1、(x-)(x-号) 1 ~(x-)(x-号), V3_(x-)(x-3)n=L(x)= =1（））（（）2（））5元sin 50°~ L0.76543218-cos5xV31四)(x一化<cos5xR,(x)=xx3!226435元<0.000770.00044<Rsin50o=0.7660444...12182th'stheactualoferrrorinterpolation2~ 0.00061Highorderinterpolation is usuallybetter than low orderon上页but not absolutely下页the better the higher返圆order hehe
上页 下页 返回 n = 2 ( )( ) ( )( ) 2 3 ( )( ) ( )( ) 2 1 ( )( ) ( )( ) 2 1 ( ) 3 6 3 4 6 4 4 6 4 3 6 3 6 4 6 3 4 3 2 − − − − + − − − − + − − − − = x x x x x x L x ) 18 5 sin 50 (2 0 L 0.76543 2 3 cos 2 1 ); 3 )( 4 )( 6 ( 3 ! cos ( ) 2 − − − − = x x R x x x x 0.00077 18 5 0.00044 2 R sin 50 = 0.7660444. the actual errror of 2th's interpolation2 0.00061 High order interpolation is usually better than low order interpolation but not absolutely the better the higher order,hehe
四、 The merits and cdrawbacks of Lagrangeinterpolation formulaUsing the interpolation function is very easy to get the Lagrangeinterpolation polynomial , the structure of which is compact. It isvery convenient in the theoretical analysis ,but ,when theinterpolation node increase,all the interpolating function I,(k = O,l,.., n)are going to change,and the formula should also change.This is notconvenient in practical computation.In order to overcome thisshortcoming,we can exchange interpolation polynomial into othereasy fomat, leading out Newton interpolation formula.上页下页返圆
上页 下页 返回 四、The meri ts and dra wbacks of Lagrange interpolation formula Using the interpolation function is very easy to get the Lagrange interpolation polynomial，the structure of which is compact. It is very convenient in the theoretical analysis ,but ,when the interpolation node increase,all the interpolating function are going to change,and the formula should also change.This is not convenient in practical computation.In order to overcome this shortcoming,we can exchange interpolation polynomial into other easy fomat, leading out Newton interpolation formula. l (k 01 n) k = ，
conclusion1.The definiton of interpolation2.The geometric significance of interpolation3.The linear interpolation4.Parabolic interpolation5.Lagrange Interpolating6.The existence and uniqueness of interpolatingpolynomial上页下页7.The interpolation remainder and the error evaluation返圆
上页 下页 返回 conclusion 2.The geometric significance of interpolation 3.The linear interpolation 4.Parabolic interpolation 1.The definiton of interpolation 5.Lagrange Interpolating 6.The existence and uniqueness of interpolating polynomial 7.The interpolation remainder and the error evaluation
s3 Newton interpolationAs we all know,the basis function of the lagrange interpolatingpolynomial isn(x-x)1,(x)=IIj= 0,1,2,...,n(x, -x,)i=0itjThe form is too complex.There is a lot of calculation,and muchdouble calculation.According to the knowledge of linear algebra,Any nth polynomial canbe expressed as:1, x-xo, (x-xo)(x-x),.", (x-xo)(x-x)..(x-xn-1)A total of n+1 linear combination of polynomialAnd then,whether can the n+1 polynomial be look as the上页interpolation basis function?下页返圆
上页 下页 返回 §3 Newton interpolation l (x) j = − − = n i j i j i i x x x x 0 ( ) ( ) j = 0,1,2, ,n As we all know,the basis function of the l agrange interpolating polynomial is The form is too complex.There is a lot of calculation,and much double calculation. According to the knowledge of linear algebra,Any nth polynomial can be expressed as: 1, , 0 x − x ( )( ), 0 1 x − x x − x ( )( ) ( ) − 0 − 1 − n−1 , x x x x x x A total of n+1 linear combination of polynomial. And then,whether can the n+1 polynomial be look as the interpolation basis function?
obviously, the polynomial group1, x-xo, (x-x)(x-x), .., (x-x)(x-x)...(x-xn-l)is linearly independent.So,they can act as interpolating basis functionSuppose that the interpolating node is x, ,the value offunction is f ,i= o,l,.,nh = max hh, = Xi+1 -X; ,i=0,1,2,...,n-1iThe interpolating condition is P(x,)= f, ,i=O,l,.".,nSuppose that the interpolating polynomial P(x) has thefollowing shape:P(x) =a, +a(x-x)+az(x-xo)(x-x)+...上页+a,(x-x,)(x-x)...(x-xn-1)下页返圆
上页 下页 返回 1, , x − x0 ( )( ), x − x0 x − x1 ( )( ) ( ) x − x0 x − x1 x − xn−1 , obviously，the polynomial group is linearly independent.So,they can act as interpolating basis function hi = xi+1 − xi , i = 0,1,2, ,n−1 i i h = max h ( )( ) ( ) ( ) ( ) ( )( ) 0 1 1 0 1 0 2 0 1 + − − − − = + − + − − + an x x x x x xn P x a a x x a x x x x Suppose that the interpolating node is ,the value of function is xi f i , i = 0,1, ,n The interpolating condition is P x f i n i i ( ) = , = 0,1, , Suppose that the interpolating polynomial has the following shape: P(x)