Microsoft Word - 4NMR2.doc
|
|
- えつみ ひろもり
- 5 years ago
- Views:
Transcription
1 4 NMR 4.1 Zeeman 1, 13 C, 19 F, 31 P NMR 1 13 C 1/ %&'- 89:;'<= 7%&':#$>?@ 89)A89B.+ " %&' 23456,#$-./01 #$( )*+! %&'!"#$ #E = h $ 2% B 0 B α β ΔE B 0 ΔE B 0 γ gyromagnetic ratio T 1 γ = T 1 s 1 13 C γ = T 1 s 1 h Planck ΔE = h(γ/2π)b 0 = hν 1
2 ΔE ν NMR Fourier FT γ NMR 4.70 T Mz 13 C 50 Mz NMR 4.70 T XX C NMR NMR 0.5 ml CDCl 3 5 mm 4.2 NMR NMR 1 m 300 Mz NMR 2
3 4.2.1 NMR 4.2 NMR NMR 4.2 2,2 1 1 NMR 1/10 6 TMS Si(C 3 ) 4 ppm (parts per million) 3
4 NMR chemical shift NMR ppm δ = TMS / TMS 10 6!"#$ 4.1 C 3 C 2 C Ar C 3 C ± ± ± ± ± ± ± ± 0.3 %&3)45* C= %&'()* 7.0~ ~ ~ ~7.4 C 2 C 2 C 2 C 2 CN 2.1 ± ± ± ± ± 0.3 N 2 C 2 C 2 C C 2 N C 2 X C ± ± ± 0.3 N C ± 0.4 X 3.3 ± 0.3 (X=F: 4.1 ± 0.1) +,-./#$ 012%&#$ R R Ar 9.6 ± ± 0.6 Ar C N 2 CN 2 2~5 4~9 9~13 1~4 5~9 4
5 4.2 2,2 1 1 NMR δ ppm 3.29 ppm C C C C Z C C (Z =, N, X) ! (ppm) C 3 C 2 C sp 2 diamagnetic anisotropy 4.4!"# ()&' ()&' ()&' $%&' $%&' $%&' (a) *+,- (b).-/- (c) *
6 ring current ppm NMR , NMR C Br C C 3 C ,1 1 NMR (270 Mz) 6
7 δ 3.47 singlet 2 δ doublet 1 δ triplet spin spin splitting spin spin coupling spin spin coupling constant J J = 5.5 z J z N N NMR C 3 CC 2 C NMR (270 Mz) 7
8 3 δ δ quartet δ 4.12 C C J = 7.0 z A!"#$ X %$&'()*+,-./0 1./ -"#$ X %$&'()*+,1./0 1"#$ X %$&'()*+,2./0 2./ J J J αα, (αβ, βα), ββ ααα, (ααβ, αβα, βαα), (αββ, βαβ, ββα), βββ (C 3 ) 2 CCl C 6 C
9 z X C Y 6~18 C C 6~8!"#$%& 8~12 2~3 2~3 0~2 X X X 6~12 12~18 7~10 2~3 0~1,-./0123 0~ ~2 0~1.5 'W()*+& < 7 geminal 2 J 3 J C J = 9~12 z J = 10~15 z Karplus A, B, C Karplus 3 J = Acos 2 + Bcos + C 3 J 4 J < 1 z NMR 1 NMR NMR 9
10 NMR 400 Mz ppm ppm 6 7 N + 1 1:6:15:20:15:6:1 6 J = 7.2 z ppm 1.11 ppm C 2.15 ppm 10
11 NMR (400 Mz) ppm J = 5.8 z ppm 6.48 ppm J trans = 14.4, J cis = 6.8, J gem = 2.4 z NMR ,2 1 NMR 11
12 3 C C* C Br a b Br ,2 1 NMR (270 Mz) C C Br Br Me Br Br Br Me a Me b a Br b a b , a 2 J
13 J J (a)!" J > 10 " A " # (b)!" J = 5 " A " # (c)!" J = 2 " A " # J J (d)!" J = 1 " A " # (e)!" J = 0 " A = " # 4.12 ν z p ppm 13
14 4.13 p 1 NMR (400 Mz) 4.13 p 1 NMR J = 8.4 z a ,2 1 C 2 C 2 14
15 C 2 CDCl 3 2,2 1 1 NMR ppm J = 6.2 z 3.29 ppm C 2 2 NMR , NMR CDCl Mz NMR J z 1~10 4 NMR 15
16 b. irradiation NMR 4.15 (a) (b) 1.9 ppm (b) (a) NMR C NMR 12 C NMR 1.11% 13 C NMR 1 16
17 1/ C NMR 1 1 broad-band proton decoupling 13 C 4.4 NE FT NMR 2 13 C 13 C 13 C 13 C NMR 200 ppm C 13 C NMR TMS (C 3 ) 4 Si R C 3 R C 2 R R R C R R R C R C NMR ppm 0 8~35 15~50 20~60 30~40 C I C Br C Cl C N C 0~40 25~65 35~80 40~60 50~80 R C Y R C 65~85 C 100~150 C 110~170 R C R R C 165~ ~ ~215 NMR 13 C 1 1/4 NMR 1/64 1/5800 NE 17
18 13 C sp 2 sp C NMR C NMR 4.17 CDCl 3 13 C NMR 10% 13 C
19 C NMR CDCl 3 13 C NMR NMR saturation relaxation 2 spin-lattice relaxation 19
20 T 1 T 2 T 1 > T 2 NMR T 1 T 1 13 C NMR NE NMR verhauser nuclear verhauser effect, NE C NMR NE 3 13 C NMR NE 1 NMR 4.5 NMR 2D NMR NMR 20
21 NMR 2D NMR 2D NMR NMR 2D NMR 1 1 CSY correlation spectroscopy 1 13 C 1 13 C 13 C 2D NMR C C C NESY nuclear verhauser effect difference spectroscopy NE 1 1 CSY x y 1 NMR CSY a b 3 C 7 3 C b' a' C 3 c c' 3!"#(3-Carene) 6 3 C 7 1 eq ax ax eq 2 C 3 5 C
22 CSY a a δ 1.7~2.3 ppm 2, 2 5, b b c c (ax), 5 (ax), 2 (eq), 5 (eq), 4 22
Nuclear Magnetic Resonance 1 H NMR spectrum PPM
Nuclear Magnetic Resonance 1 NMR spectrum PPM PPM X: X X: 1.5 1.5 1 1 1 1 : 1: 1 : 1.5 : 1.5 2 : 2 : 2 : 3 : 3 1 13 NMR spectrum BM PPM BM 13 NMR spectrum DEPT OSY a b c d a c 1 NMR spectrum PPM a-c a
More information09_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 informationChap. 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 informationkoji07-01.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> > <., 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 informationzsj2017 (Toyama) program.pdf
88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88
More information88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88
More information_170825_<52D5><7269><5B66><4F1A>_<6821><4E86><5F8C><4FEE><6B63>_<518A><5B50><4F53><FF08><5168><9801><FF09>.pdf
88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88 th Annual Meeting of the Zoological Society of Japan Abstracts 88
More informationO1-1 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 informationBD = 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 informationO x y z O ( O ) O (O ) 3 x y z O O x v t = t = 0 ( 1 ) O t = 0 c t r = ct P (x, y, z) r 2 = x 2 + y 2 + z 2 (t, x, y, z) (ct) 2 x 2 y 2 z 2 = 0
9 O y O ( O ) O (O ) 3 y O O v t = t = 0 ( ) O t = 0 t r = t P (, y, ) r = + y + (t,, y, ) (t) y = 0 () ( )O O t (t ) y = 0 () (t) y = (t ) y = 0 (3) O O v O O v O O O y y O O v P(, y,, t) t (, y,, t )
More information1 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 information2
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 information2 1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a 2 + b 2 α (norm) N(α) = a 2 + b 2 = αα = α 2 α (spure) (trace) 1 1. a R aα =
1 1 α = a + bi(a, b R) α (conjugate) α = a bi α (absolute value) α = a + b α (norm) N(α) = a + b = αα = α α (spure) (trace) 1 1. a R aα = aα. α = α 3. α + β = α + β 4. αβ = αβ 5. β 0 6. α = α ( ) α = α
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 =
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 information2 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 informationzz + 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 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
More informationax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
More informationNMRの信号がはじめて観測されてから47年になる。その後、NMRは1960年前半までPhys. Rev.等の物理学誌上を賑わせた。1960年代後半、物理学者の間では”NMRはもう死んだ”とささやかれたということであるが(1)、しかし、これほど発展した構造、物性の
8. CW-NR Bloch[]Z (longitudinal relaxation timexy (transversal relaxation timebloembergen [] Bloch Bloembergen Bloch (3.. d d d x z = ( ω ω = ωz + ( ω ω x = ω ( z x (8..a (8..b (8..c = z = z θ = ω t (
More informationlimit&derivative
- - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................
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 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( ) 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 informationNMRの信号がはじめて観測されてから47年になる。その後、NMRは1960年前半までPhys. Rev.等の物理学誌上を賑わせた。1960年代後半、物理学者の間では”NMRはもう死んだ”とささやかれたということであるが(1)、しかし、これほど発展した構造、物性の
9. I S H0 = ωii + ωss ( (9.. H A ( q (0 A = a ( os θ (9..a r ( ± ± i A =± a sinθosθe ϕ r (9..b A a sin e r ( ± ± iϕ = θ (9.. µ 0 γ I γ S a = (9..d 4π A ( q q ( q * = ( A ( q (9..e (0 = { IS ( I+ S + I
More informationn 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m
1 1 1 + 1 4 + + 1 n 2 + π2 6 x [10 n x] x = lim n 10 n n 10 k x 1.1. a 1, a 2,, a n, (a n ) n=1 {a n } n=1 1.2 ( ). {a n } n=1 Q ε > 0 N N m, n N a m a n < ε 1 1. ε = 10 1 N m, n N a m a n < ε = 10 1 N
More informationnsg04-28/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 informationuntitled
3-1 ( sit ) (stead state vibratio) (trasiet vibratio) sit(a)w s ( W s ) W / g C (b) sit ( + s ) ( + s ) c + W + sit W s si t + s + c + si t (3.1) si t (3.1) a C W b sit(respose) () 3- acost+ bsit a sit+
More information公開用:新500MHz NMRはかり方マニュアル 第2版 2
1 1-1: 10 ml 10 ml 1 H, 13 C NMR 20, 30 mg 1 H NMR 3 mg 20 mg ( 1 H NMR 30 mg 0.5 mg) 13 C NMR 20 mg 100 mg 1 H NMR 13 C NMR 1 H NMR ( 13 C 1 H ) 13 C NMR 20 mg (5 mg ) 2 1-1: NMR () 0.6 mlcdcl 3 5 ml
More information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 informationmain.dvi
SGC - 48 208X Y Z Z 2006 1930 β Z 2006! 1 2 3 Z 1930 SGC -12, 2001 5 6 http://www.saiensu.co.jp/support.htm http://www.shinshu-u.ac.jp/ haru/ xy.z :-P 3 4 2006 3 ii 1 1 1.1... 1 1.2 1930... 1 1.3 1930...
More informationr 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
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
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磁性物理学 - 遷移金属化合物磁性のスピンゆらぎ理論
email: takahash@sci.u-hyogo.ac.jp May 14, 2009 Outline 1. 2. 3. 4. 5. 6. 2 / 262 Today s Lecture: Mode-mode Coupling Theory 100 / 262 Part I Effects of Non-linear Mode-Mode Coupling Effects of Non-linear
More informationF = 0 F α, β F = t 2 + at + b (t α)(t β) = t 2 (α + β)t + αβ G : α + β = a, αβ = b F = 0 F (t) = 0 t α, β G t F = 0 α, β G. α β a b α β α β a b (α β)
19 7 12 1 t F := t 2 + at + b D := a 2 4b F = 0 a, b 1.1 F = 0 α, β α β a, b /stlasadisc.tex, cusp.tex, toileta.eps, toiletb.eps, fromatob.tex 1 F = 0 F α, β F = t 2 + at + b (t α)(t β) = t 2 (α + β)t
More information3/4/8:9 { } { } 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_0212_68<5A66><4EBA><79D1>_<6821><4E86><FF08><30C8><30F3><30DC><306A><3057><FF09>.pdf
More information
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],
More informationm(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 informationx V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R
V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x
More information(e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a
1 2 2.1 (e ) (µ ) (τ ) ( (ν e,e ) e- (ν µ,µ ) µ- (ν τ,τ ) τ- ) ( ) ( ) ( ) (SU(2) ) (W +,Z 0,W ) * 1) [ ] [ ] [ ] ν e ν µ ν τ e µ τ, e R,µ R,τ R (2.1a) L ( ) ) * 2) W Z 1/2 ( - ) d u + e + ν e 1 1 0 0
More information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 information14 4 12 10 8 6 3 2 4 2 1 0 0 100200300400500600 100200300400500600700 0 0 100200300400500600 100200300400500600700 (ppm) (ppm) 7 4 6 5 4 3 3 2 2 1 1 0 0 100200300400500600 100200300400500600700 0 0 100200300400500600
More informationuntitled
b 0 1PPm 10PPm 100PPm 1000PPm 10000PPm 0.0001% 0.001% 0.01% 0.1% 1% 10% 1PPm 10PPm 100PPm 1000PPm 10000PPm 0.0001% 0.001% 0.01% 0.1% 1% 10% 1PPm 10PPm 100PPm 1000PPm 10000PPm 0.0001% 0.001% 0.01%
More informationx () 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 informationN 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 informationTOP 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.............................
More information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
... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c
More informationD xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
More information.1 A cos 2π 3 sin 2π 3 sin 2π 3 cos 2π 3 T ra 2 deta T ra 2 deta T ra 2 deta a + d 2 ad bc a 2 + d 2 + ad + bc A 3 a b a 2 + bc ba + d c d ca + d bc +
.1 n.1 1 A T ra A A a b c d A 2 a b a b c d c d a 2 + bc ab + bd ac + cd bc + d 2 a 2 + bc ba + d ca + d bc + d 2 A a + d b c T ra A T ra A 2 A 2 A A 2 A 2 A n A A n cos 2π sin 2π n n A k sin 2π cos 2π
More informationAC 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 informationVI VI.21 W 1,..., W r V W 1,..., W r W W r = {v v r v i W i (1 i r)} V = W W r V W 1,..., W r V W 1,..., W r V = W 1 W
3 30 5 VI VI. W,..., W r V W,..., W r W + + W r = {v + + v r v W ( r)} V = W + + W r V W,..., W r V W,..., W r V = W W r () V = W W r () W (W + + W + W + + W r ) = {0} () dm V = dm W + + dm W r VI. f n
More informationω 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 information05Mar2001_tune.dvi
2001 3 5 COD 1 1.1 u d2 u + ku =0 (1) dt2 u = a exp(pt) (2) p = ± k (3) k>0k = ω 2 exp(±iωt) (4) k
More information2019 年 6 月 4 日演習問題 I α, β > 0, A > 0 を定数として Cobb-Douglas 型関数 Y = F (K, L) = AK α L β (5) と定義します. (1) F KK, F KL, F LK, F LL を求めましょう. (2) 第 1 象限のすべての点
09 年 6 月 4 日演習問題 I α, β > 0, A > 0 を定数として Cobb-Douglas 型関数 Y = F K, L) = AK α L β 5) と定義します. ) F KK, F KL, F LK, F LL を求めましょう. ) 第 象限のすべての点 K, L) R ++ に対して F KK K, L) < 0, かつ dethf )K, L) > 0 6) を満たす α,
More informationA(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 informationuntitled
71 7 3,000 1 MeV t = 1 MeV = c 1 MeV c 200 MeV fm 1 MeV 3.0 10 8 10 15 fm/s 0.67 10 21 s (1) 1fm t = 1fm c 1fm 3.0 10 8 10 15 fm/s 0.33 10 23 s (2) 10 22 s 7.1 ( ) a + b + B(+X +...) (3) a b B( X,...)
More informationuntitled
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 information06佐々木雅哉_4C.indd
3 2 3 2 4 5 56 57 3 2013 9 2012 16 19 62.2 17 2013 7 170 77 170 131 58 9 10 59 3 2 10 15 F 12 12 48 60 1 3 1 4 7 61 3 7 1 62 T C C T C C1 2 3 T C 1 C 1 T C C C T T C T C C 63 3 T 4 T C C T C C CN T C C
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 =,
[ ] 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 informationStereoelectronic 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 informationID 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 informationd-00
283-0105 298 TEL. 0475-76-0839 FAX. 0475-76-0838 400g 300 4950399167708 14 220g 300 4950399066780 Page 1 300g 200 4950399066766 100 400g 300 4950399167722 350g 160 4950399066735 100 600g 350 4950399167685
More information1 1.1 H = µc i c i + c i t ijc j + 1 c i c j V ijklc k c l (1) V ijkl = V jikl = V ijlk = V jilk () t ij = t ji, V ijkl = V lkji (3) (1) V 0 H mf = µc
013 6 30 BCS 1 1.1........................ 1................................ 3 1.3............................ 3 1.4............................... 5 1.5.................................... 5 6 3 7 4 8
More informationSFGÇÃÉXÉyÉNÉgÉãå`.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 information124
[ ] [ ] 18 123 124 ( ) 125 126 127 128 5,683,062 2,306,419 18.2 14.0 12 5389 87 13 257 88 12 22 27 56.7 41.6 1 1.7 15 129 55 38.8 13.3 15 2004 15 1 3 84.8 29.3 1 23.9 38.0 10 10 35 35 15 5 56.5.31 55.4.1
More informationA
A04-164 2008 2 13 1 4 1.1.......................................... 4 1.2..................................... 4 1.3..................................... 4 1.4..................................... 5 2
More informationd > 2 α B(y) y (5.1) s 2 = c z = x d 1+α dx ln u 1 ] 2u ψ(u) c z y 1 d 2 + α c z y t y y t- s 2 2 s 2 > d > 2 T c y T c y = T t c = T c /T 1 (3.
5 S 2 tot = S 2 T (y, t) + S 2 (y) = const. Z 2 (4.22) σ 2 /4 y = y z y t = T/T 1 2 (3.9) (3.15) s 2 = A(y, t) B(y) (5.1) A(y, t) = x d 1+α dx ln u 1 ] 2u ψ(u), u = x(y + x 2 )/t s 2 T A 3T d S 2 tot S
More information,. 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,
More information120126_RRR_jp.pptx
高磁場 NMR の利用 C 検出への期待 202 年 月 25-26 日首都大学東京秋葉原サテライトキャンパス第 回 RRR-workshop 20/2 大阪大学蛋白質研究所構造プロテオミクス研究系池上貴久 NMR の高磁場化に伴う利点 感度の上昇 B 0 3/2 磁気モーメント B 0 ラーモア周波数 B 0 ノイズレベル B 0 /2 (S/N) 950MHz / (S/N) 600MHz =
More informationgr09.dvi
.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
More informationACQUITY UPC 2 ACQUITY UPC 2 A A+D3 D ml E 10 ml K μm K μm PDA ACQUITY UPC 2 BEH mm, 1.7 μm ACQUITY UPC 2 HSS C 18 SB 3.
UPC 2 Andrew Aubin Waters Corporation, Milford, MA, USA Waters ACQUITY UPC 2TM 6 tert- UltraPerformance Convergence Chromatography TM UPC 2 ACQUITY UPC 2 1 A A D3 E D3 K1 K2 UPC 2 A E K1 ACQUITY UPC 2
More informationNMRの信号がはじめて観測されてから47年になる。その後、NMRは1960年前半までPhys. Rev.等の物理学誌上を賑わせた。1960年代後半、物理学者の間では”NMRはもう死んだ”とささやかれたということであるが(1)、しかし、これほど発展した構造、物性の
4. [] H Ψ Ψ u Ψ= cu, cc = (4..) A A m, m m < A>=< Ψ A Ψ >= c c < u A u > (4..) c A (4..)Ac m c c AA m m c c c c c m (4.. ) c c P < P m >= cmc m m, (4..3) < A >= < P m >< u A u > = Tr( PA) (4..4) m A P
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 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