APPM/MATH Problem Set 4 Solutions

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1 APPM/MATH 465 Problem Set 4 Solutions This assignment is due by 4pm on Wednesday, October 16th. You may either turn it in to me in class on Monday or in the box outside my office door (ECOT 35). 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. Let f be continuous, with two continuous derivatives, on the interval [a, b], and let s be the piecewise linear interpolant of f relative to the partition Let h = max i n 1 (x i+1 x i ). Show that a = x < x 1 < < x n 1 < x n = b. max f(x) s(x) 1 x [a,b] 8 h max f (x) x [a,b] Hint: First find a bound on the error on the interval [x i, x i+1 ] and then look at the interval [a, b]. Solution: We first notice that on an arbitrary interval [x i, x i+1 ] s (x) linearly interpolates f(x) at the nodes x i and x i+1. Thus, the error in the interpolation on the given integral is max f(x) s (x) = max f (ξ) (x x i ) (x x i+1 ) for x i < ξ < x i+1 The maximum value of (x x i ) (x x i+1 ) occurs at the midpoint between x i and x i+1, so we have max (x x i) (x x i+1 ) = = ( ) ( ) xi + x i+1 xi + x i+1 x i x i+1 ( ) ( ) xi+1 x i xi x i+1 = h i 4 Plugging this into the expression for the error on [x i, x i+1 ] and replacing ξ by x gives max f(x) s (x) = 1 8 h i max f (x) Then, we take the maximum over all subintervals in [a, b] to obtain max f(x) s(x) 1 x [a,b] 8 h max f (x) x [a,b]

2 . Consider the partition x =, x 1 =.5 and x =.1 of [,.1]. (a) Find the piecewise linear interpolating function F for f(x) = e x. (b) Approximate Solution:.1 e x dx by.1 F(x) dx and compare it to the actual value. (a) Using the general formula for a linear function on the interval [x i, x i+1 ] we have F i (x) = f(x i ) + f(x i+1) f(x i ) x i+1 x i (x x i ) = f(x i ) + f(x i+1) f(x i ) h i (x x i ) Using the given data and f(x) = e x we obtain 1 + [( e.5 1 ) /.5 ] x x.5 F(x) = e [(.1 e.1 e.5) /.5 ] x.5 x x x.5 = x.5 x.1 (b) We have.1 e x dx.1 F(x) dx = x dx = = x dx The exact value of the integral is.1 e x dx which gives a relative error of approximately

3 3. Again, consider the partition x =, x 1 =.5 and x =.1 of [,.1]. (a) Find the cubic spline s with clamped boundary conditions that interpolates f(x) = e x. (b) Approximate.1 e x dx by.1 s(x) dx and compare it to the actual value. (c) Is your approximation obtained here better or worse than in Problem? Explain your result. Solution: (a) There are numerous ways to compute the cubic spline for this problem. It is acceptable to use the code that you wrote for Problem 4. The following is one way to do it (mostly) by hand Assume that the general forms of the spline on the two subintervals are given by s (x) = a + b x + c x + d x 3 and s 1 (x) = a 1 + b 1 (x.5) + c 1 (x.5) + d 1 (x.5) 3 We now want to apply the spline conditions to determine the eight coefficients. We have Interpolation Conditions: Continuity Conditions: s () = e a = 1 (1) s (.5) = e.5 a + b (.5) + c (.5) + d (.5) 3 = e.1 () s 1 (.5) = e.5 a 1 = e.1 (3) s 1 (.1) = e.1 a 1 + b 1 (.5) + c 1 (.5) + d 1 (.5) 3 = e. (4) First note that due to the interpolation conditions in () and (3) above, the continuity condition for s(x) at the interior node is redundant. We only need to enforce the continuity conditions in the first and second derivatives. Clamped Boundary Conditions: s (.5) = s 1(.5) b + c (.5) + 3d (.5) = b 1 (5) s (.5) = s 1(.5) c 1 + 6d 1 (.5) = c 1 (6) Finally, we require that the first derivative of s(x) matches the first derivative of f at the endpoints. s () = f () b = (7) s 1(.1) = f (.1) b 1 + c 1 (.5) + 3d 1 (.5) = e. (8) We can now write (1)-(8) as a linear system and solve for the coefficients. Letting h =.5 we have 1 1 h h h h h h 3 1 h 3h 1 6h 1 1 h 3h a b c d a 1 b 1 c 1 d 1 = 1 e.1 e.1 e. e. Using your favorite linear algebra software to solve the linear system yields for the coefficients and plugging them in to the general spline form gives the following:

4 { 1 + x x 3 for x.5 s(x) = (x.5) +.58 (x.5) (x.5) 3 for.5 x.1 or s(x) = { 1 + x x x 3 for x x x x 3 for.5 x.1 (b) We have.1 e x dx.1 s(x) dx =.5 s (x) dx = = s 1 (x) dx which has a relative error of approximately when compared to the exact answer. (c) The approximation of the integral using the clamped cubic spline is significantly more accurate than the piecewise linear approximation. This is due to the use of the derivative data in the clamped cubic spline. Since we use more information about the function, it gives us a better interpolation, and this leads to a better approximation of the integral. Note: In general, it is not necessarily the case that a cubic spline interpolation rule will lead to a better integral approximation. In numerical integration, the goal is to interpolate the function as accurately as possible. With cubic spline interpolation we spend a lot of effort ensuring that the interpolant is smooth at the interior boundaries, which isn t really important for accurate integration. In fact, the natural cubic spline interpolation for this problem performs worse than the piecewise linear approximation. We will see in the chapter on number integration that our effort would be better spent constructing a Lagrange interpolating polynomial on each subinterval and using that to approximate the integral.

5 4. Use a computer to construct the cubic spline interpolant of the function f(x) = x sin (πx) on the interval [, ] using n = 1 subintervals for both natural and clamped boundary conditions. (a) Use a computer to produce the following plots on a grid with at least 5 points on [, ]: i. The interpolating cubic spline with natural boundary conditions ii. The interpolating cubic spline with clamped boundary conditions iii. The error e (x) = f(x) s(x) for the cubic spline with natural boundary conditions iv. The error e (x) = f(x) s(x) for the cubic spline with clamped boundary conditions (b) Which boundary conditions yielded a more accurate approximation? Why do you think this is? (c) Using your code, compute a reasonable approximation of the maximum error e n = max x [,] f(x) s(x) for clamped boundary conditions with n subintervals for n =, 3, 4, 5. Put your results in a table with columns h, e n, and e n /e n 1. (d) Theorem 3.13 of your textbook gives the following formula for the error in the clamped cubic spline interpolant: f(x) s(x) = h4 max x [a,b] f (x) Explain how the error ratios you computed in (c) agree with this error formula. Hint: You don t need to compute max f (x), just think about what s happening with the h 4 term. Solution: (a) The plots are as follows.5 Natural Cubic Spline.6 Error in Natural Cubic Spline

6 .5 Clamped Cubic Spline.15 Error in Clamped Cubic Spline (b) We can see from the plots that the clamped cubic spline produces an interpolation with less error than the natural cubic spline. This is because the clamped cubic spline uses derivative information from the function. More function information = better interpolation! (c) The table is as follows: h e n e n /e n e e e-6.63 According to the error formula the error should decrease like O ( h 4). This means that if we reduce the subinterval length h by a factor of we expect the error to go down by a factor of 4 = 16. The error ratios computed in part (c) appear to be converging to 1/ In fact, if we continue decreasing the node spacing by a factor of we get the following additional rows in the table, which show very clearly that the error ratios are converging to 1/16: h e n e n /e n e e e-9.65