# Module

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Module for The Newton Polynomial Background. We have seen how to expand a function in a Maclaurin polynomial about involving the powers and a Taylor polynomial about involving the powers . These polynomials have a single center . Polynomial interpolation can be used to construct the polynomial of degree that passes through the n+1 points , for . If multiple centers are used, then the result is the so called Newton polynomial. We attribute
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## Abstract Algebra

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Module   for   The Newton Polynomial   Background.  We have seen how to expand a function in a Maclaurin polynomial aboutinvolving the powers and a Taylor polynomial about involving thepowers . These polynomials have a single center . Polynomial interpolationcan be used to construct the polynomial of degree that passes through the n+1points , for . If multiple centers are used, then the result is the so called Newton polynomial. We attribute much of thefounding theory to Sir Isaac Newton (1643-1727). Theorem(Newton Polynomial). Suppose there is a known polynomial )( 1 x p n  that interpolates the data set: ),,2,1,,( ni y x ii   . When one more data point ),( 11  nn y x , which is distinct from all theold data points, is added to the data set, we can construct a new polynomial that interpolatesthe new data set. Keep also in mind that the new data point need not be at either end of theold data set. Consider the following polynomial of degree n      niinnn x xc x p x p 11 )()()( (3.2.1)where n c is an unknown constant. In the case of  1  n , we specify )( 0 x p as 10 )( y x p  ,where data point 1 need not be at the beginning of the data set.At the points of the old data set, the values of  )(  x p n are the same as those of  )( 1 x p n  .This is because the second term in Equation 3.2.1 is zero there. Since we assume )( 1 x p n   interpolates the old data set, )(  x p n does so too.At the new data point, we want 11 )(   nnn y x p . This can be accomplished by settingthe coefficient n c to be    niinnnnniinnnnn n  x x x p y x x x p x p c 1111111111 )()()()()( (3.2.2)  which definitely exists since 1  n  x is distinct from n),,21,i(x i   . Now )(  x p n is apolynomial that interpolates the new data set.For any given data set: ),,2,1,,( ni y x ii   , we can obtain the interpolatingpolynomial by a recursive process that starts from )( 0 x p and uses the above construction toget )( 1 x p , )( 2 x p ,  , )( 1 x p n  . We will demonstrate this process through the followingexample.Assume that and for are distinct values. Then,where is a polynomial that can be used to approximate ,and we write.The Newton polynomial goes through the points , i.e.for .The remainder term has the form,for some value that lies in the interval . The coefficients areconstructed using divided differences.  Definition.   Divided Differences.  The divided differences for a function are defined as follows:The divided difference formulae are used to construct the divided difference table:The coefficient of the Newton polynomial is and it is thetop element in the column of the i-th divided differences.The Newton polynomial of degree that passes through thepoints , for is  Example. Construct an interpolating polynomial for the following data set using the formulain Equations 3.2.1 and 3.2.2 i 1 2 3 4 5x 0 5 7 8 10y 0 2 -1 -2 20 Step1: for 1  i   0)( 100  yc x p  Step 2: adding point #2  xc x xc x p x p 11101 )()()(   Applying 221 )( y x p  , we get 4.0 1  c . So  x x xc x p x p 4.0)()()( 1101   Step 3: adding point #3 )5(4.0))(()()( 221212  x xc x x x x xc x p x p  Applying 332 )( y x p  , we get 2714.0 2  c . So )5(2714.04.0)( 2  x x x x p  Step 4: adding point #4 )7)(5()())()(()()( 32321323  x x xc x p x x x x x xc x p x p  Applying 443 )( y x p  , we get 0548.0 3  c . So )7)(5(0548.0)()( 23  x x x x p x p
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