IMC2014 Day2 Solutions

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  • 8/9/2019 IMC2014 Day2 Solutions

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    x n dn (x) dn (x)

    {0, 1, . . . , 9} x =

    n =1dn (x)10n 1 an

    n =1

    dn (an )

    n =1

    (an )n =1

    n0 dn (an ) = 1 n n0

    an +1 , a n +2 , . . . > 10n n n0. (1)

    an =

    k =1dk (an )10k 1 dn (an )10n 1 dn (an ) 1

    an 10n 1 m > n am 10m > 10n (1) n a1, a2, . . . , a n

    [1, 10n ] 10n n lim(10n n) =

    a1, a2, . . .

    (an ) dn (an ) = 0 n > n 0

    gn =2 dn (xn ) = 11 dn (xn ) = 1 .

    gn = dn (an ) n

    xk =k

    n =1

    gn 10n 1 k = 1, 2, . . . .

    xk +1 10k > x k (xk ) xn 0 , xn 0 +1 , xn 0 +2 , . . .

    (an ) xk = an n 1 k n0 k n dn (xk ) = gn = dn (an ) xk = an

    n > k n0 dn (xk ) = 0 = dn (an ) xk = an

    A = ( a ij )ni,j =1 n n 1, 2, . . . , n

    1 i

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    A

    n

    i=1

    a ii =n

    i=1

    i ,

    n

    i=1

    a2ii + 2i 0. g (x) < 0 0 < x < 2 x > 0

  • 8/9/2019 IMC2014 Day2 Solutions

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    sin xx

    (n )

    = dn

    dxn 1

    0 cos(xt )dt =

    1

    0

    n

    x n ( cos(xt )) d t =

    1

    0tn gn (xt )dt

    gn (u)

    sin u

    cos u

    n

    |gn | 1

    sin xx

    (n )

    1

    0tn gn (xt ) dt <

    1

    0tn dt =

    1n + 1

    .

    R n k

    k

    k

    k

    k n d(k, n) k R n k d(k, n)

    d(k, n) =k n k, n > 1k + n

    n = 1 k = 1 R n

    n + 1

    n + 1 n + 1

    k,n > 1 k n k k

    k

    n(k 1) n(k 1) k (n 1)(k 1) 1 n, k > 1

    i xi = 0 k 1 k d(k, n) kn

    d(k, n) kn k,n > 1 k k kn

    k k R n kn

    k k

  • 8/9/2019 IMC2014 Day2 Solutions

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    k nk n

    nk + 1 O

    1

    nk 1 k+1 O

    k n O k

    1, 2,...,n k O O k n

    O k

    n Dn (x1, . . . , x n ) (1, 2, . . . , n ) x j = j 1 j n 1 k n2

    ( n, k ) (x1, . . . , x n ) (1, 2, . . . , n ) xi = k + i 1 i k x j = j 1 j n

    ( n, k ) =k 1

    i=0

    k 1i

    D (n +1) (k + i )n (k + i)

    .

    a r {i1, . . . , i k } {a1, . . . , a k } ar = is s = r

    as {i1, . . . , i k } as = it x = ( x1, . . . , x n ) xi j = a j x = ( x 1, . . . , x i t 1, x

    i t +1 , x

    n )

    [n] \ { i t } xi j = a j j = t a j = a j j = s

    a s = at x = ( x1, . . . , x n ) [n] xi j = a j {i1, . . . , i k } {a1, . . . , a k } x = ( x 1, . . . , x i t 1, x

    i t +1 , x

    n )

    [n] \ { i t } xi j = a j {i1, . . . , i k } \ { i t } {a 1, . . . , a k } \ { a t }

    1 a s / {i1, . . . , i k } x = ( x1, . . . , x n ) xi j = a j x = ( x 1, . . . , x a s 1, x

    a s +1 , x

    n )

  • 8/9/2019 IMC2014 Day2 Solutions

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    [n] \ {as } xi j = a j j = s x = ( x1, . . . , x n ) [n] xi j = a j

    {i1, . . . , i k } {a1, . . . , a k } x = ( x 1, . . . , x

    a s 1, x

    a s +1 , x

    n ) [n] \ { a s } xi j = a j

    {i1, . . . , i k } \ { is } {a1, . . . , a k } \ { a s } 1 ( n,k,) = ( n 1, k 1, 1)

    ( n,k,) = ( n , k ,0).

    = 0 ( n,k, 0) 2k n k = 0 ( n, 0, 0) = Dn k 1

    ( n,k, 0) = ( n 1, k 1, 0) + ( n 2, k 1, 0).

    x = ( x1, . . . , x n ) xi j = a j xa 1 = i1 xa 1 = i1

    [n] \ { i1, a1} {i2, . . . , i k } {a2, . . . , a k } 0 ( n 2, k 1, 0)

    [n] \ { a1} {i2, . . . , i k } {a2, . . . , a k } 0 ( n 1, k 1, 0)

    k

    ( n,k, 0) =k 1

    i=0

    k 1i

    D (n +1) (k + i )n (k + i)

    , 2 2k n.

    k = 1

    ( n, 1, 0) = ( n 1, 0, 0) + ( n 2, 0, 0) = Dn 1 + Dn 2 = Dnn 1

    .

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    k 1

    ( n,k, 0) = ( n 1, k 1, 0) + ( n 2, k 1, 0)

    =k 2

    i=0

    k 2

    i

    Dn (k 1+ i )

    (n 1) (k 1 + i) +

    k 2

    i=0

    k 2

    i

    D(n 1) (k 1+ i )

    (n 2) (k 1 + i)

    =k 2

    i=0

    k 2i

    D (n +1) (k + i )n (k + i)

    +k 1

    i=1

    k 2i 1

    Dn (k + i 1)(n 1) (k + i 1)

    = D(n +1) k

    n k +

    k 2

    i=1

    k 2i

    D (n +1) (k + i )n (k + i)

    +D (n +1) (2k 1)n (2k 1)

    +k 2

    i=1

    k 2i 1

    D (n +1) (k + i )n (k + i)

    = D(n +1) k

    n k +

    k 2

    i=1

    k 2i +

    k 2i 1

    D (n +1) (k + i )n (k + i) +

    D (n +1) (2k 1)n (2k 1)

    = D(n +1) k

    n k +

    k 2

    i=1

    k 1i

    D (n +1) (k + i )n (k + i)

    + D(n +1) (2k 1)n (2k 1)

    =k 1

    i=0

    k 1i

    D (n +1) (k + i )n (k + i)

    .

    n = 2ki j = j a j = k + j j = 1 , . . . , k x = ( x 1 , . . . , x n )

    x i j = a j x = ( x k +1 , . . . , x n ) [k] x k !

    k 1i=0

    k 1i

    D k +1 ik i = k!