7/23/2019 Imc2015 Day1 Solutions

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n2 n n A, B

A1 +B1 = (A+B)1

det(A) = det(B)

(A+B)

I= (A+B)(A+B)1 = (A+B)(A1 +B1) =

=AA1 +AB1 +BA1 +BB1 =I+AB1 +BA1 +I

AB1 +BA1 +I= 0.

X=AB1

A= X B

BA1 =X1

X+X1+I= 0

(X I)X

0 = (X I)X (X+ X1 +I) = (X I) (X2 +X+ I) =X3 I.

X3 =I

(det X)3 = det(X3) = det I= 1

det X= 1

det A= det(XB) = det X det B = det B.

= 1

2(1 +i3)

/ R

3 = 1

0 = 1 ++2 = 1 ++

A= I

B

=

2

det(B)= 1

n

3

B =I

A1 =I

B1 =B

I+ B+B = 0

(A+B)1 = (B)1 =B =I+B =A1 +B1.

A

B

det A= 1= det B

7/23/2019 Imc2015 Day1 Solutions

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n

f(n)

n

n = 23

f(n)

f(23) = 8

nk=1

f(k)n2

4 .

r

k

2r1 k

7/23/2019 Imc2015 Day1 Solutions

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F(0) = 0

F(1) = 3

2 F(n) = 5

2F(n 1) F(n 2)

n2

n=0

1

F(2n)

x2 5

2x + 1 = 0

x1 = 2 x2 =

12

F(n) =a 2n + b (1

2)n

a, b

F(0) = 0

F(1) = 3

2

a+b = 0

2a+ b2

= 32

a= 1

b =1

F(n) = 2n 2n.

1

F(2n)=

22n

(22n)2 1= 1

22n 1 1

(22n)2 1= 1

22n 1 1

22n+1 1 ,

n=0

1F(2n)

=n=0

122n 1

122n+1 1

= 1

220 1= 1.

1

F(n) = 2n 2n

n=0

1

F(2n) =

n=0

1

22n 22n =n=0

(12

)2n

1 (12

)2n+1

=

n=0 (12

)2n

k=0 (12

)2n+1

k

=

n=0 (12

)2n

k=0(12

)2k2n

=n=0

k=0

(12

)2n(2k+1) =

m=1

(12

)m = 1.

m

m =

2n(2k+ 1)

n

k

1

m1, . . . , m15

15k=1

mk arctan(k) = arctan(16). (1)

m1, . . . , m15

m1, . . . , m15

z1= 1 + 16i

z2 = (1 +i)m1 (1 + 2i)m2 (1 + 3i)m3

(1 + 15i)m15.

R= z2/z1 Re z1= 1 Re z2

R

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z1 z2

(1 + 162)R2 =15k=1

(1 +k2)mk .

p = 1 + 162 = 257

p

p

k 90

n

(i, j)

1

i < j

n+ 1

vi =BAi AiBAj > 90

vi vj < 0

B

w1, . . . , wn+1>0 n+1i=1

wivi=

1, . . . , n + 1

i

j

vi vj < 0

n+ 1

n

V

W

V W ={1, 2, . . . , n+ 1}

vi vj0 iV jW

0 =

iVW

wivi

2=

iV

wivi

2+

iW

wivi

2+ 2

iV

iW

wiwj(vi vj).

iV

wivi=

iWwivi=

n

vn+1 = (1, 1, . . . , 1) vi =(0, . . . , 0

i1

,1, 0, . . . , 0 ni

)

i= 1, . . . , n vi vj