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Material Type: Exam; Class: Numerical Analysis; Subject: Mathematics; University: University of California - Berkeley; Term: Summer 2009;
Typology: Exams
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Questions (1) Fill in the table as accurately as possible using two-point formulas (including the centered difference formula) for f ′: f (−1) = 6 f (0) = 1 f (1) = 2 f ′(−1) = f ′(0) = f ′(1) = (2) Do you remember the definitions of the following quadrature methods? Newton-Cotes, composite trapezoidal, Romberg, adaptive Simpson, Gaussian. (3) Find coefficients a, b, and c so that the following quadrature rule has the best possible degree of precision: (^) ∫ 1
0
f (x) dx ≈ af (−1) + bf (0) + cf (+1).
What is the resulting rule’s degree of precision? (4) Use an easy substitution to transform the improper integral
1
1 x^2 +1 dx^ into an integral of the form
0 f^ (u)^ du.^ Transform the result into an integral which can be estimated with the following quadrature rule: (^) ∫ 1
− 1
g(t) dt ≈ f (−1) + f (+1).
(5) Use Euler’s method with h = 1 to estimate the solution to y′^ = 2t + y/t, 1 ≤ t ≤ 3, y(1) = 1. (6) Find a Lipschitz constant L for the above problem. (7) Find y′′^ in terms of t and (if necessary) y, and compute an upper bound M on |y′′(t)|. (8) Find a bound on the local truncation error τ when using Euler’s method to solve this problem. (9) Find a bound for the absolute error in estimating y(3), using the error bound hM 2 L (eL(ti−t^0 )^ − 1). (10) Write down a step (wi+1 as a function of ti, wi, h) of the second-order Taylor method for solving the IVP from problem 5. (11) Based on your answers to problem 5, estimate y(1.5) based on y(1), y′(1), y(2), and y′(2). (12) Express the following Runge-Kutta method in one line (i.e., as a single formula for wi+1):
k 1 = hf (ti, wi) k 2 = hf (ti + h, wi + k 1 ) wi+1 = wi + k 2
(13) What is the order of the Runge-Kutta method in problem 12? (14) (Skip if confused.) Consider Romberg integration with the following notation:
∫ (^) b a f^ (x)^ dx^ with step-size^ hi;
∫ (^) b a f^ (x)^ dx^ extrapolated from^ Ri,j^ and^ Ri+1,j^ , assuming^ Ri,j^ has error O(h^2 i j). Show that Ri+1,j+1 is equal to Ri+1,j + h
(^2) i h^2 i−j −h^2 i^ (Ri+1,j^ −^ Ri,j^ ).
Date: Monday 7/26. 1
Answers
(1)
f (−1) = 6 f (0) = 1 f (1) = 2 f ′(−1) ≈ − 5 f ′(0) ≈ − 2 f ′(1) ≈ 1 (2) Yes. (3) (a, b, c) = 121 (− 1 , 8 , 5); degree is 2. (4) The substitution x = 1/u produces
0
1 1+u^2 du. Then^ u^ = (t+1)/2 produces^
− 1
1 2+(t+1)^2 / 2 dt^ ≈^0 .75. (5) y(2) ≈ 4, y(3) ≈ 10. (6)
∣ (^) ∂y∂ (2t + y/t)
∣ = | 1 /t| ≤ 1. (7) y′′(t) = 2 + y′/t − y/t^2 = 2 + (2t + y/t)/t − y/t^2 = 4. We can take M = 4. (8) Euler’s method generally has τi+1 = yi+1− hw i+1= y(ti+h)−[y(ti)+hy
′(ti)] h =^
y′′(ξ)h^2 / 2 h =^
h 2 y
′′(ξ). Thus we have τ = 2h = 2, at every step. (9) 2(e^2 − 1). (Numerically, this is something smaller than 2(3^2 − 1) = 16.) The actual error is much better. (10) wi+1 = wi + h[2ti + wi/ti] + h
2 2 [4]. (11) The divided difference table below (see section 3.3) says y(t) ≈ 1+3(t−1)+0(t−1)^2 +3(t−1)^2 (t−2): t = 1 1 3 t = 1 1 0 3 3 t = 2 4 3 6 t = 2 4
Plugging in t = 1.5 gives y(1.5) ≈ 2 .125.
(12) wi+1 = wi + hf (ti + h, wi + hf (ti, wi)). (13) The method is first-order (i.e., τi+1 = O(h)). In particular, the O(h^2 ) Taylor coefficient to wi+1 is twice that of y(ti + h).
(14) This is algebra, but the important step is to realize Ri+1,j+1 = Ri+1,j − h^2 i+1j h^2 i+1j−h^2 ij^ (Ri+1,j^ −^ Ri,j^ ).
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