Midterm

Friday, July 25, 2003

Answer all below questions. You may consult only the proctor. Submit all scratch paper and full, clear reasoning; this will help us give you partial credit.

1) (34 pts) You have the three data points

(x_{1}, f(x_{1})) = (0, 0),

(x_{2}, f(x_{2})) = (\frac{π}{2} , 1)

and

(x_{3}, f(x_{3})) = (π, 0),

where f is the alledged underlying unknown function. Using the representation

p_{n−1}(x) = a_{1} + a_{2}x + ⋯ + a_{n}x^{n−1},

set up, but do not solve, the equations that give the a_{i}’s so that p_{n−1} interpolates f at the three data points. Design a 3-point interpolation function for this data based on the representation

p_{n−1}(x) = a_{1} + a_{2}(\frac{x}{π}) + a_{3}(\frac{x}{π})^{2} + ⋯ + a_{n}(\frac{x}{π})^{n−1}.

Be sure to explicitly write down your function. Explain your method’s virtues, drawbacks.

Bonus: If possible, give interpretations of the unknowns in your method.

2) (33 pts) Clearly delineate how to find x given that

x = b^{T} C^{−1}b,

where C = AA^{T}, matrices A and C are nonsingular, as efficiently as possible. Assume that pivoting is not necessary. What does the cost of your method boil down to? Suppose we formed C, formed C^{−1}, then x, and we called this method 2. What would be the savings from your method, given above, over method 2 (Assume that forming C^{−1} from C requires \frac{8}{3}n^{3})? Note that forming C^{−1} from C is usually unstable and therefore a bad idea, but we consider it for sake of comparison.

3a) (23 pts) Find w_{i}, x_{i}, so that the integration formula

∫_{−1}^{1} x^{2}f(x)dx = w_{1}f(x_{1}) + w_{2}f(x_{2})

is exact for all f ∈ Π_{3} (Hint: following the methodology in the text is one way to do it.).

3b) (10 pts) Let

I_{1} = \frac{1}{3} ∫_{a}^{b}(f − g − h)dx

and

I_{2} = \frac{1}{3} ∫_{a}^{b}(f + g + h)dx.

Clearly show how to estimate I_{i} (i = 1, 2) to within a tolerance of tol using Matlab’s quad function. Show why your approximations serve their purpose, clearly stating any assumptions you are using.

Bonus. Discuss two more reasons your method wins over a brute-force method.