回復 8# math1 的帖子
第 15 題
\(n\in N\),已知\(a_1=1\),且\(\displaystyle \frac{1}{n+1}<x\le \frac{1}{n}\)時,\(f(x)=a_n x^n\),若\(f(x)\)在區間\((0,1]\)為連續,則\(a_n=\) 。
[解答]
1/2 < x <= 1,f(x) = (a_1)x
1/3 < x <= 1/2,f(x) = (a_2)x^2
1/4 < x <= 1/3,f(x) = (a_3)x^3
:
:
由於 f(x) 在 (0,1] 連續
(a_1)(1/2) = (a_2)(1/2)^2
a_2 = 2a_1
(a_2)(1/3)^2 = (a_3)(1/3)^2
a_3 = 3a_2
a_1 = 1,a_n = n!