untitled
|
|
- しょうこ すすむ
- 5 years ago
- Views:
Transcription
1
2
3
4 1 1 Ax = b A R m m A b R m x R m A shift-and invert Lanczos -
5 LU CG
6 A = LU LU Ly = b Ux = y A LU A A = LL T 1 LU b,, Vol. 11, No. 4, pp (2006).
7 x * x (0), x (1), x (2), A Ap A #
8 x (n+1) = Cx (n) + d C d x * C d A = M N M C = M 1 N d = M 1 b 1 C - SOR ADI x (n) x (n+1) x (n)
9 A R m m A T A m 0 σ 1 (A) σ 2 (A) σ m (A) A A σ m (A) > 0 A R m m κ(a) = σ 1 (A)/σ m (A) A # r (n) = Ax (n) b x (n) #
10
11 A R m m b R m R m K n (A; b) A b n K n (A; b) = span{b, Ab, A 2 b,, A n 1 b} x (0) = 0 x (n) K n (A; b) n 1 {x (n) } # 0 K 1 (A; b) K 2 (A; b) K n (A; b) K n (A; b) x (n)
12 K n (A; b) K n+1 (A; b) K n (A; b) = K n+1 (A; b) A K n (A; b) = K n+1 (A; b) n Ax = b x * K n (A; b) A n b = c 0 b + c 1 Ab + + c n 1 A n 1 b c 1, c 2,, c n 1 c 0 = 0 A n c 0 0 c 0 A((1/c 0 )A n 1 b (c n 1 /c 0 )A n 2 b (c 1 /c 0 )b) = b x * K n (A; b)
13 K n (A; b) x (n) =d 0 b + d 1 Ab + + d n 1 A n 1 b r (n) = Ax (n) b r (n) = b + d 0 Ab + d 1 A 2 b + + d n 1 A n b = ϕ n (A)b ϕ n (z) = 1+d 0 z + d 1 z d n 1 z n 1 n K n (A; b) x (n) 1 1
14 x (n) K n (A; b) b q 1 = b / b 2 v 2 = Aq 1 v 2 q 1 q 2 v 3 = Aq 2 v 3 q 1, q 2 q 3 {v n } Arnordi
15 Arnoldi Arnoldi Arnoldi compact-wy, 2010, pp (2010).
16 Arnoldi Arnoldi Aq i q 1, q 2,, q i+1 Aq i = h 1i q 1 + h 2i q h i+1,i q i+1 m n (n+1) n AQ n = Q n+1 H n A n Arnoldi
17 Arnoldi Arnoldi AQ n = Q n+1 H n Q nt Q n n n A span{q 1, q 2,, q n } = K n (A; b) # Arnoldi
18 A Lanczos Q nt AQ n = H n H n 3 T n Anoldi h ji i j+1 0 Aq i q i q i 1 Arnoldi Lanczos
19 A Lanczos α n = h nn, β n = h n+1,n = h n,n+1 Lanczos q n q n 1 q n n
20 x (n) K n (A; b) Ritz-Galerkin Petrov-Galerkin 3
21 I r (n) = Ax (n) b x (n) K n (A; b) # e (n) = x (n) x * GMRES MINRES
22 II Ritz-Galerkin r (n) = Ax (n) b K n (A; b) x (n) Ax (n) b K n (A; b) K n (A; b) 0 K 1 (A; b) K 2 (A; b) K n (A; b) r (0) = b, r (1), r (2),, r (n 1) K n (A; b) # r (n 1) K n (A; b) n
23 II Ritz-Galerkin A φ (x) = (1/2) x T Ax x T b Ax = b φ (x) x (n) K n (A; b) x (n) = Q n y (n) y (n) R n φ (x (n) ) = (1/2) (Q n y (n) ) T AQ n y (n) (Q n y (n) ) T b y (n) Q nt (AQ n y (n) b) = 0 Ax (n) b K n (A; b) Ritz-Galerkin φ (x (n) ) x (n)
24 II Ritz-Galerkin A e (n) = x (n) x * A Ritz-Galerkin A x (n) Ritz-Galerkin CG FOM # Ritz-Galerkin
25 III Petrov-Galerkin R m L 1 L 2 L 3 L n n r (n) = Ax (n) b L n x (n) Ax (n) b L n L n 0 L n b * A T b * K n (A T ; b * ) Petrov-Galerkin Bi-CG QMR # Petrov-Galerkin Petrov-Galerkin CGS Bi-CGSTAB GPBi-CG
26 1 n P n n ϕ n (z) Ritz-Galerkin A x (n) e (n) = x (n) x * Ae (n) = Ax (n) b = r (n) min ϕ n P n ϕ n (A)b 2 e (n)t Ae (n) = r (n)t A 1 r (n) = b T ϕ n (A)A 1 ϕ n (A)b = b T A 1 ϕ n (A)Aϕ n (A)A 1 b = e (0)T ϕ n (A)Aϕ n (A)e (0) = ϕ n (A)e (0) A Ritz-Galerkin min ϕ P ϕ n (A)e (0) n n A
27
28 GMRES Generalized Minimum Residual MINRES Minimum Residual Ritz-Galerkin CG Conjugate Gradient FOM Full Orthogonalizaion Method Petrov-Galerkin Bi-CG Bi-Conjugate Gradient QMR Quasi Minimal Residual CGS Conjugate Gradient Squared Bi-CGSTAB Stabilized Bi-CG
29 GMRES K n (A; b) r (n) = Ax (n) b x (n) x (n) K n (A; b) x (n) = Q n y (n) y (n) R n 2 Arnoldi AQ n = Q n+1 H n 3 q 1 = b / b 2 4 Q n+1 Q n n+1 y (n) n y (n) 2
30 GMRES 2 2 min H n y (n) b 2 e 1 2 H n H n = U n R n U n R (n+1) n, R n R n n QR 1 R n y (n) = U nt b 2 e 1 H n (n+1) n O(n 2 ) H n 1 H n O(n) # n
31 GMRES Arnoldi K n (A; b) H n QR H n QR U n T b 2 e 1 R n y (n) = U n T b 2 e 1 x (n) = Q n y (n)
32 GMRES GMRES n # 1 n # q n q 1, q 1,, q n 1 # MINRES H n 3 1 n x (n) (κ(a)) 2 * * H. A. van der Vorst Iterative Krylov Methods for Large Linear Systems, Cambridge Univ. Press, 2003.
33 CG A r (n) = Ax (n) b K n (A; b) x (n) x (n) K n (A; b) x (n) = Q n y (n) y (n) R n Q n T (Ax (n) b) = Q n T (AQ n y (n) b) = T n y (n) Q n T b = T n y (n) e 1 = 0. Lanczos 3 T n Q n = [q 1 q 2 q n ] T n = L n U n LU x (n) y (n) = U n 1 L n 1 e 1, x (n) = Q n y (n) q 1, q 2,, q n # n
34 CG x (n) x (n) = Q n y (n) = (Q n U n 1 )(L n 1 e 1 ) [p 0 p 1 p n 1 ][v 0, v 1,, v n 1 ] T p n v n v n 1 = φ n 2 v n 2 / δ n 1
35 CG p n 1 = q n ε n 2 p n 2 2 x (n) x (n) = [p 0 p 1 p n 1 ][v 0, v 1,, v n 1 ] T = x (n 1) + v n 1 p n 1 CG
36 CG x n φ (x) = (1/2) x T Ax x T b p n 1 n p n 1 α n
37 CG CG K n (A; b) = span{b, Ab, A 2 b,, A n 1 b} = span{x 1, x 2,, x n } = span{r 0, r 1,, r n 1 } = span{p 0, p 1,, p n 1 } r nt r j = 0 (j < n) A- p nt Ap j = 0 (j < n) # p n CG A e n A = x n x * A n 1 n # 3
38 CG CG A e n A = κ A e n A P n 1 n Λ(A) A φ (λ)
39 CG CG A m * CG m * # φ (λ) m * 0 m * n = m * 0 # 1 CG Ax = b 1
40 Bi-CG CG
41 Bi-CG Breakdown r n s n 0 1 breakdown T n LU 0 breakdown 2 breakdown Bi-CG n 1 n # GMRES # 3 A A T # GMRES 2
42 Bi-CG CGS Bi-CG r n = (R n (A)) 2 b, s n = b * # s nt r n Bi-CG 2 # 2Bi-CG A A T Bi-CGSTAB S n (z) = (1 ω 0 z)(1 ω 1 z) (1 ω n 1 z) n S n (z) r n = S n (A)R(A)b, s n = b * # Bi-CG ω n CGS
43 Bi-CG GPBi-CG Bi-CGSTAB S n (z) Bi-CGSTAB QMR Bi-CG Bi-CG # Bi-CG #
44
45 K A K 1 Ax = K 1 b K 1 A A Ap K 1 Ap AK 1 y = b x = K 1 y K 1 1 AK 2 1 y = K 1 1 b x = K 2 1 y K K 1 A K K 1 x
46 LU A LU A LU K = LU ILU(0) # ILU(1) ILU(1) IC(0) # K 1 x L 1 x U 1 y # LU ILUT U 1 ILUC
47 SPAI; SParse Approximate Inverse A A 1 M M min M I AM F LU ILU D 1, D 2 A D 1 1 AD 1 2 LU
48
49 Ax = b GMRES CG Bi-CG Lis PETSc
50
51 L. N. Trefethen and D. Bau III: Numerical Linear Algebra, SIAM, Philadelphia, H. A. van der Vorst: Iterative Krylov Methods for Large Linear Systems, Cambridge University Press, Cambridge, Y. Saad: Iterative Methods for Sparse Linear Systems, PWS Publishing Company, Boston, R. Barrett et al.: Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, SIAM, Philadelphia, 1994., :,, 1996., :,, 2009.
IV (2)
COMPUTATIONAL FLUID DYNAMICS (CFD) IV (2) The Analysis of Numerical Schemes (2) 11. Iterative methods for algebraic systems Reima Iwatsu, e-mail : iwatsu@cck.dendai.ac.jp Winter Semester 2007, Graduate
More information2012年度HPCサマーセミナー_多田野.pptx
! CCS HPC! I " tadano@cs.tsukuba.ac.jp" " 1 " " " " " " " 2 3 " " Ax = b" " " 4 Ax = b" A = a 11 a 12... a 1n a 21 a 22... a 2n...... a n1 a n2... a nn, x = x 1 x 2. x n, b = b 1 b 2. b n " " 5 Gauss LU
More information橡固有値セミナー2_棚橋改.PDF
1 II. 2003 5 14 2... Arnoldi. Lanczos. Jacobi-Davidson . 3 4 Ax = x A A Ax = Mx M: M 5 Householder ln ln-1 0 l3 0 l2 l1 6 Lanczos Lanczos, 1950 Arnoldi Arnoldi, 1951 Hessenberg Jacobi-Davidson Sleijpen
More informationnumb.dvi
11 Poisson kanenko@mbkniftycom alexeikanenko@docomonejp http://wwwkanenkocom/ , u = f, ( u = u+f u t, u = f t ) 1 D R 2 L 2 (D) := {f(x,y) f(x,y) 2 dxdy < )} D D f,g L 2 (D) (f,g) := f(x,y)g(x,y)dxdy (L
More informationA11 (1993,1994) 29 A12 (1994) 29 A13 Trefethen and Bau Numerical Linear Algebra (1997) 29 A14 (1999) 30 A15 (2003) 30 A16 (2004) 30 A17 (2007) 30 A18
2013 8 29y, 2016 10 29 1 2 2 Jordan 3 21 3 3 Jordan (1) 3 31 Jordan 4 32 Jordan 4 33 Jordan 6 34 Jordan 8 35 9 4 Jordan (2) 10 41 x 11 42 x 12 43 16 44 19 441 19 442 20 443 25 45 25 5 Jordan 26 A 26 A1
More informationIDRstab(s, L) GBiCGSTAB(s, L) 2. AC-GBiCGSTAB(s, L) Ax = b (1) A R n n x R n b R n 2.1 IDR s L r k+1 r k+1 = b Ax k+1 IDR(s) r k+1 = (I ω k A)(r k dr
1 2 IDR(s) GBiCGSTAB(s, L) IDR(s) IDRstab(s, L) GBiCGSTAB(s, L) Verification of effectiveness of Auto-Correction technique applied to preconditioned iterative methods Keiichi Murakami 1 Seiji Fujino 2
More informationKrylov A04 October 8, 2010 T. Sakurai (Univ. Tsukuba) Krylov October 8, / 48
Krylov A04 October 8, 2010 T. Sakurai (Univ. Tsukuba) Krylov October 8, 2010 1 / 48 Krylov QCD, RSDFT, Shell model Block Krylov MATLAB Scilab T. Sakurai (Univ. Tsukuba) Krylov October 8, 2010 2 / 48 Krylov
More information2005 2006.2.22-1 - 1 Fig. 1 2005 2006.2.22-2 - Element-Free Galerkin Method (EFGM) Meshless Local Petrov-Galerkin Method (MLPGM) 2005 2006.2.22-3 - 2 MLS u h (x) 1 p T (x) = [1, x, y]. (1) φ(x) 0.5 φ(x)
More informationad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(
I n n A AX = I, YA = I () n XY A () X = IX = (YA)X = Y(AX) = YI = Y X Y () XY A A AB AB BA (AB)(B A ) = A(BB )A = AA = I (BA)(A B ) = B(AA )B = BB = I (AB) = B A (BA) = A B A B A = B = 5 5 A B AB BA A
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 information21 1 1 1 2 2 5 7 9 11 13 13 14 18 18 20 28 28 29 31 31 34 35 35 36 37 37 38 39 40 56 66 74 89 99 - ------ ------ -------------- ---------------- 1 10 2-2 8 5 26 ( ) 15 3 4 19 62 2,000 26 26 5 3 30 1 13
More information1 n A a 11 a 1n A =.. a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = 0 ( x 0 ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 1.1 Th
1 n A a 11 a 1n A = a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = ( x ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 11 Th9-1 Ax = λx λe n A = λ a 11 a 12 a 1n a 21 λ a 22 a n1 a n2
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 information_0212_68<5A66><4EBA><79D1>_<6821><4E86><FF08><30C8><30F3><30DC><306A><3057><FF09>.pdf
More information
( ) a C n ( R n ) R a R C n. a C n (or R n ) a 0 2. α C( R ) a C n αa = α a 3. a, b C n a + b a + b ( ) p 8..2 (p ) a = [a a n ] T C n p n a
9 8 m n mn N.J.Nigham, Accuracy and Stability of Numerical Algorithms 2nd ed., (SIAM) x x = x2 + y 2 = x + y = max( x, y ) x y x () (norm) (condition number) 8. R C a, b C a b 0 a, b a = a 0 0 0 n C n
More informationlinearal1.dvi
19 4 30 I 1 1 11 1 12 2 13 3 131 3 132 4 133 5 134 6 14 7 2 9 21 9 211 9 212 10 213 13 214 14 22 15 221 15 222 16 223 17 224 20 3 21 31 21 32 21 33 22 34 23 341 23 342 24 343 27 344 29 35 31 351 31 352
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情報処理学会研究報告 IPSJ SIG Technical Report Vol.2014-HPC-144 No /5/ CRS 2 CRS Performance evaluation of exclusive version of preconditioned ite
1 2 3 CRS 2 CRS Performance evaluation of exclusive version of preconditioned iterative method for dense matrix Abstract: As well known, only nonzero entries of a sparse matrix are stored in memory in
More information( ) 5 Reduction ( ) A M n (C) Av = λv (v 0) (11.1) λ C A (eigenvalue) v C n A λ (eigenvector) M n (R) A λ(a) A M n (R) n A λ
125 11 ( ) 5 Reduction 11.1 11.1.1 ( ) A M n (C) Av = λv (v 0) (11.1) λ C A (eigenvalue) v C n A λ (eigenvector) M n (R) A λ(a) 11.1.2 A M n (R) n A λi = 0 A C n 5 126 11 A n λ 1 (A) λ 2 (A) λ n (A) A
More informationI II III IV V
I II III IV V N/m 2 640 980 50 200 290 440 2m 50 4m 100 100 150 200 290 390 590 150 340 4m 6m 8m 100 170 250 µ = E FRVβ β N/mm 2 N/mm 2 1.1 F c t.1 3 1 1.1 1.1 2 2 2 2 F F b F s F c F t F b F s 3 3 3
More informationA
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
More informationDVIOUT-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.
More informationI , : ~/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.....................................
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
[ ]. 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> > <., 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 informationPart () () 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: : : : ) ) 1. d ij f i e i x i v j m a ij m f ij n x i =
1 1980 1) 1 2 3 19721960 1965 2) 1999 1 69 1980 1972: 55 1999: 179 2041999: 210 211 1999: 211 3 2003 1987 92 97 3) 1960 1965 1970 1985 1990 1995 4) 1. d ij f i e i x i v j m a ij m f ij n x i = n d ij
More information5 n P j j (P i,, P k, j 1) 1 n n ) φ(n) = n (1 1Pj [ ] φ φ P j j P j j = = = = = n = φ(p j j ) (P j j P j 1 j ) P j j ( 1 1 P j ) P j j ) (1 1Pj (1 1P
p P 1 n n n 1 φ(n) φ φ(1) = 1 1 n φ(n), n φ(n) = φ()φ(n) [ ] n 1 n 1 1 n 1 φ(n) φ() φ(n) 1 3 4 5 6 7 8 9 1 3 4 5 6 7 8 9 1 4 5 7 8 1 4 5 7 8 10 11 1 13 14 15 16 17 18 19 0 1 3 4 5 6 7 19 0 1 3 4 5 6 7
More information( )
7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................
More information2 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
More informationO E ( ) A a A A(a) O ( ) (1) O O () 467
1 1.0 16 1 ( 1 1 ) 1 466 1.1 1.1.1 4 O E ( ) A a A A(a) O ( ) (1) O O () 467 ( ) A(a) O A 0 a x ( ) A(3), B( ), C 1, D( 5) DB C A x 5 4 3 1 0 1 3 4 5 16 A(1), B( 3) A(a) B(b) d ( ) A(a) B(b) d AB d = d(a,
More informationy = x x R = 0. 9, R = σ $ = y x w = x y x x w = x y α ε = + β + x x x y α ε = + β + γ x + x x x x' = / x y' = y/ x y' =
y x = α + β + ε =,, ε V( ε) = E( ε ) = σ α $ $ β w ( 0) σ = w σ σ y α x ε = + β + w w w w ε / w ( w y x α β ) = α$ $ W = yw βwxw $β = W ( W) ( W)( W) w x x w x x y y = = x W y W x y x y xw = y W = w w
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 information橡魅力ある数学教材を考えよう.PDF
Web 0 2 2_1 x y f x y f f 2_2 2 1 2_3 m n AB A'B' x m n 2 1 ( ) 2_4 1883 5 6 2 2_5 2 9 10 2 1 1 1 3 3_1 2 2 2 16 2 1 0 1 2 2 4 =16 0 31 32 1 2 0 31 3_2 2 3_3 3_4 1 1 GO 3 3_5 2 5 9 A 2 6 10 B 3 7 11 C
More informationさくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a
φ + 5 2 φ : φ [ ] a [ ] a : b a b b(a + b) b a 2 a 2 b(a + b). b 2 ( a b ) 2 a b + a/b X 2 X 0 a/b > 0 2 a b + 5 2 φ φ : 2 5 5 [ ] [ ] x x x : x : x x : x x : x x 2 x 2 x 0 x ± 5 2 x x φ : φ 2 : φ ( )
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 information1 n 1 1 2 2 3 3 3.1............................ 3 3.2............................. 6 3.2.1.............. 6 3.2.2................. 7 3.2.3........................... 10 4 11 4.1..........................
More information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... (
More informationII 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 informationLINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2002 2 2 2 2 22 2 3 3 3 3 3 4 4 5 5 6 6 7 7 8 8 9 Cramer 9 0 0 E-mail:hsuzuki@icuacjp 0 3x + y + 2z 4 x + y
More information20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33
More information1 Abstract 2 3 n a ax 2 + bx + c = 0 (a 0) (1) ( x + b ) 2 = b2 4ac 2a 4a 2 D = b 2 4ac > 0 (1) 2 D = 0 D < 0 x + b 2a = ± b2 4ac 2a b ± b 2
1 Abstract n 1 1.1 a ax + bx + c = 0 (a 0) (1) ( x + b ) = b 4ac a 4a D = b 4ac > 0 (1) D = 0 D < 0 x + b a = ± b 4ac a b ± b 4ac a b a b ± 4ac b i a D (1) ax + bx + c D 0 () () (015 8 1 ) 1. D = b 4ac
More informationuntitled
- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =
More information(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
More informationNote.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 information1990 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(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
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 informationmain.dvi
3 4 2 3 4 2 3 ( ) ( ) 1 1.1 R n R n n Euclid R n R n f =(f 1 ;f 2 ;:::;f n ) T : f(x) =0 (1) x = (x 1 ;x 2 ;:::;x n ) T 2 R n 1 \T " f i (1 i n) R n R (1) f i (x) =0 (1 i n) n 2 n x =cosx f(x) :=x 0 cos
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 information1 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 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 informationx, 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
More information1 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
More information?
240-8501 79-2 Email: nakamoto@ynu.ac.jp 1 3 1.1...................................... 3 1.2?................................. 6 1.3..................................... 8 1.4.......................................
More information2 (March 13, 2010) N Λ a = i,j=1 x i ( d (a) i,j x j ), Λ h = N i,j=1 x i ( d (h) i,j x j ) B a B h B a = N i,j=1 ν i d (a) i,j, B h = x j N i,j=1 ν i
1. A. M. Turing [18] 60 Turing A. Gierer H. Meinhardt [1] : (GM) ) a t = D a a xx µa + ρ (c a2 h + ρ 0 (0 < x < l, t > 0) h t = D h h xx νh + c ρ a 2 (0 < x < l, t > 0) a x = h x = 0 (x = 0, l) a = a(x,
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 informationLCM,GCD LCM GCD..,.. 1 LCM GCD a b a b. a divides b. a b. a, b :, CD(a, b) = {d a, b }, CM(a, b) = {m a, b }... CM(a, b). q > 0, m 1, m 2 CM
LCM,GCD 2017 4 21 LCM GCD..,.. 1 LCM GCD a b a b. a divides b. a b. a, b :, CD(a, b) = {d a, b }, CM(a, b) = {m a, b }... CM(a, b). q > 0, m 1, m 2 CM(a, b) = m 1 + m 2 CM(a, b), qm 1 CM(a, b) m 1, m 2
More information1 4 1 ( ) ( ) ( ) ( ) () 1 4 2
7 1995, 2017 7 21 1 2 2 3 3 4 4 6 (1).................................... 6 (2)..................................... 6 (3) t................. 9 5 11 (1)......................................... 11 (2)
More information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
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 information1 Introduction 1 (1) (2) (3) () {f n (x)} n=1 [a, b] K > 0 n, x f n (x) K < ( ) x [a, b] lim f n (x) f(x) (1) f(x)? (2) () f(x)? b lim a f n (x)dx = b
1 Introduction 2 2.1 2.2 2.3 3 3.1 3.2 σ- 4 4.1 4.2 5 5.1 5.2 5.3 6 7 8. Fubini,,. 1 1 Introduction 1 (1) (2) (3) () {f n (x)} n=1 [a, b] K > 0 n, x f n (x) K < ( ) x [a, b] lim f n (x) f(x) (1) f(x)?
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 informationnsg02-13/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 informationmobius1
H + : ω = ( a, b, c, d, ad bc > 0) 3.. ( c 0 )... ( 5z + 2 : ω = L (*) z + 4 5z + 2 z = z =, 2. (*) z + 4 5z+ 2 6( z+) ω + = + = z+ 4 z+ 4 5z+ 2 3( z 2) ω 2 = 2= z+ 4 z+ 4 ω + z + = 2 ω 2 z 2 x + T ( x)
More informationTricorn
Triorn 016 3 1 Mandelbrot Triorn Mandelbrot Robert L DevaneyAn introdution to haoti dynamial Systems Addison-Wesley, 1989 Triorn 1 W.D.Crowe, R.Hasson, P.J.Rippon, P.E.D.Strain- Clark, On the struture
More informationnakata/nakata.html p.1/20
http://www.me.titech.ac.jp/ nakata/nakata.html p.1/20 1-(a). Faybusovich(1997) Linear systems in Jordan algebras and primal-dual interior-point algorithms,, Euclid Jordan p.2/20 Euclid Jordan V Euclid
More informationn (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz
1 2 (a 1, a 2, a n ) (b 1, b 2, b n ) A (1.1) A = a 1 b 1 + a 2 b 2 + + a n b n (1.1) n A = a i b i (1.2) i=1 n i 1 n i=1 a i b i n i=1 A = a i b i (1.3) (1.3) (1.3) (1.1) (ummation convention) a 11 x
More information, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )
81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,.,
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 informationx x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {
K E N Z OU 2008 8. 4x 2x 2 2 2 x + x 2. x 2 2x 2, 2 2 d 2 x 2 2.2 2 3x 2... d 2 x 2 5 + 6x 0 2 2 d 2 x 2 + P t + P 2tx Qx x x, x 2 2 2 x 2 P 2 tx P tx 2 + Qx x, x 2. d x 4 2 x 2 x x 2.3 x x x 2, A 4 2
More information(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)
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 information1. 1 BASIC PC BASIC BASIC BASIC Fortran WS PC (1.3) 1 + x 1 x = x = (1.1) 1 + x = (1.2) 1 + x 1 = (1.
Section Title Pages Id 1 3 7239 2 4 7239 3 10 7239 4 8 7244 5 13 7276 6 14 7338 7 8 7338 8 7 7445 9 11 7580 10 10 7590 11 8 7580 12 6 7395 13 z 11 7746 14 13 7753 15 7 7859 16 8 7942 17 8 Id URL http://km.int.oyo.co.jp/showdocumentdetailspage.jsp?documentid=
More information第8章 位相最適化問題
8 February 25, 2009 1 (topology optimizaiton problem) ( ) H 1 2 2.1 ( ) V S χ S : V {0, 1} S (characteristic function) (indicator function) 1 (x S ) χ S (x) = 0 (x S ) 2.1 ( ) Lipschitz D R d χ Ω (Ω D)
More information福岡大学人文論叢47-3
679 pp. 1 680 2 681 pp. 3 682 4 683 5 684 pp. 6 685 7 686 8 687 9 688 pp. b 10 689 11 690 12 691 13 692 pp. 14 693 15 694 a b 16 695 a b 17 696 a 18 697 B 19 698 A B B B A B B A A 20 699 pp. 21 700 pp.
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 information6 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 informationn ( (
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
More information8.1 Fubini 8.2 Fubini 9 (0%) 10 (50%) Carathéodory 10.3 Fubini 1 Introduction 1 (1) (2) {f n (x)} n=1 [a, b] K > 0 n, x f n (x) K < ( ) x [a
% 100% 1 Introduction 2 (100%) 2.1 2.2 2.3 3 (100%) 3.1 3.2 σ- 4 (100%) 4.1 4.2 5 (100%) 5.1 5.2 5.3 6 (100%) 7 (40%) 8 Fubini (90%) 2007.11.5 1 8.1 Fubini 8.2 Fubini 9 (0%) 10 (50%) 10.1 10.2 Carathéodory
More information() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.
() 6 f(x) [, b] 6. Riemnn [, b] f(x) S f(x) [, b] (Riemnn) = x 0 < x < x < < x n = b. I = [, b] = {x,, x n } mx(x i x i ) =. i [x i, x i ] ξ i n (f) = f(ξ i )(x i x i ) i=. (ξ i ) (f) 0( ), ξ i, S, ε >
More information1 α X (path) α I = [0, 1] X α(0) = α(1) = p α p (base point) loop α(1) = β(0) X α, β α β : I X (α β)(s) = ( )α β { α(2s) (0 s 1 2 ) β(2s 1) ( 1 2 s 1)
1 α X (path) α I = [0, 1] X α(0) = α(1) = p α p (base point) loop α(1) = β(0) X α, β α β : I X (α β)(s) = ( )α β { α(2s) (0 s 1 2 ) β(2s 1) ( 1 2 s 1) X α α 1 : I X α 1 (s) = α(1 s) ( )α 1 1.1 X p X Ω(p)
More informationuntitled
A = QΛQ T A n n Λ Q A = XΛX 1 A n n Λ X GPGPU A 3 T Q T AQ = T (Q: ) T u i = λ i u i T {λ i } {u i } QR MR 3 v i = Q u i A {v i } A n = 9000 Quad Core Xeon 2 LAPACK (4/3) n 3 O(n 2 ) O(n 3 ) A {v i }
More informationIA hara@math.kyushu-u.ac.jp Last updated: January,......................................................................................................................................................................................
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 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 information211 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,
More informationa n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552
3 3.0 a n a n ( ) () a m a n = a m+n () (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 55 3. (n ) a n n a n a n 3 4 = 8 8 3 ( 3) 4 = 8 3 8 ( ) ( ) 3 = 8 8 ( ) 3 n n 4 n n
More information- 2 -
- 2 - - 3 - (1) (2) (3) (1) - 4 - ~ - 5 - (2) - 6 - (1) (1) - 7 - - 8 - (i) (ii) (iii) (ii) (iii) (ii) 10 - 9 - (3) - 10 - (3) - 11 - - 12 - (1) - 13 - - 14 - (2) - 15 - - 16 - (3) - 17 - - 18 - (4) -
More information2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4 4 4 2 5 5 2 4 4 4 0 3 3 0 9 10 10 9 1 1
1 1979 6 24 3 4 4 4 4 3 4 4 2 3 4 4 6 0 0 6 2 4 4 4 3 0 0 3 3 3 4 3 2 4 3? 4 3 4 3 4 4 4 4 3 3 4 4 4 4 2 1 1 2 15 4 4 15 0 1 2 1980 8 4 4 4 4 4 3 4 2 4 4 2 4 6 0 0 6 4 2 4 1 2 2 1 4 4 4 2 3 3 3 4 3 4 4
More information