Question

Use Lagrange's formula to inverse interpolate for the number x corresponding to y = .1300 in the data of Problem 12.37.

I need help with problem 12.38

Answer 1

® The given data in 1 x 1100 .20 40 760 11.50 12.00 | 5 10.2420 0.1942 10- 1497 10.1109 10.079010.05to We are to find 2 (1.50). ing. Lagrange's interpolation formula d =1100, y = 1.80 7- 20 ) KG - 3:00 22=1.40 X3=1.60 De la f(x3), 1.- f(x) = 0.2420,- yg = f(x) = 0.6540 We know that the Lagrange interpolation poly bomial 20) is given by 4(x)=:5 li (x) y , where i PD li(x) - 64–260)(x-X;).-(--)(x-X2 +1)-(X-Xn) 7X-70) (xi-34). (X-X-1)(xi-xi xi-x1) ᎣᏍ le(x) = (x-x)(x-x2)(X-83) (X-M4) (X - 35) (x - 3) (%- Y ) (x - 13) (x - 1)(x - 1) = DADEN - 26.04 (X-120)(2-1.40)(x-1160) (2-480) (x-2:00) li(x) = 6e-60)(-x2)(x-x3)(x-Wu) (x-25). (%,- Xo)(x,-X2) (4- X3) (X.- Xy) (x2-15) = 13021(2-1:00)(x-1.40) (2-160)(x-1080)(x-2.00) 2(x) = 60-x)(x-21) (2-x3)(x-24) (X-X) TX-Xo)(x - x0) (X - x3)(x-xu) (x2-xs) - 260-42(-0)(x-120)(x-160)(x-1-80) (x-2.00)

4x) = x-Xo)(x-x4)(x-X)) (x-x4) (-x) CX3 - Xo) (x2-x) (xs-x) (x3 - Xy)(x3= X5) - 260.42(x-1)(x-1.20) (x-1-40) (2-4. 80)(x-4:00) 11(x) = (x-26) (x-2)(x-x2) (X-X3) (x-45) (24- Xo) (Xu- 24) (X4-X2) ( ^e,- X3) (xu-X 5-) -130.21(x-1)(x-120)(x-1-40)(x-1.60)(x-2.00) Is(x) - 2 - Xo)(x-2)(x-X2 ) ( x - x3)(x-x4) fs - x0) (X5 - xv) (*s - x3)(xs - x3)(xs-Ky) = 26.04(x-1)(x+1-20) (X - 140) (x-1.60)(x-1•80) Thus 120 41x)=51; (x) yi - 16(2) + 8 (1) 9,+ 2()%,+83(x) yg +86994 +15(x) ys 5-26.04(6-1:20)(x-1.40) (-1.40)/2 - 1'8+)(x-2:00) +136.21(x-1)(x-1.40) (x-1.60) (x-1180) (x-2:00) - 260.4272-1)(x-1-20)(x-/-67)(x-1.80) (x-2.00) +260.42/ X-1) (-1:20) (x-1.40) (x-1.80) (x-2:00) -130.2162-1)(x-1.20)(x-1.40) X-1.60) (x-2.00) + 26.04/2-1)(x-120)(x-1.40) (X-1•60)(x-1.80) 50) (1.50) -0.0117 -0.0976 +0.5859 +0.5859 -0.0976+0.0117 = 10000 Hence the required value in y (150) = 1.0000