餘式定理的問題疑惑?

2016-09-25 12:51 am
若f(1)=1,f(2)=2,f(3)=7,f(4)=5求f(0)=?

回答 (1)

2016-09-25 1:28 am
✔ 最佳答案
Sol
設f(x)=a(x-1)(x-2)(x-3)+b(x-1)(x-2)+c(x-1)+1
f(2)=c*1+1=2
c=1
f(x)=a(x-1)(x-2)(x-3)+b(x-1)(x-2)+x
f(3)=b*2*1+3=7
b=2
f(x)=a(x-1)(x-2)(x-3)+2x^2-5x+4
f(4)=a*3*2*1+32-20+4=5
6a=-11
a=-11/6
f(x)=-11(x-1)(x-2)(x-3)/6+2x^2-5x+4
f(0)=-11*(-1)*(-2)*(-3)/6+4=15


收錄日期: 2021-04-30 21:54:04
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https://hk.answers.yahoo.com/question/index?qid=20160924165134AArWvJF

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