December 28, 2004

 

÷
º º

December 28, 2004

 


1

1

1.1

. . . . 1

1.2

. . . . 1

1.3

. . . . 2

1

1.1

1.1

x, y

ABC



 



ABC

x, y


 





ABC

¹ ₂ .



  

A = (0, 0), B = (a, b), C = (c, d), D = (a + c, b + d)

a, b

AB

C

AB

bx−ay = bc−ad



bc − ad > 0

bc − ad > 1





bx − ay = 1

ABCD







ABC




bc − ad = 1

ABC

¹ ₂

1.2


   


1.2 (

)

P ₁ P ₂ · · · P _n

n

P 1 , P 2 , · · · , P n



 

S

n

I

n

n

P ₁ P ₂ · · · P _n

S

2 + I − 1.




n

P ₁ P ₂ · · · P _n

n




1 2

n

P ₁ P ₂ · · · P _n S, I

n

P ₁ P ₂ · · · P _n

S

2 + I − 1.

1.3 (

)

n

√

n

 

R _n

n→∞ lim R _n

n = π.








(x, y)

(x, y), (x + 1, y), (x, y + 1), (x + 1, y + 1)

 

√ n

R n


  



√ n + √ 2





√ n − √ 2



π( √ n − √

2) ² ≤ R _n ≤ π( √ n + √

2) ² .

π

1 + 2

n − 2 √

√ 2 n



≤ R _n n ≤ π

1 + 2

n + 2 √

√ 2 n



 
 

n→∞ lim R n

n = π.

1.3



x

[x]

x

[3/2] = 1, [3] = 3, [−1/5] = −1.


 

 


p q

  

- 6

O(0, 0) A(p/2, 0) C(p/2, q/2) B(0, q/2)




(1)

OACB

(p − 1)

2 · (q − 1) 2






(2)

OC

(3)

OAC

(p−1)



j=1

 j · q

p








(4)

OBC

(q−1)



j=1

 j · p

q








(5)

(p−1)



j=1

 j · q

p

 +

(q−1)



j=1

 j · p

q



= (p − 1)

2 · (q − 1) 2 .

 



 




1.1

f 1 = f 2 = 1, f n+2 = f n+1 + f n .



(0, 0), (f _n , f _n+1 ), (f _n+1 , f _n+2 )



1.2

a, b, c, d

_a ^b , ^d _c

ad − bc = 1

_a ^{b d} _c







1.3

O = (0, 0), A = (11/2, 0), B = (0, 13/2), C = (11/2, 13/2).

(1)

OAC

(2)

OBC



1.4

1

m n

 n m

 +

 2n m



+ · · · +

 (m − 1)n m



=  m n

 +

 2m n



+ · · · +

 (n − 1)m n

 .



1.5

n

√ n + √ n + 1

= √

4n + 2 .

%

 
& 
  (  

)DZ*! 
& 





m, n

(m, n) = (m + n) − mn + 2 ^m−1 

i=1

 i · n m

 .

 



P Q

Q − P = 2

(P, Q)

(3, 5), (5, 7), (11, 13), (17, 19), · · ·

 "#  


+
,
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