Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Wednesday, March 12 9.2 Series p461: 2,22,36,37 Main Idea: Achilles did catch the tortoise. a 0 2 n Key Words: -------- = a + a r + a r + ... + a r + ... 1-r 0 0 0 0 Goal: Viewing a task as an infinite sequence of smaller and smaller accomplishments is one way to solve the problem. Previous Assignment: Page 461 Problem 2 (a) Indicate whether the series converges or diverges. (b) If it converges, find its sum. Infinity -k-2 SUM (-1/4) k=1 Infinity k+2 SUM (-4) k=1 3 4 5 6 7 (-4) + (-4) + (-4) + (-4) + (-4) + ... Diverges. k k 3 1- (-4) 1-(-4) Partial Sum[k] = (-4) ------------ = -64 --------- 1-(-4) 5 Partial Sums 1 -64 2 192 3 -832 4 3264 5 -13120 6 52416 7 -209728 8 838848 9 -3355456 10 13421760 ------------------------------------ Get["font.math"]; g[k_] := (-64/5)(1-(-4)^k); f[n_] := (-1/4)^(-n-2); A = Table[{n,Sum[f[i],{i,1,n}]},{n,1, 9}]; B1 = Plot[ (-64/5)(1-(4^x)),{x,1,9}]; B2 = Plot[ (+64/5)(1-(4^x)),{x,1,9}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P2 = ParametricPlot[{0,y},{y,-10,10}]; P3 = Show[B1,B2,P1,P2, PlotLabel->"P 461 P2 SUM (-1/4)^(-k-2)",PlotRange->All]; Display["p2.ps",P3]; ------------------------------------- Page 461 Problem 22 Infinity k k Sum (-1) x -1 < x < 1 k=0 2 3 4 n = 1 - x + x - x + x (+/-)x ...... 1 = ---------- 1+x -------------------------------------------- Get["font.math"]; f[x_] := Sum[ (-1)^k x^k,{k,0,100}]; P1 = Plot[f[x],{x,-0.85,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.006]}]; P2 = Plot[1/(1+x),{x,-0.85,1},PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; P3 = Show[P2,P1,PlotLabel->"P461 P22 red = SUM (-1)^k x^k; blue = 1/(1+x)", PlotRange->All]; Display["p22.ps",P3]; -------------------------------------------- Page 461 Problem 36 Tom and Joel are good runners, both able to run at a constant speed of 10 miles per hour. Their amazing dog Trot can do even better; he runs at 20 miles per hour. Starting from towns 60 miles apart, Tom and Joel run toward each other while Trot runs back and forth between them. (a) How far does Trot run by the time the boys meet? (b) Assume that Trot started with Tom running towards Joel and that he is able to make instant turnarounds. Solve the problem two ways. (a) Use a geometric series. (b) Find a shorter way to do the problem. (2/3) d (1/3) d *-----------------------------|--------------* If the distance between the runners is d, (a) The dog will run 2/3 of the distance (b) Each runner will run 1/3 of the distance. (c) The new distance will be d minus the distance the runners ran. (d) The new distance will be (1/3) d. d d d Successive Distances d . --- . --- . ---- .... 3 2 3 3 3 The dog will run 2 2 d 2 d 2 d --- d + --- --- + --- ---- + --- ---- + .... 3 3 3 3 2 3 3 3 3 _ _ 2 | 1 1 1 | 2 1 --- d | 1 + --- + --- + --- + ,.... | = --- d ------ 3 | 3 2 3 | 3 1-1/3 |_ 3 3 _| 2d = ----------- = d = 60 miles. 3-1 (b) The runners have to run for 3 hours. The dog runs for 3 hours so he runs 60 miles. or The dog runs twice as fast as the runner. Each runner runs 30 miles, so the dog runs sixty miles. ................................................................ Page 461 Problem 37 Suppose that Peter and Paul alternate tossing a coin for which the probability of a head is 1/3 and the probability of a tail is 2/3. If they toss until someone gets a head, and Peter goes first, what is the probability that Peter wins. 2 4 6 1/3 + (2/3) (1/3) + (2/3) (1/3) + (2/3) (1/3) + ..... 2 3 (1/3) ( 1 + 4/9 + (4/9) + (4/9) + .... 1 9 (1/3) -------- = (1/3) ------ = 3/5 1-4/9 5 --------------------------------------------------------------- New Material: We now are going to start a series of techniques to determine "ball park" results about series. We simply want to know if a series converges or diverges. If a series converges, then we can next try to find the limit. If we cannot find the limit, we can give an estimate of the limit along with an error bound that tells how far we might be off. For a series a + a + a + a + .... + a + ... 1 2 3 4 n <----------------------------> The sum of an initial segment is called a partial sum. S = a + a + a + a + ... + a . We say that the n 1 2 3 4 n SERIES converges if the SEQUENCE of partial sums converges. Page 456 Definition: Geometric Series. A series of the form 2 3 4 n a + a r + a r + a r + a r + ... + a r + ... (a =/= 0 ) is called a geometric series. a The sum of a geomertic series is --------- if -1 <= r <= 1 1-r and it diverges otherwise. Page 456 Example 2. (a) Find the sum of: 4 4 4 4 --- + --- + --- + --- + ... 3 9 27 81 (b) 0.5151515151.... ans (a) = 2 ans (b) = 17/33 -------------------------------------- Get["font.math"]; f[n_] := Sum[ 4/3^i,{i,1,n}]; A = Table[{n,f[n]},{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P2 = Plot[0,{x,0,10}]; P3 = Plot[2,{x,0,10}]; P4 = Show[P1,P2,P3,PlotLabel->"P456 Ex 2, SUM[4/3^i]"]; Display["ex2x.ps",P4]; f[n_] := Sum[ 51/100^i,{i,1,n}]; A = Table[{n,f[n]},{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P2 = Plot[0,{x,0,10}]; P3 = Plot[17/33,{x,0,10}]; P4 = Show[P1,P2,P3,PlotLabel->"P456 Ex 2, SUM[4/3^i]"]; Display["ex2y.ps",P4]; ---------------------------- Page 456 Example 3 The diagram in Figure 1 represents an equliaterial triangle containing infinitely many circles tangent to the triangle and to neighboring circles, and reachinging into the corners. What fraction of the area of the triangle is occupied by the circles? Assume that the radius of the large circle is 1. Then the radii of the upward circles is 1, 1/3, 1/9, 1/27, ... Their centers are located 2 r from the top of the triangle which is at (3,0). -------------------------------------------------- Get["font.math"]; f[r_] := ParametricPlot[ {0,3-2r}+r{Cos[t],Sin[t]},{t,0,2 Pi}]; A = Table[f[1/3^n],{n,0,10}]; P1 = Show[A]; P2 = ListPlot[ {{0,0},{Sqrt[3],0},{0,3},{-Sqrt[3],0},{0,0}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]},AspectRatio->Automatic, PlotRange->All]; P3 = Show[P1,P2,PlotLabel->"P456 Ex 2, Tangent Circles", AspectRatio->Automatic,PlotRange->All]; Display["ex2.ps",P3]; --------------------------------------------------------- The area inside the circles is Pi + 3 Pi ( (1/3)^2 + (1/9)^2 + ....) = 1/9 3 11 Pi Pi ( 1 + 3 ------------ ) = Pi ( 1 + ----- ) = ------ 1-1/9 8 8 The side the triangle is s. 3/s = Sin[60] = Sqrt[3]/2 s = 6/Sqrt[3] = 2 Sqrt[3]. The area = 1/2 s 3 = 3/2 s = 3 Sqrt[3] The ratio of the two areas is (11 Pi/8)/( 3 Sqrt[3] ) = 11 Sqrt[3] Pi/72 = 0.831325 ---------------------------------------------------------- Page 457 Example 4. Suppose that Peter and Paul take turns tossing a fair coin until one of them tosses a head. If Peter goes first, what is the probability that he wins? 1/2 + (1/2)^3 + (1/2)^5 + (1/7)^7 + ...... = 1/2( 1 + 1/4 + (1/4)^2 + (1/4)^3 + ..... ) 1 = 1/2 ( ---------- ) = (1/2)(4/3) = 2/3. 1-1/4 ---------------------------------------------- Page 457 Theorem A. A general test for divergence. If the numbers don't approach zero, the SERIES cannot converge. Infinity If the series SUM a converges, then Limit a = 0 n=1 n n->Infinity n Equivalently, if Limit a =/= 0 or if Limit a does n->Infinity n n->Infinity n not exist, then the series diverges. -------------------------------------------------- Page 457 Example 5 3 Infinity n Show that SUM ------------ diverges. n=1 3 2 3 n + 2 n 3 n Limit a = Limit ------------- = 1/3 n->Infinity n n->Infinity 3 2 3 n + 2 n Since the individual terms do not approach zero, the series cannot converge. -------------------------------------------------------- Page 458 Example 6. Show that the Harmonic Series Diverges. Page 458 Example 7 Collapsing series. Show that the following series converges and find its sum. Infinity 1 SUM ---------- k=1 (k+2)(k+3) 1 1 1 ---------- = ------ - ------ (k+2)(k+3) k+2 k+3 n 1 1 Sn = SUM ----------- = 1/3 - ---- So it converges to 1/3 k=1 (k+2)(k+3) n+3 -------------------------------------------------------------- | | | | | | | | | | | | | | | 0 | | | | | | | | e | ------------------------------------------------------------- | | | | | | | | | | | | | | | 1 | | | | | | | | e | -------------------------------------------------------------- | | | | | | | | | | | | | | | 2 | | | | | | | | e | -------------------------------------------------------------- | | | | | | | | | | | | | | | 3 | | | | | | | | e | -------------------------------------------------------------- ------------------------------------------------------------- | | | | | | | | | 10 | 9 | 8 | 7 | 6 | 5 | 4 | | e | e | e | e | e | e | e | -------------------------------------------------------------- | | | | | | | | | 11 | 10 | 9 | 8 | 7 | 6 | 5 | | e | e | e | e | e | e | e | -------------------------------------------------------------- | | | | | | | | | 12 | 11 | 10 | 9 | 8 | 7 | 6 | | e | e | e | e | e | e | e | -------------------------------------------------------------- | | | | | | | | | 13 | 12 | 11 | 10 | 9 | 8 | 7 | | e | e | e | e | e | e | e | -------------------------------------------------------------- 2 3 4 5 6 7 4 5 6 7 (1 + e + e +e +e +e +e +e )(e + e + e + e ) 8 4 4 1-e e (1-e ) ----- * ---------- 1-e 1-e 2 e + e = 1 8 4 (1-e )(1-e ) < 1 thus it is impossible to jump to a corner.