Similar documents
Xray.dvi

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

#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 =

85 4

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

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 =

Ł\”ƒ-2005

第90回日本感染症学会学術講演会抄録(I)

本文/目次(裏白)


( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

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

29

Mott散乱によるParity対称性の破れを検証

( ) ( )

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e

untitled

[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

Note.tex 2008/09/19( )


08-Note2-web

LLG-R8.Nisus.pdf

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

プログラム

v_-3_+2_1.eps


untitled

5 c P 5 kn n t π (.5 P 7 MP π (.5 n t n cos π. MP 6 4 t sin π 6 cos π 6.7 MP 4 P P N i i i i N i j F j ii N i i ii F j i i N ii li i F j i ij li i i i

Z: Q: R: C: sin 6 5 ζ a, b

振動と波動

2000年度『数学展望 I』講義録

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

c 2009 i

pdf

chap1.dvi

70 : 20 : A B (20 ) (30 ) 50 1

日本内科学会雑誌第97巻第7号

日本内科学会雑誌第98巻第4号

1 7 ω ω ω 7.1 0, ( ) Q, 7.2 ( Q ) 7.1 ω Z = R +jx Z 1/ Z 7.2 ω 7.2 Abs. admittance (x10-3 S) RLC Series Circuit Y R = 20 Ω L = 100

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

抄録/抄録1    (1)V

zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {

研修コーナー

パーキンソン病治療ガイドライン2002

all.dvi

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

4‐E ) キュリー温度を利用した消磁:熱消磁

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

( ) Note (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ, µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) 3 * 2) [ ] [ ] [ ] ν e ν µ ν τ e

positron 1930 Dirac 1933 Anderson m 22Na(hl=2.6years), 58Co(hl=71days), 64Cu(hl=12hour) 68Ge(hl=288days) MeV : thermalization m psec 100

重力方向に基づくコントローラの向き決定方法

1. 4cm 16 cm 4cm 20cm 18 cm L λ(x)=ax [kg/m] A x 4cm A 4cm 12 cm h h Y 0 a G 0.38h a b x r(x) x y = 1 h 0.38h G b h X x r(x) 1 S(x) = πr(x) 2 a,b, h,π

TOP URL 1

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

PDF

grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )

吸収分光.PDF

n ( (

KENZOU

r d 2r d l d (a) (b) (c) 1: I(x,t) I(x+ x,t) I(0,t) I(l,t) V in V(x,t) V(x+ x,t) V(0,t) l V(l,t) 2: 0 x x+ x 3: V in 3 V in x V (x, t) I(x, t

日本電子News vol.44, 2012

振動工学に基礎

,.,. 2, R 2, ( )., I R. c : I R 2, : (1) c C -, (2) t I, c (t) (0, 0). c(i). c (t)., c(t) = (x(t), y(t)) c (t) = (x (t), y (t)) : (1)

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

概況

Gmech08.dvi

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B

4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t


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 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

. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e 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

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 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

untitled

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

DVIOUT-HYOU

keisoku01.dvi

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

LCR e ix LC AM m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x (k > 0) k x = x(t)

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +


CoPt 17

untitled

) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8)

= M + M + M + M M + =.,. f = < ρ, > ρ ρ. ρ f. = ρ = = ± = log 4 = = = ± f = k k ρ. k

meiji_resume_1.PDF

DVIOUT-講

高校生の就職への数学II

I 1


Transcription:

29 1 6 1

1 1.1 1.1 1.1( ) 1.1( ) 1.1: 2

1.2 1.2( ) 4 4 1 2,3,4 1 2 1 2 1.2: 1,2,3,4 a 1 2a 6 2 2,3,4 1,2,3,4 1.2( ) 4 1.2( ) 3

1.2( ) 1.3 1.3 1.3: 4

1.4 1.4 1.4: 1.5 1.5 1 2 1 a a R = l a l 5

R = l a + const. a 1.5: a, b 1.2 6

1.6 7 14 R = l a+m b+n c l, m, n a, b, c α, β, γ 7 14 1.6 1.9 2.1 1.6: 7 14 1/4 7

1.7: 7 14 2/4 1.8: 7 14 3/4 8

1.9: 7 14 4/4 2 2.1 7 14 a, b, c α, β, γ 1.2 a 360 90 180 n n = 1, 2, 4 2.1 32 9

2.1: 5 a 7 2.2 1.4 7 14 32 14 230 230 10

International Tables for Crystallography vol. A 2.2: 2.2 P 2 1 /c 2.3 2.3 a, b, c [u, v, w] R = u a + v b + w c [uvw] [111] [222] [ 1 1 1 2 2 ] 2 [111] a, b, c [111] [111] < 111 > 11

2.3: a, b, c a, b, c (hkl) h k l i 2.4 2.4 2.5 ABCABC ABAB < 111 > < 001 > 2.5 (hkl) (h k l ) [uvw] [uvw] 2.6 [uvw] 12

2.4: 2.5: 13

(hkl) [uvw] hu + kw + lv = 0 2.6: 2.6 2.7 14

2.7: 2.8 2.7 2.9 (001) 2.9 001 2.9 100 111 011 111 100 [011] (2.5 ) (hkl) (h k l ) (phqh pkqk plql ) p, q 15

2.8: 2.9: 001 16

3 Ewald 3.1 3.1 (010) 010 (hkl) g hkl 3.1: 17

3.2 3.2 3.2: 3.3 3.4 3.2 3.5 1.2 3.6 18

3.3: 3.4: (1/2) 19

3.5: (2/2) 3.6: 20

3.3 3.7 A, B k 0 k k 0 k λ k 0 = k = 1 λ R Ψ A Ψ B 3.7: Ψ A Ψ B 3.8 l λ l A B λ r r k k 0 k k 0 k 0 k r q = 3.9 e iθ = cosθ + isinθ n 21

3.8: 3.4 r n q 3.10 R pqr = p a + q b + r c r n = R pqr g hkl R pqr = q = g hkl q = g hkl 2sinθ = 1 λ d hkl λ = 2dsinθ 3.5 Ewald Ewald 3.11 22

3.9: 3.10: 23

λ 1 λ k 0 0 Ewald 0 k 0 = k = 1 λ k 0 k q k k 0 k 0 k g hkl q 3.11: Ewald Ewald 24

4 4.1 t T 0 x(t) x(t) = a 0 2 + {a n cos(nω 0 t) + b n sin(nω 0 t)} (4.1) n=1 ω 0 T 0 f 0 ω 0 = 2πf 0 = 2π T 0 (4.2) (4.1) ω 0 nω 0 a n b n a n b n a n = 2 T 0 T0 2 b n = 2 T 0 T0 2 T 0 2 x(t)cos(nω 0 t)dt (4.3) T 0 2 x(t)sin(nω 0 t)dt (4.4) 4.2 (e iθ = cosθ + jsinθ) x(t) = n= c n e jnω 0t where, c n = 1 T 0 T0 2 T 0 2 x(t)e jnω 0t dt (4.5) nω 0 ω 0, 2ω 0, 3ω 0 ω 0 0 ω X(ω) = c n T 0 X(ω) = 25 x(t)e jwt dt (4.6)

(4.6) x(t) t s ω s 1 4.1 4.2 4.1 f(t)=sin(t) g(t)=sint(2t)+ 1sin( t 2 2 ) f(t) t(s) sin g(t) sin(t) ( ) 4.2 f sin(t) 2,000(s)(0.5 mhz) sin(t)( ) 0.5 mhz 2t t 0.25 mhz 2 1.0 mhz 1.0 0.5 sin(t) 0.0-0.5-1.0 0 2000 4000 t / s 6000 8000 4.1: f(t) g(t) 4.3 x(t) x( ) 26

Power 0.0 0.2 0.4 0.6 0.8 1.0 f / Hz 1.2 1.4 1.6 1.8 2.0x10-3 4.2: f(t) g(t) 4.3 4.4 4.5 4.6 4.3 4.4 27

4.3: 2 4.4: 2 28

4.5: 2 4.6: 2 29

5 5.1 5.1 X 5.2 Mo 5.1 5.1: 30

5.2: l l = ±1 31

5.3 5.3 K α1, K α2 α1,α2 K α1 K α2 K α1 K α2 K α1 K α2 K α1 K α2 K α 5.3: K α X K β K α K α X K α K β 5.4 5.5 32

5.4: X 5.2 5.6 ω θ θ 0 o ω θ θ θ 33

5.5: 5.6: 34

( 1 + cos I = F 2 2 ) 2θ 1 2µt p (1 e 2sin 2 sinθ )e 2M θcosθ 2µ (5.1) F 2 p 1 2µt (1 e sinθ ) 2µ - e 2M (5.1) 1. hkl N F hkl = f j e 2πi(hu j+kv j +lw j ) j=1 f j j u j, v j, w j j N 2. 3. {100} (100), (010), (001), (100), (010), (001) 4. 5.7 2θ 35

2θ 5.7: 5. 5.8 γ = β = θ 1 2µt (1 e sinθ ) 2µ t 1 2µ 6. - u 2 2 M = 8π 2 u 2 ( sinθ 2 λ ) = B( sinθ λ 2 ) (5.2) 36

5.8: 5.3 5.9 (5.1) 4sin 2 θ λ 2 = 1 d 2 = h2 + k 2 + l 2 a 2 (5.3) θ 5.2 h 2 + k 2 + l 2 a (5.3) 5.2 2 1 0.1828 0.1828 1 2, 1 3 0.1828 1 3 37

5.9: JCPDS(Joint Committee on Powder Diffraction Standards) 38

6 6.1 6.1 6.2 6.2 6.1: 6.2 39

6.2: 6.2.1 1. a, b, c a = b c a (b c), b = c a b (c a), c = a b c (a b) (6.1) a, b, c bc ca ab a, b, c a, b, c a, b, c a = 1 a, b = 1 b, c = 1 c (6.2) 40

2. 1. (hkl) F (hkl) F (hkl) 2 F (hkl) = n f n e 2πi(hxn+kyn+lzn) (6.3) f n n (x n, y n, z n ) n (6.3) n 6.3 6.3: 41

6.3 1. R d Rd = Lλ (6.4) 6.4 6.4: 2. 42

100kV 1 100 6.5 6.5: [double reflection] 6.6 Nd 2 Fe 1 4B 6.6(b) h + l = 43

2n + 1(n ) h + l = 2n + 1 6.6(a),(c) h + l = 2n + 1 300, 003 6.6(b) (a),(c) (b) h + l = 2n + 1 (a),(c) 6.6: [systematic reflections] h00 h = 0, +1, +2... 6.4 1. h, k, l h + k + l 44

1 1 1 4 4 4 [110] [110] 2. 6.7 Cu 3 Au CsCl ZnS 6.7: 45

3. Incommensurate Structure [Incommensurate Structure] 6.8(a) Cu 3 Pd M 6.8 6.8(b) Tl- a 5.9 (a) (b) 6.8: 46

4. SiC 6.9(a) 6.9: 6.9(b) c 47

7 7.1 7.1 d λ n 7.1: d = 0.61λ (7.1) n sinα α n sinα 1.5 λ 0.5 µm 0.2 µm (7.1) 48

d e CS λ e d e = 0.65C 1 4 s λ 3 4 e (7.2) 300kV 0.00197 nm 0.60 mm (7.2) 0.17 nm 7.2 7.2 kev 0.8 7.1 A-B 7.3 θ 49

7.2: 0.1µmϕ nmϕ 7.3 7.4(a) 7.4(b) 50

7.3: 7.5(a) (b) 7.5(b) g 7.6(a) 7.6(b) (c) 7.6(b) g 3g g 51

7.4: 7.5: 52

7.6: 7.6(c) g 7.6(b) g g-3g 3g 7.7 g = 220 g 3g nm 7.4(c) (7.2) 7.8 53

7.7: 54

7.8: 55