Re: [中學] 多項式
※ 引述《g418 (我有問題)》之銘言：
: 令f(x)為四次的多項式 滿足f(42)=f(69)=f(96)=f(123)=13 且f(165)=20
: 則f(1)-f(2)+f(3)-f(4)+....+f(165)=
: 這該怎麼設呢? 我只觀察到42 69 96 123都相差27 不會算...
令 g(x) = f(x) - 13
=> g(42) = g(69) = g(96) = g(123) = 0, g(165) = 7
(x-42)(x-69)(x-96)(x-123)
且 g(x) = ----------------------------------- * 7
(165-42)(165-69)(165-96)(165-123)
所求 = g(1)-g(2)+g(3)-...+g(165) + 13
(-41) * (-68) * (-95) * (-122)
觀察 f(1) = -------------------------------- * 7
42 * 69 * 96 * 123
41 * 68 * 95 * 122
f(164) = --------------------- * 7 = f(1)
42 * 69 * 96 * 123
故所求 = g(165) + 13 = 20
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