Question

Using Lagrange's interpolation formula find y(0.25) from the following table: x0.2 0.4 0.6 0.8 y 0.12 0.48 0.12 Write 3-decimal plates

Answer 1

Given Xo = 0.2, x1 = 0.4, x2 = 0.6, xz = 0.8 yo = 0.12, y1 = 0.48, yz =0., y; = 2 Lagrange's Interpolation formula is given by -XY. + -XY (x- x)(x-x)(x– x;) (x- x)(x– x3)(x – x3) y=f(x)= ( *o - x)( *o - x)(x -xz) (x; - x)(x; – x3)(x; – x3) (x-xo)(x- x)(x– x3) (x-xo)(x x)(x – x2) (x3 - x)(xz - x)(xz -xz) (x3 - x)(x3 - x)(x3 - x2) + Xy, + XY y= (x-0.4)(x-0.6) (x-0.8) (x-0.2)(x-0.6)(x-0.8) -0.12 + x 0.48 (0.2-0.4)(0.2-0.6) (0.2-0.8) (0.4-0.2)(0.4-0.6) (0.4-0.8) (x-0.2)(x-0.4)(x-0.8) (x-0.2)(x-0.4)(x-0.6) x 0.1+ x2 (0.6-0.2)(0.6-0.4)(0.6-0.8) (0.8 -0.2) (0.8 -0.4)(0.8-0.6) + y=((62.9167(x – 4)+3.33333)(x– 0.2)+3.1333)(x-0.8)+2 y(x) = 62.9167x -84.75x2 +35.0333x-4

Therefore y(0.25) is given by y(0.25) = 62.9167 (0.25) – 84.75(0.25) +35.0333(0.25)-4 y(0.25)= 0.44453 Therefore y(0.25)=0.44453

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