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One of the useful application of Lagrange interpolation is to develop a look-up table. Create such a cubic Lagrange function to fit the data below and determine the function value at x = 1.4. (Answers should be in 5 decimal places) хо x1 x2 x3 -6 3 4 f(x) 54 5.4 0.8 f(x0) f(x1) f(x2) f(x3) Lolx)f(xo) = Blank 1 (Blank 2x3 + Blank 3x2 + Blank 4x + Blank 5) L1(x) f(x1) = Blank 6 (Blank 7x3 + Blank 8x2 + Blank 9x + Blank 10) L2(x) f(x2) = Blank 11 (Blank 12x3 + Blank 13x + Blank 14x + Blank 15) L3(x) f(x3) = Blank 16 (Blank 17x3 + Blank 18x + Blank 19x + Blank 20) f(1.4) = Blank 21

One of the useful application of Lagrange interpolation is to develop a look-up table. Create such a cubic Lagrange function to fit the data below and determine the function value at x = 1.4. (Answers should be in 5 decimal places) хо x1 x2 x3 -6 3 4 f(x) 54 5.4 0.8 f(x0) f(x1) f(x2) f(x3) Lolx)f(xo) = Blank 1 (Blank 2x3 + Blank 3x2 + Blank 4x + Blank 5) L1(x) f(x1) = Blank 6 (Blank 7x3 + Blank 8x2 + Blank 9x + Blank 10) L2(x) f(x2) = Blank 11 (Blank 12x3 + Blank 13x + Blank 14x + Blank 15) L3(x) f(x3) = Blank 16 (Blank 17x3 + Blank 18x + Blank 19x + Blank 20) f(1.4) = Blank 21

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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One of the useful application of Lagrange interpolation is to develop a look-up table. Create such a cubic Lagrange function to fit the data below and determine the function value at x = 1.4. (Answers should be in 5 decimal places) хо x1 x2 x3 -6 3 4 f(x) 54 5.4 0.8 f(x0) f(x1) f(x2) f(x3) Lo(x)f(xo) = Blank 1 (Blank 2x3 + Blank 3x2 + Blank 4x + Blank 5) L1(x) f(x1) = Blank 6 (Blank 7x3 + Blank 8x2 + Blank 9x + Blank 10) L2(x) f(x2) = Blank 11 (Blank 12x + Blank 13x + Blank 14x + Blank 15) L3(x) f(x3) = Blank 16 (Blank 17x3 + Blank 18x2 + Blank 19x + Blank 20) %3D f(1.4) = Blank 21 %3D
Transcribed Image Text:One of the useful application of Lagrange interpolation is to develop a look-up table. Create such a cubic Lagrange function to fit the data below and determine the function value at x = 1.4. (Answers should be in 5 decimal places) хо x1 x2 x3 -6 3 4 f(x) 54 5.4 0.8 f(x0) f(x1) f(x2) f(x3) Lo(x)f(xo) = Blank 1 (Blank 2x3 + Blank 3x2 + Blank 4x + Blank 5) L1(x) f(x1) = Blank 6 (Blank 7x3 + Blank 8x2 + Blank 9x + Blank 10) L2(x) f(x2) = Blank 11 (Blank 12x + Blank 13x + Blank 14x + Blank 15) L3(x) f(x3) = Blank 16 (Blank 17x3 + Blank 18x2 + Blank 19x + Blank 20) %3D f(1.4) = Blank 21 %3D
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