0104.pages

0104.pages

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I No. sin cos sine, cosine : trigonometric function π : π =.4 : n =, ±, ±, sin + nπ = sin cos + nπ = cos sin = sin : cos = cos :. sin. sin. sin + π si

I 8 No. : No. : No. : No.4 : No.5 : No.6 : No.7 : No.8 : No.9 : No. : I No. sin cos sine, cosine : trigonometric function π : π =.4 : n =, ±, ±, sin + nπ = sin cos + nπ = cos sin = sin : cos = cos :. sin.

sin cos No. sine, cosine : trigonometric function π : π = 3.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin : odd cos = cos : even.

08 No. : No. : No.3 : No.4 : No.5 : No.6 : No.7 : No.8 : No.9 : No.0 : No. : sin cos No. sine, cosine : trigonometric function π : π = 3.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin

I No. sin cos sine, cosine : trigonometric function π : π =.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin : odd cos = cos : even.

I 0 No. : No. : No. : No.4 : No.5 : No.6 : No.7 : No.8 : No.9 : No.0 : I No. sin cos sine, cosine : trigonometric function π : π =.4 : n = 0, ±, ±, sin + nπ = sin cos + nπ = cos : parity sin = sin : odd

. sinh x sinh x) = e x e x = ex e x = sinh x 3) y = cosh x, y = sinh x y = e x, y = e x 6 sinhx) coshx) 4 y-axis x-axis : y = cosh x, y = s

. 00 3 9 [] sinh x = ex e x, cosh x = ex + e x ) sinh cosh 4 hyperbolic) hyperbola) = 3 cosh x cosh x) = e x + e x = cosh x ) . sinh x sinh x) = e x e x = ex e x = sinh x 3) y = cosh x, y = sinh x y =

di-problem.dvi

005/05/05 by. I : : : : : : : : : : : : : : : : : : : : : : : : :. II : : : : : : : : : : : : : : : : : : : : : : : : : 3 3. III : : : : : : : : : : : : : : : : : : : : : : : : 4 4. : : : : : : : : : :

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数学の基礎訓練I

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4) 07.3.7 ) Poincaré) Poincaré disk) hyperboloid) [] [, 3, 4] [] y 0 L hyperboloid) K Klein disk) J hemisphere) I Poincaré disk) : hyperboloid) L Klein disk) K hemisphere) J Poincaré) I y 0 x + y z 0 z

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9 8 3............................................. 3.......................................... 4.3............................................ 4 5 3 6 3..................................................

高等学校学習指導要領

高等学校学習指導要領

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,

2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a

009 I II III 4, 5, 6 4 30. 0 α β α β l 0 l l l l γ ) γ αβ ) α β. n n cos k n n π sin k n π k k 3. a 0, a,..., a n α a 0 + a x + a x + + a n x n 0 ᾱ 4. [a, b] f y fx) y x 5. ) Arcsin 4) Arccos ) ) Arcsin

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/

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( ) 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) (

6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

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Chap10.dvi

=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n

さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =

Chap2

逆三角関数の微分 Arcsin の導関数を計算する Arcsin I. 初等関数の微積分 sin [, ], [π/, π/] cos sin / (Arcsin ) 計算力の体力をつけよう π/ π/ E. II- 次の関数の導関数を計算せよ () Arccos () Arctan E. I- の解答 不定積分あれこれ () Arccos n log C C (n ) n e e C log (log

(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {

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( ; ) 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 =

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

f f x, y, u, v, r, θ r > = x + iy, f = u + iv C γ D f f D f f, Rm,. = x + iy = re iθ = r cos θ + i sin θ = x iy = re iθ = r cos θ i sin θ x = + = Re, y = = Im i r = = = x + y θ = arg = arctan y x e i =

LLG-R8.Nisus.pdf

d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a

1 1 1.1...................................... 1 1.2................................... 5 1.3................................... 7 1.4............................. 9 1.5....................................

2 Part Part

Contents 7 7 Part 1. 8 1. 8 2. 11 Part 2. 13 3. 13 3.1. 13 3.2. 14 4. 15 5. 16 Part 3. 17 6. 17 6.1. 17 6.2. 18 6.3. 19 7. 20 7.1. 20 7.2. 21 7.3. 23 7.4. 25 Date:. 1 2 Part 4. 27 8. 27 8.1. 27 8.2. 2

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3

Design of highly accurate formulas for numerical integration in weighted Hardy spaces with the aid of potential theory 1 Ken ichiro Tanaka 1 Ω R m F I = F (t) dt (1.1) Ω m m 1 m = 1 1 Newton-Cotes Gauss

di-problem.dvi

III 005/06/6 by. : : : : : : : : : : : : : : : : : : : : :. : : : : : : : : : : : : : : : : : : : : : : : : : : 3 3. : : : : : : : : : : : : : : 4 4. : : : : : : : : : : : : : : : : : : : : : : 5 5. :

no35.dvi

p.16 1 sin x, cos x, tan x a x a, a>0, a 1 log a x a III 2 II 2 III III [3, p.36] [6] 2 [3, p.16] sin x sin x lim =1 ( ) [3, p.42] x 0 x ( ) sin x e [3, p.42] III [3, p.42] 3 3.1 5 8 *1 [5, pp.48 49] sin

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

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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,

RIMS98R2.dvi

RIMS Kokyuroku, vol.084, (999), 45 59. Euler Fourier Euler Fourier S = ( ) n f(n) = e in f(n) (.) I = 0 e ix f(x) dx (.2) Euler Fourier Fourier Euler Euler Fourier Euler Euler Fourier Fourier [5], [6]

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

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

(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n

3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4

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 (

1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

項別超微分

13 項別超微分本章では 2 階以上の高階導関数を簡単な一般式で表すことが困難な関数について これら を級数に展開した上項別に超微分するものである 従って 12 超微分 で扱った e x, logx, sinx, cosx, sinhx, coshx の各関数は本章では扱わない 13 1 三角関数 双曲線関数の項別超微分 公式 13 1 1 ベルヌイ数とオイラー数をそれぞれ B 0 =1, B 2

36 3 D f(z) D z f(z) z Taylor z D C f(z) z C C f (z) C f(z) f (z) f(z) D C D D z C C 3.: f(z) 3. f (z) f 2 (z) D D D D D f (z) f 2 (z) D D f (z) f 2 (

3 3. D f(z) D D D D D D D D f(z) D f (z) f (z) f(z) D (i) (ii) (iii) f(z) = ( ) n z n = z + z 2 z 3 + n= z < z < z > f (z) = e t(+z) dt Re z> Re z> [ ] f (z) = e t(+z) = (Rez> ) +z +z t= z < f(z) Taylor

Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0

Tips KENZOU 28 7 6 P no problem 2 w = f(z) z w w z w = (z z ) b b w = log (z z ) z = z 2π 2 z = z w = z /2 z = re iθ θ (z = ) 2π 4π 2 θ θ 2π 4π z r re iθ re i2π = r re i4π = r w r re iθ/2 re iπ = r re

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z

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

4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

[ ] 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 =

I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10

1 2007.4.13. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 0. 1. 1. 2. 3. 2. ɛ-δ 1. ɛ-n

08 p Boltzmann I P ( ) principle of equal probability P ( ) g ( )g ( 0 ) (4 89) (4 88) eq II 0 g ( 0 ) 0 eq Taylor eq (4 90) g P ( ) g ( ) g ( 0

08 p. 8 4 k B log g() S() k B : Boltzmann T T S k B g g heat bath, thermal reservoir... 4. I II II System I System II II I I 0 + 0 const. (4 85) g( 0 ) g ( )g ( ) g ( )g ( 0 ) (4 86) g ( )g ( 0 ) 0 (4

4 14 4 14 4 1 1 4 1.1................................................ 4 1............................................. 4 1.3................................................ 5 1.4 1............................................

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 =

arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =

arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927

fx-260A_Users Guide_J

fx-260a http://edu.casio.jp J 1 5 2 Fl SD F0 COMP F4 DEG F5 RAD F6 GRA 3 F7 FIX F8 SCI F9 NORM COMP DEG, RAD, GRA COMP SD F0 SD SC FIX F9 SD DEG, RAD, GRA t SD COMP DEG RAD GRA COMP 23 4.5 53 23 + 4.5,

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)

4.6 (E i = ε, ε + ) T Z F Z = e βε + e β(ε+ ) = e βε (1 + e β ) F = kt log Z = kt log[e βε (1 + e β )] = ε kt ln(1 + e β ) (4.18) F (T ) S = T = k = k

4.6 (E i = ε, ε + ) T Z F Z = e ε + e (ε+ ) = e ε ( + e ) F = kt log Z = kt loge ε ( + e ) = ε kt ln( + e ) (4.8) F (T ) S = T = k = k ln( + e ) + kt e + e kt 2 + e ln( + e ) + kt (4.20) /kt T 0 = /k (4.20)

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

1 1 Gnuplot gnuplot Windows gnuplot gp443win32.zip gnuplot binary, contrib, demo, docs, license 5 BUGS, Chang

Gnuplot で微分積分 2011 年度前期 数学解析 I 講義資料 (2011.6.24) 矢崎成俊 ( 宮崎大学 ) 1 1 Gnuplot gnuplot http://www.gnuplot.info/ Windows gnuplot 2011 6 22 4.4.3 gp443win32.zip gnuplot binary, contrib, demo, docs, license 5

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5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =

( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED)

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