Estimating Error in 2nd Order Polynomial Interpolation

1. The University of JordanMechatronics Engineering DepartmentDr. Osama M. Al-Habahbeh2012

2. Example: Estimate the error of the 2nd order polynomial inter polation Where x0=1 x1=4 x2=6 x3=5 and f(x0)=f(1)=0 f(x1)=f(4)=1.386 f(x2)=f(6)=1.792 f(x3)=f(5)=1.609

3. Solution: R2 = f [x3,x2,x1,x0](x-x0)(x-x1)(x-x2) 3rd divided difference found in previous example as 0.00787 =b3 R2=0.00787(x-1)(x-4)(x-6) At x=2 R2=0.00787(2-1)(2-4)(2-6) R2=0.0629 Rn=fn=1(x)-fn (x) (difference between (nth&nth+1)) Fn+1(x)= fn(x) +Rn

4. Lagrange interpolating polynomial It is a reformulation of the Newton polynomial avoids the computation of divided differences. Represented as Fn(x)= Where Li(x)= When x=xi Li=0 Where design "product of "

5. For n=1 f1(x)=L0(x)f(x)+L1 1st order polynomial

6. Estimated error of the Lagrange polynomial is Rn =f[x,xn,xn-1,…,x0] Lagrange polynomial is preferred when the order of the polynomial is known in advance, where a Newton polynomial is preferred when the order of the polynomial is unknown in advance higher-order polynomial tend to be ill-Condi-tioned.

7. 2nd order polynomial X0,x1,x2 X-Data points

8. Coefficient of an interpolating polynomial To represent the interpolating polynomial in the form F(x)=a0+a1x+a2x2 for example three data points are required [x0,f(x0)] ,[x1,f(x1)] and [x2,f(x2)] Each points is substituted into f(x) to give: F(x0)=a0+a1x0+a2x02 F(x1)=a0+a1x1+a2x12 F(x2)=a0+a1x2+a2x22 But a0,a1,a2 are unknowns and x0,x1,x2 are known's Solve the a's to get f(x)=a0+a1x+a numerical differentiation and integration.