TEAM PROJECT. Interpolation and Extrapolation. (a) Lagrange practical error estimate

Chapter 19, Problem 19.1.67

(choose chapter or problem)

Interpolation and Extrapolation.

(a) Lagrange practical error estimate (after Theorem 1). Apply this to \(p_{1}(9.2)\) and \(p_{2}(9.2)\) for the data \(x_{0}=9.0, x_{1}=9.5, x_{2}=11.0, f_{0}=\ln x_{0}\). \(f_{1}=\ln x_{1}, f_{2}=\ln x_{2}\) (6S-values).

(b) Extrapolation. Given \(\left(x_{j}, f\left(x_{j}\right)\right)=(0.2,0.9980)\), (0.4,0.9686), ( 0.6,0.8443), (0.8,0.5358), (1.0,0)). Find f(0.7) from the quadratic interpolation polynomials based on \((\alpha) 0.6,0.8,1.0,(\beta) 0.4,0.6\), \(0.8,(\gamma) 0.2,0.4,0.6\). Compare the errors and comment. [Exact \(f(x)=\cos \left(\frac{1}{2} \pi x^{2}\right), f(0.7)=0.7181\) (4S).]

(c) Graph the product of factors \(\left(x-x_{j}\right)\) in the error formula (5) for \(n=2, \cdots, 10\) separately. What do these graphs show regarding accuracy of interpolation and extrapolation?

(d) Central differences. Show that

\(\delta^{2} f_{m}=f_{m+1}-2 f_{m}+f_{m-1}\), and, furthermore

\(\delta^{3} f_{m+1 / 2}=f_{m+2}-3 f_{m+1}+3 f_{m}-f_{m-1}\),

\(\delta^{n} f_{m}=\Delta^{n} f_{m-n / 2}=\nabla^{n} f_{m+n / 2}\).

(e) Everett's interpolation formula

\(f(x) \approx (1-r) f_{0}+r f_{1}\)

(20)       \(+\frac{(2-r)(1-r)(-r)}{3 !} \delta^{2} f_{0}\)

\(+\frac{(r+1) r(r-1)}{3 !} \delta^{2} f_{1}\)

is an example of a formula involving only even-order differences. Use it to compute the Bessel function \(J_{0}(x)\) for x = 1.72 from \(J_{0}(1.60)=0.4554022\) and \(J_{0}(1.7)\), \(J_{0}(1.8), J_{0}(1.9)\) in Example 6.

Text Transcription:

p_1(9.2)

p_2(9.2)

x_0 = 9.0, x_1 = 9.5, x_{2} = 11.0, f_{0} = ln x_0

f_1 = ln x_1, f_{2} = ln x_2

(x_j, f (x_j)) = (0.2, 0.9980)

(alpha) 0.6,0.8,1.0,(beta) 0.4,0.6,

0.8, (gamma) 0.2,0.4,0.6

f(x) = cos (1 / 2 pi x^2), f(0.7) = 0.7181

(x - x_j)

n = 2, cdots, 10

delta^{2} f_{m} = f_{m + 1} - 2f_{m} + f_{m-1}

delta^{3} f_{m + 1/2} = f_{m + 2} - 3f_{m + 1} + 3f_{m} - f_{m - 1}

delta^{n} f_{m} = Delta^{n} f_{m - n/2} = nabla^{n} f_{m + n/2}

f(x) approx (1 - r) f_{0} + rf_1

+ (2 - r) (1 - r) (-r) / 3! delta^{2} f_0

+ (r + 1) r(r - 1) / 3! delta^{2} f_1

J_0(x)

J_0(1.60) = 0.4554022

J_0(1.7)

J_0(1.8), J_{0}(1.9)