Re: [微積] 二項展開式和線性近似之關係
※ 引述《saltlake (SaltLake)》之銘言:
: 請問下列二者之間的大小關係怎樣確定?
: A = 1 - (1 - a)^n
: B = n*a
: n 是自然數而 a 乃實數
: A < B? 還是 A > B?
: 倘若 (1) 0 <= a <= 1; (2) 1 <= a
f(a) = 1 - (1 - a)^n - na
f(0) = 0
f'(a) = -n[1 - (1 - a)^(n-1)]
當0 <= a <= 1:
f'(a) <= 0 => f(a) <= 0 => A <= B
當a > 1:
f'(a) = -n[1 + (-1)^n |a - 1|^(n-1)]
若n為偶數
則f'(a) = -n[1 + |a - 1|^(n-1)] < 0 => f(0) <= 0 => A <= B
若n為奇數
則f'(a) = -n[1 - |a - 1|^(n-1)]
當0 <= a <= 2:f'(a) <= 0
當a > 2:f'(a) > 0,存在一點a'使得f(a') = 0
所以
當0 <= a <= a',A < B
當a > a',A > B
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