The Intermediate Value Theorem states that if a function g is continuous, then for any number M satisfying. g(x 1 ) M g(x 2 )

Transcription

1 APPM/MATH 450 Problem Set 5 s This assignment is due by 4pm on Friday, October 25th. You may either turn it in to me in class or in the box outside my office door (ECOT 235). Minimal credit will be given for incomplete solutions or solutions that do not provide details on how the solution is found. For problems that require you to implement a method, you may use code from our textbook website faires/numerical-analysis/programs/, although they may require some modification. For your own personal understanding of the methods involved, it is highly recommended that you attempt your own implementations. 1. Consider the derivation of the second-order accurate Centered Difference Formula using Taylor Expansion. If we re very particular about the handling of the error terms we might obtain the following formula f (x) = f(x + h) f(x h) h2 12 [f (ξ 1 ) + f (ξ 2 )] where ξ 1 (x, x + h), ξ 2 (x h, x) Assuming that f is continuous, use a common theorem from calculus to show that we are justified in writing the formula with a simplified error term as f (x) = f(x + h) f(x h) h2 f (ξ) where ξ (x h, x + h) We need to show that there exists some ξ (x h, x + h) such that 1 2 [f (ξ 1 ) + f (ξ 2 )] = f (ξ) The Intermediate Value Theorem states that if a function g is continuous, then for any number M satisfying g(x 1 ) M g(x 2 ) there exists a c in [x 1, x 2 ] such that g(c) = M. Here, we notice that [f (ξ 1 ) + f (ξ 2 )] /2 is the average of f (x 1 ) and f (x 2 ). Thus, if we (without loss of generality) assume that f (ξ 1 ) f (ξ 2 ), then we have f (ξ 1 ) 1 2 [f (ξ 1 ) + f (ξ 2 )] f (ξ 2 ) Then, by the Intermediate Value Theorem, we have that there must exist a ξ in [ξ 1, ξ 2 ] such that 1 2 [f (ξ 1 ) + f (ξ 2 )] = f (ξ) Finally, since ξ 1 (x h, x) and ξ 2 (x, x + h), we have that ξ (x h, x + h)

2 2. The following second-order accurate Backward Difference Formula formula was discussed in class f (x n ) f(x n ) 4f(x n h) + 3f(x n ). (a) Derive the formula and its corresponding truncation error using Taylor Expansion. We begin by expanding f(x n h) and f(x n ) about x n. We have f(x n h) = f(x n ) hf (x n ) + h2 2 f (x n ) h3 f (ξ) (1) f(x n ) = f(x n ) f (x n ) + 2 f (x n ) 4h3 3 f (ξ) (2) We notice that if we take 4 times the first equation and subtract the second equation the O ( h 2) terms will cancel out. Solving for f (x n ) we have 4f(x n 2) f(x n ) = 3f(x n ) f (x n ) f (ξ) f (x n ) = f(x n ) 4f(x n h) + 3f(x n ) + h2 3 f (ξ) for some ξ (x n, x n + ) (b) Derive the formula and its corresponding truncation error using Lagrange interpolation. Since this is a 3-point formula, we interpolate f on the nodes located at (x n ), (x n h), and x n, which we now relabel as x n 2, x n 1 and x n, respectively. The interpolating polynomial at these nodes, along with its error term, is give by f(x) = n j=n 2 L 2,j (x) f(x j ) + f (ξ(x)) (x x n 2 ) (x x n 1 ) (x x n ) Taking the derivative we have f (x) = n j=n 2 L 2,j(x) f(x j ) + d [ f ] (ξ(x)) (x x n 2 ) (x x n 1 ) (x x n ) dx The first term will give us the finite difference approximation and the second term will give us the error. Let s deal with the approximation first. We need to compute the derivatives of the Lagrange basis functions. We have L 2,n 2 (x) = L 2,n 1 (x) = (x x n 1 ) (x x n ) (x n 2 x n 1 ) (x n 2 x n ) (x x n 2 ) (x x n ) (x n 1 x n 2 ) (x n 1 x n ) L 2,n (x) = (x x n 2) (x x n 1 ) (x n x n 2 ) (x n x n 1 ) L 2,n 2(x) = (x x n 1) + (x x n ) (x n 2 x n 1 ) (x n 2 x n ) L 2,n 1(x) = (x x n 2) + (x x n ) (x n 1 x n 2 ) (x n 1 x n ) L 2,n(x) = (x x n 2) + (x x n 1 ) (x n x n 2 ) (x n x n 1 )

3 Recalling the definitions of x n 1 and x n 2 in terms of h, and evaluating the basis functions at x n we have L 2,n 2(x n ) = h ( h) ( ) = 1 L 2,n 1(x n ) = (h) ( h) = 2 h L 2,n(x n ) = + h () (h) = 3 Then the approximation term becomes n j=n 2 L 2,j(x n ) f(x j ) = f(x n 2) 2f(x n 1) h + 3f(x n) = f(x n ) 4f(x n h) + 3f(x n ) Now for the error term. We differentiate using the product rule [ d f ] (ξ(x)) (x x n 2 ) (x x n 1 ) (x x n ) = d dx dx + f (ξ(x)) [ f (ξ(x)) ] (x x n 2 ) (x x n 1 ) (x x n ) [(x x n 1 ) (x x n ) + (x x n 2 ) (x x n ) + (x x n 2 ) (x x n 1 )] Plugging in x = x n and letting ξ = ξ(x n ) we see that the first term is zero, and the second term is f (ξ) [ () (h)] = h2 3 f (ξ) Note that this is identical to the results of the Taylor Expansion derivation in part (a).

4 3. (a) Using Taylor Expansion, derive the following four-point Forward Difference approximation to the second derivative along with the corresponding truncation error: f (x 0 ) 2f(x 0) 5f(x 0 + h) + 4f(x 0 + ) f(x 0 + 3h) h 2 Expanding the three relevant terms in the formula using Taylor Expansion, we have f(x 0 + h) = f(x 0 ) + hf (x 0 ) + h2 2 f (x 0 ) + h3 f (x) + h4 24 f (ξ) (3) f(x 0 + ) = f(x 0 ) + f (x 0 ) + f (x 0 ) + 4h3 3 f (x) f (ξ) (4) f(x 0 + 3h) = f(x 0 ) + 3hf (x 0 ) + 9h2 2 f (x 0 ) + 9h3 2 f (x) + 27h4 8 f (ξ) (5) Then 2f(x 0 ) 5f(x 0 + h) + 4f(x 0 + ) f(x 0 + 3h) = h 2 f (x 0 ) h4 f (ξ) Dividing by h 2 and solving for f (x 0 ) gives f (x 0 ) = 2f(x 0) 5f(x 0 + h) + 4f(x 0 + ) f(x 0 + 3h) h h2 f (ξ) Notice that the error term is O ( h 2), so the given difference formula is second-order accurate. (b) Using step sizes of h = 2 n for n = 1,..., 8, verify numerically that the given approximation has second-order accuracy by approximating f (1) for f(x) = 1/x. Clearly explain your method of verification and include all relevant numerical results. To verify that the method is second-order accurate we will approximate f (1) for step sizes that decrease by a factor of 2 and look at the ratios of the error. Since the method is second-order we expect that reducing the step size by 2 should result in a decrease in the error by a factor of 4. Thus, we should have e n 1 e n 4 or e n e n We have the following table h f (1) e n e n 1 /e n e n /e n

5 4. In this problem you will use Richardson s Extrapolation to compute higher-order approximations of f (x). Let N 1 (h) be the second-order centered difference approximation N 1 (h) = f(x + h) f(x h) (a) Use Richardson s Extrapolation to derive the difference formula N 2 (h) (in terms of N 1 ) and give the corresponding order of accuracy. Let M = f (x). Upon close examination of the Taylor Expansion derivation of the centered difference approximation to f (x) we see that the only error terms that survive the subtraction have even powers (e.g. h 2,, etc). Then, we can write the approximation and it s error terms using step-size h and h/2: M = N 1 (h) + K 1 h 2 + K 2 + K 3 + () M = N 1 (h/2) + K 1 h K K (7) where the K j s are some coefficients that we could find from the Taylor Expansion, but are unimportant here. Multiplying the second expression by 4 and subtracting the first will eliminate the h 2 term, so we have Dividing through by 3 we have 3M = [4N 1 (h/2) N 1 (h)] K K M = [4N 1(h/2) N 1 (h)] 3 K 2 4 K Letting N 2 (h) = [4N 1 (h) N 1 (h)] /3 we have M = N 2 (h) K 2 4 K From this we can see that N 2 (h) approximates f (x) with fourth-order accuracy. (b) Extrapolate one more time to derive the difference formula N 3 (h) (in terms of N 2 ) and give the corresponding order of accuracy. From part (a) we see that our extrapolated difference formula N 2 (h) has an error expansion that looks like M = N 2 (h) + E 1 + K 2 + (8) M = N 2 (h/2) + E E (9) where again E 1 and E 2 are some coefficients. Multiplying the second equation by 1 and subtracting the first we have

6 15M = [1N 2 (h/2) N 2 (h)] E Dividing through by 15 and defining N 3 (h) = [1N 2 (h/2) N 2 (h)] /15 we have M = N 3 (h) E So the difference operator defined by N 3 (h) has sixth-order accuracy! (c) Compute the extrapolation table that approximates f (2) for f(x) = ln x using the difference formulas N 1 (h), N 2 (h) and N 3 (h). Use mesh sizes of h = 2 n for n = 1,...,. Make sure to display enough decimal places to accurately show the approximation. Note that the exact answer is f (2) = 1/2. The extrapolation table is as follows h N 1 N 2 N (d) Compute the absolute error in the N 3 approximation and display the relevant error ratios. Do your results agree with what you predicted in part (b)? We found in part (c) that the third extrapolation method N 3 was sixth-order accurate. This suggests that when we divide the step size h by 2 we expect the error to go down by a factor of 2 = 4. The relevant errors and error ratios are given in the following table. h e n e n 1 /e n