Homework and Computer Solutions for Math*2130 (W17). MARCUS R. GARVIE 1 December 21, 2016 1 Department of Mathematics & Statistics, University of Guelph STOP! Before looking at the answers the best study strategy is to: 1. First read your lecture notes for the relevant section. 2. Then attempt the questions without first looking at the answers. 3. Finally, if you are stuck, look at the general approach in the relevant answer and try again. Remember, some struggle is often necessary for effective learning. 1 Chapter 1 Basic Tools 1.4 Taylor’s Theorem 1. What is the third-order Taylor polynomial for f (x) = x0 = 0? Solution: P3 (x) = f (0) + xf 0 (0) + x3 x2 00 f (0) + f 000 (0) 2! 3! f (0) = 1 1 1 1 (x + 1)− 2 so f 0 (0) = f 0 (x) = 2 2 3 1 1 f 00 (x) = − (x + 1)− 2 so f 00 (0) = − 4 4 5 3 3 f 000 (x) = (x + 1)− 2 so f 000 (0) = 8 8 Substituting in (1.1) yields 1 1 1 P3 (x) = 1 + x − x2 + x3 . 2 8 16 2. Given that R(x) = 2 √ |x|6 ξ e 6! x + 1, about (1.1) for x ∈ [−1, 1], where ξ is between x and 0, find and upper bound for |R|, valid for all x ∈ [−1, 1], that is independent of x and ξ. Solution: Note first that x ∈ [−1, 1] ⇐⇒ |x| ≤ 1. so |x|6 · eξ 1 · e1 e ≤ = ≈ 0.00378. 6! 6! 6! We also used the fact that the function eξ is a monotonic increasing function of ξ. 3. Given that |x|4 −1 R(x) = 4! 1 + ξ 1 1 for x ∈ − , , where ξ is between x and 0, find and upper bound 2 2 1 1 for |R|, value for all x ∈ − , , that is independent of x and ξ. 2 2 Solution: We use the following facts 1 1 1 (i) x ∈ − , ⇐⇒ |x| ≤ 2 2 2 1 (ii) for 0 < z < , y = z 4 is increasing, thus for z1 < z2 , we have that 2 4 1 z14 < z24 , e.g., z 4 < 2 1 1 1 for ξ ∈ − , is maximized for ξ that minimizes |1 + ξ|, (iii) |1 + ξ| 2 2 1 i.e. when ξ = − . Thus 2 |x|4 4! −1 1+ξ 1 4 1 4 1 4 |x|4 2 2 2 = ≤ ≤ ≤ = 4!|1 + ξ| 4!|1 + ξ| 4! 21 4! 1 + − 21 3 1 3 2 4 = 1 . 192 4. What is the fourth-order Taylor polynomial for 1 about x0 = 0? x+1 Solution: P4 (x) = f (0) + xf 0 (0) + f (0) f 0 (x) f 00 (x) f 000 (x) f (iv) (x) = = = = = x3 x4 x2 00 f (0) + f 000 (0) + f (iv) (0) 2! 3! 4! (1.2) 1 −(x + 1)−2 so f 0 (0) = −1 2(x + 1)−3 so f 00 (0) = 2 −6(x + 1)−4 so f 000 (0) = −6 24(x + 1)−5 so f (iv) (0) = 24 Substituting in (1.2) yields P4 (x) = 1 − x + x2 − x3 + x4 . 5. Find the Taylor polynomial of third-order for sin(x) using x0 = π . 2 Solution: 2 3 π x − π2 π 0 π x − π2 00 π 000 π P3 (x) = f + x− f + f + f 2 2 2 2! 2 3! 2 (1.3) f (π/2) f 0 (x) f 00 (x) f 000 (x) = = = = 1 cos(x) so f 0 (π/2) = 0 − sin(x) so f 00 (π/2) = −1 − cos(x) so f 000 (π/2) = 0 Substituting into (1.3) yields x − π2 P3 (x) = 1 − 2 4 2 . 6. For the function below construct the third-order Taylor polynomial approximation, using x0 = 0, and then estimate the error by computing an upper bound on the remainder, over the given interval: f (x) = ln(1+x), 1 1 x∈ − , . 2 2 Solution: P4 (x) = f (0) + xf 0 (0) + x3 x4 x2 00 f (0) + f 000 (0) + f (iv) (ξ) 2! 3! 4! (1.4) for ξ between 0 and x. f (x) = ln(1 + x) so f (0) = ln(1) = 0 1 f 0 (x) = so f 0 (0) = 1 1+x f 00 (x) = −(x + 1)−2 so f 00 (0) = −1 f 000 (x) = 2(x + 1)−3 so f 000 (0) = 2 6 6 f (iv) (x) = −6(x + 1)−4 = − so f (iv) (ξ) = − 4 (1 + x) (1 + ξ)4 Substituting into (1.4) yields ln(1 + x) = x − x2 x3 x4 + − . 2 3 4(1 + ξ)4 x2 x3 So P3 (x) = x − + . For the error bound, first note that x ∈ 2 3 1 1 1 − , ⇐⇒ |x| ≤ . 2 2 2 1 4 1 4 1 4 |x|4 1 2 2 2 |R3 (x)| ≤ ≤ ≤ 4 ≤ 4 = 4 4 1 1 4(1 + ξ) 4(1 + ξ) 4 4 1 + −2 4 2 √ 7. Find the Taylor Series expansion of f (x) = 3 x of degree 2 about x0 = 8. Using the remainder term estimate the maximum absolute error on the interval [7, 9]. 5 Solution: Calculate f (x) = x1/3 =⇒ f (8) = 81/3 = 2, 1 1 f 0 (x) = x−2/3 =⇒ f 0 (8) = , 3 12 2 −5/3 1 00 00 f (x) = − x =⇒ f (8) = − , 9 144 10 f 000 (x) = x−8/3 . 27 Thus the 2nd degree polynomial is (x − 8)2 00 f (8) 2! (x − 8) (x − 8)2 1 =2+ − 12 2! 144 (x − 8) (x − 8)2 − . =2+ 12 288 Using the Lagrange form of the remainder T2 (x) = f (8) + (x − 8)f 0 (8) + (x − 8)3 10 −8/3 (x − 8)3 000 f (ξ) = ξ R2 (x) = 3! 3! 27 5(x − 8)3 = , ξ between x and 8. 81ξ 8/3 I.e., we maximize |R2 (x)| = 5|x − 8|3 , 81|ξ|8/3 ξ between x and 8. 1 Now note that as x is between 7 and 9 so is ξ. Thus as ξ8/3 is decreasing 1 1 on [7, 9], it follows that |ξ|8/3 is maximized on [7, 9] by 78/3 . Also, powers of x (or powers of (x − c), for some c) are maximized at the end points. Just think of the general shape of x2 or x3 . So |x − 8|3 is maximized on [7, 9], at x = 7 or x = 9. Here, we get the same answer for both values. Thus 5|9 − 8|3 5|7 − 8|3 or 81 · 78/3 81 · 78/3 = 29/84238 = 3.4426 × 10−4 |R2 (x)| ≤ 6 (4 d.p.). 8. Construct a Taylor polynomial approximation that is accurate to within 10−3 , over the indicated interval, for the following function, using x0 = 0: f (x) = e−x , x ∈ [0, 1]. Solution: As in an example from the lecture notes, we need n such that |Rn (x)| ≤ 10−3 for all x ∈ [0, 1]. n+1 n+1 x 1 1 · e0 −ξ −ξ = |x| e e = ≤ (n + 1)! (n + 1)! (n + 1)! (n + 1)! 1 ≤ 10−3 , we find that n = 6 (n + 1)! is the minimum value to do this (get about 1.984 × 10−4 ), thus we must find P6 (x). Since f (n) (x) = (−1)n e−x for all n we have that f (n) (0) = (−1)n for all n. Thus Plugging in values of n to make P6 (x) = 1 − x + x2 x3 x4 x5 x6 − + − + . 2 3! 4! 5! 6! 9. For each function below, use the Mean Value Theorem to find a value M such that |f (x1 ) − f (x2 )| ≤ M |x1 − x2 | that is valid for all x1 , x2 in the stated interval: (a) f (x) = ln(1 + x) on the interval [−1, 1], (b) f (x) = sin(x) on the interval [0, π]. Solution: From the notes, |f (x2 ) − f (x1 )| ≤ |f 0 (ξ)| · |x2 − x1 |. For (a): [x1 , x2 ] = [−1, 1] and ξ ∈ [−1, 1]. 1 1+ x 1 1 0 =∞ =⇒ |f (ξ)| = ≤ 1 + ξ 1 − 1 f (x) = ln(1 + x) =⇒ f 0 (x) = 7 So M does not exist. For (b): [x1 , x2 ] = [0, π] and ξ ∈ [0, π]. f (x) = sin(x) =⇒ f 0 (x) = cos(x) =⇒ |f 0 (ξ)| = | cos(x)| ≤ 1 = M 10. A function is called monotone on an interval if its derivative is strictly positive or strictly negative on the interval. Suppose f is continuous and monotone on the interval [a, b], and f (a)f (b) < 0; prove that there is exactly one value α ∈ [a, b] such that f (α) = 0. Solution: If f is monotone on [a, b] and f (a)f (b) < 0 then either f (a) < 0 and f (b) > 0 or f (a) > 0 and f (b) < 0. Thus, as f ∈ C([a, b]), by the Intermediate Value Theorem ∃α ∈ [a, b] such that f (α) = 0. 11. (MATLAB) Use the program GRAPH_PRES to estimate the (true) maximum absolute error in using the Taylor polynomial of Question 7 to √ 3 approximate x on [7, 9]. √ Solution: Just plot the exact error given by |T2 (x)− 3 x| over the interval [7, 9]. Visually estimate the maximum absolute error and compare this value with what you estimated was the maximum theoretical error using R2 (x). So from the graph below: 8 we see that the true maximum error on [7, 9] is about 2.6 × 10−4 . The theoretical maximum error is estimated to be ≤ 3.4 × 10−4 , which is not bad. 12. (MATLAB) Use the program GRAPH_MANY_PRES to graphically investigate how well the Taylor series Tn (x) (for n = 1, 2, . . . ) approximates the functions f (x) near x = x0 (see Questions 1, 4, 5, 6, 7, 8). Solution: On a given interval [a, b] containing x0 plot Tn (x), n = 1, 2, . . . , with y = f (x), and observe how the fit between the Taylor series near x0 gets better as n increases. An example (see Question 4) is given below: 9 1.5 Error and Asymptotic Error 1. Use Taylor’s Theorem to show that √ 1 1 + x = 1 + x + O(x2 ), 2 for x sufficiently small. Solution: x0 = 0 and x2 00 f (0) + ... 2 = f (0) + xf 0 (0) + O(x2 ) f (x) = f (0) + xf 0 (0) + Calculating coefficients: 1 f (x) = (1 + x) 2 1 1 f 0 (x) = (1 + x) 2 2 Substituting into (1.5) yields √ 1+x=1+ 10 =⇒ f (0) = 1 1 =⇒ f 0 (0) = 2 1 x + O(x2 ). 2 (1.5) 2. Use Taylor’s Theorem to show that ex = 1 + x + O(x2 ), for x sufficiently small. Solution: x0 = 0 and f (x) = f 0 (x) = ex =⇒ f (0) = f 0 (0) = 1 Thus, as in #1, ex = 1 + x + O(x2 ). 3. Use Taylor’s Theorem to show that 1 1 − cos(x) = x + O(x3 ), x 2 for x sufficiently small. Solution: x0 = 0 and 1 2 1 4 x + x − ... 2! 4! 1 2 1 4 =⇒ 1 − cos(x) = x − x + ... 2! 4! 1 − cos(x) 1 1 3 =⇒ = x − x + ... x 2! 4! 1 = x + O(x3 ). 2 cos(x) = 1 − 4. Show that sin(x) = x + O(x3 ). Solution: x0 = 0 and sin(x) = x − 1 3 x + ... = x + O(x3 ). 3! 11 5. Recall the summation formula 2 3 n 1 + r + r + r + ... + r = n X rk = k=0 1 − rn+1 , 1−r r 6= 1. Use this to prove that for |r| < 1 n X k=0 rk = 1 + O(rn+1 ), 1−r Hint: What is the definition of O notation? Solution: n X 1 − rn+1 1 rn+1 r = = − ,. 1−r 1−r 1−r k=0 k Need to show that rn+1 = O(rn+1 ), for n large. 1−r Using the definition of O notation and the fact that |r| < 1, rn+1 1 1−r = 1 < ∞. lim n+1 = lim n→∞ 1 − r n→∞ r 1−r 6. Use the above result to show that 9 terms are all that is needed to compute ∞ X S= e−k k=0 −4 to within 10 absolute accuracy. Solution: n X k=0 rk = 1 − rn+1 1 rn+1 rn+1 = − =S− , for |r| < 1. 1−r 1−r 1−r 1−r 12 Need to show that n+1 n n+1 X r −4 k S − = r = r 1 − r 1 − r < 10 , k=0 for r = 1 and n = 9. e rn+1 e−10 = = 7.2 × 10−5 < 10−4 . 1−r 1 − e−1 7. Recall the summation formula S= n X k= k=1 n(n + 1) . 2 Use this to show that n X 1 k = n2 + O(n). 2 k=0 Solution: n X k=0 k= n(n − 1) n2 n 1 = + = n2 + O(n). 2 2 2 2 8. Use the definition of O to show that if y = yh + O(hp ), then hy = hyh + O(hp+1 ). y − yp Solution: Given y = yh + O(h ), i.e. lim p = C < ∞. But h→0 h hy − hyh y − yh = lim = C < ∞, lim h→0 h→0 hp hp+1 p i.e. hy = hyh + O(hp+1 ). 13 9. Show that if an = O(np ) and bn = O(nq ) then an bn = O(np+q ). Solution: Given that an = O(np ) and bn = O(nq ), i.e. an bn lim p = C1 < ∞, and lim q = C2 < ∞, n→∞ n n→∞ n then an b n an bn lim p+q = lim p · lim q = C1 C2 < ∞, n→∞ n n→∞ n n→∞ n i.e. an bn = O(np+q ). 10. (MATLAB) Recall that the infinite (classic) geometric series is S = 1 + r + r2 + r3 + · · · = 1 , 1−r |r| < 1. In fact we saw in Question 5 that S = 1 + r + r2 + r3 + . . . rn + O(rn+1 ), i.e., the error in using only n + 1 terms in the approximation of S is O(rn+1 ). The MATLAB program GEOMETRIC_PRES sums this finite n X rk < tol, where ‘tol’ series for a given r until the absolute error S − k=0 is a tolerance supplied by the user. Use this program with tol = 1×10−6 and (a) r = 0.001 and (b) r = 0.999. Explain the difference in results for these two cases. Solution: Running the MATLAB code yields that the number of terms needed to achieve the required accuracy is 3 for case (a) and 20713 for case (b). The series with r = 0.999 clearly converged much more slowly than the series with r = 0.001 as the error term for the former is O(0.999n+1 ) while for the latter it is O(0.001n+1 ), and an order 0.001n+1 term goes to zero much faster than an order 0.999n+1 term as n → ∞. 14 1.6 Computer Arithmetic 1. Perform the indicated computations in each of three ways: (i) Exactly; (ii) Using 3 digit decimal arithmetic, with chopping; (iii) Using 3 digit decimal arithmetic, with rounding. For both approximations, compute the absolute error and the relative error. 1 1 1 1 1 + (b) + + (a) 6 10 7 10 9 Solution: (a) (i) 1 1 + = 0.26̇ 6 10 (ii) 1 1 fl fl + fl = 6 10 = = = f l(f l(0.16̇) + f l(0.1)) f l(0.166 + 0.100) f l(0.266) 0.266 (chopping) absolute error = |0.26̇ − 0.266| = 0.0006̇ 0.0006̇ = 2.5 × 10−3 relative error = 0.26̇ (2 significant figures) (iii) 1 1 fl fl + fl = 6 10 = = = f l(f l(0.16̇) + f l(0.1)) f l(0.167 + 0.100) f l(0.267) 0.267 (rounding) absolute error = |0.26̇ − 0.267| = 3.3̇ × 10−4 0.0003̇ relative error = = 1.25 × 10−3 (2 significant figures) 0.26̇ 15 (b) (i) 1 1 + 7 10 + 1 = 0.3539682 9 (ii) 1 1 fl fl + fl = f l(f l(0.142) + 0.100) 7 10 = f l(0.242) = 0.242 so 1 f l 0.142 + f l = f l(0.142 + 0.111) 9 = f l(0.353) = 0.353 (chopping) absolute error = |0.3539682 − 0.353| = 9.682 × 10−4 9.683... × 10−4 relative error = = 2.735 × 10−3 (4 significant figures) 0.3539682 1 (iii) Rounding obtained via the same method, but as f l = 7 0.143, this leads to the answer 0.354. Thus, absolute error = |0.3539682 − 0.354| = 3.175 × 10−5 3.175... × 10−5 relative error = = 8.969 × 10−5 (4 significant figures) 0.3539682 ex − 1 , how many terms in the Taylor expansion are needed x 1 to get single precision accuracy (7 decimal digits) for all x ∈ 0, ? 2 How many terms are needed for double precision accuracy (14 decimal digits) over this same range? 2. For f (x) = Solution: Recall ex = 1 + x + x2 xn xn+1 ξ + ... + + e, 2! n! (n + 1)! 16 1 for ξ between 0 and x, and x ∈ 0, . Rearranging: 2 x x2 xn−1 xn ex − 1 =1+ + + ... + + eξ . x 2! 3! n! (n + 1)! | {z } Rn (x) For single precision (i.e., rounding to t = 7 decimal places) find n such that (see your lecture notes for the formula we use) |Rn (x)| < 5 × 10−(t+1) = 5 × 10−8 , 1 for all x ∈ 0, . Thus 2 1 n 21 n ξ xn e |x| e ξ 2 = e ≤ ≤ 1.78 × 10−8 (n + 1)! (n + 1)! (n + 1)! if n = 8 (and > 5 × 10−8 if n = 7). So we need 9 terms or n = 8. For double precision we need n such that (see your lecture notes for the formula we use) |Rn (x)| < 5 × 10−(t+1) = 5 × 10−15 , where t = 14. Same procedure as before leads to |Rn (x)| ≤ 2.308 × 10−15 if n = 13. So we need n = 14 terms or n = 13. 3. Using 3−digit decimal arithmetic, find three values a, b, and c, such that (a + b) + c 6= a + (b + c). Solution: Let a = 1 × 10−20 , b = 1 and c = −1. Then (a + b) + c = (1 × 10−20 + 1) − 1, which with 3−digit decimal arithmetic yields 0. However, a + (b + c) = 1 × 10−20 + (1 − 1) with 3−digit decimal arithmetic yields 1 × 10−20 . 17 4. Assume we are using 3−digit decimal arithmetic. For = 0.0001 and a1 = 5, compute 1 a2 = a0 + a1 for a0 equal to each of 1, 2, and 3. Comment. Solution: a0 = 1 1 fl = 1 × 104 0.0001 f l(5 × 104 ) = 50000 f l(1 + 5 × 104 ) = f l(50001) = 50000. a0 = 2 . . . f l(2 + 5 × 104 ) = f l(50002) = 50000 a0 = 3 . . . f l(3 + 5 × 104 ) = f l(50003) = 50000 a2 is the same in all three cases. 5. What is the machine epsilon for a computer that uses binary arithmetic, 24 bits for the fraction, and rounds? What if it chops? Solution: β = 2 and t = 24: From the lecture notes 1 1 δ = β 1−t = × 21−24 = 0.596 × 10−7 2 2 δ = β 1−t = ×21−24 = 0.119 × 10−7 (rounding) (chopping) 6. What is the machine epsilon for a computer that uses binary arithmetic, 24 bits for the fraction, and rounds? What if it chops? 18 7. (MATLAB) Perform the following calculation in MATLAB: >> 1000.00000000000000005-1000 What does MATLAB give as the answer? What is the exact answer? Explain the discrepancy. Solution: MATLAB gives 0 as an answer, however the actual answer is 5 × 10−17 . The discrepancy is due to the fact that MATLAB works in double precision (rounding numbers to about 16 significant figures) and so MATLAB stores the number 1000.00000000000000005 as 1000. This is a classic problem of loss of significance due to subtracting two nearly equal numbers. To be precise: f l(f l(1000.00000000000000005) − f l(1000)) = f l(1000 − 1000) = f l(0) = 0. 8. (MATLAB) Perform the following calculation in MATLAB: >> format long >> 0.12345678-0.12345677 What does MATLAB give as the answer? What is the exact answer? Explain the discrepancy. Solution: MATLAB gives 1.000000000861423e − 08 as an answer (i.e. 1.000000000861423 × 10−8 ). The actual answer is of course 1 × 10−8 . What is puzzling about this is that both numbers in the calculation only involve 8 significant figures which one might argue should be stored exactly in MATLAB (as it uses double precision). However, MATLAB works internally in binary arithmetic and even the number 0.1 has an infinite binary expansion (which is 0.000110011 . . .). Thus loss of significance is again due to finite precision (in the binary system). 19 9. (MATLAB) Open the MATLAB program EPS_CALC_PRES by typing: >> open eps_calc_pres Explain what the program does. If necessary type >> help while Also run the program and compare the answer you get with the value of machine epsilon in MATLAB, obtained by simply entering the commands >> format long >> eps Solution: We approximate machine epsilon M using the definition that M is the smallest positive number x such that f l(1 + x) > 1. We start with the number 1 and keep halving it until the condition is satisfied. The final value is multiplied by 2 as the test 1 + x > 1 failed, indicating that we cut x in half once to often. When we run the program we get the output 2.220446049250313e − 16, which is the same answer we get if we type ‘eps’ in MATLAB. 10. (MATLAB) Recall the well-known expression for the number e: x 1 = 2.718281828459045 . . . . e = lim 1 + x→∞ x The program EXP_CALC_PRES uses this result with integer x values to approximate e. Run this program and explain the pattern in the errors. What is the optimal x value? Solution: Clearly from the tabulated results the absolute errors decrease to a minimum value for n (i.e., x) around 108 . After that the errors increase to a maximum around n ≥ 1016 . This is because for n ≥ 1016 MATLAB calculates 1/n as 0 (due to finite precision), yielding the approximation (1 + 1/n)n = 1. Thus the absolute error becomes |e − 1| = 1.718281828459 . . . , which is what we see. 20 Chapter 2 A Survey of Simple Methods and Tools 2.1 Horner’s Rule 1. Write the following polynomial in nested form: x3 + 3x + 2. Solution: 2 + 3x + x3 = 2 + x(3 + x2 ). 2. Write the following polynomial in nested form, but this time take advantage of the fact that they involve only even powers of x to minimize the 1 1 computations: 1 − x2 + x4 . 2 24 Solution: 1 1 1 − x2 + x4 = 1 + x2 2 24 1 − 2 1 2 + x . 24 1 1 3. Write the following polynomial in nested form: 1 − x2 + x4 − x6 . 2 6 Solution: 1 4 1 6 1 2 1 4 2 1 − x + x − x = 1 + x (−1) + x − x 2 6 2 6 1 1 = 1 + x2 (−1) + x2 − x2 2 6 2 21 Note: it takes longer to work out 1 1 2 . − x 1 + x x −1 + x x 2 6 4. Consider the polynomial 1 1 p(x) = 1 + (x − 1) + (x − 1)(x − 2) + (x − 1)(x − 2)(x − 4). 6 7 This can be written in “nested-like” form by factoring out each binomial term as far as it will go, thus: 1 1 p(x) = 1 + (x − 1) 1 + (x − 2) + (x − 4) . 6 7 Write the following polynomial in this kind of nested form: 6 1 5 1 1 1 p(x) = −1 + x− − x− (x − 4) + x− (x − 4)(x − 2). 7 2 21 2 7 2 Solution: 6 1 5 1 1 1 p(x) = −1 + x− − x− (x − 4) + x− (x − 4)(x − 2) 7 2 21 2 7 2 6 5 1 1 − (x − 4) + (x − 4)(x − 2) = −1 + x − 2 7 21 7 1 6 5 1 = −1 + x − + (x − 4) − + (x − 2) 2 7 21 7 2.2 Difference Approximations 1. Compute, by hand, approximations to f 0 (1) for each of the following functions, with h = 1/16 using the Forward and Centered Difference Approximations: (a) f (x) = arctan(x), (b) f (x) = e−x . Solution: For (a) f (x) = arctan(x) 22 Forward Difference Approximation arctan 1 + f (x + h) − f (x) = h 0.8157 − π4 = 1 1 16 1 16 − arctan(1) 16 = 0.485 ≈ f 0 (1)∗∗ (3 decimal places) Centered Difference Approximation 1 1 − arctan(1 − 16 ) arctan 1 + 16 f (x + h) − f (x − h) = 2 2h 16 = 0.500 (3 decimal places)...which is better. Observe that f 0 (x) = x2 1 so f 0 (1) = 0.5. +1 For (b) f (x) = e−x Forward Difference Approximation 1 f (x + h) − f (x) e−(1+ 16 ) − e−1 = 1 h 16 = −0.356 (3 decimal places) Centered Difference Approximation 1 1 f (x + h) − f (x − h) e−(1+ 16 ) + e−(1− 16 ) = 2 2h 16 = −0.368 (3 decimal places)...which is better. Observe that f 0 (x) = −e−x so f 0 (1) = −e−1 ≈ −0.367 to 3 decimal places. 2. Use the Forward Difference Approximation to approximate f 0 (1), the Centered Difference Approximation to approximate f 0 (1.1), and the Backward Difference Approximation to approximate f 0 (1.2), using the data in the table below: 23 x f (x) 1.00 1.0000000000 1.10 0.9513507699 1.20 0.9181687424 Solution: Using h = 0.1 Forward Difference Approximation 0.9513... − 1.00 f (1 + 0.1) − f (1) = = −0.487 ≈ f 0 (1) 0.1 0.1 Centered Difference Approximation (3 s.f.) f (1.1 + 0.1) − f (1.1 − 0.1) 0.91816... − 1.00 = = −0.409 ≈ f 0 (1.1) 2(0.1) 0.2 (3 s.f.) Backward Difference Approximation f (1.2) − f (1.2 − 0.1) 0.91816... − 0.951350 = = −0.332 ≈ f 0 (1.2) 0.1 0.1 (3 s.f.) 3. Use the Centered Difference Approximation to the first derivative to prove that this approximation will be exact for any quadratic polynomial. Solution: Let p(x) = a + bx + cx2 so p0 (x) = b + 2cx. Using the centered difference approximation [a + b(x + h) + c(x + h)2 ] − [a + b(x − h) + c(x − h)2 ] p(x + h) − p(x − h) = 2h 2h 2bh + c(x + h)2 − c(x − h)2 = 2h c = b + [(x + h)2 − (x − h)2 ] 2h c = b + [(x2 + 2xh + h2 ) − (x2 − 2xh + h2 )] 2h 4cxh = b+ 2h = b + 2cx = p0 (x). 24 4. Find coefficients A, B, and C so that (a) f 0 (x) = Af (x) + Bf (x + h) + Cf (x + 2h) + O(h2 ) (hard!). Hint: Use Taylor’s theorem. Solution: f (x + h) = f (x) + hf 0 (x) + h2 00 h3 f (x) + f 000 (c1 ), 2! 3! (2.1) for c1 between x and x + h. f (x + 2h) = f (x) + 2hf 0 (x) + (2h)3 000 (2h)2 00 f (x) + f (c2 ), 2! 3! (2.2) for c2 between x and x + 2h. Now computing 4×(2.1)-(2.2) gives h3 000 [4f (c1 ) − 8f 000 (c2 )] 6 2h3 000 = 3f (x) + 2hf 0 (x) + [f (c1 ) − 2f 000 (c2 )]. 3 4f (x + h) − f (x + 2h) = 3f (x) + 2hf 0 (x) + So dividing by 2h yields: 2 1 3 f (x + h) − f (x + 2h) = f (x) + f 0 (x) + O(h2 ) h 2h 2h or 3 2 1 f (x) + f (x + h) − f (x + 2h) + O(h2 ). 2h h 2h 2 1 3 Thus A = − , B = , and C = − . 2h h 2h f 0 (x) = − 5. Use Taylor’s Theorem to show that the approximation f 0 (x) ≈ 8f (x + h) − 8f (x − h) − f (x + 2h) + f (x − 2h) 12h is O(h4 ) (hard!). Again, use Taylor’s theorem. Solution: Expand to powers of h4 (with remainder term O(h5 ) as we then divide by 12h: f (x + h) = f (x) + hf 0 (x) + h2 00 h3 h4 h5 f (x) + f 000 (x) + f (iv) (x) + f (v) (c1 ) 2! 3! 4! 5! 25 f (x − h) = f (x) − hf 0 (x) + h2 00 h3 h4 h5 f (x) − f 000 (x) + f (iv) (x) − f (v) (c2 ) 2! 3! 4! 5! (2h)3 000 (2h)4 (iv) (2h)5 (v) (2h)2 00 f (x) + f (x) + f (x) + f (c3 ) 2! 3! 4! 5! 4h3 000 2h4 (iv) 4h5 (v) = f (x) + 2hf 0 (x) + 2h2 f 00 (x) + f (x) + f (x) + f (c3 ) 3 3 15 f (x + 2h) = f (x) + 2hf 0 (x) + f (x−2h) = f (x)−2hf 0 (x)+2h2 f 00 (x)− 2h4 (iv) 4h5 (v) 4h3 000 f (x)+ f (x)− f (c4 ) 3 3 15 Thus, 8f (x + h) − 8f (x − h) − f (x + 2h) + f (x − 2h) = 16hf 0 (x) − 4hf 0 (x) + ĉh5 = 12hf 0 (x) + ĉh5 , where ĉ depends on f (v) (ci ) for i = 1, 2, 3, 4. So dividing by 12h yields 8f (x + h) − 8f (x − h) − f (x + 2h) + f (x − 2h) = f 0 (x) + ch4 , 12h where c = ĉ . Rearranging 12 f 0 (x) = 8f (x + h) − 8f (x − h) − f (x + 2h) + f (x − 2h) + O(h4 ). 12h 6. Use the derivative approximation from the previous problem with h = 0.1 to evaluate f 0 (1.2), using the data in the following table: x 1.00 1.10 1.20 1.30 1.40 f (x) 1.0000000000 0.9513507699 0.9181687424 0.8974706963 0.8872638175 26 Solution: 8f (1.2 + 0.1) − 8f (1.2 − 0.1) − f (1.2 + 0.2) + f (1.2 − 0.2) 12(0.1) 8f (1.3) − 8f (1.1) − f (1.4) + f (1.0) = 1.2 8(0.89747...) − 8(0.95135...) − (0.88726...) + 1.0 = 1.2 = −0.265 (3 decimal places) f 0 (1.2) = 7. Use Taylor expansions for f (x ± h) to derive an O(h2 ) accurate approximation to f 00 (x) using f (x) and f (x ± h). Provide all the details of the error estimate. Hint: Go out as far as the fourth derivative term, and then add the two expansions. Solution: See class notes. 8. (MATLAB) Use the programs FDA_PRES, BDA_PRES and CDA_PRES to investigate how the errors in difference approximations change with a decrease in the step sizes h. Explain what you see. (Note: the above programs compute approximations to the derivative of a function y = f (x) at some point x0 using the Forward, Backward and Centered Difference Approximations respectively.) Solution: Just run the program and see. They all show pretty much the same thing so we just show the output for a single example. We looked at the example of f (x) = ln(x) (entered as log(x) in MATLAB) with x0 = 1. The Forward Difference Approximation was used to approximate the derivative of this function. N was chosen as 16: h approx |error| 1/10 0.95310180 0.0468982019567507 1/100 0.99503309 0.0049669146831908 1/1000 0.99950033 0.0004996669165768 1/10000 0.99995000 0.0000499966670269 1/100000 0.99999500 0.0000049999601159 1/1000000 0.99999950 0.0000005000819332 27 1/10000000 0.99999995 0.0000000494161295 1/100000000 0.99999999 0.0000000110774709 1/1000000000 1.00000008 0.0000000822403710 1/10000000000 1.00000008 0.0000000826903710 1/100000000000 1.00000008 0.0000000827353710 1/1000000000000 1.00008890 0.0000889005818410 1/10000000000000 0.99920072 0.0007992778374091 1/100000000000000 0.99920072 0.0007992778373641 1/1000000000000000 1.11022302 0.1102230246251559 1/10000000000000000 0.00000000 1.0000000000000000 As expected, initially as we decrease h the accuracy of the approximation decreases until we obtain the least error around h = 10−8 , after which the error begins to rise again due to rounding error. 28 2.3 Euler’s Method 1. Use Euler’s method with h = 0.25 to compute approximate solution values for the initial value problem y 0 = sin(t + y), y(0) = 1. You should eventually get y4 = 1.851566895 (be sure that your calculator is set to radians). Solution: yn+1 = yn + h sin(tn + yn ) tn = nh = 0.25n So yn+1 = yn = 0.25 sin(0.25n + yn ) y0 = y(0) = 1 n=0 y1 = y0 + 0.25 sin(0 + y0 ) = 1 + 0.25 sin(1) = 1.2103677462...(≈ y(0.25)) n=1 y2 = y1 + 0.25 sin(0.25 + y1 ) = 1.45884498566...(≈ y(0.5)) n=2 y3 = y2 + 0.25 sin(0.5 + y2 ) = 1.6902572796...(≈ y(0.75)) n=3 y4 = y3 + 0.25 sin(0.75 + y3 ) = 1.85156689533...(≈ y(1)) 29 2. Use Euler’s method with h = 0.25 to compute approximate solution values for y 0 = et−y , y(0) = −1. What approximate value do you get for y(1) = 0.7353256638? Solution: yn+1 = yn + hf (tn , yn ) = yn + 0.25etn −yn = yn + 0.25e0.25n−yn , with y0 = y(0) = −1. n=0 y1 = y0 + 0.25e0−y0 = −1 + 0.25e1 = −0.320429542885...(≈ y(0.25)) n=1 y2 = y1 + 0.25e0.25−y1 = 0.121827147394...(≈ y(0.5)) n=2 y3 = y2 + 0.25e0.5−y2 = 0.486730952236...(≈ y(0.75)) n=3 y4 = y3 + 0.25e0.75−y3 = 0.812025139875...(≈ y(1)) 3. Repeat the above with h = 0.125. What value do you now get for y8 ≈ y(1)? 30 Solution: yn+1 = yn + hf (tn , yn ) = yn + 0.125etn −yn = yn + 0.125e0.125n−yn , with y0 = y(0) = −1. n=0 y1 = y0 + 0.125e0−y0 = −0.660214771443...(≈ y(0.125)) n=1 y2 = y1 + 0.125e0.125−y1 = −0.38610503923...(≈ y(0.25)) n=2 y3 = y2 + 0.125e0.25−y2 = −0.149966473291...(≈ y(0.375)) n=3 y4 = y3 + 0.125e0.375−y3 = 0.061333798435...(≈ y(0.5)) n=4 y5 = y4 + 0.125e0.5−y4 = 0.255163498508...(≈ y(0.625)) n=5 y6 = y5 + 0.125e0.625−y5 = 0.436100739954...(≈ y(0.75)) 31 n=6 y7 = y6 + 0.125e0.75−y6 = 0.607194720154...(≈ y(0.875)) n=7 y8 = y7 + 0.125e0.875−y7 = 0.770581294899...(≈ y(1)) 4. (MATLAB) Use the program EULER_PRES to compute approximations to each of the following initial value1 problems on [0, 1], using M = 2, 4, 8, 16 steps. (i) f (t, y) = −y + sin(t), y(0) = 1, where y(t) = (3/2) exp(−t) + (sin(t) − cos(t))/2, (ii) f (t, y) = t − y, y(0) = 2, where y(t) = 3 exp(−t) + t − 1, (iii) f (t, y) = exp(t − y), y(0) = −1, where y(t) = log(exp(t) − 1 + exp(−1)), (iv) f (t, y) = −y log(y), y(0) = 3, where y(t) = exp(log(3) exp(−t)). Where is the maximum error on [0, 1] for each value of h = 1/M ? How does the approximation change as h is reduced? Solution: Just run the program and see. In general we expect the maximum error to be close to t = 1 and the errors to decrease as h → 0. However, as we saw in our lecture notes Euler’s Method converges quite slowly. An example of the program output with the graph is given for example (iii) below with h = 1/4: t 0.000 0.250 0.500 0.750 1.000 1 exact -1.000000 -0.427857 0.016464 0.395334 0.735326 approx -1.000000 -0.320430 0.121827 0.486731 0.812025 In MATLAB log means natural log. 32 |error| 0.000000 0.107427 0.105363 0.091397 0.076699 2.4 Linear Interpolation 1. Use linear interpolation to find an approximation to erf(0.34), where f (x) = erf(x) is the error function, using the data in the following table: x f (x) 0.3 0.32862675945913 0.4 0.42839235504667 Also, give an upper bound on the error in the approximation. Solution: Choose x0 = 0.3 and x1 = 0.4 so that [a, b] = [0.3, 0.4]. x − x1 x − x0 P1 (x) = f (x0 ) + f (x1 ). x0 − x1 x1 − x0 With x = 0.34, P1 (0.34) = 0.6(0.3286...)+0.4(0.4283...) = 0.3685 33 (4 significant figures). For the error in the approximation we need the 2nd derivative of f (x) = erf(x), which is given by 4x 2 f 00 (x) = − √ e−x . π Without resorting to a graphical approach this is difficult to bound. A crude upper bound is 4(0.4) 2 |f 00 (x)| ≤ √ e−(0.3) = 0.8250087277, π max (numerator) and the fact that y1 = x min (denominator) 2 is increasing on [0.3, 0.4] and y2 = e−x is decreasing on [0.3, 0.4]. Thus from the error bound formula M (x1 − x0 )2 |f (x) − P1 (x)| ≤ 8 0.250087277 = (0.4 − 0.3)2 8 = 0.001031 (4 significant figures), where we have bounded using where M = max |f 00 (x)| for all x ∈ [x0 , x1 ]. 2 Check: using erf(x) = √ π get Z x 2 e−t dt on my TI-89 graphing calculator I 0 Z 2 |f (x) − P1 (x)| = √ π 0 0.34 −t2 e dt − 0.3685329977 = |0.369364529345... − 0.3685329...| = 0.000831..., consistent with our result above. 2. The gamma function, denoted by Γ(x), occurs in a number of applications, most notably probability theory and the solution of certain differential equations. It is basically the generalization of the factorial function to non-integer values, in that Γ(n + 1) = n!. The table below gives values of Γ(x) for x between 1 and 2. Use linear interpolation to approximate Γ(1.005) = 0.99713853525101.... 34 x 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 Γ(x) 1.0000000000 0.9513507699 0.9181687424 0.8974706963 0.8872638175 0.8862269255 0.8935153493 0.9086387329 0.9313837710 0.9617658319 1.000000000 Solution: Choose x0 = 1.00 and x1 = 1.10. Set f (x) = Γ(x). x − x0 x − x1 f (x0 ) + f (x1 ). P1 (x) = x0 − x1 x1 − x0 With x = 1.005 P1 (1.005) = 0.95(1.00)+0.05(0.9513507699) = 0.9976 (4 significant figures). 3. Construct a linear interpolating polynomial to the function f (x) = x−1 using x0 = 1/2 and x1 = 1 as nodes. What is the upper bound on the error over the interval [1/2, 1], according to the error estimate? Solution: x − x1 x − x0 P1 (x) = f (x0 ) + f (x1 ) x0 − x 1 x1 − x0 −1 x − 12 x−1 1 = + 1−1 1 1 2 − 1 1 − 2 2 = 3 − 2x. For the error bound we need f 00 (x): f 0 (x) = −x−2 and 35 f 00 (x) = 2x−3 = 2 . x3 1 We need an upper bound for f (x) on , 1 . Noting that x3 is increasing 2 1 2 1 on , 1 =⇒ 3 is decreasing on , 1 and positive throughout, thus 2 x 2 1 −3 00 −3 |f (x)| = 2|x | ≤ 2 = 2 (2−1 )−3 = 2(8) = 16 = M. 2 00 So using the error bound formula 16 M (x1 − x0 )2 = |f (x) − P1 (x)| ≤ 8 8 2 1 1 1 1− =2 = . 2 4 2 1 4. Repeat the above for f (x) = x 3 , using the interval [1/8, 1]. Solution: x − x1 x − x0 P1 (x) = f (x0 ) + f (x1 ) x0 − x 1 x1 − x0 13 x − 81 1 1 x−1 + 13 = 1 1 8 −1 1− 8 8 4 3 1 = x+ or (4x + 3). 7 7 7 Now 1 2 f 0 (x) = x− 3 3 2 5 2 f 00 (x) = − x− 3 = − 5 . 9 9x 3 1 Where is |f 00 (x)| a maximum on , 1 ? Now (positive) powers of x and 8 their reciprocals are maximized or minimized at one of the end-points of the interval. Thus the absolute values of these functions are maximized at one of the end-points. To find which end-point leads to a maximum we check each and see. 1 00 f = −7.1̇ and f 00 (1) = −0.2̇ 8 and 36 =⇒ |f 00 (x)| ≤ 7.1̇ = M . Thus from the error bound formula 7.1̇ M (x1 − x0 )2 = |f (x) − P1 (x)| ≤ 8 8 2 1 1− = 0.6806 8 (4 significant figures) Note: alternatively, to find the maximum value of |f 00 | we could have argued as follows. Because we took absolute values we can ignore the sign of f 00 . As x5/2 is increasing, f 00 is decreasing, and so the function is maximized at the left end-point. See the graph below: 5. (MATLAB) Apply the program INTEROL_PRES to Question 3. Solution: Running the program with a = 1/2, b = 1 with 2 nodes yields: 37 2.5 Trapezoid Rule 1 1. Apply the trapezoid rule with h = , to approximate the integral 8 Z 1 1 √ dx = 0.92703733865.... I= 1 + x4 0 How small does h have to be to get an error less that 10−3 ? 10−6 ? Solution: f (x) = √ 1 , 1 + x4 1 h= , 8 [a, b] = [0, 1], 38 n= b−a 1−0 = 1 = 8. h 8 i xi 0 0 1 18 2 14 3 38 4 12 5 58 6 34 7 78 8 1 f (xi ) 1 0.99987... 0.99805... 0.99025... 0.97014... 0.93145... 0.87157... 0.79400... 0.70710... 1 h(f (x0 ) + 2f (x1 ) + ... + 2f (x7 ) + f (x8 )) 2 1 1 = (1 + 2(0.99987...) + 2(0.99805...) + ... + 2(0.79400...) + 0.70710...) 2 8 = 0.926115180158... T8 (f ) = In order to estimate the error we need the 2nd derivative of f (x) = 1 (1 + x4 )− 2 . 3 3 1 f 0 (x) = − (1 + x4 )− 2 · 4x3 = −2x3 (1 + x4 )− 2 . 2 Using the product rule we can show that f 00 (x) = 6x2 (x4 − 1) 5 (x4 + 1) 2 . This is a complicated function to get an upper bound on. We would use a combination of calculus and curve sketching techniques to deduce the following diagram: 39 Alternatively, you could get an upper bound by maximizing and minimizing the numerator and denominator in f 00 respectively. (Note: I wouldn’t give you one as involved as this in our tests, but the principle is still important.) Now recall from lecture notes that to get an error ≤ 10−3 we require |I(f ) − Tn (f )| ≤ h2 b−a 2 h M = (1.39275) ≤ 10−3 , 12 12 (2.3) where M = max |f 00 (x)| for all x ∈ [a, b] (as indicated in the diagram). Rearranging, we require h2 ≤ 0.08616047... or h ≤ 0.0928227... But h must divide b − a = 1, so as n= b−a h or h= b−a 1 = ≤ 0.0928227... n n =⇒ n ≥ 10.77 40 so choose n = 11 =⇒ h = h2 (1.39275) ≤ 10−6 12 1 = 0.09. As in (2.3) we seek h such that n h2 ≤ 8.616047×10−6 or h ≤ 0.0029353104. so As before h= 1 ≤ 0.0029353104... =⇒ n ≥ 340.679... n 1 = 0.00293255132... n π 2. Use the trapezoid rule with h = , to approximate the integral 4 Z π 2 I= sin(x) dx = 1. so with n = 341, =⇒ h = 0 How small does h have to be to get an error less that 10−3 ? 10−6 ? Solution: f (x) = sin(x), h= π , 4 h πi [a, b] = 0, , 2 i xi 0 0 1 π 4 2 π 2 n= f (xi ) 0 1 √ 2 1 1 h(f (x0 ) + 2f (x1 ) + f (x2 )) 2 2 1 π = 0+ √ +1 2 4 2 ! √ 2+1 π ≈ 0.948059... = 8 T2 (f ) = 41 b−a = h π 2 −0 π 4 = 2. (Note: exact answer is 1.) In order to estimate the error we need the 2nd derivative of f (x) = sin(x). f 0 (x) = cos(x) and f 00 (x) = − sin(x) h πi Thus |f 00 (x)| = | sin(x)| ≤ 1 = M on 0, . 2 From the error bound formula we need π b−a 2 π |I(f ) − Tn (f )| ≤ h M = 2 h2 · 1 = h2 ≤ 10−3 12 12 24 =⇒ h2 ≤ 0.007639437... (2.4) h ≤ 0.08740387... so π π b−a But h must divide b − a = , so noting that h = = 2 ≤ 2 n n 0.08740387... =⇒ n ≥ 17.97. So choose n = 18 =⇒ h = π 2 = 0.087266.... n From (2.4) we also seek h such that π 2 h ≤ 10−6 24 =⇒ h2 ≤ 7.639... × 10−6 =⇒ h ≤ 0.002763953... As before h= π 2 n ≤ 0.002763953... Choosing n = 569 =⇒ h = π 2 569 =⇒ n ≥ 568.315... = 0.0027606... 1 3. Apply the trapezoid rule with h = , to approximate the integral 8 Z 1 1 I= x(1 − x2 ) dx = . 4 0 How small does h have to be to get an error less than 10−3 ? 10−6 ? 42 Solution: f (x) = x(1−x2 ), 1 h= , 8 [a, b] = [0, 1], i xi 0 0 1 18 2 14 3 38 4 12 5 58 6 34 7 78 8 1 n= b−a 1−0 = 1 = 8. h 8 f (xi ) 0 63 512 15 64 165 512 3 8 195 512 21 64 105 512 0 1 h(f (x0 ) + 2f (x1 ) + ... + 2f (x7 ) + f (x8 )) 2 1 1 63 15 105 = (0 + 2 +2 + ... + 2 + 0) 2 8 512 64 512 63 = 256 = 0.2461 (4 significant figures) 1 . In order to estimate the error we need the Note: exact answer is 4 2nd derivative of f (x) = x(1 − x2 ). T8 (f ) = f 0 (x) = 1 − 3x2 =⇒ f 00 (x) = −6x. So |f 00 (x)| = 6|x| ≤ 6 = M on [0, 1]. From the error bound formula we need h such that |I(f ) − Tn (f )| ≤ b−a 2 h2 h M = (6) ≤ 10−3 , 12 12 Rearranging, we require h2 ≤ 0.002 or 43 h ≤ 0.0447213595... (2.5) But h must divide b − a = 1, so as 1 b−a = ≤ 0.0447213595... h= n n =⇒ n ≥ 22.36... 1 so choose n = 23 =⇒ h = = 0.04347826. For an error less than 23 10−6 we need h such that h2 b−a 2 h M = (6) ≤ 10−6 |I(f ) − Tn (f )| ≤ 12 12 Rearranging, we require h2 ≤ 2 × 10−6 or h ≤ 0.00141421356... As before 1 ≤ 0.00141421356... =⇒ n ≥ 707.106... n 1 = 0.00141242937... so with n = 708, =⇒ h = 708 h= 4. Apply the trapezoid rule with h = 81 , to approximate the integral Z 1 I= ln(1 + x) dx = 2 ln(2) − 1. 0 How small does h have to be to get an error less than 10−3 ? 10−6 ? Solution: f (x) = ln(1+x), 1 h= , 8 [a, b] = [0, 1], i xi 0 0 1 18 2 14 3 38 4 12 5 58 6 34 7 78 8 1 44 f (xi ) 0 ln 98 ln 54 ln 11 8 ln 32 ln 13 8 ln 74 ln 15 8 ln(2) n= b−a 2−1 = 1 = 8. h 8 1 h(f (x0 ) + 2f (x1 ) + ... + 2f (x7 ) + f (x8 )) 2 1 1 9 5 15 = 0 + 2 ln + 2 ln + ... + 2 ln + ln(2) 2 8 8 4 8 = 0.3856 (4 significant figures) T8 (f ) = (Note: exact answer is 0.3863 to 4 significant figures). In order to estimate the error we need the 2nd derivative of f (x) = ln(1 + x). f 0 (x) = 1 1 =⇒ f 00 (x) = − . 1+x (1 + x)2 1 1 ≤ = 1 = M on [0, 1]. From the error 2 (1 + x) (1 + 0)2 bound formula we need h such that So |f 00 (x)| = |I(f ) − Tn (f )| ≤ b−a 2 h2 h M = (1) ≤ 10−3 , 12 12 (2.6) Rearranging, we require h2 ≤ 0.012 h ≤ 0.10954451150... or But h must divide b − a = 1, so as b−a 1 = ≤ 0.10954451150... n n =⇒ n ≥ 9.12... 1 so choose n = 10 =⇒ h = = 0.1. For an error less than 10−6 we 10 need h such that h= |I(f ) − Tn (f )| ≤ b−a 2 h2 h M = (1) ≤ 10−6 12 12 Rearranging, we require h2 ≤ 1.2 × 10−5 or h ≤ 0.003464101615... As before 1 ≤ 0.003464101615... =⇒ n ≥ 288.67... n 1 so with n = 289, =⇒ h = = 0.0034602076... 289 h= 45 5. (MATLAB) Apply the program TRAPEZOID_PRES to numerically verify any of the above results. For example, what is the actual error incurred for Question 3 when using h = 1/23? Solution: Running the program yields that the approximate area is 0.249483. The exact answer is 1/4, thus the error is about |1/4 − 0.249483| = 5.17 × 10−4 , which is consistent with the theoretical result (|error| ≤ 10−3 ). The approximation is illustrated below: 2.6 Solution of Tridiagonal Linear Systems 1. Use the tridiagonal algorithm to solve the following system of equations: π 4 2 0 0 x1 √9 1 4 1 0 x2 3 √2 0 1 4 1 x3 = 3 2 0 0 2 4 x4 − π9 46 Solution: 4 2 1 4 0 1 0 0 0 1 4 2 π √9 3 √2 3 2 −π 9 0 0 1 4 r2 − 41 r1 → r2 4 0 0 0 r3 − 27 r2 → r3 r4 − 7 r 13 3 → r4 2 0 1 4 2 7 2 1 0 4 0 0 0 0 0 26 7 2 0 1 4 0 0 0 2 7 2 7 2 0 0 0 0 1 4 0 1 2 26 7 0 √ π 9 3 − π 2 √ 36 3 2 − π9 π 9 0 0 1 4 0 0 1 π 9 √ 3 π − = 0.778759 36 √2 5 3 + π = 0.643523 14 126 − 5√3 − 3π = −0.695578 26 26 45 13 √ + 2x2 7 x 2 2 + x3 26 x 7 3 + x4 45 x 13 4 π − 36 π + 126 − π9 π = 9 = 0.778759 = 0.643523 = −0.695578 Back substitution yields: x1 = 0.008495 x2 = 0.1575 x3 = 0.2274 x4 = −0.2009 (each to 4 significant figures). 2. Use the tridiagonal algorithm to solve the following system of equations: 1 12 0 0 x1 2 1 1 1 0 x2 2 2 31 41 1 0 x3 = 53 4 5 6 30 15 0 0 61 17 x4 14 47 3 √2 5 3 14 (each to 6 significant figures). Keeping 6 significant figures for the right-hand side values yields the following associated linear system (see lecture notes): 4x1 . Solution: 1 12 0 1 1 1 2 31 14 0 4 5 0 0 16 2 2 53 30 15 0 0 1 6 1 7 1 0 0 0 r2 − 21 r1 → r2 14 1 2 1 12 1 4 0 1 r3 − 3r2 → r3 0 0 0 r4 + 10 r 33 3 → r4 1 0 0 0 0 1 4 1 5 1 6 1 2 1 12 0 0 1 2 1 12 0 0 0 0 1 6 1 7 2 1 53 30 15 14 0 1 4 − 11 20 1 6 0 1 4 − 11 20 0 0 0 1 6 1 7 2 1 − 37 1530 0 0 1 6 134 693 + 1 x 2 2 1 x 12 2 + 1 x 4 3 11 − 20 x3 + 1 x 6 4 134 x 693 4 2 1 − 37 96730 = 2 = 1 = − 37 30 967 = 1386 Back substitution yields: x1 = 1.004 x2 = 1.993 x3 = 3.336 x4 = 3.608 (each to 4 significant figures). 3. Is the following system diagonally dominant? Use the triadiagonal algorithm to find the solution. 1 10 61 0 0 x1 2 21 42 1 1 1 x2 179 0 4 31 13 = 156 1 1 0 x3 563 5 4 21 420 1 1 13 0 0 5 x4 6 10 48 14 The associated linear system is: x1 1386 Solution: Checking for diagonal dominance: 1 10 > = 0.4761...X 2 21 1 1 1 17 0.3̇ = > + = = 0.3269...X 3 4 13 52 1 1 1 26 0.25 = > + = = 0.247...X 4 5 21 105 1 1 < ...× 0.16̇ = 6 5 Hence the system is not diagonally dominant. Nevertheless, Gaussian elimination may still work: 1 10 1 10 0 0 61 0 0 61 2 21 42 2 21 42 1 1 1 1 1 0 2 179 115 0 r − r → r 0 2 1 2 2 21 13 4 31 13 156 273 1 1 563 1 1 563 0 0 1 5 4 21 420 5 4 21 420 1 13 1 13 0 0 51 0 0 15 6 10 6 10 0.5 = r3 − 21 r 10 2 → r3 0 0 0 r4 − 52 r 23 3 → r4 1 2 1 2 0 0 0 10 21 2 21 0 0 10 21 2 21 0 0 0 1 13 23 260 1 5 0 1 13 23 260 0 The associated linear system is: 1 x 2 1 + 10 x 21 2 2 x 21 2 + 1 x 13 3 23 x 260 3 + 1 x 21 4 19 x 322 4 = = = = 61 42 115 273 2489 5460 1301 4830 Back substitution yields: x1 = 0.7661 x2 = 2.246 (each to 4 significant figures). 49 x3 = 2.696 x4 = 4.565 0 0 1 21 1 6 0 0 1 21 19 322 61 42 115 273 2489 5460 13 10 61 42 115 273 2489 5460 1301 4830 4. (MATLAB) Use MATLAB to verify any of the solutions you have found by using the ‘backslash’ command. That is, with the system Ax = b to find x just type >> A\b. See any tutorial to see how to enter vectors and matrices in MATLAB. Solution: The MATLAB commands for Question 2 are given below: >> A=[1,1/2,0,0;1/2,1/3,1/4,0;0,1/4,1/5,1/6;0,0,1/6,1/7] A = 1 1/2 0 0 1/2 1/3 1/4 0 0 1/4 1/5 1/6 0 0 1/6 1/7 >> b=[2;2;53/30;15/14] b = 2 2 53/30 15/14 >> A\b ans = 1.0037 1.9925 3.3358 3.6082 5. (MATLAB) We can use some simple commands in MATLAB to implement the individual row operations in the tridiagonal algorithm. The commands applied to a hypothetical augmented matrix B are as follows (ri refers to the ith row of B and k is a nonzero scalar): 50 To do kri → ri (multiply row i by k) type >> B(i,:) = k*B(i,:) To do ri − krj → ri (subtract k times row j from row i) type >> B(i,:) = B(i,:) - k*B(j,:) To do ri ↔ rj (swap rows i and j) type >> B([i,j],:) = B([j,i],:) Solution: The MATLAB commands for the first step of Question 2 are given below (I am assuming you have already constructed the augmented matrix called B - see previous solution): >> format rat % Comment: so calculations done with fractions >> B(2,:)=B(2,:)-(1/2)*B(1,:) B = 1 0 0 0 2.7 1/2 1/12 1/4 0 0 1/4 1/5 1/6 0 0 1/6 1/7 Boundary Value Problems 1. Solve, by hand, the following two-point BVP: −u00 + u = f (x), x ∈ [0, 1], u(0) = u(1) = 0, where f (x) = x, using h = 1/4. Write out the linear system explicitly prior to a solution. You should get the following 3 × 3 system: 2.0625 -1 0 U1 0.015625 -1 2.0625 -1 U2 = 0.03125 0 -1 2.0625 U3 0.046875 51 2 1 53/30 15/14 Solution: −u00 + u = x, u(0) = u(1) = 0 0≤x≤1 With uk ≈ u(xk ) and xk = kh, we replace the above BVP with uk−1 −2uk +uk+1 + u k = xk , − h2 u0 = un = 0 Multiplying the DE through by h2 and simplifying yields: −uk−1 + (2 + h2 )uk − uk+1 = h2 xk 1 which with h = , so that n = 11 = 4 yields 4 4 2 1 33 −uk−1 + 16 uk − uk+1 = 14 k = 4 u0 = u4 = 0, 1 3 4 (2.7) k, (2.8) for k = 1, 2, 3. (2.8) leads to 3 linear equations: k=1 33 −u0 + u1 − u2 = 16 3 1 33 1 =⇒ u1 − u2 = 4 16 64 (2.9) 3 1 1 (2) = 4 32 (2.10) k=2 33 −u1 + u2 − u3 = 16 52 k=3 −u2 + 33 3 u3 = 16 64 (2.11) Writing (2.9),(2.10) and (2.11) as a single matrix equation: 33 1 −1 0 u 1 16 64 −1 33 −1 u2 = 1 16 32 33 3 0 −1 16 u3 64 (Note: 33/16 = 2.0625). Gaussian elimination and back substitution yield 266 u1 0.0349 7625 u2 = 65 = 0.0563 1154 68 u3 0.0500 1359 to 4 decimal places. 2. Repeat the above problem, but this time using h = 1/5. What does the new system become? Solution: −uk−1 + 51 u 25 k u0 = u5 = 0, − uk+1 = 1 2 5 1 k 5 = 1 3 5 k, (2.12) for k = 1, 2, 3, 4. k=1 51 −u0 + u1 − u2 = 25 3 1 51 1 =⇒ u1 − u2 = 5 25 125 (2.13) 3 1 2 (2) = 5 125 (2.14) k=2 51 −u1 + u2 − u3 = 25 k=3 51 −u2 + u3 = 25 53 3 1 3 (3) = 5 125 (2.15) k=4 51 −u3 + u4 − u5 = 25 Writing (2.13)-(2.16) as a 51 −1 25 −1 51 25 0 −1 0 0 3 1 51 4 (4) =⇒ −u3 + u4 = 5 25 125 single matrix equation: 0 0 u1 −1 0 u2 = 51 −1 u3 25 −1 51 u4 25 1 125 2 125 3 125 4 125 (2.16) 3. Approximate the solution of the two-point BVP −u00 + u0 + u = 1, x ∈ [0, 1], u(0) = u(1) = 0. Assuming h < 2 determine if the resulting coefficient matrix is diagonally dominant? Hint: For the approximation of the 2nd derivative see your lecture notes 1 . Solution: −u00 + u0 + u = 1, u(0) = u(1) = 0 x ∈ [0, 1] with uk ≈ u(xk ) and xk = hk, we replace the above BVP with ( −uk−1 k +uk+1 − uk−1 −2u + uk+12h + uk = 1, 2 h u0 = un = 0. Multiplying the DE by h2 yields or h −uk−1 + 2uk − uk+1 + (uk+1 − uk−1 ) + h2 uk = h2 , 2 − h2 + 1 uk−1 + (2 + h2 )uk + h2 − 1 uk+1 = h2 u0 = un = 0, k = 1, 2, ..., n − 1 1 Note: the problem of determining diagonal dominance is harder than I would give you in an exam, but shows you the sort of calculations one might have to do in practice. However, the initial part is fair game! 54 As k runs from 1 to n−1, we get the following tridiagonal linear system in matrix form, after using in the first and last equation that u0 = 0 and un = 0, respectively: 2 + h2 − h2 + 1 0 h 2 −1 2 + h2 − h2 + 1 .. . 0 h −1 2 2 + h2 .. . − h 2 +1 0 .. . 2 + h2 − h2 + 1 h − 1 2 2 + h2 u1 u2 u3 u4 u5 u6 = h2 h2 h2 h2 h2 h2 Is the coefficient matrix diagonally dominant? h First equation: Is |2 + h2 | > − 1? 2 Recall the triangle inequality, namely |a + b| ≤ |a| + |b|, so h h − 1 = + (−1) ≤ h + 1 < 2 < 2 + h2 = |2 + h2 | 2 2 2 X (We used the fact that h < 2 by assumption.) h Middle equations: Is |2 + h2 | ≥ − h2 + 1 + 2 − 1? First observe h h h h − 1 = 1 − , after that − 1 < 0 as h < 2, so − 1 = − 2 2 2 2 recalling that x, x ≥ 0 |x| = −x, x < 0 So h h h h − + − 1 = + − 1 = h +1+1− h = 2 < 2+h2 = |2+h2 | + 1 + 1 2 2 2 2 2 2 h Last equation: Is |2 + h2 | > − + 1 ? 2 − h + 1 = h + 1 = h + 1 < 2 < 2 + h2 = |2 + h2 | X 2 2 2 The coefficient matrix is diagonally dominant for all h. 55 X 4. (MATLAB) Use the program BVP_SIMPLE_PRES to numerically solve the BVP −u00 + u = f (x), u(0) = u(1) = 0, x ∈ [0, 1] where f and the exact solutions u are given by (a) f (x) = (π 2 + 1) sin(πx), u(x) = sin(πx); (b) f (x) = 4 exp(−x) − 4x exp(−x), u(x) = x(1 − x) exp(−x); (c) f (x) = π(π sin(πx) + 2 cos(πx)) exp(−x), u(x) = exp(−x) sin(πx); (d) f (x) = 3 − 1/x − (x2 − x − 2) ln(x), u(x) = x(1 − x) ln(x). Using h = 1/2n where n = 2, 3, 4, . . . numerically investigate the accuracy of the approximations. Recall that with n equal subintervals we have the approximations uk ≈ u(xk ), k = 1, 2, . . . , n − 1 where the nodes are xk = hk, h = 1/n. As a measure of the accuracy choose the error to be E(h) := max |u(xk ) − uk |, k k = 1, 2, . . . , n − 1. Solution: We used the MATLAB program to solve problem (a). The output and graph are displayed below for n = 4, i.e. h = 1/4: 56 xk 0.000000 0.250000 0.500000 0.750000 1.000000 Approx 0.000000 0.740989 1.047917 0.740989 0.000000 Exact 0.000000 0.707107 1.000000 0.707107 0.000000 |Approx - Exact| 0.000000 0.033882 0.047917 0.033882 0.000000 Clearly the maximum error is (to 6 d.p.) 0.047917. We ran the program again with h = 1/8, h = 1/16 and h = 1/32 and summarized our findings in the table below. We have also shown the ratio of consecutive h 1/4 1/8 1/16 1/32 E(h) E(h)/E(h/2) 0,047917 4.0789 0.011745 4.0195 0.002922 4.0027 0.000730 −−− errors. From lecture notes we know that with the assumption that E(h) = Chp (and so E(h)/E(h/2) = hp ) the data suggests that hp = 4 i.e. p = 2. That is we have shown numerically that E(h) = O(h2 ) (as expected). 57 Chapter 3 Root - Finding 3.1 Bisection Method 1. Do three iterations of the bisection method, applied to f (x) = x3 − 2 using a = 0 and b = 2. Solution: f (0) = −2 f (2) = 8 − 2 = 6 so by the IMVT a root of f exists in [0, 2]. Set x1 = f (1) = 13 − 2 = −1 f (2) = 8 − 2 = 6 0+2 = 1. 2 so by the IMVT a root of f exists in [1, 2]. Set x2 = f (1.5) = (1.5)3 − 2 = 1.375 f (1) = −1 1+2 = 1.5. 2 so by the IMVT a root of f exists in [1, 1.5]. Set x3 = 1 + 1.5 = 1.25. 2 2. (a) How many iterations of the bisection method are needed to find 2 the root of f (x) = x − e−x on [0, 1] to an accuracy of 0.1? (b) Apply the method to find the root to the specified accuracy. (c) If the actual 58 root is 0.6529186... verify that your root does indeed have the required accuracy. Solution: (a) Need k such that |α − xk | ≤ b−a 1−0 < 0.1 =⇒ < 0.1 k 2 2k =⇒ 1 < (0.1)2k =⇒ ln(1) < ln(0.1) + k ln(2) ln(0.1) = 3.321... =⇒ k > − ln(2) Thus we need 4 iterations (bisections) to achieve the desired accuracy. (b) f (0) = 0 − e0 = −1 f (1) = 0.632... so by the IMVT a root of f exists in [0, 1]. Set x1 = f (0.5) = −0.278... f (1) = 0.632... 0+1 = 0.5. 2 so by the IMVT a root of f exists in [0.5, 1]. Set x2 = 0.75. f (0.75) = 0.180... f (0.5) = −0.278... so by the IMVT a root of f exists in [0.5, 0.75]. Set x3 = 0.625. f (0.625) = −0.0516... f (0.75) = 0.180... 0.5 + 1 = 2 0.75 + 0.5 = 2 so by the IMVT a root of f exists in [0.625, 0.75]. Set x4 = 0.625 + 0.75 = 0.6875. 2 59 (c) The actual root is 0.6529186... so the absolute error is: |0.6529186... − 0.6875| = 0.0345... < 0.1.X 3. (MATLAB) Use the program BISECT_PRES to apply the bisection method to the following test problems: (a) f (x) = 2 − exp(x) on [0, 1], (b) f (x) = x3 − 2 on [0, 2], (c) f (x) = 5 − 1/x on [0.1, 0.25], (d) f (x) = x2 − sin(x) on [0.5, 1]. Can you check the exact roots of the problems with the answers given by the program? Use an appropriate error bound to check a priori the minimum number of steps needed for the roots to be accurate within 10−6 . Solution: We show output for problem (a). The iterations and graph of how xn varies with increasing n is shown below: x1 = 0.5000000000000000 x2 = 0.7500000000000000 x3 = 0.6250000000000000 etc. x19 = 0.6931476593017578 x20 = 0.6931467056274414 After 20 iterations the root is approximately 0.6931467056274414 with an |error|< 1.000000e-06 60 The root is clearly worked out to be ln(2). With a tolerance of 10−6 we require (see your lecture notes) b−a ≤ 10−6 . 2n With [a, b] = [0, 1] we solve for n yielding n ≥ ln(106 )/ ln(2) = 19.9316 . . .. Thus the minimum number of steps is 20. Similar results are obtained for the other examples. The roots are easily found (exactly on paper) √ 3 for problems (b) and (c), to be 2 and 1/5 respectively. The root for problem (d) cannot be found exactly on paper. 3.2 Newton’s Method 1. (a) Write down Newton’s method applied to the function f (x) = x3 −2. Simplify the computation as much as possible. (b) What has been accomplished if we find the root of this function? Solution: (a) f (x) = x3 − 2 =⇒ f 0 (x) = 3x2 . 61 Thus f (xn ) f 0 (xn ) (x3 − 2) = xn − n 2 3xn 3 2(xn + 1) = , 3x2n xn+1 = xn − where x0 is given. (b) If we find a root of f we have found the cube root of 2 (do you see why?). 2. (a) Write down Newton’s method applied to the function f (x) = a − x−1 . Simplify the resulting computation as much as possible. (b) What has been accomplished if we find the root of this function? Solution: (a) f (x) = a − x−1 = a − 1 1 =⇒ f 0 (x) = x−2 = 2 . x x Thus xn+1 = xn − = xn − f (xn ) f 0 (xn ) (a − x1n ) 1 x2n = xn (2 − axn ), where x0 is given. (b) A root of f corresponds to a = 1 1 , or x = , a 6= 0. x a 3. (a) Do three iterations of Newton’s method for f (x) = 3 − ex , using x0 = 1. Repeat, with x0 = 2, 4, 8, 16, but using as many iterations as 62 needed in order to see a pattern. (b) Comment on your results. Solution: (a) f (x) = 3 − ex =⇒ f 0 (x) = −ex . Thus f (xn ) f 0 (x ) n xn 3−e = xn − −exn = ((xn − 1)exn + 3) e−xn , xn+1 = xn − x0 = 1 x1 = = = x2 = x3 = (1 − 1)e1 + 3 e−1 3e−1 1.103... 1.098... 1.0986122... x0 = 2 x1 = (2 − 1)e2 + 3 e−2 = 1.406... x2 = 1.141... x3 = 1.0995133... x0 = 4 x1 = (4 − 1)e4 + 3 e−4 = 3.054... x2 = 2.196... x3 = 1.529... 63 We need to go further to see the pattern: x4 = 1.179... x5 = 1.101... x6 = 1.098617... x0 = 8 x1 = (8 − 1)e8 + 3 e−8 = 7.00... x2 = 6.00... x3 = 5.011... We need to go further to see the pattern: x4 x5 x6 x7 x8 x9 x10 = = = = = = = 4.031... 3.084... 2.22... 1.546... 1.185... 1.102... 1.09861... x0 = 16 x1 = = x2 = .. . (16 − 1)e16 + 3 e−16 15.00... 14.00.00... .. . x18 = 1.09861... (b) The actual root is ln(3) = 1.09861228... (convince yourself of this!). For this particular problem, all chosen starting values converge to the root, but the further the initial guess is from the root the more iterations are needed to achieve a given accuracy. However, in general we must be sufficiently close to the root for the iterative procedure to converge. 64 4. Figure 3.1 shows the geometry of a planetary orbit around the sun. The position of the sun is given by S, the position of the planet is given by P . Let x denote the angle defined by P0 OA, measured in radians. The dotted line is a circle concentric to the ellipse and having a radius equal to the major axis of the ellipse. Let T be the total orbital period of the planet, and let t be the time required for the planet to go from A to P . Then Kepler’s equation from orbital mechanics, relating x and t, is 2πt . x − sin(x) = T Here is the eccentricity of the elliptical orbit (the extent to which it deviates from a circle). For an orbit of eccentricity = 0.01 (roughly equivalent to that of the Earth), what is the value of x corresponding to T T t = ? What is the value of x corresponding to t = ? Use Newton’s 4 8 method to solve the required equation. Figure 3.1: Orbital geometry 65 Solution: = 0.01 and t = T leads to 4 x − 0.01 sin(x) = 2π T4 π = . T 2 Set f (x) = x − 0.01 sin(x) − π2 , then f (0) = − π2 < 0 f (π) = π − 0 − π 2 = π 2 so by the IMVT a root α ∈ [0, π]. Set x0 = >0 π . Newton’s method is: 2 f (xn ) f 0 (xn ) xn − 0.01 sin(xn ) − π2 = xn − (1 − 0.01 cos(xn )) xn cos(xn ) − sin(xn ) − 50π = cos(xn ) − 100 xn+1 = xn − Iterating with x0 = π : 2 n = 0 n = 1 n = 2 The method for t = x1 = 1.5807963267... x2 = 1.5807958268... x3 = 1.5807958268... T is done in the same way. 8 5. (MATLAB) Use the program NEWTON_RAPH_PRES to apply Newton’s method to the following test problems: (a) f (x) = x2 − 4 on [1, 6.5], x0 = 6, tol = 1e − 06; (b) f (x) = (x − 2)2 on [1.5, 3.2], x0 = 3, tol = 1e − 03; (c) f (x) = 6(x − 2)5 on [2, 3], x0 = 3, tol = 1e − 03; (d) f (x) = (4/3) exp((2 − x/2)(1 + ln(x)/x)) on [0, 20], x0 = 2.5, f 0 (x) = (4/3) exp((2 − x/2)(1 + ln(x)/x)) × ((2 − x/2)(1 − ln(x))/x2 − (1 + ln(x)/2)/2); 66 (e) f (x) = x3 − 2x + 2 on [−0.5, 1.5], x0 = 0; (f) f (x) = 1 − x2 on [−1, 1], x0 = 0. The number ‘tol’ refers to the tolerance that the program uses in it’s stopping criterion for Newton’s method (default is 10−6 ). In each case: (i) explain the behaviour of Newton’s method that you see, and (ii) for those cases where we have convergence, explain how the rate of convergence is related to the multiplicity of the roots. Solution: (i): We have convergence for problems (a) - (c). In case (d) we have divergence because the iterates move further and further away away from the root (eventually becoming unbounded). In case (e) we have divergence because the iterates simple oscillate between 0 and 1. In problem (f) the initial guess lands on the graph where the slope is horizontal (thus the tangent line never intersects with the x-axis). In Newton’s method we attempt to divide by zero, which is impossible. (ii): In case (a) we have the simple roots ±2. Thus from a Theorem in our lecture notes we know that provided the initial guess is chosen sufficiently close to the root the method is a quadratically convergent method (i.e., the number of correct decimal places we obtain in the iterates is approximately doubled every step). We see this in the data output from the Matlab program (see below). In case (b) we have a double root (+2, twice) and so we expect a rate of convergence less than quadratic (which is verified if you look at the digits in the iterates output by the Matlab program). In Case (c) we have a root (+2) of multiplicity 5, thus we expect an even slower rate of convergence. We show the Matlab out put for problem (a) below: Iteration Iteration Iteration Iteration Iteration Iteration Iteration x0 x1 x2 x3 x4 x5 x6 is is is is is is is 6.0000000000000000 3.3333333333333335 2.2666666666666666 2.0156862745098039 2.0000610360875868 2.0000000009313226 2.0000000000000000 67 Iterations stop because |f(x6)|+|x6-x5|<0.000001 3.3 How to Stop Newton’s Method 1. Approximate the root of f (x) = ex + x using Newton’s Method. Stop the method using an appropriate Stopping Criterion with a tolerance of 10−4 . (see your lecture notes). (Hint: to start, show using the IMVT that there is a root in [−1, 0].) Solution: Following the hint, f (0) = e0 + 0 = 1, f (−1) = e−1 − 1 = −0.632 . . . , thus by the IMVT there is a root in the interval [−1, 0]. Hence it is 68 sensible to choose our initial guess as x0 = −0.51 . Newton’s Method is (exn + xn ) f (xn ) = x − n f 0 (xn ) (exn + 1) exn (xn − 1) = , exn + 1 xn+1 = xn − (3.1) after simplifying. We start iterating with (3.1) as follows ex0 (x0 − 1) = −0.566311 . . . , ex0 + 1 ex1 (x1 − 1) = −0.567143165034862 . . . , x2 = ex1 + 1 ex2 (x2 − 1) = −0.567143290409781 . . . , x3 = ex2 + 1 x1 = etc, to fill in the the table of values: n xn 0 −0.5 1 −0.566311 . . . 2 −0.567143165034862 . . . 3 −0.567143290409781 . . . |xn − xn−1 | — 0.066 . . . 8.321 · · · × 10−4 1.253 · · · × 10−7 |f (xn )| 0.106 . . . 0.0013 . . . 1.96 · · · × 10−7 4.551 · · · × 10−15 Clearly after three iterations we have |f (x3 )| + |x3 − x2 | < 10−4 as required, thus we stop iterating. 2. (MATLAB) Use the program NEWTON_RAPH_PRES to check your written answer in Problem 1. Solution: Just run the Matlab program with the data used in the solution of Problem 1 and a tolerance of 10−4 . 1 Of course this is just one step of the Bisection Method 69 3.4 Secant Method 1. Do three steps of the secant method for f (x) = x3 − 2, using x0 = 0, x1 = 1. Solution: We approximate the cube root of 2 since f (x) = 0 =⇒ x = √ 3 2. The secant method is: xn − xn−1 xn+1 = xn − f (xn ) f (xn ) − f (xn−1 ) xn − xn−1 3 = xn − (xn − 2) (x3 − 2) − (x3n−1 − 2) n xn − xn−1 3 = xn − (xn − 2) (3.2) x3n − x3n−1 Using (3.2) to do apply the Bisection Method is not a good idea. If xn is very close to xn−1 then both the numerator and denominator in the fractional part will lead to loss of significance (see the section on Computer Arithmetic). So we will rearrange this expression to avoid the problem. Now b 3 − a3 = b2 + ab + a2 , b−a so with b = xn and a = xn−1 we have x3n − x3n−1 = x2n + xn xn−1 + x2n−1 xn − xn−1 so 1 xn − xn−1 = 2 . 3 3 xn − xn−1 xn + xn xn−1 + x2n−1 Substituting this into (3.2) yields: xn+1 (x3n − 2) = xn − 2 xn + xn xn−1 + x2n−1 (3.3) Ideally, we would do some further rearranging to get a single expression that avoids the (x3n − 2) term. This is because once we get very close to the root x3n will be very nearly equal to 2, again leading to loss of significance. However, unless we are needing a VERY accurate answer this shouldn’t make much of a difference. Doing 3 steps of (3.3) (with x0 = 0 and x1 = 1) yields 70 n=1 (x31 − 2) (x21 + x0 x1 + x20 ) (13 − 2) = 1− 2 (1 + (0)(1) + 02 ) = 2 x 2 = x1 − n=2 x3 (x32 − 2) = x2 − 2 (x2 + x1 x2 + x21 ) (23 − 2) = 2− 2 (2 + (1)(2) + 12 ) 8 = or 1.14286 (6 significant figures) 7 n=3 (x33 − 2) (x23 + x2 x3 + x22 ) 8 3 − 2 7 8 − 2 = 8 8 7 2 + (2) + 2 7 7 x4 = x3 − = Note: √ 3 75 62 or 1.20968 (6 significant figures) 2 ≈ 1.259921... 2. Repeat the above using x0 = 1, x1 = 0. Comment on how the algorithm is performing compared to the last example. Solution: Using the same method we get n = 1 n = 2 n = 3 x2 = 2 1 x3 = 2 6 x4 = 7 or 0.857143 71 (6 significant figures) The method appears to be converging slower with x0 = 1 and x1 = 0 than with x0 = 0 and x1 = 1. 3. (MATLAB) Use the program SECANT_PRES to apply the Secant method to the following test problems: (a) f (x) = x2 − 4 on [2, 6], x0 = 6, x1 = 5.5, tol = 1e − 03; (b) f (x) = (x − 2)2 on [2, 3], x0 = 3, x1 = 2.7, tol = 1e − 06; (c) f (x) = 6(x − 2)5 on [2, 3], x0 = 3, x1 = 2.95, tol = 1e − 03; (d) f (x) = x2 − 4 on [−3, 3], x0 = −1, x1 = 1; (e) f (x) = (4/3) exp((2 − x/2)(1 + ln(x)/x)) on [2, 20], x0 = 2.5, x1 = 3.5. The number ‘tol’ refers to the tolerance that the program uses in it’s stopping criterion for the Secant method (default is 10−6 ). In each case: (i) explain the behaviour of the Secant method that you see, and (ii) for those cases where we have convergence, explain how the rate of convergence is related to the multiplicity of the roots. Solution: (i): We have convergence for problems (a) - (c). In case (d) we have divergence because the slope of the first secant line is horizontal, and thus never intersects with the x-axis (we divide by zero in the formula). In case (e) we have divergence because the iterates move further and further away from the root (becoming unbounded as x tends to infinity). (ii): Although we don’t cover this in the lecture notes, the convergence theorem for the Secant Method is almost the same as for Newton’s Method. The difference being that the optimal rate of convergence for the Secant Method is a little less than 2, but bigger than 1 (‘superlinear’). It’s interesting that this optimal rate of convergence is equal √ to the Golden Mean (namely, (1 + 5)/2 ≈ 1.62). When you run the program you should notice that the rate of convergence for problem (a) is faster than for either problem (b) or (c) (as problem (a) has simple roots, while problems (b) and (c) have roots of multiplicity 2 and 5 72 respectively). We show (some of) the Matlab output for problem (e) below (program interupted): Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Iteration Operation x1 is 2.5000000000000000 x2 is 3.5000000000000000 x2 is 4.5156679582822132 x3 is 5.5514402736656114 x4 is 6.6339592833128247 x5 is 7.7514171147633677 x6 is 8.9020232758648721 x7 is 10.0794196823085791 x8 is 11.2797803442435374 x9 is 12.4992630213283906 x10 is 13.7349726174479319 x11 is 14.9844470660306381 x12 is 16.2457144330682333 terminated by user during secant_pres (line 99) 73 3.5 Fixed Point Iteration 1. Do three steps of the following fixed point iteration problem: xn+1 = ln(1 + xn ), x0 = 12 . Solution: n=0 x1 1 = ln 1 + 2 = 0.405465108108... n=1 x2 = ln(1 + 0.405465108108...) = 0.340368285804... n=2 x3 = ln(1 + 0.340368285804...) = 0.292944416354... 2. Let y0 and h = 1/8 be fixed numbers. Do three steps of the following fixed point iteration 1 yn+1 = y0 + h(−y0 ln(y0 ) − yn ln(yn )), 2 1 1 and h = this becomes 2 8 1 1 1 1 = + − ln − yn ln(yn ) . 2 16 2 2 Solution: With y0 = yn+1 y0 = 1/2. Iterating 74 n=0 y1 = = = = 1 1 1 1 + − ln − y0 ln(y0 ) 2 16 2 2 1 1 1 1 1 1 + − ln − ln 2 16 2 2 2 2 1 1 1 + ln 2 16 2 0.543321698785... n=1 y2 1 1 1 1 = + − ln − y1 ln(y1 ) 2 16 2 2 = 0.542376812253... n=2 y3 1 1 1 1 = + − ln − y2 ln(y2 ) 2 16 2 2 = 0.54239978931... 3. Consider the fixed point iteration xn+1 = 1 + e−xn . Show that this iteration converges for any x0 ∈ [1, 2]. How many iterations does the theory predict it will take to achieve 10−5 accuracy? Solution: Observe that with g(x) = 1 + e−x g(1) = 1.367... ∈ [1, 2], g(2) = 1.135... ∈ [1, 2], and as g(x) is monotonically decreasing on [1, 2] we know that 1 ≤ g(x) ≤ 2 for all x ∈ [1, 2], i.e. g : [1, 2] ⊂ [1, 2]. To show the second condition of the fixed point theorem (see your lecture notes) consider |g 0 (x)| = |e−x | ≤ |e−1 | = 0.367 < 1, for all x ∈ (1, 2), as e−x is decreasing. Thus by the fixed point theorem we see that the iteration converges for all x0 ∈ [1, 2]. We apply the error bound formula n L |xn − α| ≤ (x1 − x0 ), 1−L 75 where L = 1 = 0.367... and e x0 = x1 = = = 1.5 1 + e−x0 1 + e−1.5 1.22313016015..., so we seek n such that 1 1 n e − 1e ! (1.2231... − 1.5) < 10−5 or e−n < 2.2830965... × 10−5 so − n < ln(2.2830965... × 105 ) or n > 10.687.. so take n = 11. 4. For each function listed below, find an interval [a, b] such that g([a, b]) ⊂ [a, b]. Draw a graph of y = g(x) and y = x over this interval, and confirm that a fixed point exists there. Estimate (by eye) the value of the fixed point, and use this as a starting value for a fixed point iter 2 1 x+ , ation. Does the iteration converge? Explain: (a) g(x) = 2 x (b) g(x) = 1 + e−x . Solution: (a) Not straightforward. To clarify things we need to do some curve sketching. Notice first that we can work the fixed points out exactly, as √ 1 1 1 1 g(x) = x + = x =⇒ = x or 2 = x2 or x = ± 2. 2 x x 2 √ √ √ Focus on the fixed point at x = 2. Notice g( 2) = 2 tells us that √ √ ( 2, 2) is on the graph of g. Now √ 1 1 g 0 (x) = − 2 = 0 =⇒ x = ± 2 2 x 76 are critical points. Maxima, minima, or inflection points? Look at the second derivative. √ 2 2 g 00 (x) = 3 =⇒ g 00 ( 2) = 3 > 0 x 22 p √ so ( (2), 2) is a local minimum. Notice also that lim g(x) = +∞ x→+∞ and lim g(x) = −∞. x→−∞ This information leads to the following graph: Since g(1) = 1.5 = g(2), thus √ 2 ≤ g(x) ≤ 1.5 1 ≤ g(x) ≤ 1.5 for all x ∈ [1, 2] for all x ∈ [1, 2] Thus, condition (i) in the key fixed point theorem is satisfied with [a, b] = [1, 2]. To satisfy the second condition we need to show that 1 1 |g 0 (x)| < 1 for all x ∈ (1, 2). Now g 0 (x) = − 2 , so 2 x |g 0 (1)| = 0.5 and 77 g 0 (2) = 0.25, √ and as g has a local minimum at x = 2 ≈ 1.414, the slope of g, i.e. g 0 (x), must be less than 1 throughout (1, 2) X (b) g(x) = 1 + e−x . Now g(x) = x =⇒ 1 + e−x = x or e−x = x − 1 illustrated below: So visually we see that the fixed point is between 1 and 2. Now g(1) = 1 + e−1 ≈ 1.136... g(2) = 1 + e−2 ≈ 1.135... and as g(x) is decreasing, 1 ≤ g(x) ≤ 2 for all x ∈ [1, 2] 78 X i.e. g : [1, 2] ⊂ [1, 2] and so condition (i) in the fixed point theorem is satisfied. To satisfy condition (ii) in this theorem, consider g 0 (x) = −e−x so |g 0 (x)| = e−x , which again is a decreasing function. And as |g 0 (1)| = e−1 ≈ 0.367... =⇒ |g 0 (x)| < 1 for all x ∈ (1, 2) |g 0 (2)| = e−2 ≈ 0.135... 5. Let h(x) = 1 − x2 /4. Show that this function has a root at x = α = 2. Using x0 = 1/2, apply fixed point iteration with xn+1 = h(xn ) to approximate the fixed point of h. Comment on your results. Solution: Consider the iterative process xn+1 = 1 − n = n = n = n = n = .. . n = x2n , 4 1 x0 = . 2 x20 = 0.9375 0 : x1 = 1 − 4 x2 1 : x2 = 1 − 1 = 0.780273... 4 x22 2 : x3 = 1 − = 0.847793... 4 x2 3 : x4 = 1 − 3 = 0.820311... 4 x24 4 : x5 = 1 − = 0.83177... 4 .. . x229 = 0.8284271... 29 : x30 = 1 − 4 Can we check our answer? The fixed points of h(x), i.e., the roots of x2 h(x) = 1 − = x, correspond to the roots of 4 √ −4 ± 32 2 x +4x−4 = 0 so x= (usual quadratic formula). 2 Just keeping the positive solution yields x = 0.8284271... 79 X X 6. (MATLAB) Use the program FIXED_POINT_ITER_PRES to apply Fixed Point iteration to the following iteration functions: (a) g(x) = sin(x) on [0, 2.5], x0 = 2, tol = 1e − 03; (b) g(x) = 1 + 2/x on [0, 3.5], x0 = 1, tol = 1e − 03; (c) g(x) = x2 /4 + x/2 on [2, 20], x0 = 2.3. In each case (i) work out by hand (if possible) the fixed points, (ii) explain what you see, and state whether the iterative process converges or diverges. Solution: (i): (a) solving sin(x) = x yields α = 0 (just recall the graph of sin(x) on [0, π]), (b) solving 1 + 2/x = x yields (after application of the quadratic formula) that α = 2 (α = −1 is not in the specified interval); solving x2 /4 + x/2 = x yields α = 2 (α = 0 is not in the specified interval). (ii): (a) We see a ‘stair’ shaped diagram. Convergence is quite slow. At best fixed point iteration is linearly convergent. (b) We see a ‘cobweb’ shaped diagram. The iterates are convergent. (c) We see a ‘stair’ shaped diagram. The iterates are diverging from the root. We give some MATLAB output for problem (a). Iteration Iteration Iteration Iteration Iteration etc .. Iteration Iteration Iteration x0 x1 x2 x3 x4 is is is is is 2.000000 0.9092974268256817 0.7890723435728884 0.7097000402345258 0.6516062636498291 x15 is 0.4008516296070669 x16 is 0.3902026038089464 x17 is 0.3803757974172909 Iterations stop because |g(x16)|-x16|<0.010000 80 81 Chapter 4 Interpolation and Approximation 4.1 Lagrange Interpolation 1. Find the polynomial of degree 2 that interpolates y = x3 at the nodes x0 = 0, x1 = 1 and x2 = 2. Plot y = x3 and the interpolating polynomial over the interval [0, 2]. Solution: To find P2 (x) first find the Lagrange basis functions: (x − x1 )(x − x2 ) (x − 1)(x − 2) 1 = = (x − 1)(x − 2), (x0 − x1 )(x0 − x2 ) (0 − 1)(0 − 2) 2 (x − x0 )(x − x2 ) (x − 0)(x − 2) L1 (x) = = = −x(x − 2), (x1 − x0 )(x1 − x2 ) (1 − 0)(1 − 2) (x − 0)(x − 1) 1 (x − x0 )(x − x1 ) L2 (x) = = = x(x − 1). (x2 − x0 )(x2 − x1 ) (2 − 0)(2 − 1) 2 L0 (x) = Thus, P2 (x) = f (x0 )L0 (x) + f (x1 )L1 (x) + f (x2 )L2 (x) 1 = −x(x − 2) + (8) x(x − 1) 2 = x(3x − 2) The sketch is difficult to draw as both graphs match well over the specified interval: 82 2. Construct the quadratic polynomial that interpolates y = nodes x0 = 1/4, x1 = 9/16, and x2 = 1. √ x at the Solution: To find P2 (x) first find the Lagrange basis functions: 9 x − 16 (x − 1) 65 9 (x − x1 )(x − x2 ) 1 = = 1 x− (x − 1), L0 (x) = 9 (x0 − x1 )(x0 − x2 ) 15 16 − 16 −1 4 4 x − 41 (x − 1) (x − x0 )(x − x2 ) 256 1 9 =− = 9 L1 (x) = x− (x − 1), (x1 − x0 )(x1 − x2 ) 35 4 − 14 16 −1 16 9 x − 14 x − 16 64 (x − x0 )(x − x1 ) 1 9 = = L2 (x) = x− x− . 9 (x2 − x0 )(x2 − x1 ) 21 4 16 1 − 14 1 − 16 83 Thus, P2 (x) = f (x0 )L0 (x) + f (x1 )L1 (x) + f (x2 )L2 (x) 1 64 9 3 256 1 = x− (x − 1) + − x− (x − 1) 2 15 16 4 35 4 1 9 64 x− x− +(1) 21 4 16 = after much tedious simplification 32 22 9 2 = − x + x+ . 105 21 35 3. Find the polynomial of degree 3 that interpolates y = x3 at the nodes x0 = 0, x1 = 1, x2 = 2, and x3 = 3. (Simplify your interpolating polynomial as much as possible.) Hint: This is easy if you think about the implications of the uniqueness of the interpolating polynomial. Solution: From our lecture notes we know that for distinct nodes, the interpolating polynomial is unique. Thus the cubic polynomial that interpolates the cubic x3 at x0 = 0, x1 = 1, x2 = 2 and x3 = 3 must be x3 ! We will do the calculation nonetheless. (x − 1)(x − 2)(x − 3) (x − x1 )(x − x2 )(x − x3 ) = L0 (x) = (x0 − x1 )(x0 − x2 )(x0 − x3 ) (0 − 1)(0 − 2)(0 − 3) 1 = − (x − 1)(x − 2)(x − 3), 6 (x − x0 )(x − x2 )(x − x3 ) (x − 0)(x − 2)(x − 3) L1 (x) = = (x1 − x0 )(x1 − x2 )(x1 − x3 ) (1 − 0)(1 − 2)(1 − 3) 1 = x(x − 2)(x − 3), 2 (x − x0 )(x − x1 )(x − x3 ) (x − 0)(x − 1)(x − 3) L2 (x) = = (x2 − x0 )(x2 − x1 )(x2 − x3 ) (2 − 0)(2 − 1)(2 − 3) 1 = − x(x − 1)(x − 3), 2 (x − x0 )(x − x1 )(x − x2 ) (x − 0)(x − 1)(x − 2) L3 (x) = = (x3 − x0 )(x3 − x1 )(x3 − x2 ) (3 − 0)(3 − 1)(3 − 2) 1 = x(x − 1)(x − 2). 6 84 Thus, after setting f (x) = x3 P3 (x) = f (x0 )L0 (x) + f (x1 )L1 (x) + f (x2 )L2 (x) + f (x3 )L3 (x) 1 1 x(x − 1)(x − 3) = 0 + x(x − 2)(x − 3) + (8) − 2 2 1 + (27) x(x − 1)(x − 2) 6 = x3 (after much simplification). 4. What is the error in quadratic interpolation to f (x) = equally spaced nodes on the interval [1/4, 1]? √ x, using Solution: From lecture notes we have h3 |f (x) − P2 (x)| ≤ √ M, 9 3 √ where M = max |f 000 (x)|. Here f (x) = x and x∈(x0 ,x2 ) h= so 1 x0 = , 4 x1 = 1− 2 1 4 = 3 1 3 · = , 4 2 8 1 3 5 + = , 4 8 8 x2 = 5 3 + = 1. 8 8 Also 1 1 1 f (x) = x 2 =⇒ f 0 (x) = x− 2 2 1 3 00 =⇒ f (x) = − x− 2 4 3 −5 000 =⇒ f (x) = x 2 , 8 1 which is a decreasing function on , 1 . Thus 4 3 |f (x)| ≤ 8 000 − 52 5 1 3 3 = 42 = (32) = 12 = M, 4 8 8 85 1 for all x ∈ , 1 . Thus 4 √ 3 3 |f (x) − P2 (x)| ≤ √ (12) = = 0.0406 128 9 3 3 3 8 5. Repeat the above for f (x) = x −1 (3 significant figures). 1 on ,1 . 2 Solution: Proceed as in the problem above. 1 − 12 1 = , so f (x) = x−1 and h = 2 4 1 x0 = , 2 3 x1 = , 4 x2 = 1. Also f (x) = x−1 =⇒ f 0 (x) = −x−2 =⇒ f 00 (x) = 2x−3 =⇒ f 000 (x) = −6x−4 . So 6 |f (x)| = 4 = 96 = M, |x| 000 1 for all x ∈ , 1 . 2 Thus using the error estimate given in the last problem we have √ 1 3 3 4 |f (x) − P2 (x)| ≤ √ (96) = = 0.0962 (3 significant figures). 18 9 3 6. (MATLAB) Use the program LAGRANGE_PRES to construct and plot the Lagrange interpolant Pn (x) for the following functions f (x) with n nodes on the specified intervals [a, b]: (a) f (x) = exp(x) on [−2, 2] with n = 2, 3, 4; (b) f (x) = x3 on [0, 2] with n = 2, 3, 4; (c) f (x) = tan(x) on [−1.5, 1.5] with n = 4, 5, 6, 7, 8; 86 (d) f (x) = 1/(1 + 25x2 ) on [−1, 1] with n = 4, 8, 16, 32. We can measure the fit between f (x) and the interpolating polynomial Pn (x) using the maximum norm kf − Pn k∞ := max |f (x) − Pn (x)|, x∈[a,b] which is just the maximum absolute error between f and Pn for all x ∈ [a, b]. We say that the sequence of Lagrange interpolation polynomials Pn converges to f if kf − Pn k∞ → 0 as n → ∞. For each problem above decide in which cases the Lagrange interpolation polynomials are convergent to the given functions. If the Lagrange polynomials are divergent explain why based on the plots. Solution: Running the MATLAB program reveals that we have convergence for cases (a) - (c), as measured by the maximum norm (just assess whether the maximum absolute error on [a, b] tends to zero as n → ∞). However, in case (d) we have divergence. As we increase the degree of the interpolating polynomial the maximum error increases (near the end points of the domain). This is a famous problem illustrating nonconvergence called ‘Runge’s phenomena’. So not every function can be adequately approximated using Lagrange interpolation. The ‘fix’ for this problem is to either use piecewise Lagrange interpolation (e.g., piecewise linear interpolation), or to relax the assumption that the nodes are equally spaced (in some clever way!). We show the plot from the MATLAB program for Problem (c) with n = 5 below: 87 The program also outputs the Lagrange polynomial in traditional algebraic form, namely P (x) = 4.83486x3 − 1.47748x. Note: the highest power shown here is x3 because the coefficient of x4 was either zero or almost zero. 4.2 Hermite Interpolation √ 1. Construct the cubic Hermite interpolant to f (x) = 1 + x using the nodes a = 0 and b = 1. Plot the interpolant and the function together. Where is the approximation error the greatest? 88 1 1 1 Solution: f (x) = (1 + x) 2 =⇒ f 0 (x) = (1 + x)− 2 . Then 2 x − x2 x−1 = =1−x x1 − x2 0−1 x − x1 x−0 L2 (x) = = =x x2 − x1 1−0 L1 (x) = with L01 (x) = −1 and L02 (x) = 1. Now h1 (x) = = = h2 (x) = = = h̃1 (x) = = = h̃2 (x) = = [1 − 2(x − x1 )L01 (x1 )]L21 (x) [1 − 2(x − 0)(−1)](1 − x)2 (1 + 2x)(1 − x)2 [1 − 2(x − x2 )L02 (x2 )]L22 (x) [1 − 2(x − 1)(1)]x2 (3 − 2x)x2 (x − x1 )L21 (x − 0)(1 − x)2 x(1 − x)2 (x − x2 )L22 (x) (x − 1)x2 (4.1) (4.2) (4.3) (4.4) Using (4.1)-(4.4) we have H(x) = f (x1 )h1 (x) + f 0 (x1 )h̃1 (x) + f (x2 )h2 (x) + f 0 (x2 )h̃2 (x) 1 1 1 1 1 = 1 2 (1 + 2x)(1 − x)2 + x(1 − x)2 + 2 2 (3 − 2x)x2 + 2− 2 (x − 1)x2 2 2 = (after simplifying) = 0.0251x3 − 0.111x2 + 0.5x + 1 (coefficients to ≤ 3 significant figures) Based on the graph (see below) we see that the error is greatest at the end of the interval. 89 Figure 4.1: Graph of f (x) and the interpolant 90 2. Construct the cubic Hermite interpolant to f (x) = sin(x) using the nodes a = 0 and b = π. Plot the error between the interpolant and the function. Where is the error greatest? Solution: f (x) = sin(x) =⇒ f 0 (x) = cos(x). Then x−π π−x x − x2 = = x1 − x2 0−π π x − x1 x−0 x L2 (x) = = = x2 − x1 π−0 π L1 (x) = with L01 (x) = − 1 1 and L02 (x) = . Now π π h1 (x) = [1 − 2(x − x1 )L01 (x1 )]L21 (x) (π − x)2 1 = 1 − 2(x − 0) − π π2 2 (π − x)2 = 1+ x π π2 h2 (x) = [1 − 2(x − x2 )L02 (x2 )]L22 (x) 2 x 1 = 1 − 2(x − π) π π2 2 2 x = 3− x π π2 h̃1 (x) = (x − x1 )L21 (π − x)2 = (x − 0) π2 2 (π − x) = x π2 h̃2 (x) = (x − x2 )L22 (x) x2 = (x − π) 2 π 91 (4.5) (4.6) (4.7) (4.8) Using (4.5)-(4.8) we have H(x) = f (x1 )h1 (x) + f 0 (x1 )h̃1 (x) + f (x2 )h2 (x) + f 0 (x2 )h̃2 (x) 2 x (π − x)2 2 x2 x = 0 + 1x + 0 3 − + (−1)(x − π) π2 π π2 π2 = (after simplifying) 1 3 = 0x − x2 + x + 0 π The maximum error occurs near the centre of the interval. Figure 4.2: Error between f (x) and the interpolant 92 1 3. Construct the cubic Hermite interpolatant to f (x) = x 3 on the interval 1 , 1 . What is the maximum error as predicted by theory? 8 1 1 2 Solution: f (x) = x 3 =⇒ f 0 (x) = x− 3 . Then 3 x−1 x−1 x − x2 8 = 1 = 7 = (1 − x) x1 − x2 7 −1 −8 8 1 x− 8 x − x1 1 8 L2 (x) = = x− = x2 − x1 7 8 1 − 18 L1 (x) = with L01 (x) = − 8 8 and L02 (x) = . Now 7 7 h1 (x) = [1 − 2(x − x1 )L01 (x1 )]L21 (x) 2 8 8 1 − (1 − x)2 = 1−2 x− 8 7 7 5 16 64 x+ = (1 − x)2 7 7 49 h2 (x) = [1 − 2(x − x2 )L02 (x2 )]L22 (x) 2 2 8 1 8 x− = 1 − 2(x − 1) 7 7 8 2 23 16 64 1 = − x x− 7 7 49 8 h̃1 (x) = (x − x1 )L21 1 8 = x− (1 − x)2 8 7 h̃2 (x) = (x − x2 )L22 (x) 2 64 1 = (x − 1) x− 49 8 93 (4.9) (4.10) (4.11) (4.12) Using (4.9)-(4.12) we have H(x) = f (x1 )h1 (x) + f 0 (x1 )h̃1 (x) + f (x2 )h2 (x) + f 0 (x2 )h̃2 (x) 13 16x 5 1 64 = + (1 − x)2 8 7 7 49 − 23 1 8 1 1 (1 − x)2 + x− 3 8 8 7 2 1 23 16 64 1 + (1 3 ) − x x− 7 7 49 8 2 1 1 64 − 23 (1 )(x − 1) x− + 3 49 8 = (after simplifying) = 0.466x3 − 1.26x2 + 1.46x + 0.336 (coefficients to ≤ 3 significant figures) From the lecture notes, the interpolation error is f (4) (ξx ) 4! so 2 M 1 |f (x) − H(x)| ≤ x − (x − 1)2 · , 24 8 1 where M is an upper bound for |f (4) (x)| on ,1 . 8 f (x) − H(x) = (x − x1 )2 (x − x2 )2 (4.13) 1 1 2 f (x) = x 3 =⇒ f 0 (x) = x− 3 3 2 5 =⇒ f 00 (x) = − x− 3 9 8 10 =⇒ f 000 (x) = x− 3 27 80 11 =⇒ f (4) (x) = − x− 3 81 so |f (4) 1 (x)| = which is a decreasing function on , 1 , thus 8 − 113 80 1 80 20736 (4) |f (x)| ≤ = (2048) = = M. 81 8 81 10 11 80 |x|− 3 , 81 94 2 1 1 2 (x − 1) on , 1 . The graph of We also need to maximize x − 8 8 2 1 h(x) = x − (x − 1)2 is: 8 We use Calculus to find the local maximum point. After some simplifying we obtain: 1 (x − 1)(8x − 1)(16x − 9). 32 1 9 Thus local extrema occur at x = 1, x = and x = . Thus h(x) ≤ 8 16 9 2401 h 16 = 65536 . So from (4.13) h0 (x) = 1 |x 3 − H(x)| ≤ 2401 20736 1 · · = 3.165 65536 10 24 (predicted). Note: if you plot the actual error between H(x) and f (x) the actual maximum error is closer to 0.016. So the theoretical (predicted) maximum error (≤ 3.165) is quite a bit different from this (but still consistent with the prediction). 4. (MATLAB) Use the program HERMITE_PRES to construct and plot the Hermite interpolant H(x) for the following functions f (x) with n nodes on the specified intervals [a, b]: 95 (a) f (x) = exp(x) on [−1, 1] with n = 2, 3 (b) f (x) = 1/x on [1/2, 4] with n = 2, 3, 4 (c) f (x) = x8 + 1 on [−1, 1] with n = 2, 3, 4, 5; (d) f (x) = 1/(1 + 25x2 ) on [−1, 1] with n = 2, 3, 4, 5, 6; We can measure the fit between f (x) and the interpolating polynomial H(x) using the maximum norm kf − Hk∞ := max |f (x) − H(x)|, x∈[a,b] which is just the maximum absolute error between f and H for all x ∈ [a, b]. We say that the sequence of Hermite interpolation polynomials H converges to f if kf − Hk∞ → 0 as n → ∞. For each problem above decide in which cases the Hermite interpolation polynomials are convergent to the given functions. If the Hermite polynomials are divergent explain why based on the plots. How do the graphs of the Hermite interpolating polynomials differ from the graphs of the Lagrange interpolating polynomials? Solution: Basically, the situation is the same as for the Lagrange case. For cases (a) - (c) we have convergence as n → ∞. Divergence in problem (d) is due to the same ‘Runge’s phenomena’ as in the Lagrange case. The main difference here is that it is clear from the plots that not only do the function values between f (x) and H(x) match, but so do the slopes. We show the plot for problem (c) with n = 3 below: 96 The program also outputs the Hermite polynomial in traditional algebraic form, namely H(x) = 3x4 − 2x2 + 1. It is clear that in this case we need greater than 3 nodes to get a good approximation. Recall that the degree of the Hermite interpolant is given by 2n − 1, thus if if n = 5 (thus the degree of H(x) would be 9) the approximation would be exact (i.e., f (x) = H(x)). This is because we are fitting polynomials to polynomials, and if you use a higher degree than necessary the coefficients of the higher (unnecessary) powers of x will be zero. 97

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