Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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5. The interpolating polynomial P3 (x) Xi P3(x) = 37(x + 1) + 8(x + 1)x − 6(x + 1)x(x - 1). is obtained using the Newton Divided Difference table given as below. Find the correct value for each blank box. Please show necessary work. f(x₂) 1st divided difference 2nd divided difference 3rd divided difference ☐☐☐ 6. Prove the claim on Lecture Notes3.3 Page 6: x - P₂ (x) = q (x) + Xn - Xo where P(x) is the polynomial of degree at most k that interpolates f at the nodes Xo, X₁, X, and q(x) is the polynomial of degree at most n - 1 that interpolates f at the nodes x₁,x₂,.,Xn- -(q (x) - Pn-1(x)) (Hint: Notice that both LHS and RHS are polynomials of degree at most n, show that LHS=RHS at n + 1 points: Xo, X₁, X, therefore by the uniqueness of interpolation polynomial, LHS=RHS.)
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Transcribed Image Text:5. The interpolating polynomial P3 (x) Xi P3(x) = 37(x + 1) + 8(x + 1)x − 6(x + 1)x(x - 1). is obtained using the Newton Divided Difference table given as below. Find the correct value for each blank box. Please show necessary work. f(x₂) 1st divided difference 2nd divided difference 3rd divided difference ☐☐☐ 6. Prove the claim on Lecture Notes3.3 Page 6: x - P₂ (x) = q (x) + Xn - Xo where P(x) is the polynomial of degree at most k that interpolates f at the nodes Xo, X₁, X, and q(x) is the polynomial of degree at most n - 1 that interpolates f at the nodes x₁,x₂,.,Xn- -(q (x) - Pn-1(x)) (Hint: Notice that both LHS and RHS are polynomials of degree at most n, show that LHS=RHS at n + 1 points: Xo, X₁, X, therefore by the uniqueness of interpolation polynomial, LHS=RHS.)
Expert Solution
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Step 1

Given : x4=2   ,  P3=3-7x+1+8x+1x-6x+1xx-1

To Find : Value for Box

Note : According to the guideline we are suppose to answer only one question at a time . kindly repost the remaining question and mention it .

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