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UCB MATH 128A-2, SUMMER 2009: FINAL EXAM SOLUTIONS
JUSTIN BLANCHARD
- (a) Prove that the bisection method on [1, 2] will converge to a root of x^3 − x − 1. (b) Find a bound on the absolute error in approximating this root after n iterations. (c) How many iterations are necessary to ensure the relative error is at most 2−^5? (Your answer does not have to be exact.) Solution: (a) This function is continuous. Its values at the endpoints are − 1 < 0 and 5 > 0. Therefore, the interval brackets a root and the bisection method will converge. (b) On the nth iteration, the root is known to lie in an interval of width (^22) n−−^11. Its distance from this interval’s midpoint can be at most 1 / 2 n^. (The answer 1/ 2 n+1^ is also acceptable, depending on how the phrase “after n iterations” is interpreted.) (c) Since the root lies in [1, 2], the relative error is at most equal to the absolute error. By (b), we must use n = 5 iterations.
- (a) Find an algebraic expression for the unique root (in R) of f (x) = x^2 − 2 /x. (b) Newton’s method searches for this root using the iteration pn+1 = g(pn). Find g(x). Simplify your answer. (c) Is the convergence linear? Is it at least quadratic? (d) Find the exact order of convergence. Solution: (a) x^2 − 2 /x = 0 ⇐⇒ x^3 − 2 = 0 ⇐⇒ x = 2^1 /^3.
(b) g(x) = x − (^) ff ′^ ((xx)) = x − x
(^2) − 2 /x 2 x+2/x^2 =^ x^ −^ x^
x^3 − 2 2 x^3 +2 =^
x(x^3 + 4) 2(x^3 + 1) (c) The convergence is at least quadratic, for either of the following reasons: 2^1 /^3 is a simple root of f (x), as f ′(2^1 /^3 ) = 2 · 21 /^3 + 2/ 22 /^3 6 = 0. Alternatively, g′(2^1 /^3 ) = 0 after the calculation g′(x) = (4x
(^3) +4)2(x (^3) +1)−x(x (^3) +4)(6x (^2) ) 22 (x^3 +1)^2 =^
8(x^3 +1)^2 − 6 x^3 (x^3 +4) 22 (x^3 +1)^2. (d) (This was intended to be a harder problem.) By definition, the order of convergence is α iff limn→∞
pn+1− 21 /^3 pn− 21 /^3
∣ =^ λ^6 = 0.^ If^ pn =^ x, this fraction’s numerator is g(x) − 21 /^3 = x
(^4) − 24 / (^3) x (^3) +4x− 2 2(x^3 +1).^ Factor out (x^ −^2
1 / (^3) ) as many times
as possible: (x−^2
1 / (^3) ) (^3) (x+2 1 / (^3) ) 2(x^3 +1). (Alternatively, Taylor-expand^ x
(^4) − 24 / (^3) x (^3) + 4x − 2 with center
x = 2^1 /^3 .) It follows that the order of convergence is 3
- (a) Estimate y(1/2) with polynomial interpolation given the data y(0) = 1, y(1) = 2. (b) Repeat part (a) with Hermite interpolation, if y(t) is a solution to y′^ = y. Solution: (a) Polynomial interpolation in this case is linear interpolation, so y(0.5) ≈ 1. 5 (b) Hermite interpolation, Newton form:
Date: Thursday 8/13. 1
x 0 = 0 f [x 0 ] = 1 f [x 0 , x 1 ] = 1 x 1 = 0 f [x 1 ] = 1 f [x 0 , x 1 , x 2 ] = 0 f [x 1 , x 2 ] = 1 f [x 0 , x 1 , x 2 , x 3 ] = 1 x 2 = 1 f [x 2 ] = 2 f [x 0 , x 1 , x 2 ] = 1 f [x 2 , x 3 ] = 2 x 3 = 1 f [x 3 ] = 2
Using the co-
efficients above, we can read off the polynomial P (x) = 1+1(x−0)+0(x−0)^2 +1(x−0)^2 (x−1) = 1 + x − x^2 + x^3. Plugging in x = 0.5 yields 1. 375
- (a) Transform
− 3
1 t^2 +1 dt^ to an integral of the form^
− 1 f^ (x)^ dx. (b) Find the coefficients below for the three-point Gaussian quadrature rule: ∫ (^1)
− 1
f (x) dx ≈ af
(c) What is this rule’s degree of precision? (d) Estimate the integral from (a) using this rule. Express your answer as a fraction. Solution:
(a) Let t = 3x to get
− 1
9 x^2 + 1 dx
(b) Method one:
f (x) = x^0 :
− 1
x^0 dx =
= +a
=⇒ a + b + c = 2
f (x) = x^1 :
− 1
x^1 dx =0 = −a
=⇒ a = c
f (x) = x^2 :
− 1
x^2 dx =
= +a
=⇒ a = c =
=⇒ b =
Method two: Integrate Lagrange polynomials: for instance, a =
− 1
(x−0)(x−
3 /5) (−
3 / 5 −0)(−
3 / 5 −
3 /5) dx, etc. (This strategy leads to a mess, and will not be elaborated here.) (c) Method one: It’s 5 because Gaussian quadrature achieves degree of precision 2(# nodes) − 1. Method two: Continue the tests above. The first one that fails is f (x) = x^6 , so the degree is 5.
(d) (^59) 9(3/^3 5)+1 + 89 (3) + (^59) 9(3/^3 5)+1 = (^92) · 32 ·^5 ·/^35 + 89 ·^3 = (^325) · 16 + 83 = 15348 =
- (a) Find the local truncation error for the method wi+1 = 4wi − 3 wi− 1 − 2 hf (ti− 1 , wi− 1 ). (b) Classify this multistep method as strongly stable, weakly stable, or unstable. Solution: (a) Note: This may be clearer if written out, long-hand. Let y(t) =
cn(t − ti)n. Then y′(t) =
ncn(t − ti)n−^1 , and
τi+1 =
y(ti + h) − wi+ h
= − 4 c 0 h−^1 +
cn[1 + 3(−1)n^ + 2n(−1)n−^1 ]hn−^1 = 0 + 0 + 0 + 4c 3 h^2 +...
Thus, the local truncation error is O(h^2 ) (b) The characteristic polynomial for this method is λ^2 − 4 λ + 3 = (λ − 3)(λ − 1). Since | 3 | > 1, the method is unstable.
Find a Cholesky (LL∗) factorization for this matrix: A =
Solution: If you remember (or re-derive, as described in class) a Cholesky factorization algorithm, you can practically write down the answer. If not, there is still hope: A = LL∗^ must take the form
a 0 0 0 b c 0 0 d e f 0 g h i j
a b d g 0 c e h 0 0 f i 0 0 0 j
a^2 ab ad ag ???? ???? ????
Without further computation, you can find a = 1, b = 0, d = 1, g = 1, and write
0 c 0 0 1 e f 0 1 h i j
0 c e h 0 0 f i 0 0 0 j
0 c^2 ce ch ???? ????
Write down c = 1, e = 0, h = 2, and
1 0 f 0 1 2 i j
0 0 f i 0 0 0 j
1 0 1 + f 2 1 + f i ????
Write down f = 1, i = 0, and
1 2 0 j
0 0 0 j
1 2 1 5 + j^2
Finally, j = 1, and the desired factorization has
L =
