Here's a sample of how to do one of the problems in section 8.1. Below, I look at the first 50 elements in the sequence. > n := 50; n := 50 > for i from 1 by 1 to n > do > a[i] := evalf(8 + log(i)/i); > od; a[1] := 8. a[2] := 8.346573590 a[3] := 8.366204096 a[4] := 8.346573590 a[5] := 8.321887582 a[6] := 8.298626578 a[7] := 8.277987164 a[8] := 8.259930193 a[9] := 8.244136064 a[10] := 8.230258509 a[11] := 8.217990479 a[12] := 8.207075554 a[13] := 8.197303797 a[14] := 8.188504095 a[15] := 8.180536680 a[16] := 8.173286795 a[17] := 8.166659609 a[18] := 8.160576209 a[19] := 8.154970473 a[20] := 8.149786614 a[21] := 8.144977259 a[22] := 8.140501930 a[23] := 8.136325836 a[24] := 8.132418910 a[25] := 8.128755033 a[26] := 8.125311405 a[27] := 8.122068032 a[28] := 8.119007304 a[29] := 8.116113649 a[30] := 8.113373246 a[31] := 8.110773781 a[32] := 8.108304247 a[33] := 8.105954775 a[34] := 8.103716486 a[35] := 8.101581373 a[36] := 8.099542193 a[37] := 8.097592376 a[38] := 8.095725952 a[39] := 8.093937478 a[40] := 8.092221986 a[41] := 8.090574928 a[42] := 8.088992134 a[43] := 8.087469770 a[44] := 8.086004310 a[45] := 8.084592500 a[46] := 8.083231335 a[47] := 8.081918034 a[48] := 8.080650021 a[49] := 8.079424904 a[50] := 8.078240460 Okay, they ask us to plot the sequence. I can't figure out how to do that using maple. But I can at least plot the function and see how it behaves... I do this below. > plot({8,8 + log(x)/x}, x=1..2500); > Looking at the sequence and the plot, it looks like it's decreasing to 8. So I take L=8 and try to answer part b) First I need to find N so that if n > N then |a_n-L| < 0.01. Also, I want to find N so that if n > N then |a_n - L| < 0.0001. I do this by blindly guessing values of N... 50 isn't big enough... > n := 50: evalf((8+log(n)/n)-8); .07824046010 100 isn't big enough... > n := 100: evalf((8+log(n)/n)-8); .04605170186 None of the following have n big enough to have a_n within 0.01 of L... > n := 200: evalf((8+log(n)/n)-8); .02649158684 > n := 300: evalf((8+log(n)/n)-8); .01901260825 > n := 400: evalf((8+log(n)/n)-8); .01497866137 > n := 500: evalf((8+log(n)/n)-8); .01242921620 Aha! If n > 800 then a_n is within 0.01 of L. So I can take N = 800. If I looked at n between 500 and 800 I might find a smaller value of N that works. But the point is that there's some N that works and we found one. > n := 800: evalf((8+log(n)/n)-8); .008355764660 Now we look for N so that if n > N then |a_n - L| < .0001 3000 isn't big enough... > n := 3000: evalf((8+log(n)/n)-8); .002668789189 5000 isn't big enough... > n := 5000: evalf((8+log(n)/n)-8); .001703438638 > n := 10000: evalf((8+log(n)/n)-8); .0009210340372 > n := 50000: evalf((8+log(n)/n)-8); .0002163955656 > n := 70000: evalf((8+log(n)/n)-8); .0001593750075 > n := 90000: evalf((8+log(n)/n)-8); .0001267507217 > n := 100000: evalf((8+log(n)/n)-8); .0001151292546 Aha! If n > 300000 then |a_n - 8| < .0001. So we can take N = 300000. We could find a smaller value for N, but that's not really the point... > n := 300000: evalf((8+log(n)/n)-8); .00004203845916 Now I'll do problem 79 in section 8.2 You want to find x so that x - sin(x) = 1. I do this by Picard iteration. If x - sin(x) = 1, then sin(x) + 1 = x. So I'll define x_(n+1) = sin(x_n) + 1. If x_n is converging to some number, call it a, then a will satisfy a=sin(a)+1, as desired. To get a good first guess, I plot x and sin(x) + 1 on the same plot to see where they seem to intersect. > plot({x, sin(x)+1},x=0..3); So 2 looks like a good first guess. > n := 25; a[1] := 2; n := 25 a[1] := 2 > for i from 2 by 1 to n > do > a[i] := evalf(sin(a[i-1]) + 1); > od; > a[2] := 1.909297427 a[3] := 1.943253470 a[4] := 1.931435991 a[5] := 1.935671295 a[6] := 1.934168384 a[7] := 1.934703616 a[8] := 1.934513246 a[9] := 1.934580987 a[10] := 1.934556886 a[11] := 1.934565461 a[12] := 1.934562410 a[13] := 1.934563496 a[14] := 1.934563109 a[15] := 1.934563247 a[16] := 1.934563198 a[17] := 1.934563215 a[18] := 1.934563209 a[19] := 1.934563211 a[20] := 1.934563211 a[21] := 1.934563211 a[22] := 1.934563211 a[23] := 1.934563211 a[24] := 1.934563211 a[25] := 1.934563211 So to the number of significant digits shown, the answer is 1.934563211. If I'd used evalf(sin(a[i-1])+1,20), then I'd see that a_24 and a_25 are actually different.