], N 1 N 1 z[n] = (τ m X[k])e jωnk = (X[k m])e jωnk k=0 k=0 N 1 = (X[k m])e jωn(k m) e jωnm = e jωnm x[n] k=0 : 44
45 ( 3.2 4,5 ) x 2 x 2 ( ) X 1 X 2 ( ). N 1 (x 1 x 2 )[n] = x 1 [m]x 2 [n m], m=0 N 1 (X 1 X 2 )[n] = X 1 [m]x 2 [n m] m=0 : 45
46 ,,. : F DFT [x 1 x 2 ] = X 1 X 2,FDFT[ 1 1 X ] N 1 X 2 = x1 x , 5,. : 46
47 N 1 (F DFT [x 1 x 2 ])[n] = (x 1 x 2 )[k]e jωnk k=0 N 1 N 1 = x 1 [m]x 2 [k m]e jωnk k=0 m=0 N 1 N 1 = x 1 [m]e jωnm x 2 [k m]e jωn(k m) k=0 m=0 N 1 N 1 = x 1 [m]e jωnm x 2 [k m]e jωn(k m) m=0 k=0 = X 1 [n]x 2 [n] : 47
48 ( [ ]) 1 F 1 DFT N X 1 X 2 [n] = 1 N 1 ( 1 N N X 1 X 2 )[k]e jωnk = 1 N 2 = 1 N 2 N 1 N 1 k=0 X 1 [m]x 2 [k m]e jωnk k=0 m=0 N 1 N 1 X 1 [m]e jωnm X 2 [k m]e jωn(k m) k=0 m=0 N 1 = 1 X 1 [m]e jωnm 1 N 1 X 2 [k m]e jωn(k m) N N m=0 = x 1 [n]x 2 [n] k=0 : 48
49 , ( ), N. x[n],, x x, x[n]. x n x[n]. x, n x[n] = x[n],, : 49
50 N 1 X[ n] = X[ n] = x[m]e jωm( n) = = N 1 m=0 m=0 x[n]e jωmn = (F DFT [ x])[n] N 1 x[m]e jωmn m=0 x[ n] = x[ n] = 1 N 1 X[m]e N jωm( n = 1 N 1 X[m]e jωmn N = 1 N N 1 m=0 m=0 X[m]e jωmn = ( F 1 DFT [ X] ) [n] m=0 : 50
51 x N 1 X[ n] = X[ n] = m=0 x[m]e jωm( n) N 1 = x[m]e jωmn = X[n]. m= : 51
52 (Discretetime Fourier Transform, DTFT) ( ),. : 52
53 ( T) f(t) m Z C m = 1 T C m e j 2πm T t T 0 f(t)e 2πm T t dt, m Z T = 2π ( )., m n. : 53
54 ( 2π) f(t) C n e jnt n Z C m = 1 2π f(t)e nt dt, n Z 2π 0, e jnt e jnt, (e jnt ) n Z,. : 54
55 ( 2π), f(t) n Z C n = 1 2π C n e jnt 2π 0 f(t)e nt dt, n Z : t f C n ω X x[n]. : 55
56 X(ω) x[n]e jnω n Z x[n] = 1 2π X(ω)e nω dω, n Z 2π 0 X( f) x( C n )., x X., x. : 56
57 x x l 1 (x l 1 x[n] <. n= x, [0,2π) ( ; X(e jω ) ). X(ω) = n Z x[n]e jnω : 57
58 t < 0 t 2π, f 2π. x[n] (n Z) X.. x[n] = 1 2π X(ω)e nω dω. 2π 0 : 58
59 ,. X(ω) = x[n]e jnω n Z x[n] = 1 2π X(ω)e nω dω, n Z 2π 0,. : 59
60 Logz, Y(z) = (X Log)(z), Y(e jω ) = X(Log(e jω )) = X(ω), X(ω) Y(e jω ), Y(e jω ) = n Z x[n]e jnω, x[n] = 1 2π Y(e jω )e nω dω (n Z). 2π 0,.,. : 60
61 , x l 1, x, l 1 2π., l 2 (l 1 l 2 ),.. : 61
62 ( ),, MATLAB,, 2000.,, 1993.,, A. V. Oppenheim and R. W. Schafter, Discrete-Time Signal Processing, International Ed., Pearson, M. Mandal and A. Asif, Continuous and Discrete Time Signals and Systems, Cambridge University Press, K. B. Howell, Principles of Fourier Analysis, 2/e, CRC Press, 2017.,.. : 62
main.dvi
3 Discrete Fourie Transform: DFT DFT 3.1 3.1.1 x(n) X(e jω ) X(e jω )= x(n)e jωnt (3.1) n= X(e jω ) N X(k) ωt f 2π f s N X(k) =X(e j2πk/n )= x(n)e j2πnk/n, k N 1 (3.2) n= X(k) δ X(e jω )= X(k)δ(ωT 2πk
[] 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 )
![[] 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 )](/thumbs/85/91744595.jpg "[] 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 )")
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)
(5 B m e i 2π T mt m m B m e i 2π T mt m m B m e i 2π T mt B m (m < 0 C m m (6 (7 (5 g(t C 0 + m C m e i 2π T mt (7 C m e i 2π T mt + m m C m e i 2π T
2.6 FFT(Fast Fourier Transform 2.6. T g(t g(t 2 a 0 + { a m b m 2 T T 0 2 T T 0 (a m cos( 2π T mt + b m sin( 2π mt ( T m 2π g(t cos( T mtdt m 0,, 2,... 2π g(t sin( T mtdt m, 2, 3... (2 g(t T 0 < t < T
() 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
main.dvi
6 FIR FIR FIR FIR 6.1 FIR 6.1.1 H(e jω ) H(e jω )= H(e jω ) e jθ(ω) = H(e jω ) (cos θ(ω)+jsin θ(ω)) (6.1) H(e jω ) θ(ω) θ(ω) = KωT, K > 0 (6.2) 6.1.2 6.1 6.1 FIR 123 6.1 H(e jω 1, ω
main.dvi
4 DFT DFT Fast Fourier Transform: FFT 4.1 DFT IDFT X(k) = 1 n=0 x(n)e j2πkn (4.1) 1 x(n) = 1 X(k)e j2πkn (4.2) k=0 x(n) X(k) DFT 2 ( 1) 2 4 2 2(2 1) 2 O( 2 ) 4.2 FFT 4.2.1 radix2 FFT 1 (4.1) 86 4. X(0)
09 8 9 3 Chebyshev 5................................. 5........................................ 5.3............................. 6.4....................................... 8.4...................................
, 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
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 (
[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a
![[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a](/thumbs/85/91634532.jpg "[1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin 4πt = a")
13/7/1 II ( / A: ) (1) 1 [] (, ) ( ) ( ) ( ) etc. etc. 1. 1 [1] 1.1 x(t) t x(t + n ) = x(t) (n = 1,, 3, ) { x(t) : : 1 [ /, /] 1 x(t) = a + a 1 cos πt + a cos 4πt + + a n cos nπt + + b 1 sin πt + b sin
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,
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
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 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
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 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
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
(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
1 Fourier Fourier Fourier Fourier Fourier Fourier Fourier Fourier Fourier analog digital Fourier Fourier Fourier Fourier Fourier Fourier Green Fourier
Fourier Fourier Fourier etc * 1 Fourier Fourier Fourier (DFT Fourier (FFT Heat Equation, Fourier Series, Fourier Transform, Discrete Fourier Transform, etc Yoshifumi TAKEDA 1 Abstract Suppose that u is
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 ),
30 I .............................................2........................................3................................................4.......................................... 2.5..........................................
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
4 14 4 14 4 1 1 4 1.1................................................ 4 1............................................. 4 1.3................................................ 5 1.4 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..............................
IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α
18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t
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..................................
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
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
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................
36 3 D f(z) D z f(z) z Taylor z D C f(z) z C C f (z) C f(z) f (z) f(z) D C D D z C C 3.: f(z) 3. f (z) f 2 (z) D D D D D f (z) f 2 (z) D D f (z) f 2 (
3 3. D f(z) D D D D D D D D f(z) D f (z) f (z) f(z) D (i) (ii) (iii) f(z) = ( ) n z n = z + z 2 z 3 + n= z < z < z > f (z) = e t(+z) dt Re z> Re z> [ ] f (z) = e t(+z) = (Rez> ) +z +z t= z < f(z) Taylor
n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.
1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,
II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k
II 231017 1 1.1. R n k +1 v 0,, v k k v 1 v 0,, v k v 0 1.2. v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ kσ dimσ = k 1.3. k σ {v 0,...,v k } {v i0,...,v il } l σ τ < τ τ σ 1.4.
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
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)
X G P G (X) G BG [X, BG] S 2 2 2 S 2 2 S 2 = { (x 1, x 2, x 3 ) R 3 x 2 1 + x 2 2 + x 2 3 = 1 } R 3 S 2 S 2 v x S 2 x x v(x) T x S 2 T x S 2 S 2 x T x S 2 = { ξ R 3 x ξ } R 3 T x S 2 S 2 x x T x S 2
[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt
![[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt](/thumbs/101/152334931.jpg "[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt")
3.4.7 [.] =e j(t+/4), =5e j(t+/3), 3 =3e j(t+/6) ~ = ~ + ~ + ~ 3 = e j(t+φ) =(e 4 j +5e 3 j +3e 6 j )e jt = e jφ e jt cos φ =cos 4 +5cos 3 +3cos 6 =.69 sin φ =sin 4 +5sin 3 +3sin 6 =.9 =.69 +.9 =7.74 [.]
- II
- II- - -.................................................................................................... 3.3.............................................. 4 6...........................................
Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)
![Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)](/thumbs/94/118373946.jpg "Z[i] Z[i] π 4,1 (x) π 4,3 (x) 1 x (x ) 2 log x π m,a (x) 1 x ϕ(m) log x 1.1 ( ). π(x) x (a, m) = 1 π m,a (x) x modm a 1 π m,a (x) 1 ϕ(m) π(x)")
3 3 22 Z[i] Z[i] π 4, (x) π 4,3 (x) x (x ) 2 log x π m,a (x) x ϕ(m) log x. ( ). π(x) x (a, m) = π m,a (x) x modm a π m,a (x) ϕ(m) π(x) ϕ(m) x log x ϕ(m) m f(x) g(x) (x α) lim f(x)/g(x) = x α mod m (a,
untitled
1 SS 2 2 (DS) 3 2.1 DS................................ 3 2.2 DS................................ 4 2.3.................................. 4 2.4 (channel papacity)............................ 6 2.5........................................
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
2 4 202 9 202 9 6 3................................................... 3.2................................................ 4.3......................................... 6.4.......................................
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],
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
Z: Q: R: C: sin 6 5 ζ a, b
Z: Q: R: C: 3 3 7 4 sin 6 5 ζ 9 6 6............................... 6............................... 6.3......................... 4 7 6 8 8 9 3 33 a, b a bc c b a a b 5 3 5 3 5 5 3 a a a a p > p p p, 3,
2 2 L 5 2. L L L L k.....
L 528 206 2 9 2 2 L 5 2. L........................... 5 2.2 L................................... 7 2............................... 9. L..................2 L k........................ 2 4 I 5 4. I...................................
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
(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fou
(Bessel) (Legendre).. (Hankel). (Laplace) V = (x, y, z) n (r, θ, ϕ) r n f n (θ, ϕ). f n (θ, ϕ) n f n (θ, ϕ) z = cos θ z θ ϕ n ν. P ν (z), Q ν (z) (Fourier) (Fourier Bessel).. V ρ(x, y, z) V = 4πGρ G :.
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
006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................
,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
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
ft. ft τfτdτ = e t.5.. fx = x [ π, π] n sinnx n n=. π a π a, x [ π, π] x = a n cosnx cosna + 4 n=. 3, x [ π, π] x 3 π x = n sinnx. n=.6 f, t gt n 3 n
![ft. ft τfτdτ = e t.5.. fx = x [ π, π] n sinnx n n=. π a π a, x [ π, π] x = a n cosnx cosna + 4 n=. 3, x [ π, π] x 3 π x = n sinnx. n=.6 f, t gt n 3 n](/thumbs/92/108735706.jpg "ft. ft τfτdτ = e t.5.. fx = x [ π, π] n sinnx n n=. π a π a, x [ π, π] x = a n cosnx cosna + 4 n=. 3, x [ π, π] x 3 π x = n sinnx. n=.6 f, t gt n 3 n")
[ ]. A = IC X n 3 expx = E + expta t : n! n=. fx π x π. { π x < fx = x π fx F k F k = π 9 s9 fxe ikx dx, i =. F k. { x x fx = x >.3 ft = cosωt F s = s4 e st ftdt., e, s. s = c + iφ., i, c, φ., Gφ = lim
chap1.dvi
1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f
1 9 v.0.1 c (2016/10/07) Minoru Suzuki T µ 1 (7.108) f(e ) = 1 e β(e µ) 1 E 1 f(e ) (Bose-Einstein distribution function) *1 (8.1) (9.1)
1 9 v..1 c (216/1/7) Minoru Suzuki 1 1 9.1 9.1.1 T µ 1 (7.18) f(e ) = 1 e β(e µ) 1 E 1 f(e ) (Bose-Einstein distribution function) *1 (8.1) (9.1) E E µ = E f(e ) E µ (9.1) µ (9.2) µ 1 e β(e µ) 1 f(e )
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
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
) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4
![) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4](/thumbs/92/107866707.jpg ") ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4")
1. k λ ν ω T v p v g k = π λ ω = πν = π T v p = λν = ω k v g = dω dk 1) ) 3) 4). p = hk = h λ 5) E = hν = hω 6) h = h π 7) h =6.6618 1 34 J sec) hc=197.3 MeV fm = 197.3 kev pm= 197.3 ev nm = 1.97 1 3 ev
untitled
0. =. =. (999). 3(983). (980). (985). (966). 3. := :=. A A. A A. := := 4 5 A B A B A B. A = B A B A B B A. A B A B, A B, B. AP { A, P } = { : A, P } = { A P }. A = {0, }, A, {0, }, {0}, {}, A {0}, {}.
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,
214 March 31, 214, Rev.2.1 4........................ 4........................ 5............................. 7............................... 7 1 8 1.1............................... 8 1.2.......................
29 21 3 1 1 2 1.1............................ 2 1.2................................... 2 2 3 2.1 Neumann...................... 3 2.2............................. 5 2.3.................................
213 March 25, 213, Rev.1.5 4........................ 4........................ 6 1 8 1.1............................... 8 1.2....................... 9 2 14 2.1..................... 14 2.2............................
(yx4) 1887-1945 741936 50 1995 1 31 http://kenboushoten.web.fc.com/ OCR TeX 50 yx4 e-mail: yx4.aydx5@gmail.com i Jacobi 1751 1 3 Euler Fagnano 187 9 0 Abel iii 1 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.
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)
数値計算：フーリエ変換
( ) 1 / 72 1 8 2 3 4 ( ) 2 / 72 ( ) 3 / 72 ( ) 4 / 72 ( ) 5 / 72 sample.m Fs = 1000; T = 1/Fs; L = 1000; t = (0:L-1)*T; % Sampling frequency % Sample time % Length of signal % Time vector y=1+0.7*sin(2*pi*50*t)+sin(2*pi*120*t)+2*randn(size(t));
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
( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,
[ ] 8 IC. y d y dx = ( dy dx ( p = dy p y dx ( ( ( 8 ( s8. 3 A A = ( A ( A (3 A P A P AP.3 π y(x = { ( 8 ( s8 x ( π < x x ( < x π y(x π π O π x ( 8 ( s83.4 f (x, y, z grad(f ( ( ( f f f grad(f = i + j
untitled
2 : n =1, 2,, 10000 0.5125 0.51 0.5075 0.505 0.5025 0.5 0.4975 0.495 0 2000 4000 6000 8000 10000 2 weak law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ),i=1, 2,,n 3. σi 2 = V (X i ) σ 2,i=1, 2,,n ɛ>0
1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2
1 1.1 R(x) = 0 y + P (x)y + Q(x)y = R(x)...(1) y + P (x)y + Q(x)y = 0...(2) 1 2 u(x) v(x) c 1 u(x)+ c 2 v(x) = 0 c 1 = c 2 = 0 c 1 = c 2 = 0 2 0 2 u(x) v(x) u(x) u (x) W (u, v)(x) = v(x) v (x) 0 1 1.2
z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy
z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z
main.dvi
2 e jωt, 0 ωt < 2π 2.1 2.1.1 x(n) X(z) = x(n)z n (2.1) z z z = e σ+jωt = re jωt, r = e σ (2.2) ωt 0 2π σ e σ+jωt = re jωt r = e σ 2.1 0
B
B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T
TOP URL 1
TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................
II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
![II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re](/thumbs/94/118770263.jpg "II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re")
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
Microsoft Word - 章末問題
1906 R n m 1 = =1 1 R R= 8h ICP s p s HeNeArXe 1 ns 1 1 1 1 1 17 NaCl 1.3 nm 10nm 3s CuAuAg NaCl CaF - - HeNeAr 1.7(b) 2 2 2d = a + a = 2a d = 2a 2 1 1 N = 8 + 6 = 4 8 2 4 4 2a 3 4 π N πr 3 3 4 ρ = = =
#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/
211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
1. 1 BASIC PC BASIC BASIC BASIC Fortran WS PC (1.3) 1 + x 1 x = x = (1.1) 1 + x = (1.2) 1 + x 1 = (1.
Section Title Pages Id 1 3 7239 2 4 7239 3 10 7239 4 8 7244 5 13 7276 6 14 7338 7 8 7338 8 7 7445 9 11 7580 10 10 7590 11 8 7580 12 6 7395 13 z 11 7746 14 13 7753 15 7 7859 16 8 7942 17 8 Id URL http://km.int.oyo.co.jp/showdocumentdetailspage.jsp?documentid=
4 R f(x)dx = f(z) f(z) R f(z) = lim R f(x) p(x) q(x) f(x) = p(x) q(x) = [ q(x) [ p(x) + p(x) [ q(x) dx =πi Res(z ) + Res(z )+ + Res(z n ) Res(z k ) k
f(x) f(z) z = x + i f(z). x f(x) + R f(x)dx = lim f(x)dx. R + f(x)dx = = lim R f(x)dx + f(x)dx f(x)dx + lim R R f(x)dx Im z R Re z.: +R. R f(z) = R f(x)dx + f(z) 3 4 R f(x)dx = f(z) f(z) R f(z) = lim R
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)
Gmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r
( ) ( ) ( ) 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
4................................. 4................................. 4 6................................. 6................................. 9.................................................... 3..3..........................
A 2008 10 (2010 4 ) 1 1 1.1................................. 1 1.2..................................... 1 1.3............................ 3 1.3.1............................. 3 1.3.2..................................
A
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
, 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)
構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
2
16 1050026 1050042 1 2 1 1.1 3 1.2 3 1.3 3 2 2.1 4 2.2 4 2.2.1 5 2.2.2 5 2.3 7 2.3.1 1Basic 7 2.3.2 2 8 2.3.3 3 9 2.3.4 4window size 10 2.3.5 5 11 3 3.1 12 3.2 CCF 1 13 3.3 14 3.4 2 15 3.5 3 17 20 20 20
[Ver. 0.2] 1 2 3 4 5 6 7 1 1.1 1.2 1.3 1.4 1.5 1 1.1 1 1.2 1. (elasticity) 2. (plasticity) 3. (strength) 4. 5. (toughness) 6. 1 1.2 1. (elasticity) } 1 1.2 2. (plasticity), 1 1.2 3. (strength) a < b 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
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
5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (
![5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (](/thumbs/92/107866626.jpg "5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (")
5 5.1 [ ] ) d f(t) + a d f(t) + bf(t) : f(t) 1 dt dt ) u(x, t) c u(x, t) : u(x, t) t x : ( ) ) 1 : y + ay, : y + ay + by : ( ) 1 ) : y + ay, : yy + ay 3 ( ): ( ) ) : y + ay, : y + ay b [],,, [ ] au xx
IA
IA 31 4 11 1 1 4 1.1 Planck.............................. 4 1. Bohr.................................... 5 1.3..................................... 6 8.1................................... 8....................................
振動と波動
Report JS0.5 J Simplicity February 4, 2012 1 J Simplicity HOME http://www.jsimplicity.com/ Preface 2 Report 2 Contents I 5 1 6 1.1..................................... 6 1.2 1 1:................ 7 1.3