learnerlilia

高中多項式問題(拉格朗日插值法相關)

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三次多項式 f(x) 滿足 f(1)=1,  f(2)=4,  f(3)=9
 f(x) 除以 (x-1)(x-2)(x-3) 的餘式為1‧(x-2)(x-3)/(1-2)(1-3)  +4‧(x-1)(x-3)/(2-1)(2-3)  +9‧(x-1)(x-2)/(3-1)(3-2) ???

若設 f(x)=a(x-1)(x-2)(x-3)+  b(x-1)(x-2)+  c(x-1)+  d

帶入可得f(x)=a(x-1)(x-2)(x-3)+  1(x-1)(x-2)+  3(x-1)+  1
暴力法可得餘式式x平方沒錯

基本上三次式要由四個點確定,但這個只有三個點,所以a是不確定的,用拉格朗日法不是也要四個點才能寫出f(x)=.......
我有自己再找一個三次式試試看是否也可以這麼做,卻發現也可以,我想請問有人證明過這個嗎?我從高中龍騰課本第一冊2-2習題找的問題

這樣的找餘式方法有一般性嗎?3次式,找到f(x1)=y1  f(x2)=y2  f(x3)=y3  
即可得f(x)除以(x-x1)(x-x2)(x-x3)的餘式=y1‧(x-x2)(x-x3)/(x1-x2)(x1-x3)  +y2‧(x-x1)(x-x3)/(x2-x1)(x2-x3)  +y3‧(x-x1)(x-x2)/(x3-x1)(x3-x2)

 

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  請問第一個問題是要求什麼?是f(1)=1,f(2)=4,f(3)=9,然後問f(x)除以(x-1)(x-2)(x-3)的餘式嗎? 如果是的話答案應該是x^2-x+3 沒有計算錯誤的話啦XD

(不過我不確定喔,畢竟是上學期了orz

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被除式f(x)和商式q(x)是三次多項式,但你要求的是餘式r(x)
餘式的次數比商式少一次,所以是二次多項式
找r(x)要用到的條件是r(1)=1,r(2)=4,r(3)=9
這些條件是從f(x)=(x-1)(x-2)(x-3)q(x)+r(x)得到的

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