Nuclear Magnetic Resonance 1 H NMR spectrum PPM

Size: px
Start display at page:

Download "Nuclear Magnetic Resonance 1 H NMR spectrum PPM"

Transcription

1 Nuclear Magnetic Resonance 1 NMR spectrum PPM

2 PPM X: X X:

3

4 : 1: 1 : 1.5 : : 2 : 2 : 3 : 3 1

5 13 NMR spectrum BM PPM BM

6 13 NMR spectrum DEPT

7 OSY a b c d a c 1 NMR spectrum PPM a-c a b c d d c b a 1 NMR spectrum PPM

8 - OSY a a b c d c 1 NMR spectrum PPM a-c d c b a 13 NMR spectrum PPM

9 OSY

10 - OSY

11 O + 3 O + ( O O O O

12 IR O

13 MS

14 NMR

15 NMR : 1, 12, 13, 14 N, 15 N, 19 F, 31 P,... : I 0, I = 1/2, 1, 3/2, 2, 5/2, 3,...

16 or

17 : : ΔE N /N = e E/kT : B0 = 2.35 T 1 N > N N - N = 10-5

18 NMR NMR ΔE ΔE = hν

19 ΔE = 2µB 0, γhi µ = 2π (μ:, : ( ), B0:, h: ) I = 1/2 ( 1, 13,...) ν [z] = ΔE = h γb 0 2π B0 = 2.35 T 1 ΔE = 3.99 x 10-2 (J mol -1 ) ν = 100 x 10-6 (z)

20 : Beff = B0 Bind Bind = σb0 Beff = (1-σ)B0 B0 Bind γb ν [z] = 0 (1-σ) 2π σ:

21 σ = σdia + σpara + σ σdia: ( σpara: p (1/ E, p, 1/r 3 ) σ : (

22 z z ν χ ν standard = ν Δν ν [ppm] (perts per million) Mz singlet 3 3 Si Si D 2 D 2 O 2 Na 3

23 deuterated solvent formula mp ( ) bp ( ) δ δ c (multiplicity) acetone (D 3 ) 2 O (7) acetonitrile D 3 N (7) benzene 6 D (3) chloroform Dl (3) dichloromethane D 2 l (5) dimethylsulfoxide (D 3 ) 2 SO (7) methanol D 3 OD (7) pyridine 6 D 5 N (3) (3) (3) water D 2 O

24 X X = F: 4.30, 1.68, 0.97 X = l: 3.30, 1.61, 0.89 N( 3 ) 2 O 3 < < o 6.59 o 6.80 < 7.27 N + ( 3 ) 2 o 7.98

25 < < < B0

26 [10] 0.51 ppm 2.63 ppm

27 [18]アヌレン 9.28 ppm 非遮蔽領域 低磁場シフト ppm 遮蔽領域 高磁場シフト

28 B0 Alkenes arbonyl Groups

29 1.30 ppm 4.93/4.97 ppm 1.94 ppm B0 Alkynes

30 ax. eq. δ (ax) > δ (eq) ( 0.5 ppm) - bond B0

31 RO aromatic R R OR l NR 2

32 : 4-methylenzaldehyde D A B B O 3

33 : benzaldehyde A B D B D O

34 : 2-hydroxy-4-methoxy-benzaldehyde A B F D E D E O 3 O O

35 a b Jab Jba a b Jab = Jba : z

36 a b a(β)b(α) a b a(α)b(α) a b a(β)b(β) a b a(α)b(β) a b

37 a ν0 ν1 a ν0 ν2 B0 b β b α b a (Jab = Jba)

38 a b ν 0 J ab J ba ν 1 ν 2 or :

39 singlet 2 doublet Mz 200 Mz (1 ppm = 10 cm) 1 cm 2 cm doublet singlet 2 cm

40 100 Mz 200 Mz 10 z 10 z 10 z 10 z ~1.985 ppm (m, 2) 2.00 ppm (t, 1, J = 10.0 z) 1.99 ppm (d, 1, J = 10.0 z)

41 b b ' a a b b ' α α β β α β α β β α β α β α β α up-field shift no shift down-field shift b b : ( + 1 )

42 1 singlet 1 1 doublet (1 : 1) triplet (1 : 2 : 1) quartet (1 : 3 : 3 : 1) quintet (1 : 4 : 6 : 6 : 4 : 1) sextet (1 : 5 : 10 : 10 : 5 : 1)

43 ' ' ' geminal ( 2 J) 9~15 z vicinal ( 3 J) 0~18 z : 7~8 z long range ( 4 J) 0~3 z φ φ ' Karplus 3 J = A cos 2 φ + B cos φ +

44 ax ax eq eq eq eq ax ax 3 J (~12 z) >> 3 J (2~3 z) ~ー 2 J (2~3 z) cis geminal trans 3 Jtrans (11~18 z) > 3 Jcis (6~14 z) >> 2 Jgeminal (~3 z)

45 o m : 3 J (6~9 z) : 3 J (1~3 z) : 2 J (~0 z) p 4 J = 2~4 z

46 O

47 l O 2 3

48

49

50 a O 3 b AnMmXl A (m+1) (l+1) JAM = JAX (m+l+1) a b dd: double doublet (AMX) t: triplet (AX2)

51

52 Allyl Ethyl Ether 2 O 2 3

53 ~17 z ~10 z ~6 z

54 AA XX a b Br O a ' b ' νa = νa νb = νb Ja,b Ja,b Ja,b Ja,b

55 90 Mz AA BB Br Br l l Br l 1 2 ν1 = ν2 enantiomers p1 p2 p3 Br 3 4 B B' Br l 3 B B' 4 Br l A A 1 2 l 4 A 3 A' A' A' B' B (P87) J1,3 = p1jab + p2jad + p3ja J1,4 = p1jab + p2ja + p3jad J1,3 J1,4

56 JAB JAB JAB JAB ν1 ν2 ν3 ν4 νa νb νa - νb > 10 JAB ν1 ν2 ν3 ν4 νa νb νa - νb < 10 JAB

57 A2X2 90 Mz O N 2.61 ppm 3 J = 7 z AA XX

58 ax. eq. eq. very fast at rt slow below 100 ax. ax. eq. ax. eq.

59 (ABX 3.08 ppm JAX = 8.6 z O 2 JAB = 17.1 z A B X 4.52 ppm 2.89 ppm Br O 2 JBX = 6.4 z A O 2 B O 2 O 2 A B O 2 O 2 Br X Br X Br O 2 B A X A B

60 O O 3 N 3 N 3 E/Z 3 cf. 3 N 3 3

61 1 { 1 }NMR A cf. - OSY X hν A X A X

62 13 { 1 }NMR (~ 1%) (99.99%) J(, ) = 125~250 z 2 J(, ) = 70 ~ 20 z 3 J(, ) = <15 z cf. 1 J(, X): X = D, 19 F, 31 P 3 BM

63 O δ R O < Ar O < R OO O + fast O 2 O + 2 O quartet (α) O a singlet b + fast O a b a + fast O b (β)

64 O cf. in DMSO cf. dq O a t 3 O - S+ 3 O O

65 N 14 N: I = 1, 3 δ 1.98 O δ 6.7 N δ z 7.2 z δ 1.14

66 S 7.7 z 7.7 z S t 7.7 z q sex t

67 O fast + D O D 2 O + DO 3 O N fast O + D 2 O + DO 3 N D

68 O 2 3 O O 3 3

69 O

70 13 NMR spectrum % γ 1 1/4 γ J(, ), 2 J(, ), 3 J(, ) 400 Mz ( 1 NMR vs. 100 Mz ( 13 NMR) 1. FT-NMR 2. 1 NOE J(, ) : 1 J(, ) : DEPT -OSY, MQ, LR--OSY, MB

71 13 NMR spectrum O O O : 0~220 ppm : 0~50 ppm : ppm sp 2 : 100~160 ppm =O: 150~220 ppm : singlet

72 13 NMR OR R 3 X R 2 X R 2 X aldehydes aromatics alkynes X 3 ketones carboxylates alkenes NR

Microsoft Word - 4NMR2.doc

Microsoft Word - 4NMR2.doc 4 NMR 4.1 Zeeman 1, 13 C, 19 F, 31 P NMR 1 13 C 1/2 4.1 7%&'- 89:;'

More information

09_organal2

09_organal2 4. (1) (a) I = 1/2 (I = 1/2) I 0 p ( ), n () I = 0 (p + n) I = (1/2, 3/2, 5/2 ) p ( ), n () I = (1, 2, 3 ) (b) (m) (I = 1/2) m = +1/2, 1/2 (I = 1/2) m = +1/2, 1/2 I m = +I, +(I 1), +(I 2) (I 1), I ( )

More information

koji07-01.dvi

koji07-01.dvi 2007 I II III 1, 2, 3, 4, 5, 6, 7 5 10 19 (!) 1938 70 21? 1 1 2 1 2 2 1! 4, 5 1? 50 1 2 1 1 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 3 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k,l m, n k,l m, n kn > ml...?

More information

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

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 3 5 5 5 3 3 7 5 33 5 33 9 5 8 > e > f U f U u u > u ue u e u ue u ue u e u e u u e u u e u N cos s s cos ψ e e e e 3 3 e e 3 e 3 e 3 > A A > A E A f A A f A [ ] f A A e > > A e[ ] > f A E A < < f ; >

More information

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.   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)

More information

Chap. 1 NMR

Chap. 1   NMR β α β α ν γ π ν γ ν 23,500 47,000 ν = 100 Mz ν = 200 Mz ν δ δ 10 8 6 4 2 0 δ ppm) Br C C Br C C Cl Br C C Cl Br C C Br C 2 2 C C3 3 C 2 C C3 C C C C C δ δ 10 8 6 4 δ ppm) 2 0 ν 10 8 6 4 δ ppm) 2 0 (4)

More information

keisoku01.dvi

keisoku01.dvi 2.,, Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 4 Mon, 2006, 401, SAGA, JAPAN Dept. of Mechanical Engineering, Saga Univ., JAPAN 5 Mon, 2006, 401, SAGA, JAPAN Dept.

More information

( ) ,

( ) , II 2007 4 0. 0 1 0 2 ( ) 0 3 1 2 3 4, - 5 6 7 1 1 1 1 1) 2) 3) 4) ( ) () H 2.79 10 10 He 2.72 10 9 C 1.01 10 7 N 3.13 10 6 O 2.38 10 7 Ne 3.44 10 6 Mg 1.076 10 6 Si 1 10 6 S 5.15 10 5 Ar 1.01 10 5 Fe 9.00

More information

2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 1, 2 1, 3? , 2 2, 3? k, l m, n k, l m, n kn > ml...? 2 m, n n m

2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 1, 2 1, 3? , 2 2, 3? k, l m, n k, l m, n kn > ml...? 2 m, n n m 2009 IA I 22, 23, 24, 25, 26, 27 4 21 1 1 2 1! 4, 5 1? 50 1 2 1 1 2 1 4 2 2 2 1?? 2 1 1, 2 1, 2 1, 2, 3,... 1, 2 1, 3? 2 1 3 1 2 1 1, 2 2, 3? 2 1 3 2 3 2 k, l m, n k, l m, n kn > ml...? 2 m, n n m 3 2

More information

SO(2)

SO(2) TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

More information

The Physics of Atmospheres CAPTER :

The Physics of Atmospheres CAPTER : The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(

More information

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......................

More information

( ) 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

( ) 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 ( ) Note 3 19 12 13 8 8.1 (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 µ τ, e R, µ R, τ R (1a) L ( ) ) * 3) W Z 1/2 ( - )

More information

Note.tex 2008/09/19( )

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..................................

More information

研修コーナー

研修コーナー l l l l l l l l l l l α α β l µ l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l l

More information

Part () () Γ Part ,

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

More information

25 7 18 1 1 1.1 v.s............................. 1 1.1.1.................................. 1 1.1.2................................. 1 1.1.3.................................. 3 1.2................... 3

More information

( )

( ) 18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................

More information

, 1. x 2 1 = (x 1)(x + 1) x 3 1 = (x 1)(x 2 + x + 1). a 2 b 2 = (a b)(a + b) a 3 b 3 = (a b)(a 2 + ab + b 2 ) 2 2, 2.. x a b b 2. b {( 2 a } b )2 1 =

, 1. x 2 1 = (x 1)(x + 1) x 3 1 = (x 1)(x 2 + x + 1). a 2 b 2 = (a b)(a + b) a 3 b 3 = (a b)(a 2 + ab + b 2 ) 2 2, 2.. x a b b 2. b {( 2 a } b )2 1 = x n 1 1.,,.,. 2..... 4 = 2 2 12 = 2 2 3 6 = 2 3 14 = 2 7 8 = 2 2 2 15 = 3 5 9 = 3 3 16 = 2 2 2 2 10 = 2 5 18 = 2 3 3 2, 3, 5, 7, 11, 13, 17, 19.,, 2,.,.,.,?.,,. 1 , 1. x 2 1 = (x 1)(x + 1) x 3 1 = (x 1)(x

More information

H22環境地球化学4_化学平衡III_ ppt

H22環境地球化学4_化学平衡III_ ppt 1 2 3 2009年度 環境地球化学 大河内 温度上昇による炭酸水の発泡 気泡 温度が高くなると 溶けきれなくなった 二酸化炭素が気泡として出てくる 4 2009年度 環境地球化学 圧力上昇による炭酸水の発泡 栓を開けると 瓶の中の圧力が急激に 小さくなるので 発泡する 大河内 5 CO 2 K H CO 2 H 2 O K H + 1 HCO 3- K 2 H + CO 3 2- (M) [CO

More information

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B 2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :

More information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

nsg02-13/ky045059301600033210

nsg02-13/ky045059301600033210 φ φ φ φ κ κ α α μ μ α α μ χ et al Neurosci. Res. Trpv J Physiol μ μ α α α β in vivo β β β β β β β β in vitro β γ μ δ μδ δ δ α θ α θ α In Biomechanics at Micro- and Nanoscale Levels, Volume I W W v W

More information

(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

(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 1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45

More information

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

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 . 4cm 6 cm 4cm cm 8 cm λ()=a [kg/m] A 4cm A 4cm cm h h Y a G.38h a b () y = h.38h G b h X () S() = π() a,b, h,π V = ρ M = ρv G = M h S() 3 d a,b, h 4 G = 5 h a b a b = 6 ω() s v m θ() m v () θ() ω() dθ()

More information

) 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)

) 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) 4 4 ) 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) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7

More information

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

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 1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67

More information

.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 =,

.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 =, [ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b

More information

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e 7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z

More information

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

More information

SFGÇÃÉXÉyÉNÉgÉãå`.pdf

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

More information

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)

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

More information

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

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 i j ij i j ii,, i j ij ij ij (, P P P P θ N θ P P cosθ N F N P cosθ F Psinθ P P F P P θ N P cos θ cos θ cosθ F P sinθ cosθ sinθ cosθ sinθ 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

More information

2

2 th 37 ICh Theoretical Examination - - - - 5 - - : - ( ): ( ) - : 279 - - - - - - - G D L U C K 1 2 1 amu = 1.6605 10-27 kg N = 6.02 10 23 mol -1 k = 1.3806503 10-23 J K -1 e = 1.6022 10-19 C F = 9.6485

More information

(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

(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 Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν 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 O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

More information

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....................

More information

直交座標系の回転

直交座標系の回転 b T.Koama x l x, Lx i ij j j xi i i i, x L T L L, L ± x L T xax axx, ( a a ) i, j ij i j ij ji λ λ + λ + + λ i i i x L T T T x ( L) L T xax T ( T L T ) A( L) T ( LAL T ) T ( L AL) λ ii L AL Λ λi i axx

More information

1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1

1: *2 W, L 2 1 (WWL) 4 5 (WWL) W (WWL) L W (WWL) L L 1 2, 1 4, , 1 4 (cf. [4]) 2: 2 3 * , , = , 1 I, A 25 8 24 1 1.1 ( 3 ) 3 9 10 3 9 : (1,2,6), (1,3,5), (1,4,4), (2,2,5), (2,3,4), (3,3,3) 10 : (1,3,6), (1,4,5), (2,2,6), (2,3,5), (2,4,4), (3,3,4) 6 3 9 10 3 9 : 6 3 + 3 2 + 1 = 25 25 10 : 6 3 + 3 3

More information

『共形場理論』

『共形場理論』 T (z) SL(2, C) T (z) SU(2) S 1 /Z 2 SU(2) (ŜU(2) k ŜU(2) 1)/ŜU(2) k+1 ŜU(2)/Û(1) G H N =1 N =1 N =1 N =1 N =2 N =2 N =2 N =2 ĉ>1 N =2 N =2 N =4 N =4 1 2 2 z=x 1 +ix 2 z f(z) f(z) 1 1 4 4 N =4 1 = = 1.3

More information

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 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

More information

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 + ( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n

More information

. 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

. 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 003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........

More information

limit&derivative

limit&derivative - - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................

More information

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds

128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds = 0 (3.4) S 1, S 2 { B( r) n( r)}ds 127 3 II 3.1 3.1.1 Φ(t) ϕ em = dφ dt (3.1) B( r) Φ = { B( r) n( r)}ds (3.2) S S n( r) Φ 128 3 II S 1, S 2 Φ 1, Φ 2 Φ 1 = { B( r) n( r)}ds S 1 Φ 2 = { B( r) n( r)}ds (3.3) S 2 S S 1 +S 2 { B( r) n( r)}ds

More information

AC Modeling and Control of AC Motors Seiji Kondo, Member 1. q q (1) PM (a) N d q Dept. of E&E, Nagaoka Unive

AC Modeling and Control of AC Motors Seiji Kondo, Member 1. q q (1) PM (a) N d q Dept. of E&E, Nagaoka Unive AC Moeling an Control of AC Motors Seiji Kono, Member 1. (1) PM 33 54 64. 1 11 1(a) N 94 188 163 1 Dept. of E&E, Nagaoka University of Technology 163 1, Kamitomioka-cho, Nagaoka, Niigata 94 188 (a) 巻数

More information

漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト

漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,

More information

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2

6 2 T γ T B (6.4) (6.1) [( d nm + 3 ] 2 nt B )a 3 + nt B da 3 = 0 (6.9) na 3 = T B V 3/2 = T B V γ 1 = const. or T B a 2 = const. (6.10) H 2 = 8π kc2 1 6 6.1 (??) (P = ρ rad /3) ρ rad T 4 d(ρv ) + PdV = 0 (6.1) dρ rad ρ rad + 4 da a = 0 (6.2) dt T + da a = 0 T 1 a (6.3) ( ) n ρ m = n (m + 12 ) m v2 = n (m + 32 ) T, P = nt (6.4) (6.1) d [(nm + 32 ] )a

More information

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C 0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

More information

B

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

More information

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

70 : 20 : A B (20 ) (30 ) 50 1 70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................

More information

(MRI) 10. (MRI) (MRI) : (NMR) ( 1 H) MRI ρ H (x,y,z) NMR (Nuclear Magnetic Resonance) spectrometry: NMR NMR s( B ) m m = µ 0 IA = γ J (1) γ: :Planck c

(MRI) 10. (MRI) (MRI) : (NMR) ( 1 H) MRI ρ H (x,y,z) NMR (Nuclear Magnetic Resonance) spectrometry: NMR NMR s( B ) m m = µ 0 IA = γ J (1) γ: :Planck c 10. : (NMR) ( 1 H) MRI ρ H (x,y,z) NMR (Nuclear Magnetic Resonance) spectrometry: NMR NMR s( B ) m m = µ 0 IA = γ J (1) γ: :Planck constant J: Ĵ 2 = J(J +1),Ĵz = J J: (J = 1 2 for 1 H) I m A 173/197 10.1

More information

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 +

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 + 2.6 2.6.1 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 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j

More information

機器分析化学 3.核磁気共鳴(NMR)法(1)

機器分析化学 3.核磁気共鳴(NMR)法(1) 機器分析化学 3. 核磁気共鳴 (NMR) 法 (1) 2011 年度 5. 核磁気共鳴スペクトル法 (Nucler Mgnetic Resonnce:NMR) キーワード原子核磁気共鳴 ⅰ) 原子核 ( 陽子 + 中性子 ) 原子番号 (= 陽子数 ) 質量数 (= 陽子数 + 中性子数 ) もし原子番号も質量数も偶数の場合その原子核はスピンを持たない そうでない場合 ( どちらか あるいは一方が奇数

More information

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

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 63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5

More information

[ ] 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

[ ] 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 [ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =

More information

II

II II 16 16.0 2 1 15 x α 16 x n 1 17 (x α) 2 16.1 16.1.1 2 x P (x) P (x) = 3x 3 4x + 4 369 Q(x) = x 4 ax + b ( ) 1 P (x) x Q(x) x P (x) x P (x) x = a P (a) P (x) = x 3 7x + 4 P (2) = 2 3 7 2 + 4 = 8 14 +

More information

nsg04-28/ky208684356100043077

nsg04-28/ky208684356100043077 δ!!! μ μ μ γ UBE3A Ube3a Ube3a δ !!!! α α α α α α α α α α μ μ α β α β β !!!!!!!! μ! Suncus murinus μ Ω! π μ Ω in vivo! μ μ μ!!! ! in situ! in vivo δ δ !!!!!!!!!! ! in vivo Orexin-Arch Orexin-Arch !!

More information

高校生の就職への数学II

高校生の就職への数学II II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................

More information

6.1 (P (P (P (P (P (P (, P (, P.

6.1 (P (P (P (P (P (P (, P (, P. (011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.

More information

II (No.2) 2 4,.. (1) (cm) (2) (cm) , (

II (No.2) 2 4,.. (1) (cm) (2) (cm) , ( II (No.1) 1 x 1, x 2,..., x µ = 1 V = 1 k=1 x k (x k µ) 2 k=1 σ = V. V = σ 2 = 1 x 2 k µ 2 k=1 1 µ, V σ. (1) 4, 7, 3, 1, 9, 6 (2) 14, 17, 13, 11, 19, 16 (3) 12, 21, 9, 3, 27, 18 (4) 27.2, 29.3, 29.1, 26.0,

More information

85 4

85 4 85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V

More information

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

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 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

O1-1 O1-2 O1-3 O1-4 O1-5 O1-6

O1-1 O1-2 O1-3 O1-4 O1-5 O1-6 O1-1 O1-2 O1-3 O1-4 O1-5 O1-6 O1-7 O1-8 O1-9 O1-10 O1-11 O1-12 O1-13 O1-14 O1-15 O1-16 O1-17 O1-18 O1-19 O1-20 O1-21 O1-22 O1-23 O1-24 O1-25 O1-26 O1-27 O1-28 O1-29 O1-30 O1-31 O1-32 O1-33 O1-34 O1-35

More information

untitled

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}, {}.

More information

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r 2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2

More information

Stereoelectronic Effect

Stereoelectronic Effect node anti bonding M ( σ* ) A A : bonding M ( σ ) A: atomic orbital M: molecular orbital node anti bonding M filled orbital of molecular 1 M bonding M vacant orbital of molecular 2 LUM LUM (lowest unoccupied

More information

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

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 ),

More information

( )

( ) 7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................

More information

1

1 GL (a) (b) Ph l P N P h l l Ph Ph Ph Ph l l l l P Ph l P N h l P l .9 αl B βlt D E. 5.5 L r..8 e g s e,e l l W l s l g W W s g l l W W e s g e s g r e l ( s ) l ( l s ) r e l ( s ) l ( l s ) e R e r

More information

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq 49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r

More information

統計的データ解析

統計的データ解析 ds45 xspec qdp guplot oocalc (Error) gg (Radom Error)(Systematc Error) x, x,, x ( x, x,..., x x = s x x µ = lm = σ µ x x = lm ( x ) = σ ( ) = - x = js j ( ) = j= ( j) x x + xj x + xj j x + xj = ( x x

More information

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

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 6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

More information

1 911 9001030 9:00 A B C D E F G H I J K L M 1A0900 1B0900 1C0900 1D0900 1E0900 1F0900 1G0900 1H0900 1I0900 1J0900 1K0900 1L0900 1M0900 9:15 1A0915 1B0915 1C0915 1D0915 1E0915 1F0915 1G0915 1H0915 1I0915

More information

) ] [ 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 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

More information

第86回日本感染症学会総会学術集会後抄録(I)

第86回日本感染症学会総会学術集会後抄録(I) κ κ κ κ κ κ μ μ β β β γ α α β β γ α β α α α γ α β β γ μ β β μ μ α ββ β β β β β β β β β β β β β β β β β β γ β μ μ μ μμ μ μ μ μ β β μ μ μ μ μ μ μ μ μ μ μ μ μ μ β

More information

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P 9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)

More information

ID POS F

ID POS F 01D8101011L 2005 3 ID POS 2 2 1 F 1... 1 2 ID POS... 2 3... 4 3.1...4 3.2...4 3.3...5 3.4 F...5 3.5...6 3.6 2...6 4... 8 4.1...8 4.2...8 4.3...8 4.4...9 4.5...10 5... 12 5.1...12 5.2...13 5.3...15 5.4...17

More information

newmain.dvi

newmain.dvi 数論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/008142 このサンプルページの内容は, 第 2 版 1 刷発行当時のものです. Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published

More information

春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,

春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an

More information

0.6 A = ( 0 ),. () A. () x n+ = x n+ + x n (n ) {x n }, x, x., (x, x ) = (0, ) e, (x, x ) = (, 0) e, {x n }, T, e, e T A. (3) A n {x n }, (x, x ) = (,

0.6 A = ( 0 ),. () A. () x n+ = x n+ + x n (n ) {x n }, x, x., (x, x ) = (0, ) e, (x, x ) = (, 0) e, {x n }, T, e, e T A. (3) A n {x n }, (x, x ) = (, [ ], IC 0. A, B, C (, 0, 0), (0,, 0), (,, ) () CA CB ACBD D () ACB θ cos θ (3) ABC (4) ABC ( 9) ( s090304) 0. 3, O(0, 0, 0), A(,, 3), B( 3,, ),. () AOB () AOB ( 8) ( s8066) 0.3 O xyz, P x Q, OP = P Q =

More information

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

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

6.1 (P (P (P (P (P (P (, P (, P.101

6.1 (P (P (P (P (P (P (, P (, P.101 (008 0 3 7 ( ( ( 00 1 (P.3 1 1.1 (P.3.................. 1 1. (P.4............... 1 (P.15.1 (P.15................. (P.18............3 (P.17......... 3.4 (P................ 4 3 (P.7 4 3.1 ( P.7...........

More information

ρ /( ρ) + ( q, v ) : ( q, v ), L < q < q < q < L 0 0 ( t) ( q ( t), v ( t)) dq ( t) v ( t) lmr + 0 Φ( r) dt lmr + 0 Φ ( r) dv ( t) Φ ( q ( t) q ( t)) + Φ ( q+ ( t) q ( t)) dt ( ) < 0 ( q (0), v (0)) (

More information

chap9.dvi

chap9.dvi 9 AR (i) (ii) MA (iii) (iv) (v) 9.1 2 1 AR 1 9.1.1 S S y j = (α i + β i j) D ij + η j, η j = ρ S η j S + ε j (j =1,,T) (1) i=1 {ε j } i.i.d(,σ 2 ) η j (j ) D ij j i S 1 S =1 D ij =1 S>1 S =4 (1) y j =

More information

3/4/8:9 { } { } β β β α β α β β

3/4/8:9 { } { } β β β α β α β β α β : α β β α β α, [ ] [ ] V, [ ] α α β [ ] β 3/4/8:9 3/4/8:9 { } { } β β β α β α β β [] β [] β β β β α ( ( ( ( ( ( [ ] [ ] [ β ] [ α β β ] [ α ( β β ] [ α] [ ( β β ] [] α [ β β ] ( / α α [ β β ] [ ] 3

More information