Numerical Analysis

Numerical Analysis

[Exercise Set 3.1, Q6] Use appropriate Lagrange interpolating polynomials of degrees one, two, and three to approximate each of the following (by hand): (b) !(0) if!(−5) = 1.93750, !(−0.25) = 1.33203, !(0.25) = 0.800781, !(0.5) = 0.687500 (d) !(0.25) if!(−1) = 0.86199480, !(−0.5) = 0.95802009, !(0) = 1.0986123, !(0.5) = 1.2943767 2. [Exercise Set 3.2, Q6] Neville’s method is used to approximate!(0.5), giving the following table. Determine12 = !(0.7). 34 = 0 14 = 0 35 = 0.4 15 = 2.8 14,5 = 3.5 32 = 0.7 12 15,2 14,5,2 = 27/7 3. [Exercise Set 3.3, Q15] The Newton forward divided-difference formula is used to approximate !(0.3) given the following data. x 0.0 0.2 0.4 0.6 f(x) 15.0 21.0 30.0 51.0 Suppose it is discovered that !(0.4) was understated by 10 and !(0.6) was overstated by 5. By what amount should the approximation to !(0.3) be changed? 4. [Exercise Set 3.4, Q2] (a) Use Theorem 3.9 or Algorithm 3.3 to construct an approximating polynomial (by hand) for the following data. x f(x) f’(x) 0.1 -0.29004996 -2.8019975 0.2 -0.56079734 -2.6159201 0.3 -0.81401972 -2.9734038 (b) Use Theorem 3.9 or Algorithm 3.3 to construct an approximating polynomial (by MAPLE or other software, program) for the data listed in above table. 5. [Exercise Set 3.5, Q3 & Q5] (a) Construct the natural cubic spline for the following data. x f(x) 8.3 17.56492 8.6 18.50515 (b) The data in above table were generated using !(3) = 3 ln (3). Use the cubic splines constructed in 5(a) for the given value of x to approximate !(0.9) and !′(0.9), and calculate the actual error.

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