(7) u 1 θ A {u 1, u, u 3 } U = (u 1, u, u 3 ) A = UT (θ) + tu t UAU = T (θ) + () θ x z cos θ 0 sin θ cos θ sin θ 0 X(θ) = 0 cos θ sin θ, Y (θ) =

December 28, 2018

e-mail : kigami@i.kyoto-u.ac.jp December 28, 28 Contents 2............................. 3.2......................... 7.3..................... 9.4................ 4.5............. 2.6.... 22 2 36 2..........................

1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x

. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +

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

2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =

) ] [ 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

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

1 1.1 sin 2π [rad] 3 ft 3 sin 2t π 4 3.1 2 1.1: sin θ 2.2 sin θ ft t t [sec] t sin 2t π 4 [rad] sin 3.1 3 sin θ θ t θ 2t π 4 3.2 3.1 3.4 3.4: 2.2: sin θ θ θ [rad] 2.3 0 [rad] 4 sin θ sin 2t π 4 sin 1 1

( )

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

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 {

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

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

(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

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,

(, Goo Ishikawa, Go-o Ishikawa) ( ) 1

(, Goo Ishikawa, Go-o Ishikawa) ( ) 1 ( ) ( ) ( ) G7( ) ( ) ( ) () ( ) BD = 1 DC CE EA AF FB 0 0 BD DC CE EA AF FB =1 ( ) 2 (geometry) ( ) ( ) 3 (?) (Topology) ( ) DNA ( ) 4 ( ) ( ) 5 ( ) H. 1 : 1+ 5 2

pdf

http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

TOP URL 1

TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

I , : ~/math/functional-analysis/functional-analysis-1.tex

I 1 2004 8 16, 2017 4 30 1 : ~/math/functional-analysis/functional-analysis-1.tex 1 3 1.1................................... 3 1.2................................... 3 1.3.....................................

量子力学 問題

3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

i I II I II II IC IIC I II ii 5 8 5 3 7 8 iii I 3........................... 5......................... 7........................... 4........................ 8.3......................... 33.4...................

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

数学Ⅱ演習（足助・09夏）

II I 9/4/4 9/4/2 z C z z z z, z 2 z, w C zw z w 3 z, w C z + w z + w 4 t R t C t t t t t z z z 2 z C re z z + z z z, im z 2 2 3 z C e z + z + 2 z2 + 3! z3 + z!, I 4 x R e x cos x + sin x 2 z, w C e z+w

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

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

9 7 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 x x B } B C y C y + x B y C x C C x C y B = A

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

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

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) 3 R ij R ik = δ jk (4) i=1 δ ij Kronecker δ ij = { 1 (i = j) 0 (i

1. 1 1.1 1.1.1 1.1.1.1 v v = v 1 v 2 v 3 (1) R = (R ij ) (2) R (R 1 ) ij = R ji (3) R ij R ik = δ jk (4) δ ij Kronecker δ ij = { 1 (i = j) 0 (i j) (5) 1 1.1. v1.1 2011/04/10 1. 1 2 v i = R ij v j (6) [

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy

DVIOUT-HYOU

() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

Part2 47 Example 161 93 1 T a a 2 M 1 a 1 T a 2 a Point 1 T L L L T T L L T L L L T T L L T detm a 1 aa 2 a 1 2 + 1 > 0 11 T T x x M λ 12 y y x y λ 2 a + 1λ + a 2 2a + 2 0 13 D D a + 1 2 4a 2 2a + 2 a

B ver B

B ver. 2017.02.24 B Contents 1 11 1.1....................... 11 1.1.1............. 11 1.1.2.......................... 12 1.2............................. 14 1.2.1................ 14 1.2.2.......................

,. 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,

さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n

1 1.1 1.1.1 A 2 P Q 3 R S T R S T P 80 50 60 Q 90 40 70 80 50 60 90 40 70 8 5 6 1 1 2 9 4 7 2 1 2 3 1 2 m n m n m n n n n 1.1 8 5 6 9 4 7 2 6 0 8 2 3 2 2 2 1 2 1 1.1 2 4 7 1 1 3 7 5 2 3 5 0 3 4 1 6 9 1

24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x

24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),

2019 1 5 0 3 1 4 1.1.................... 4 1.1.1......................... 4 1.1.2........................ 5 1.1.3................... 5 1.1.4........................ 6 1.1.5......................... 6 1.2..........................

(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y

(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b

熊本県数学問題正解

00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (

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

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

行列代数2010A

a ij i j 1) i +j i, j) ij ij 1 j a i1 a ij a i a 1 a j a ij 1) i +j 1,j 1,j +1 a i1,1 a i1,j 1 a i1,j +1 a i1, a i +1,1 a i +1.j 1 a i +1,j +1 a i +1, a 1 a,j 1 a,j +1 a, ij i j 1,j 1,j +1 ij 1) i +j a

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

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 ),

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

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 =

IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (

IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1

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 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N m 1.

1.1 1. 1.3.1..3.4 3.1 3. 3.3 4.1 4. 4.3 5.1 5. 5.3 6.1 6. 6.3 7.1 7. 7.3 1 1 variation 1.1 imension unit L m M kg T s Q C QT 1 A = C s 1 MKSA F = ma N N = kg m s 1.1 J E = 1 mv W = F x J = kg m s 1 = N

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

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =

IA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................

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

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

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) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37

4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin

13 0 1 1 4 11 4 12 5 13 6 2 10 21 10 22 14 3 20 31 20 32 25 33 28 4 31 41 32 42 34 43 38 5 41 51 41 52 43 53 54 6 57 61 57 62 60 70 0 Gauss a, b, c x, y f(x, y) = ax 2 + bxy + cy 2 = x y a b/2 b/2 c x

Z: Q: R: C:

0 Z: Q: R: C: 3 4 4 4................................ 4 4.................................. 7 5 3 5...................... 3 5......................... 40 5.3 snz) z)........................... 4 6 46 x

2.2 ( y = y(x ( (x 0, y 0 y (x 0 (y 0 = y(x 0 y = y(x ( y (x 0 = F (x 0, y(x 0 = F (x 0, y 0 (x 0, y 0 ( (x 0, y 0 F (x 0, y 0 xy (x, y (, F (x, y ( (

(. x y y x f y = f(x y x y = y(x y x y dx = d dx y(x = y (x = f (x y = y(x x ( (differential equation ( + y 2 dx + xy = 0 dx = xy + y 2 2 2 x y 2 F (x, y = xy + y 2 y = y(x x x xy(x = F (x, y(x + y(x 2

II 1 II 2012 II Gauss-Bonnet II

II 1 II 212 II Gauss-Bonnet II 1 1 1.1......................................... 1 1.2............................................ 2 1.3.................................. 3 1.4.............................................

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

n ( (

1 2 27 6 1 1 m-mat@mathscihiroshima-uacjp 2 http://wwwmathscihiroshima-uacjp/~m-mat/teach/teachhtml 2 1 3 11 3 111 3 112 4 113 n 4 114 5 115 5 12 7 121 7 122 9 123 11 124 11 125 12 126 2 2 13 127 15 128

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

..3. Ω, Ω F, P Ω, F, P ). ) F a) A, A,..., A i,... F A i F. b) A F A c F c) Ω F. ) A F A P A),. a) 0 P A) b) P Ω) c) [ ] A, A,..., A i,... F i j A i A

.. Laplace ). A... i),. ω i i ). {ω,..., ω } Ω,. ii) Ω. Ω. A ) r, A P A) P A) r... ).. Ω {,, 3, 4, 5, 6}. i i 6). A {, 4, 6} P A) P A) 3 6. ).. i, j i, j) ) Ω {i, j) i 6, j 6}., 36. A. A {i, j) i j }.

(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z

B 4 24 7 9 ( ) :,..,,.,. 4 4. f(z): D C: D a C, 2πi C f(z) dz = f(a). z a a C, ( ). (ii), a D, a U a,r D f. f(z) = A n (z a) n, z U a,r, n= A n := 2πi C f(ζ) dζ, n =,,..., (ζ a) n+, C a D. (iii) U a,r

2 R U, U Hausdorff, R. R. S R = (S, A) (closed), (open). (complete projective smooth algebraic curve) (cf. 2). 1., ( ).,. countable ( 2 ) ,,.,,

15, pp.1-13 1 1.1,. 1.1. C ( ) f = u + iv, (, u, v f ). 1 1. f f x = i f x u x = v y, u y = v x.., u, v u = v = 0 (, f = 2 f x + 2 f )., 2 y2 u = 0. u, u. 1,. 1.2. S, A S. (i) A φ S U φ C. (ii) φ A U φ

II K116 : January 14, ,. A = (a ij ) ij m n. ( ). B m n, C n l. A = max{ a ij }. ij A + B A + B, AC n A C (1) 1. m n (A k ) k=1,... m n A, A k k

: January 14, 28..,. A = (a ij ) ij m n. ( ). B m n, C n l. A = max{ a ij }. ij A + B A + B, AC n A C (1) 1. m n (A k ) k=1,... m n A, A k k, A. lim k A k = A. A k = (a (k) ij ) ij, A k = (a ij ) ij, i,

1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =

1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i

meiji_resume_1.PDF

β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

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

Gmech08.dvi

145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

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 (

II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) 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 x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )

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

II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =

all.dvi

5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0

1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2

filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin

(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y

(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K

II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F

( ) 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 ( - )

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

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

量子力学Ａ

c 1 1 1.1....................................... 1 1............................................ 4 1.3.............................. 6 10.1.................................. 10......................................

16 B

16 B (1) 3 (2) (3) 5 ( ) 3 : 2 3 : 3 : () 3 19 ( ) 2 ax 2 + bx + c = 0 (a 0) x = b ± b 2 4ac 2a 3, 4 5 1824 5 Contents 1. 1 2. 7 3. 13 4. 18 5. 22 6. 25 7. 27 8. 31 9. 37 10. 46 11. 50 12. 56 i 1 1. 1.1..

OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P

4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e

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)

338 7 7.3 LCR 2.4.3 e ix LC AM 7.3.1 7.3.1.1 m k x m x x > 0 x < 0 F x > 0 x < 0 F = k x k > 0 k 5.3.1.1 x = xt 7.3 339 m 2 x t 2 = k x 2 x t 2 = ω 2 0 x ω0 = k m ω 0 1.4.4.3 2 +α 14.9.3.1 5.3.2.1 2 x

> > <., 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 >

「図形の数理」テキストfinal.dvi

008 9 5 .. O A, B AB AB A, B. A, B A, B AB O A D B AB AB AB. AB A, B.4 AB, D E AE : EB E D E A B D A : B = AE : EB E AEB AD : AB = AE : EB ED AEB EB = EF EBF = EFB E FB EB = EBF, GE = EFB F G E A B D.

) 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

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

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

08-Note2-web

r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t)

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,

p = mv p x > h/4π λ = h p m v Ψ 2 Ψ

II p = mv p x > h/4π λ = h p m v Ψ 2 Ψ Ψ Ψ 2 0 x P'(x) m d 2 x = mω 2 x = kx = F(x) dt 2 x = cos(ωt + φ) mω 2 = k ω = m k v = dx = -ωsin(ωt + φ) dt = d 2 x dt 2 0 y v θ P(x,y) θ = ωt + φ ν = ω [Hz] 2π

d ϕ i) t d )t0 d ϕi) ϕ i) t x j t d ) ϕ t0 t α dx j d ) ϕ i) t dx t0 j x j d ϕ i) ) t x j dx t0 j f i x j ξ j dx i + ξ i x j dx j f i ξ i x j dx j d )

23 M R M ϕ : R M M ϕt, x) ϕ t x) ϕ s ϕ t ϕ s+t, ϕ 0 id M M ϕ t M ξ ξ ϕ t d ϕ tx) ξϕ t x)) U, x 1,...,x n )) ϕ t x) ϕ 1) t x),...,ϕ n) t x)), ξx) ξ i x) d ϕi) t x) ξ i ϕ t x)) M f ϕ t f)x) f ϕ t )x) fϕ

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

July 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i

July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac

,2,4

2005 12 2006 1,2,4 iii 1 Hilbert 14 1 1.............................................. 1 2............................................... 2 3............................................... 3 4.............................................

2 (2016 3Q N) c = o (11) Ax = b A x = c A n I n n n 2n (A I n ) (I n X) A A X A n A A A (1) (2) c 0 c (3) c A A i j n 1 ( 1) i+j A (i, j) A (i, j) ã i

[ ] (2016 3Q N) a 11 a 1n m n A A = a m1 a mn A a 1 A A = a n (1) A (a i a j, i j ) (2) A (a i ca i, c 0, i ) (3) A (a i a i + ca j, j i, i ) A 1 A 11 0 A 12 0 0 A 1k 0 1 A 22 0 0 A 2k 0 1 0 A 3k 1 A rk

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

5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4

... A F F l F l F(p, 0) = p p > 0 l p 0 P(, ) H P(, ) P l PH F PF = PH PF = PH p O p ( p) + = { ( p)} = 4p l = 4p (p 0) F(p, 0) = p O 3 5 5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 =