4 (induction) (mathematical induction) P P(0) P(x) P(x+1) n P(n) 4.1 (inductive definition) A A (basis ) ( ) A (induction step ) A A (closure ) A clos

Size: px
Start display at page:

Download "4 (induction) (mathematical induction) P P(0) P(x) P(x+1) n P(n) 4.1 (inductive definition) A A (basis ) ( ) A (induction step ) A A (closure ) A clos"

Transcription

1 4 (induction) (mathematical induction) P P(0) P(x) P(x+1) n P(n) 4.1 (inductive definition) A A (basis ) ( ) A (induction step ) A A (closure ) A closure 81 3 A 3 A x A x + A A ( A. ) 3 closure A N 1, closure 8 N. 40

2 0 N. n N n + 1 N 83 A = 1, 3, 7, 1, 31,...} 1 A. x A x + 1 A (list) (sequence) ( ) : A A. A List A List A = x 1,..., x n n 0 i.(1 i n x i A)} :, 1,,, 3,, 1 0 nil cons, head, tail. cons(x, x 1,..., x n ) = x, x 1,..., x n head( x 1,..., x n ) = x 1 tail( x 1,..., x n ) = x,..., x n head tail (List A ) 8 head( a, b, c ) = a tail( a, b, c ) = b, c cons(a, ) = a cons(a, b, c ) = a, b, c x A L List A x = head(cons(x, L)) L = tail(cons(x, L)). L L = cons(head(l), tail(l)). cons cons : A A List A List A 41

3 x A L List A cons(x, L) List A cons(x, L) :: x :: L a :: (b :: (c :: ))) = cons(a, cons(b, cons(c, ))) = cons(a, cons(b, c )) = cons(a, b, c ) = a, b, c S 0 S. 1 S. L S head(l) = 0 cons(1, L) S. L S head(l) = 1 cons(0, L) S S = 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1,...} 1,, 3 Lisp Scheme : (1 3) ML : [1,, 3] 4.1. (string) abc 1 0 Λ. ( ) Σ Σ Σ ( ). 87 a, b}. Λ, a, b, aa, ab, ba, bb, Σ Σ (Σ 1) Λ Σ x Σ s Σ xs Σ (Σ ) 1 abc 4

4 Λ Σ x Σ s Σ sx Σ (Σ 3) Λ Σ x Σ x Σ s Σ t Σ st Σ 3 1 Σ bab a, b} Λ a, b}. b a, b} b a, b}. a a, b} ab a, b}. b a, b} bab a, b}. bab a, b} 3 s, t, u Σ 3 stu ( ) s Σ t Σ st Σ u Σ stu Σ t Σ u Σ tu Σ s Σ stu Σ (ab)c a(bc) 88 A = 0, 1} 0 L 0 L. s L 1s L. 89 S = a, b, ab, ba, aab, bba, aaab, bbba,...} a S, b S. s S bs S as S as S bs S (s = a ) (s = b ) (s a strhead(s) = a ) (s b strhead(s) = b ) strhead ( ) A B C (A B) C 43

5 4.1.3 (binary Tree) ( ) 3 [ BinTree] BinTree (L BinTree R BinTree) L, R BinTree. L BinTree L BinTree. 1 ( 0 1 ) L R L, R L R L, R L, R} L, R L R ( ) A ( ) A 1 BinTree A (x A L BinTree A R BinTree A ) L, x, R BinTree A. x L R. (x A L BinTree A ) L, x BinTree A BinTree N ( ), 1,,,, 1,,, 1, 44

6 L, x, R Tree(L, x, R) L, x Tree(L, x). 91 Tree(Tree(,, ), 1, ) Tree(Tree(, ), 1, ) head, tail. label(tree(t 1, x, t )) = x label(tree(t 1, x)) = x left(tree(t 1, x, t )) = t 1 left(tree(t 1, x)) = t 1 right(tree(t 1, x, t )) = t right 1 9 BinTree A T A T A. x A t T A Tree(t, x, t) T A ( ) 93 0, 1} 0, 1 Opps.,, 0,,, 1, Opps. x 0, 1} t Opps Tree(t, x, Tree(right(t), 1, left(t))) Opps, if label(t) = 0 Tree(t, x, Tree(right(t), 0, left(t))) Opps, if label(t) = ( ) 4

7 4.1.4 BNF ( ) BNF (Backus-Naur Form Backus Normal Form) 1 BNF BNF 94 [ ] a, b, c,, z} 0, 1,,, 9} BNF (expression) ::= + - (basis induction step) BNF x y x+y x y x-y abc+31-0 BNF 1 ::= := begin end while do if then else ::= ; while (statement) x:=3; y:=0; while x do begin y:=y+3; x:=x-1 end 4. head, tail ( ) 4 [ f : A B A ] 46

8 (basis) A basis x f(x) B (induction step) A induction step a A b A f(a) f(b) B A f f : A B N f : N B f 0 N f(0) n N f(n) f(n + 1) (n = 0 ) f(n) = f(m) (n = m + 1 ) n = m + 1 f(m) f ( ) f(n + 1) 0 (n = 0 ) f(n) = f(n + 1) + 3 (n = m + 1 ) 3 9 f : N N, n N 1 n + 1 f(n) n 1 f(n) = ((n 1) + 1) + (n + 1) = f(n 1) + (n + 1) f 1 (n = 0 ) f(n) = f(m) + n + 1 (n = m + 1 ) 96 add : N N N. N m (n = 0 ) add(n, m) = add(p, m) + 1 (n = p + 1 ) add(n, m) = n add(n, p) + 1 (m = 0 ) (m = p + 1 ) ( ) 3 f(n) = f(n + 1) f f (recursive function) (inductively defined function) f 47

9 97 times : N N N. 1 0 (n = 0 ) times(n, m) = times(p, m) + m (n = p + 1 ) 98 factorial(n). factorial(n) = 1 (n = 0 ) n factorial(m) (n = m + 1 ) 99 f : N List N f(n) = n,..., 1, 0 0 (n = 0 ) f(n) = cons(n, f(p)) (n = p + 1 ) f(3) = cons(3, f()) = cons(3, cons(, f(1))) = cons(3, cons(, cons(1, f(0)))) = cons(3, cons(, cons(1, 0 ))) = 3,, 1, 0 n = 0 n 1 n = 0, 1,, k n > k f(n) m < n f(m) 100 Fibonacci fib(n) 1 (n = 1, ) fib(n) = fib(n 1) + fib(n ) (n > ) 101 head; List A A, tail; List A List A. (x = ) head(x) = y (x = cons(y, L) ) (x = ) tail(x) = L (x = cons(y, L) ) 10 length : List A N. 0 (x = ) length(x) = 1 + length(l) (x = cons(y, L) ) 1 + length(tail(x)) 103 (concatenation) : List A List A List A. y (x = ) x y = cons(z, L y) (x = cons(z, L) ) 48

10 104 nodes : BinTree A N. 0 (x = ) nodes(x) = 1 + nodes(l) + nodes(r) (x = Tree(L, y, R) ) 10 : Σ Σ Σ. Σ 1 t (s = Λ ) s t = x (u t) (s = xu ) s, t, u Σ x Σ 1 x Σ u Σ s = xu s s (t = Λ ) s t = (s u) x (t = ux ) 4.3 A A P (induction) x, P (x) (base case) P (0). (induction step) x N P (x) 4 P (x + 1) n ( ) n(n+1)(n+1) 6 : P (n) P (0) 0 = P (n) 0 n + 1 n(n + 1)(n + 1) 6 + (n + 1) = (n + 1)(n + )(n + 3) 6 P (n + 1) n N P (n) n N.P (n) 4 (induction hypothesis) 49

11 n 0 : n k P (n) base case P (k) induction step n > k n P (n) P (n + 1) n : P (n) base case n = 1 induction step P (k + 1) P (k + 3) : induction step P (n 1) 0 k < n k P (k) P (n) 107 fib n > 0 (( ) n ( fib(n) = g(n) g(n) = 1 (( 1 + ) n ( 1 ) n ) 1 ) n ) n n.(n > 0 fib(n) = g(n)). n = 1 n = g(1) = 1 (( g() = 1 (( 1 + ) ( 6 + ) ( 4 1 )) = 1 = fib(1) 6 )) = 1 = fib() 4 n > k < n k fib(k) = g(k). n n 1 k = n k = n 1 fib(n) = fib(n ) + fib(n 1) = g(n ) + g(n 1) ( = ) ( 1 + ) n ( (( = ) n ( 1 ) n ) = g(n) 3 ) ( 1 ) n 0

12 4.3. L List A, P (L). P ( ). L List A a A P (L) P (a::l). 108 x, y List A length(x y) = length(x) + length(y) ( ) x x List A y List A.length(x y) = length(x) + length(y) P (x) (base case) x =. y List A P (x) length( y) = length(y) = length( ) + length(y) (induction step) x = a :: z. y List A ( ) length((a :: z) y) = length(a :: (z y)) ( ) = length(z y) + 1 (length ) = length(z) + length(y) + 1 ( P (z) ) ( ) length(a :: z) + length(y) = length(z) length(y) (length ) P (a :: z) x List A P (x) 6 x y x y 6 1

13 4.3.3 ( ) x BinTree A, P (x) P ( ). x, y BinTree A a A P (x), P (y) P (Tree(x, a, y)). (structural induction)

1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 5 3. 4. 5. A0 (1) A, B A B f K K A ϕ 1, ϕ 2 f ϕ 1 = f ϕ 2 ϕ 1 = ϕ 2 (2) N A 1, A 2, A 3,... N A n X N n X N, A n N n=1 1 A1 d (d 2) A (, k A k = O), A O. f

More information

1. A0 A B A0 A : A1,...,A5 B : B1,...,B

1. A0 A B A0 A : A1,...,A5 B : B1,...,B 1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A B f : A B 4 (i) f (ii) f (iii) C 2 g, h: C A f g = f h g = h (iv) C 2 g, h: B C g f = h f g = h 4 (1) (i) (iii) (2) (iii) (i) (3) (ii) (iv) (4)

More information

A, B, C. (1) A = A. (2) A = B B = A. (3) A = B, B = C A = C. A = B. (3)., f : A B g : B C. g f : A C, A = C. 7.1, A, B,. A = B, A, A A., A, A

A, B, C. (1) A = A. (2) A = B B = A. (3) A = B, B = C A = C. A = B. (3)., f : A B g : B C. g f : A C, A = C. 7.1, A, B,. A = B, A, A A., A, A 91 7,.,, ( ).,,.,.,. 7.1 A B, A B, A = B. 1), 1,.,. 7.1 A, B, 3. (i) A B. (ii) f : A B. (iii) A B. (i) (ii)., 6.9, (ii) (iii).,,,. 1), Ā = B.. A, Ā, Ā,. 92 7 7.2 A, B, C. (1) A = A. (2) A = B B = A. (3)

More information

1. A0 A B A0 A : A1,...,A5 B : B1,...,B

1. A0 A B A0 A : A1,...,A5 B : B1,...,B 1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A, B Z Z m, n Z m n m, n A m, n B m=n (1) A, B (2) A B = A B = Z/ π : Z Z/ (3) A B Z/ (4) Z/ A, B (5) f : Z Z f(n) = n f = g π g : Z/ Z A, B (6)

More information

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

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

More information

NOTE P, A,. A P ( A, P ),,.,. P A. (set) (set), (). (element), (element).,.,. ( A, B, X, Y, P ), { } (),..3 (union) A, B,A B, A B (union),. A B = {x x

NOTE P, A,. A P ( A, P ),,.,. P A. (set) (set), (). (element), (element).,.,. ( A, B, X, Y, P ), { } (),..3 (union) A, B,A B, A B (union),. A B = {x x 2. (set)............... 2.2,.... 2.3 (union)............ 2.4 (intersection)......... 2.5 (power set)......... 3.6 (cartesian product set)... 3 2, 3 2. (length)........ 3 2.2 (epsilon)............ 3 2.3

More information

II Time-stamp: <05/09/30 17:14:06 waki> ii

II Time-stamp: <05/09/30 17:14:06 waki> ii II waki@cc.hirosaki-u.ac.jp 18 1 30 II Time-stamp: ii 1 1 1.1.................................................. 1 1.2................................................... 3 1.3..................................................

More information

Chap9.dvi

Chap9.dvi .,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim

More information

2

2 2011.11.11 1 2 MapReduce 3 4 5 6 Functional Functional Programming 7 8 9 10 11 12 13 [10, 20, 30, 40, 50] 0 n n 10 * 0 + 20 * 1 + 30 * 2 + 40 * 3 + 50 *4 = 400 14 10 * 0 + 20 * 1 + 30 * 2 + 40 * 3 + 50

More information

2/ 土 :30 11:20 似通った科目名がありますので注意してください. 受験許可されていない科目を解答した場合は無効 整理番号と科目コードは受験許可証とよく照合し正確に記入

2/ 土 :30 11:20 似通った科目名がありますので注意してください. 受験許可されていない科目を解答した場合は無効 整理番号と科目コードは受験許可証とよく照合し正確に記入 2/ 土 28 9 10 10:30 11:20 似通った科目名がありますので注意してください. 受験許可されていない科目を解答した場合は無効 整理番号と科目コードは受験許可証とよく照合し正確に記入 30 10 11 12 00011 00016 01101 02607 02703 (1) AB AB 100 cm 2 3.00 cm 2 9.80 m/s 2 AB A B A 10.0 kg A

More information

II

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

More information

:30 12:00 I. I VI II. III. IV. a d V. VI

:30 12:00 I. I VI II. III. IV. a d V. VI 2017 2017 08 03 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF X [ S ] a S S ; X X X, S [, a, ], ; BNF X (parse tree) (1) [a;a] (2) [[a]] (3) [a;[a]] (4) [[a];a] : [a] X 2 222222

More information

ohp02.dvi

ohp02.dvi 172 2017.7.16 1 ? X A B A X B ( )? IBMPL/I FACOM PL1 ( ) X 2 ( ) 3 2-0 ( ) ( ) ( ) (12) ( ) (112) 31) 281 26 1 4 (syntax) (semantics) ( ) 5 BNF BNF(Backus Normal Form) Joun Backus (grammer) English grammer

More information

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 = 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 information

Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1) (2) ( ) BASIC BAS

Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1) (2) ( ) BASIC BAS Appendix A BASIC BASIC Beginner s All-purpose Symbolic Instruction Code FORTRAN COBOL C JAVA PASCAL (NEC N88-BASIC Windows BASIC (1 (2 ( BASIC BASIC download TUTORIAL.PDF http://hp.vector.co.jp/authors/va008683/

More information

「計算と論理」 Software Foundations その3

「計算と論理」  Software Foundations   その3 Software Foundations 3 cal17@fos.kuis.kyoto-u.ac.jp October 24, 2017 ( ) ( 3) October 24, 2017 1 / 47 Lists.v ( ) ( ) ( ) ( 3) October 24, 2017 2 / 47 ( ) Inductive natprod : Type := pair : nat nat natprod.

More information

( )

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

More information

# let st1 = {name = "Taro Yamada"; id = };; val st1 : student = {name="taro Yamada"; id=123456} { 1 = 1 ;...; n = n } # let string_of_student {n

# let st1 = {name = Taro Yamada; id = };; val st1 : student = {name=taro Yamada; id=123456} { 1 = 1 ;...; n = n } # let string_of_student {n II 6 / : 2001 11 21 (OCaml ) 1 (field) name id type # type student = {name : string; id : int};; type student = { name : string; id : int; } student {} type = { 1 : 1 ;...; n : n } { 1 = 1 ;...; n = n

More information

org/ghc/ Windows Linux RPM 3.2 GHCi GHC gcc javac ghc GHCi(ghci) GHCi Prelude> GHCi :load file :l file :also file :a file :reload :r :type expr :t exp

org/ghc/ Windows Linux RPM 3.2 GHCi GHC gcc javac ghc GHCi(ghci) GHCi Prelude> GHCi :load file :l file :also file :a file :reload :r :type expr :t exp 3 Haskell Haskell Haskell 1. 2. 3. 4. 5. 1. 2. 3. 4. 5. 6. C Java 3.1 Haskell Haskell GHC (Glasgow Haskell Compiler 1 ) GHC Haskell GHC http://www.haskell. 1 Guarded Horn Clauses III - 1 org/ghc/ Windows

More information

「計算と論理」 Software Foundations その4

「計算と論理」  Software Foundations   その4 Software Foundations 4 cal17@fos.kuis.kyoto-u.ac.jp http://www.fos.kuis.kyoto-u.ac.jp/~igarashi/class/cal/ November 7, 2017 ( ) ( 4) November 7, 2017 1 / 51 Poly.v ( ) ( 4) November 7, 2017 2 / 51 : (

More information

³ÎΨÏÀ

³ÎΨÏÀ 2017 12 12 Makoto Nakashima 2017 12 12 1 / 22 2.1. C, D π- C, D. A 1, A 2 C A 1 A 2 C A 3, A 4 D A 1 A 2 D Makoto Nakashima 2017 12 12 2 / 22 . (,, L p - ). Makoto Nakashima 2017 12 12 3 / 22 . (,, L p

More information

r02.dvi

r02.dvi 172 2017.7.16 1 1.1? X A B A X B ( )? IBMPL/I FACOM PL1 ( ) X ( ) 1.2 1 2-0 ( ) ( ) ( ) (12) ( ) (112) (131) 281 26 1 (syntax) (semantics) ( ) 2 2.1 BNF BNF(Backus Normal Form) Joun Backus (grammer) English

More information

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. 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),

More information

ad 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(

ad 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

:30 12:00 I. I VI II. III. IV. a d V. VI

:30 12:00 I. I VI II. III. IV. a d V. VI 2018 2018 08 02 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF N N y N x N xy yx : yxxyxy N N x, y N (parse tree) (1) yxyyx (2) xyxyxy (3) yxxyxyy (4) yxxxyxxy N y N x N yx

More information

( ) ( ) 1729 (, 2016:17) = = (1) 1 1

( ) ( ) 1729 (, 2016:17) = = (1) 1 1 1729 1 2016 10 28 1 1729 1111 1111 1729 (1887 1920) (1877 1947) 1729 (, 2016:17) 12 3 1728 9 3 729 1729 = 12 3 + 1 3 = 10 3 + 9 3 (1) 1 1 2 1729 1729 19 13 7 = 1729 = 12 3 + 1 3 = 10 3 + 9 3 13 7 = 91

More information

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 =

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 =

More information

N88 BASIC 0.3 C: My Documents 0.6: 0.3: (R) (G) : enterreturn : (F) BA- SIC.bas 0.8: (V) 0.9: 0.5:

N88 BASIC 0.3 C: My Documents 0.6: 0.3: (R) (G) : enterreturn : (F) BA- SIC.bas 0.8: (V) 0.9: 0.5: BASIC 20 4 10 0 N88 Basic 1 0.0 N88 Basic..................................... 1 0.1............................................... 3 1 4 2 5 3 6 4 7 5 10 6 13 7 14 0 N88 Basic 0.0 N88 Basic 0.1: N88Basic

More information

iii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................

More information

ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University

ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2004 1 1 1 2 2 1 3 3 1 4 4 1 5 5 1 6 6 1 7 7 1 8 8 1 9 9 1 10 10 1 E-mail:hsuzuki@icu.ac.jp 0 0 1 1.1 G G1 G a, b,

More information

¥¢¥ë¥´¥ê¥º¥à¥¤¥ó¥È¥í¥À¥¯¥·¥ç¥ó ÎØ¹Ö #1

¥¢¥ë¥´¥ê¥º¥à¥¤¥ó¥È¥í¥À¥¯¥·¥ç¥ó ÎØ¹Ö #1 #1 id:motemen August 27, 2008 id:motemen 1-3 1-5 6-9 10-14 1 2 : n < a 1, a 2,..., a n > a 1 a 2 a n < a 1, a 2,..., a n > : Google: insertion sort site:youtube.com 1 : procedure Insertion-Sort(A) for

More information

つくって学ぶプログラミング言語 RubyによるScheme処理系の実装

つくって学ぶプログラミング言語 RubyによるScheme処理系の実装 Ruby Scheme 2013-04-16 ( )! SICP *1 ;-) SchemeR SICP MIT * 1 Structure and Interpretaion of Computer Programs 2nd ed.: 2 i SchemeR Ruby Ruby Ruby Ruby & 3.0 Ruby ii https://github.com/ichusrlocalbin/scheme_in_ruby

More information

5 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

5 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

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

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

More information

25 II :30 16:00 (1),. Do not open this problem booklet until the start of the examination is announced. (2) 3.. Answer the following 3 proble

25 II :30 16:00 (1),. Do not open this problem booklet until the start of the examination is announced. (2) 3.. Answer the following 3 proble 25 II 25 2 6 13:30 16:00 (1),. Do not open this problem boolet until the start of the examination is announced. (2) 3.. Answer the following 3 problems. Use the designated answer sheet for each problem.

More information

ケミカルエンジニアのためのExcelを用いた化学工学計算法

ケミカルエンジニアのためのExcelを用いた化学工学計算法 VBA 7.1 f ()= ( f ( )) y = f ()(1) y = f ( )( ) + f ( ) (1) 1 1 1 = f ( ) f ( ) (2) 1 n = = y = f() y = () 1 n+1 = n (f( n )f( n )) 1.7 1 2 log()2 145 7.2 f ( ) 1-1 -1 f ( ) (3) (4) 34 5 1 1) ( ) ( ) '(

More information

2 1/2 1/4 x 1 x 2 x 1, x 2 9 3x 1 + 2x 2 9 (1.1) 1/3 RDA 1 15 x /4 RDA 1 6 x /6 1 x 1 3 x 2 15 x (1.2) (1.3) (1.4) 1 2 (1.5) x 1

2 1/2 1/4 x 1 x 2 x 1, x 2 9 3x 1 + 2x 2 9 (1.1) 1/3 RDA 1 15 x /4 RDA 1 6 x /6 1 x 1 3 x 2 15 x (1.2) (1.3) (1.4) 1 2 (1.5) x 1 1 1 [1] 1.1 1.1. TS 9 1/3 RDA 1/4 RDA 1 1/2 1/4 50 65 3 2 1/15 RDA 2/15 RDA 1/6 RDA 1 1/6 1 1960 2 1/2 1/4 x 1 x 2 x 1, x 2 9 3x 1 + 2x 2 9 (1.1) 1/3 RDA 1 15 x 1 + 2 1/4 RDA 1 6 x 1 1 4 1 1/6 1 x 1 3

More information

1 1 [1] ( 2,625 [2] ( 2, ( ) /

1 1 [1] ( 2,625 [2] ( 2, ( ) / [] (,65 [] (,3 ( ) 67 84 76 7 8 6 7 65 68 7 75 73 68 7 73 7 7 59 67 68 65 75 56 6 58 /=45 /=45 6 65 63 3 4 3/=36 4/=8 66 7 68 7 7/=38 /=5 7 75 73 8 9 8/=364 9/=864 76 8 78 /=45 /=99 8 85 83 /=9 /= ( )

More information

Microsoft PowerPoint L03-Syntex and Semantics-1-students ( )

Microsoft PowerPoint L03-Syntex and Semantics-1-students ( ) プログラミング言語論 A (Concepts on Programming Languages) 趙建軍 (Jianjun Zhao) http://stap.ait.kyushu-u.ac.jp/~zhao/course/2018/concepts of Programming Languages.html 1 第 3 回 構文と意味 (1) (Syntax and Semantics) 2017.04.26

More information

( 9 1 ) 1 2 1.1................................... 2 1.2................................................. 3 1.3............................................... 4 1.4...........................................

More information

入試の軌跡

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

More information

i

i i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,

More information

r3.dvi

r3.dvi 2012 3 / Lisp(2) 2012.4.19 1 Lisp 1.1 Lisp Lisp (1) (setq) (2) (3) setq defun (defun (... &aux...)...) ( ) ( nil ) [1]> (defun sisoku (x y &aux wa sa sho seki) (setq wa (+ x y)) (setq sa (- x y)) (setq

More information

Akito Tsuboi June 22, T ϕ T M M ϕ M M ϕ T ϕ 2 Definition 1 X, Y, Z,... 1

Akito Tsuboi June 22, T ϕ T M M ϕ M M ϕ T ϕ 2 Definition 1 X, Y, Z,... 1 Akito Tsuboi June 22, 2006 1 T ϕ T M M ϕ M M ϕ T ϕ 2 Definition 1 X, Y, Z,... 1 1. X, Y, Z,... 2. A, B (A), (A) (B), (A) (B), (A) (B) Exercise 2 1. (X) (Y ) 2. ((X) (Y )) (Z) 3. (((X) (Y )) (Z)) Exercise

More information

2012 A, N, Z, Q, R, C

2012 A, N, Z, Q, R, C 2012 A, N, Z, Q, R, C 1 2009 9 2 2011 2 3 2012 9 1 2 2 5 3 11 4 16 5 22 6 25 7 29 8 32 1 1 1.1 3 1 1 1 1 1 1? 3 3 3 3 3 3 3 1 1, 1 1 + 1 1 1+1 2 2 1 2+1 3 2 N 1.2 N (i) 2 a b a 1 b a < b a b b a a b (ii)

More information

r3.dvi

r3.dvi / 94 2 (Lisp ) 3 ( ) 1994.5.16,1994.6.15 1 cons cons 2 >(cons a b) (A. B).? Lisp (S ) cons 2 car cdr n A B C D nil = (A B C D) nil nil A D E = (A (B C) D E) B C E = (A B C D. E) A B C D B = (A. B) A nil.

More information

) 9 81

) 9 81 4 4.0 2000 ) 9 81 10 4.1 natural numbers 1, 2, 3, 4, 4.2, 3, 2, 1, 0, 1, 2, 3, integral numbers integers 1, 2, 3,, 3, 2, 1 1 4.3 4.3.1 ( ) m, n m 0 n m 82 rational numbers m 1 ( ) 3 = 3 1 4.3.2 3 5 = 2

More information

Microsoft PowerPoint - IntroAlgDs-06-8.ppt

Microsoft PowerPoint - IntroAlgDs-06-8.ppt アルゴリズムとデータ構造入門 2006 年 11 月 21 日 アルゴリズムとデータ構造入門 2. データによる抽象の構築 2.2 階層データ構造と閉包性 奥乃博大学院情報学研究科知能情報学専攻知能メディア講座音声メディア分野 http://winnie.kuis.kyoto-u.ac.jp/~okuno/lecture/06/introalgds/ okuno@i.kyoto-u.ac.jp 12

More information

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

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

More information

記号と準備

記号と準備 tbasic.org * 1 [2017 6 ] 1 2 1.1................................................ 2 1.2................................................ 2 1.3.............................................. 3 2 5 2.1............................................

More information

Programming D 1/15

Programming D 1/15 プログラミング D ML 大阪大学基礎工学部情報科学科中田明夫 nakata@ist.osaka-u.ac.jp 教科書 プログラミング言語 Standard ML 入門 6 章 2005/12/19 プログラミング D -ML- 1 2005/12/19 プログラミング D -ML- 2 補足 : 再帰関数の作り方 例題 : 整数 x,y( ただし x

More information

(2000 )

(2000 ) (000) < > = = = (BC 67» BC 1) 3.14 10 (= ) 18 ( 00 ) ( ¼"½ '"½ &) ¼ 18 ¼ 0 ¼ =3:141596535897933846 ¼ 1 5cm ` ¼ = ` 5 = ` 10 () ` =10¼ (cm) (1) 3cm () r () () (1) r () r 1 4 (3) r, 60 ± 1 < > µ AB ` µ ±

More information

Basic Math. 1 0 [ N Z Q Q c R C] 1, 2, 3,... natural numbers, N Def.(Definition) N (1) 1 N, (2) n N = n +1 N, (3) N (1), (2), n N n N (element). n/ N.

Basic Math. 1 0 [ N Z Q Q c R C] 1, 2, 3,... natural numbers, N Def.(Definition) N (1) 1 N, (2) n N = n +1 N, (3) N (1), (2), n N n N (element). n/ N. Basic Mathematics 16 4 16 3-4 (10:40-12:10) 0 1 1 2 2 2 3 (mapping) 5 4 ε-δ (ε-δ Logic) 6 5 (Potency) 9 6 (Equivalence Relation and Order) 13 7 Zorn (Axiom of Choice, Zorn s Lemma) 14 8 (Set and Topology)

More information

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

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

More information

c-all.dvi

c-all.dvi III(994) (994) from PSL (9947) & (9922) c (99,992,994,996) () () 2 3 4 (2) 2 Euler 22 23 Euler 24 (3) 3 32 33 34 35 Poisson (4) 4 (5) 5 52 ( ) 2 Turbo 2 d 2 y=dx 2 = y y = a sin x + b cos x x = y = Fortran

More information

I: 2 : 3 +

I: 2 : 3 + I: 1 I: 2008 I: 2 : 3 + I: 3, 3700. (ISBN4-00-010352-0) H.P.Barendregt, The lambda calculus: its syntax and semantics, Studies in logic and the foundations of mathematics, v.103, North-Holland, 1984. (ISBN

More information

数値計算:常微分方程式

数値計算:常微分方程式 ( ) 1 / 82 1 2 3 4 5 6 ( ) 2 / 82 ( ) 3 / 82 C θ l y m O x mg λ ( ) 4 / 82 θ t C J = ml 2 C mgl sin θ θ C J θ = mgl sin θ = θ ( ) 5 / 82 ω = θ J ω = mgl sin θ ω J = ml 2 θ = ω, ω = g l sin θ = θ ω ( )

More information

2002.N.x.h.L.......g9/20

2002.N.x.h.L.......g9/20 1 2 3 4 5 6 1 2 3 4 5 8 9 1 11 11 12 13 k 14 l 16 m 17 n 18 o 19 k 2 l 2 m 21 n 21 o 22 p 23 q 23 r 24 24 25 26 27 28 k 28 l 29 m 29 3 31 34 42 44 1, 8, 6, 4, 2, 1,2 1, 8 6 4 2 1, 8, 6, 4, 2, 1,2 1, 8

More information

D 24 D D D

D 24 D D D 5 Paper I.R. 2001 5 Paper HP Paper 5 3 5.1................................................... 3 5.2.................................................... 4 5.3.......................................... 6

More information

Parametric Polymorphism

Parametric Polymorphism ML 2 2011/04/19 Parametric Polymorphism Type Polymorphism ? : val hd_int : int list - > int val hd_bool : bool list - > bool val hd_i_x_b : (int * bool) list - > int * bool etc. let hd_int = function (x

More information

CALCULUS II (Hiroshi SUZUKI ) f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b)

CALCULUS II (Hiroshi SUZUKI ) f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b) CALCULUS II (Hiroshi SUZUKI ) 16 1 1 1.1 1.1 f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b) lim f(x, y) = lim f(x, y) = lim f(x, y) = c. x a, y b

More information

1

1 2 章 1 整数を一つ読み込み, その階乗を計算する RAM プログラムを書け f (n) = n! ( n 0) 何でもよい ( n

More information

Mathematical Logic I 12 Contents I Zorn

Mathematical Logic I 12 Contents I Zorn Mathematical Logic I 12 Contents I 2 1 3 1.1............................. 3 1.2.......................... 5 1.3 Zorn.................. 5 2 6 2.1.............................. 6 2.2..............................

More information

untitled

untitled () 1 / 11 2 / 11 ( ) X X X 3 / 11 26 ()2 1 (d)5 (f)6 芦 4 / 11 5 / 11 6 / 11 (a) (b) 7 / 11 (c) 8 / 11 9 / 11 2011 1 2 1 30 20 20 Y A B Y B A Y B Y Y A B Y BA B A Y A A Y Y 4 X A B2 Y X A B 2/3 3 A B X

More information

15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = N N 0 x, y x y N x y (mod N) x y N mod N mod N N, x, y N > 0 (1) x x (mod N) (2) x y (mod N) y x

15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = N N 0 x, y x y N x y (mod N) x y N mod N mod N N, x, y N > 0 (1) x x (mod N) (2) x y (mod N) y x A( ) 1 1.1 12 3 15 3 9 3 12 x (x ) x 12 0 12 1.1.1 x x = 12q + r, 0 r < 12 q r 1 N > 0 x = Nq + r, 0 r < N q r 1 q x/n r r x mod N 1 15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = 3 1.1.2 N N 0 x, y x y N x y

More information

waseda2010a-jukaiki1-main.dvi

waseda2010a-jukaiki1-main.dvi November, 2 Contents 6 2 8 3 3 3 32 32 33 5 34 34 6 35 35 7 4 R 2 7 4 4 9 42 42 2 43 44 2 5 : 2 5 5 23 52 52 23 53 53 23 54 24 6 24 6 6 26 62 62 26 63 t 27 7 27 7 7 28 72 72 28 73 36) 29 8 29 8 29 82 3

More information

haskell.gby

haskell.gby Haskell 1 2 3 Haskell ( ) 4 Haskell Lisper 5 Haskell = Haskell 6 Haskell Haskell... 7 qsort [8,2,5,1] [1,2,5,8] "Hello, " ++ "world!" "Hello, world!" 1 + 2 div 8 2 (+) 1 2 8 div 2 3 4 map even [1,2,3,4]

More information

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ

More information

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

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

More information

VI 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

VI 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 information

04年度LS民法Ⅰ教材改訂版.PDF

04年度LS民法Ⅰ教材改訂版.PDF ?? A AB A B C AB A B A B A B A A B A 98 A B A B A B A B B A A B AB AB A B A BB A B A B A B A B A B A AB A B B A B AB A A C AB A C A A B A B B A B A B B A B A B B A B A B A B A B A B A B A B

More information

R R 16 ( 3 )

R R 16   ( 3 ) (017 ) 9 4 7 ( ) ( 3 ) ( 010 ) 1 (P3) 1 11 (P4) 1 1 (P4) 1 (P15) 1 (P16) (P0) 3 (P18) 3 4 (P3) 4 3 4 31 1 5 3 5 4 6 5 9 51 9 5 9 6 9 61 9 6 α β 9 63 û 11 64 R 1 65 13 66 14 7 14 71 15 7 R R 16 http://wwwecoosaka-uacjp/~tazak/class/017

More information

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

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

More information

3 1 5 1.1........................... 5 1.1.1...................... 5 1.1.2........................ 6 1.1.3........................ 6 1.1.4....................... 6 1.1.5.......................... 7 1.1.6..........................

More information

II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k

II R n k +1 v 0,, v k k v 1 v 0,, v k v v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ k σ dimσ = k 1.3. k II 231017 1 1.1. R n k +1 v 0,, v k k v 1 v 0,, v k v 0 1.2. v 0,, v k R n 1 a 0,, a k a 0 v 0 + a k v k v 0 v k k k v 0,, v k σ kσ dimσ = k 1.3. k σ {v 0,...,v k } {v i0,...,v il } l σ τ < τ τ σ 1.4.

More information

1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA appointment Cafe D

1W II K =25 A (1) office(a439) (2) A4 etc. 12:00-13:30 Cafe David 1 2 TA  appointment Cafe D 1W II K200 : October 6, 2004 Version : 1.2, kawahira@math.nagoa-u.ac.jp, http://www.math.nagoa-u.ac.jp/~kawahira/courses.htm TA M1, m0418c@math.nagoa-u.ac.jp TA Talor Jacobian 4 45 25 30 20 K2-1W04-00

More information

(1) (2) (3) (4) (1) (a) (b) (c) (d) kg 9.8 N 5.0 kg 19.6 m/s kg m/s 8.0kg (2) 1 r=1.0m ABC QA =1

(1) (2) (3) (4) (1) (a) (b) (c) (d) kg 9.8 N 5.0 kg 19.6 m/s kg m/s 8.0kg (2) 1 r=1.0m ABC QA =1 2/ 土 28 6 11 10:30 11:20 似通った科目名がありますので注意してください. 受験許可されていない科目を解答した場合は無効 整理番号と科目コードは受験許可証とよく照合し正確に記入 30 10 11 12 00011 00016 2 01101 02607 4 (1) (2) (3) (4) 02703 (1) (a) (b) (c) (d) 1 5.0 kg 9.8 N 5.0

More information

Int Int 29 print Int fmt tostring 2 2 [19] ML ML [19] ML Emacs Standard ML M M ::= x c λx.m M M let x = M in M end (M) x c λx.

Int Int 29 print Int fmt tostring 2 2 [19] ML ML [19] ML Emacs Standard ML M M ::= x c λx.m M M let x = M in M end (M) x c λx. 1, 2 1 m110057@shibaura-it.ac.jp 2 sasano@sic.shibaura-it.ac.jp Eclipse Visual Studio ML Standard ML Emacs 1 ( IDE ) IDE C C++ Java IDE IDE IDE IDE Eclipse Java IDE Java Standard ML 1 print (Int. 1 Int

More information

I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) + x * x + x x (4) * *

I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) + x * x + x x (4) * * 2015 2015 07 30 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) +

More information

(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n

(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n . 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n

More information

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

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. +

More information

koji07-01.dvi

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

More information

13 ACG110CBF-30K20A000.book

13 ACG110CBF-30K20A000.book ZOOMER-X d w c P. 2 P. 12 P. 30 P. 54 P. 58 P. 67 P. 70 P. 3 P. 4 P. 5 P. 9 P. 10 a 3 a w 4 u u u u a 5 a 1. 2. u 3. 4. 2 P. 18 6 a c 7 a w 8 w a 9 10a u a 11 (P31) (P23) 12b (P16) (P17) (P18) (P18) (P19)

More information

1

1 005 11 http://www.hyuki.com/girl/ http://www.hyuki.com/story/tetora.html http://www.hyuki.com/ Hiroshi Yuki c 005, All rights reserved. 1 1 3 (a + b)(a b) = a b (x + y)(x y) = x y a b x y a b x y 4 5 6

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 information

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

More information

Jacques Garrigue

Jacques Garrigue Jacques Garrigue Garrigue 1 Garrigue 2 $ print_lines () > for i in $1; do > echo $i > done $ print_lines "a b c" a b c Garrigue 3 Emacs Lisp (defun print-lines (lines) (dolist (str lines) (insert str)

More information

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P 1 1.1 (population) (sample) (event) (trial) Ω () 1 1 Ω 1.2 P 1. A A P (A) 0 1 0 P (A) 1 (1) 2. P 1 P 0 1 6 1 1 6 0 3. A B P (A B) = P (A) + P (B) (2) A B A B A 1 B 2 A B 1 2 1 2 1 1 2 2 3 1.3 A B P (A

More information

# let rec sigma (f, n) = # if n = 0 then 0 else f n + sigma (f, n-1);; val sigma : (int -> int) * int -> int = <fun> sigma f n ( : * -> * ) sqsum cbsu

# let rec sigma (f, n) = # if n = 0 then 0 else f n + sigma (f, n-1);; val sigma : (int -> int) * int -> int = <fun> sigma f n ( : * -> * ) sqsum cbsu II 4 : 2001 11 7 keywords: 1 OCaml OCaml (first-class value) (higher-order function) 1.1 1 2 + 2 2 + + n 2 sqsum 1 3 + 2 3 + + n 3 cbsum # let rec sqsum n = # if n = 0 then 0 else n * n + sqsum (n - 1)

More information

橡CompSimmAllcpct.PDF

橡CompSimmAllcpct.PDF 3 1 M dx 1 dx/ 1/ x P 0 (x) x dx x P 0 (x) x+dx P 0 (x+dx) x dx P 0 (x+dx)=p 0 (x)(1-dx/ dx/) x P 0 (x) P 0 (x+dx)=p 0 (x)(1-dx/ dx/) Tayler (dx) P 0 (x)+(dp 0 /dx)dx (dp 0 (x)/dx)dx= -P 0 (x) dx/

More information

Part () () Γ Part ,

Part () () Γ Part , Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

More information

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b) 2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................

More information

1 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 = 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

.. ( )T p T = p p = T () T x T N P (X < x T ) N = ( T ) N (2) ) N ( P (X x T ) N = T (3) T N P T N P 0

.. ( )T p T = p p = T () T x T N P (X < x T ) N = ( T ) N (2) ) N ( P (X x T ) N = T (3) T N P T N P 0 20 5 8..................................................2.....................................3 L.....................................4................................. 2 2. 3 2. (N ).........................................

More information