For ODD number questions you can find the solutions at the back of text book. You need add more details following the hints below. _________________________________________________________________ _________________________________________________________________ Sec 1.1, no 10: f(x) = (x+1)^(1/2) f(0) = 1 f'(x) = (1/2)(x+1)^(-1/2) f'(0) = 1/2 f''(x) = (-1/4)(x+1)^(-3/2) f''(0) = -1/4 f'''(x) = (3/8)(x+1)^(-5/2) f'''(0) = 3/8 so P_3(x) = 1 +(1/2) x - (1/8)x^2 + (1/16)x^3. sqrt(0.50) is approx P_3(-0.5) = 0.7109375 The actual value is 0.70710678... so the error is 0.00383072. (The relative error is 0.0054.. or slightly under one-half percent). The other parts are just as easy. ________________________________________________________________ Sec 1.1, no 11(a): f(x) = e^x cos x f(0) = 1 f'(x) = e^x cos x - e^x sin x f'(0) = 1 f''(x) = - 2 e^x sin x f''(0) = 0 f'''(x) = - 2 e^x (sin x + cos x) Hence P_2(x) = 1+x and e^x cos x = 1 + x + R_2(x) where R_2(x) = -(1/3)e^c(sin c + cos c)x^3 and c is some number between 0 and x. Thus |f(0.5) - P_2(0.5)| = |R_2(0.5)| < = (2/3)e^(0.5)0.5^3 = 0.1373... (Here we have used |sin c| < =1 and |cos c| <= 1.) The actual error is much smaller. It is 1 + 0.5 - e^.5 cos .5 = 0.0531... (b) |f(x) - P_2(x)| = |R_2(x)| = | (1/3)e^c(sin c + cos c)x^3 | <= (1/3) e (1+1) 1^3 <= 1.812 (Since 0 < c <1, so e^c <= e, and sin c <=1, cos c <=1.) ____________________________________________________________________ Sec 1.1, no 16: According to Taylor's theorem, for suitable f, f(b) = f(a) + (b-a)f'(a) + (1/2)(b-a)^2 f''(c) where c is a number between a and b. Now choose f(x) = sin x, a = 0, b = pi/180, so f'(x) = cos x, f''(x) = -sin x, so we get sin (pi/180) = sin 0 + (pi/180 - 0) cos 0 + (1/2)(pi/180)^2[-sin(c)] where c is a number between 0 and pi/180. Thus sin (pi/180) = 0.0174553 - 0.0001523 sin (c) So sin (pi/180) is approximately 0.0174553 with an error no bigger in absolute value than 0.0001523 (Note that |sin (c)| <= 1.) Another way to say this is that sin (pi/180) is between 0.0174553 - 0.0001523 and 0.0174553 + 0.0001523 (Check out its "actual" value) _________________________________________________________________ Sec 1.1, no 17: 42 degrees = 45 degrees - 3 degrees = pi/4 - pi/60 radians. By Taylor's theorem cos (pi/4 - pi/60) = cos (pi/4) + (-pi/60) (-sin(pi/4)) + (1/2!)(-pi/60)^2 (-cos(pi/4)) + (1/3!)(-pi/60)^3 (sin(pi/4)) + (1/4!)(-pi/60)^4 (cos(pi/4)) + (1/5!)(-pi/60)^5 (-sin(c)), where c is some number between pi/4 - pi/60 and pi/4. Since |sin x| <=1 for all x, the last term is no bigger in absolute value than (1/5!)(pi/60)^5 = 3.2795 times 10^(-9) so if we consider the other five terms only, we will have the required accuracy. The other five terms give the approximate value 0.7431448 for cos (42 degrees) - we use sin (pi/4) = cos(pi/4) = 1/sqrt(2) __________________________________________________________________ Sec 1.2, no. 3: a. Rel err. <= 10^-3, so abs err < = 150*10^-3 = 0.15, so p* must be between 148.5 and 151.5. _________________________________________________________________ Sec 1.2, no. 15: (a) (1/3)x^2 - (123/4)x + 1/6 = 0 a = 1/3, b = -123/4. c = 1/6 exact solution: x_1=92.24457963, x_2 = 0.00542037285 Using four digit to get approximation solution: b^2-4ac = (123/4)^2 - 4/(3*6) = 945.2 sqrt(b^2-4ac) = 30.74 x_1approx = (-b + sqrt(b^2 - 4ac))/(2a) = (123/4 + 30.74)/0.6666 = 92.24 x_2approx = 2c/(-b + sqrt(b^2 - 4ac)) = 0.3332/(123/4 + 30.74) = 0.005418 abs errors: | x_1 -x_1approx | = 4.6 x 10^{-3} | x_2 -x_2approx | = 2.4 x 10^{-6} Rel errors: | x_1 -x_1approx |/|x_1| = 5.0 x 10^{-5} | x_2 -x_2approx |/|x_2| = 4.4 x 10^{-4} __________________________________________________________________ Sec 1.2, no. 21: a. easy b. Using the first method, we first calculate x0*y1 to be 6.24 and x1*y0 to be 6.25. (The actual values are 6.2356 and 6.2532.) The difference is -0.0100. (Actual is -0.0176.) We have made a big error here. The basic reason is that we are subtracting nearly equal numbers the "6.2" s cancel in the two numbers and with only three digits in total all our information about the two products has to be included in the one remaining digit. This method gives a bad result. The other one is reasonably satisfactory even with three digits. ___________________________________________________________________ Sec 1.2, no. 23. a) use the fomulas given in the question, and four-digit rounding step by step it gets the solution x = 2.451, y = - 1.635 ____________________________________________________________________ Sec 1.3, nos. 1: 1a. The actual answer is 1.5497... Using 3 -digit arithmetic we get 1 + 1/4 = 1.25 Then 1+1/4 + 1/9 = 1.25 + .111 = 1.36 * Adding 1/16 =0.0625 we get 1.42 * ' 1/25 = 0.0400 " 1.46 1/36 = 0.0277 1.48 * 1/49 = 0.0204 1.50 * 1/64 = 0.0156 1.51 * 1/81 = 0.0123 1.52 * 1/100 = 0.0100 1.53 Using the other method we start with 1/100 + 1/81 = 0.0100 + 0.0123 = 0.0223 adding 1/64 = 0.0156, we get 0.0379 adding 1/49 = 0.0204, we get 0.0583 adding 1/36 = 0.0277, 0.0860 adding 1/25 = 0.0400 0.126 * adding 1/16 = 0.0625 0.188 * adding 1/9 = .111 0.299 adding 1/4 = 0.250 0.549 adding 1/1 = 1.00 1.54 * The second method gives the better result. There are two kinds of errors arising here. First, numbers like 1/49 are not dealt with exactly but as 0.0204, for example. But this should lead to about the same inaccuracy in both methods. Second, when we add two 3-digit numbers and get a number with four or more digits, we lose accuracy in the chopping. This happens six times (indicated by an asterisk) in the first method but only three times in the second method _______________________________________________________________ Sec. 1.3, no 3: It is not easy to write the general form of the nth derivative for arctan (x) and so to get the remainder in Taylor's theorem. However, we can use here the fact that, for an alternating series whose term decrease in absolute value to zero, the error involved in stopping at any stage is less, in absolute value, than the first term omitted. Here we use pi = 4 arctan (1) = 4 P_n(1) + 4(-1)^(n+2)/(2n+1)+ ..., where P_n(1) = 1 - 1/3 + 1/5 - ... + (-1)^(n+1)/(2n-1). so |4 P_n(1) - pi| <= 4/(2n+1). So, in part (a), we have to find the smallest value of n for which 4/(2n+1) is less than 10^(-3). So we want 2n + 1 > 4000 or n > = 2000. In part (b), we want 4/(2n+1) < 10^(-10), that is, 2n+1 > 40 000 000 000 or n >= 20 000 000 000. _________________________________________________________________ Sec. 1.3, no 4: Again, as in no 3, we use the fact that we are dealing with alternating series. In the series for arctan(1/2), if we include 5 non-zero terms, the first (non-zero) term omitted will be (1/2)^11/11 so the error committed in stopping there will be no more than 0.000044389; if we include three non-zero terms in the series for arctan(1/3). The error will be no more than (1/3)^7/7 = 0.0000654. Thus, we can be sure that the total error involved in taking pi/4 = 1/2 - (1/2)^3/3 + (1/2)^5/5 -(1/2)^7/7 + (1/2)^9/9 + 1/3 - (1/3)^3/3 + (1/3)^5/5 is no more than 4 (0.00004439 + 0.0000654) = 0.000439 < 10^(-3). You have to do soem trial and error to see how many terms to take in each series. (If we actually do the calculation, we find that we get 3.141979902... so the actual error is 0.000387.) ___________________________________________________________________ 1.3, no 5: Just taking one term in the series for arctan(1/239) causes an error of no more than (1/239)^3/3 = 0.000000024. Taking three terms in the series for arctan(1/4) causes an error of no more than (1/5)^7/7 = 0.000001828. So if we approximate pi/4 by 4[1/5 -(1/5)^3/3 +(1/5)^5/5] - 1/239, the error in pi will be no more than 16(0.000001828) + 4(0.000000024) = 0.0000293... < 10^(-3). (If we actually do the calculation, we find that we get 3.141620932 so the actual error is 0.0000282.) ____________________________________________________________________ Sec. 1.3, no 7 (a)(b): a. sin h /h = 1 - h^2/6 + O(h^4) so the rate is O(h^2). b. (1/h)(1 - cos h) = h/2 + O(h^3), so the order is O(h). Note: FOR more detailed answers, see examples given in class. _____________________________________________________________________