HMMT 二月 2011 · ALGCALC 赛 · 第 19 题

英文解析

19. Let
    F ( x ) = ,1
    5 2011
    (2 − x − x )
    ∞
    ∑
    and note that F may be expanded as a power series so that F ( x ) = a x . Find an ordered pair ofnnn =0
    positive real numbers ( c, d ) such that lim = c .nadn →∞
    1 n
    Answer: ( , 2010) First notice that all the roots of 2 − x − x that are not 1 lie strictly 5
    2011
    6 2010!
    outside the unit circle. As such, we may write 2 − x − x as 2(1 − x )(1 − r x )(1 − r x )(1 − r x )(1 − r x )5
    1 2 3 4
    (1 − x )
    1 b b b
    0 1 4
    where | r | < 1, and let = + + . . . + . We calculate b as lim =
    i 5 0 x → 1 5
    (2 − x − x ) (1 − x ) (1 − r x ) (1 − r x ) (2 − x − x )
    1 4
    ( − 1)
    lim = .1
    x → 14
    ( − 1 − 5 x ) 6
    Now raise the equation above to the 2011 th power.
    ( )
    2011
    1 1 / 6 b b
    1 4 = + + . . . +
    5 2011
    (2 − x − x ) (1 − x ) (1 − r x ) (1 − r x )
    1 4
    Expand the right hand side using multinomial expansion and then apply partial fractions. The result − k − kwill be a sum of the terms (1 − x ) and (1 − r x ) , where k ≤ 2011.
    − ki
    Since | r | < 1, the power series of (1 − r x ) will have exponentially decaying coefficients, so we onlyi i
    ( )
    n + k − 1 − k n − kneed to consider the (1 − x ) terms. The coefficient of x in the power series of (1 − x ) is ,
    k − 1
    which is a ( k − 1)th degree polynomial in variable n . So when we sum up all coefficients, only the − 2011 2010
    power series of (1 − x ) will have impact on the leading term n .
    − 2011 20111
    The coefficient of the (1 − x ) term in the multinomial expansion is ( ) . The coefficient of
    ( )6
    n +2010
    n − 2011 1 2010
    the x term in the power series of (1 − x ) is = n + . . . . Therefore, ( c, d ) =
    2010 2010!
    ( , 2010).1
    2011
    6 2010!
    √
    2 2