My Math Forum Neville algorithm

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 June 6th, 2017, 09:12 AM #1 Newbie   Joined: Jun 2017 From: France Posts: 3 Thanks: 0 Neville algorithm Hello, I post this message because in the following exercise I am asked to use the Neville algorithm, and I do not know exactly how to use it and whether I have the right formula. Here is this exercise: Let f be a real function whose value is known in 5 points: 0 <= i <= 4 xi = {- 1,1,1,2,3}, f (xi) is equivalent to fi = {8, -2,2,1,5} And I'm asked to use the Neville algorithm to compute an approximate value of f (1/2) and f (-1/2). The formula that I find is not first of this site "http://www.uvt.rnu.tn/resources-uvt/cours/analyse_num/chap4/node4.html" The P represents a polynomial normally. Can anyone help me please? Last edited by skipjack; June 6th, 2017 at 10:21 PM.
June 6th, 2017, 10:33 PM   #2
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Quote:
 Originally Posted by Joan94 0 <= i <= 4 xi = {- 1,1,1,2,3}, f (xi) is equivalent to fi = {8, -2,2,1,5}
Should you have typed "xi = {- 1,0,1,2,3}" instead of "xi = {- 1,1,1,2,3}"?

 June 7th, 2017, 03:59 AM #3 Newbie   Joined: Jun 2017 From: France Posts: 3 Thanks: 0 Yes,sorry it's xi={-1,0,1,2,3}!
 June 17th, 2017, 04:11 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,171 Thanks: 869 So you have f(-1)= 8, f(0)= -2, f(1)= 2, f(2)= 1, f(3)= 5. "Neville's algorithm" is basically a "divided difference" method using Newton's difference formula. $\Delta f(-1)= f(0)- f(-1)= -2- 8= -10$, $\Delta f(0)= f(1)- f(0)= 2- (-2)= 4$, $\Delta f(1)= f(2)- f(1)= 1- 2= -1$, and $\Delta f(2)= 5- 2= 3$. Then $\Delta^2(-1)= \Delta f(0)- \Delta f(-1)= 4- (-10)= 14$, $\Delta^2(0)= \Delta f(1)- \Delta f(0)= -1- 4= -5$, and $\Delta^2(1)= \Delta f(2)- \Delta f(1)= 3- (-1)= 4$. $\Delta^3(-1)= \Delta^2(0)- \Delta^2(-1)= -5- 14= -19$, and $\Delta^3(0)= \Delta^2(1)- \Delta^2(0)= 4- (-5)= 9$. $\Delta^4(-1)= \Delta^3(0)- \Delta^3(-1)= -19- 9= -28$. "Newton's difference formula" says that we can approximate f(x) by $f(-1)+ \Delta f(-1) (x+ 1)+ \frac{\Delta^2 f(-1)}{2}(x+ 1)(x)+ \frac{\Delta^3 f(-1)}{3!} (x+ 1)(x)(x- 1)+ \frac{\Delta^4(x)}{4!}(x+1)(x)(x- 1)(x- 2)= 8- 10(x+ 1)+ 7(x+ 1)x- (19/6)(x+ 1)(x)(x- 1)- (28/24)(x+1)(x)(x- 1)(x- 2)$. Thanks from topsquark

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