Use appropriate Lagrange interpolating polynomials of degree

Use appropriate Lagrange interpolating polynomials of degrees one, two, and three

to approximate f(0.9) if f(0.6) = -0.17694460, f(0.7) = 0:01375227, f(0.8) =

0:22363362, f(1.0) = 0:65809197.

2

The above data were generated with f(x) = sin(e^x-2). Use the error

formula to nd a bound for the error, and compare the bound to the actual error for

the cases n = 1 and n = 2.

Solution

LAGRANGE INTERPOLATION POLYNOMIALS TO FIND F[0.9]..GIVEN F[0.6]= -0.17694 F[0.7]= 0.013752 F[0.8]= 0.223634 F[1.0]= 0.658092 NO. OF DATA POINTS = N+1..[(Xo,Yo) , (X1,Y1) , …..(Xn,Yn) ] N = 3 PROPERTY OF LAGRANGE POLYNOMIAL ………. Ln,k [ Xj] = 1 …WHEN …J=K………..AND ……. Ln,k [ Xj] = 0 …WHEN …J IS NOT EQUAL TO K L3,0(X) = [(X-X1)(X-X2)(X-X3)] / [ (Xo-X1)(Xo-X2)(Xo-X3)] L3,1(X) = [(X-X0)(X-X2)(X-X3)] / [ (X1-X0)(X1-X2)(X1-X3)] L3,2(X) = [(X-X0)(X-X1)(X-X3)] / [ (X2-Xo)(X2-X1)(Xo-X3)] L3,3(X) = [(X-X0)(X-X1)(X-X2)] / [ (X3-Xo)(X3-X1)(X3-X2)] THEN N TH DEGREE POLYNOMIAL IS GIVEN BY …. L[X] = SIGMA [ Yk*Ln,k(X)] ..FOR K=0 TO N IN OUR CASE WE HAVE …. N = 3 ….NO.OF DATA POINTS = 3+1 = 4 N X Y 0 0.6 -0.17694 1 0.7 0.013752 2 0.8 0.223634 3 1 0.658092 L3,0(X) = [(X-X1)(X-X2)(X-X3)] / [ (Xo-X1)(Xo-X2)(Xo-X3)]= [(X-0.7)(X-0.8)(X-1)] / -0.008 L3,1(X) = [(X-X0)(X-X2)(X-X3)] / [ (X1-X0)(X1-X2)(X1-X3)]= [(X-0.6)(X-0.8)(X-1)] / 0.003 L3,2(X) = [(X-X0)(X-X1)(X-X3)] / [ (X2-Xo)(X2-X1)(Xo-X3)]= [(X-0.6)(X-0.7)(X-1)] / -0.004 L3,3(X) = [(X-X0)(X-X1)(X-X2)] / [ (X3-Xo)(X3-X1)(X3-X2)]= [(X-0.6)(X-0.7)(X-0.8)] / 0.024 L1[X] = Yo*L3,o(X)] + Y1*L3,1(X) L2[X] = Yo*L3,o(X)] + Y1*L3,1(X) + Y2*L3,2(X) L3[X] = Yo*L3,o(X)] + Y1*L3,1(X) + Y2*L3,2(X)+Y3*L3,3(X) L3,0(X) = -0.002         / -0.008         = 0.25 L3,1(X) = -0.003         / 0.003         = -1 L3,2(X) = -0.006         / -0.004         = 1.5 L3,3(X) = 0.006         / 0.024         = 0.25 L1[0.9] = -0.05799 ……………….ANSWER L2[0.9] = 0.277462 ……………….ANSWER ……………….ANSWER L3[0.9] = 0.441985 ……………….ANSWER ……………….ANSWER