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. Monday, March 10 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: 453:(2),(22),(36),(37) Page 453 Problem 2. (a) Write the first five terms of {a }, n (b) Determine whether the sequence converges or diverges, (c) If it converges, find Limit a n-> Infinity n 3n + 2 a = ---------- n n + 1 5 8 11 14 17 (a) -, -, --, --, --, 2 3 4 5 6 3n+2 (b) Limit --------- = 3 n->Infinity n+1 Notice that the terms do seem to be approaching 3. 3n+2 1 In fact ----- = 3 - ----- so the terms approach 3 from below. n+1 n+1 -------------------------------------------------------------------- Get["font.math"]; f[n_] := (3n+2)/(n+1); {f[1],f[2],f[3],f[4],f[5]}; Limit[f[n],{n->Infinity}]; A = Table[{n,f[n]},{n,1,20}]; P1 = ListPlot[A,PlotStyle->{PointSize[0.02]}]; P2 = Plot[3,{x,0,20},PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P3 = Plot[0,{x,0,20}]; P4 = Show[P1,P2,P3,PlotLabel->"P453 P2 an = (3n+2)/(n+1)",PlotRange->All]; Display["p2.ps",P4]; -------------------------------------------------------------- Page 453 Problem 22 (a) Find an explicit formula a = _______________ for each sequence n (b) Determine whether the sequence converges or diverges. (c) If it converges, find Limit a n->Infinity n 1 2 3 4 ... ---, ----, ---- , ----, 2 3 4 5 2 2 2 2 n (a) a = ------ n n+1 2 n 1 (b) Limit a = Limit -------- = Limit --------- = 0 n->Infinity n n->Infinity n+1 n+1 2 2 ln(2) -------------------------------------------------------------------- Get["font.math"]; f[n_] := n/2^(n+1); t = {f[1],f[2],f[3],f[4],f[5]}; Limit[f[n],{n->Infinity}]; A = Table[{n,f[n]},{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P2 = Plot[0,{x,1,10},PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P3 = ParametricPlot[{0,y},{y,0,1}]; P4 = Plot[f[x],{x,0,10},PlotStyle->{Thickness[0.005]}]; P5 = Show[P4,P3,P2,P1,PlotLabel->"P453 P22, an = n/2^(n+1)",PlotRange->All]; Display["p22.ps",P5]; -------------------------------------------------------------------- Page 453 Problem 36 (a) Write the first four terms of the sequence {a }. n (b) Then use Theorem D to show that the sequence converges. Theorem D is the Monotonic Sequence Theorem. _ _ | 2 | a = 2; a = 1/2 | a + ----- | 1 n+1 | n a | |_ n _| 3 17 577 665857 (a) 2, -, --, ---, ------ 2 12 408 470832 2, 1.5, 1.41667, 1.41422, 1.41421} It looks like the sequence is decreasing to Sqrt[2]. How can we show this. (a) We first show that the sequence stays larger than Sqrt[2]. Write a = Sqrt[2] + x n 2 a = 1/2 ( a + ---- ) n+1 n a n Sqrt[2] + x 1 a = ------------- + ----------- n+1 2 Sqrt[2]+x 2 2 + 2 Sqrt[2] x + x + 2 a = ---------------------------- n+1 2 (Sqrt[2]+x) 2 4 + 2 Sqrt[2] x + x a = ----------------------- n+1 2 (Sqrt[2]+x) 2 2 Sqrt[2] Sqrt[2] + 2 Sqrt[2] x + x a = -------------------------------------------- n+1 2 (Sqrt[2]+x) 2 2 Sqrt[2]( Sqrt[2] + x ) + x a = ------------------------------- n+1 2 (Sqrt[2]+x) 2 x a = Sqrt[2] + ------------- n+1 2 (Sqrt[2]+x) 2 x Then a = Sqrt[2] + --------------- n+1 2( Sqrt[2] + x ) Therefore if a is larger than then Sqrt[2], then a is also n n+1 larger than Sqrt[2]. They continue larger than Sqrt[2] forever. (b) It is also clear that it is decreasing because the new distance above Sqrt[2] is less than half of the old distance. (* once x < 1 *) The Sequence is decreasing, but remains above Sqrt[2]. By the monotone convergence theorem the sequence converges. Now taking limits assuming that a -----> a n _ _ | 2 | Limit a = Limit 1/2 | a + ---- | n->Infinity n+1 n->Infinity | n a | |_ n _| a = 1/2( a + 2/a) 2 2 2 a = a + 2 a = Sqrt[2]. So the only possible limit is Sqrt[2]. Since it does have a limit, it must converge to Sqrt[2]. ------------------------------------------------------- Get["font.math"] f[x_] := 1/2 (x + 2/x); A = Table[0,{i,1,5}]; A[[1]] := {1,2}; Do[A[[i+1]] = {i+1,f[A[[i,2]]]},{i,1,4}]; P1 = ListPlot[A,PlotRange->All,PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P2 = Plot[Sqrt[2],{x,1,5},PlotStyle->{RGBColor[0,1,0]}]; P3 = Plot[0,{x,1,5}]; P4 = ParametricPlot[{0,y},{y,0,1}]; P5 = Show[P1,P2,P3,P4,PlotLabel-> "P453 P36 an = 1/2(an + 2/an)"]; Display["p36x.ps",P5]; A = Table[ Graphics[ Line[ { {1,j/20}, {2,f[j/20]}, {3,f[f[j/20]]}, {4,f[f[f[j/20]]]}, {5,f[f[f[f[j/20]]]]}, {6,f[f[f[f[f[j/20]]]]]}, {7,f[f[f[f[f[f[j/20]]]]]]}, {8,f[f[f[f[f[f[f[j/20]]]]]]]} } ] ],{j,30,100,10}]; P6 = Plot[0,{x,0,10}]; P7 = Show[A,P6, PlotLabel-> "P453 P36 an = 1/2(an + 2/an), black > Sqrt[2]", PlotRange->All,Axes->True]; Display["p36y.ps",P7]; B = Table[ Graphics[{RGBColor[1,0,0], Line[ { {1,j/20}, {2,f[j/20]}, {3,f[f[j/20]]}, {4,f[f[f[j/20]]]}, {5,f[f[f[f[j/20]]]]}, {6,f[f[f[f[f[j/20]]]]]}, {7,f[f[f[f[f[f[j/20]]]]]]}, {8,f[f[f[f[f[f[f[j/20]]]]]]]} } ] } ],{j, 2 ,20}]; P8 = Show[B,P7, PlotLabel-> "P453 P36 an = 1/2(an + 2/an) red < Sqrt[2]", PlotRange->All]; Display["p36z.ps",P8]; ------------------------------------------------ Page 453 Problem 37 Assuming that u = Sqrt[3] and u = Sqrt[3+u ] determine a 1 n+1 n convergent sequence, find Lim u n->Infinity n Assuming that u ->u, then n u = Sqrt[3+u] 2 u = 3+u 2 u - u - 3 = 0 1 +/- Sqrt[1+12] u = ------------------ 2 Since we are taking the positive square root 1 + Sqrt[1+12] u = ---------------- 2 Check. 7 + Sqrt[13] 14 + 2 Sqrt[13] Sqrt[3+u] = Sqrt[ -------------- ] = Sqrt[ ---------------] 2 4 1 + 2 Sqrt[13] + 13 1 + Sqrt[13] = Sqrt[ ---------------------] = ----------------- = u 4 2 ----------------------------------------------------------------- Get["font.math"]; f[x_] := Sqrt[3+x]; A = Table[0,{i,1,10}]; A[[1]] := {1,Sqrt[3]}; Do[A[[i+1]] = {i+1,f[A[[i,2]]]},{i,1,9}]; P1 = ListPlot[A,PlotRange->All,PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P2 = Plot[(1+Sqrt[13])/2,{x,1,10},PlotStyle->{RGBColor[0,1,0]}]; P3 = Plot[0,{x,0,10}]; P4 = ParametricPlot[{0,y},{y,0,1}]; P5 = Show[P1,P2,P3,P4,PlotLabel-> "P453 P37, un+1 = Sqrt[3 + un]"]; Display["p37x.ps",P5]; f[x_] := Sqrt[3+x]; A = Table[0,{i,1,10}]; g[t_] := ( A[[1]] = {1,t}; Do[ A[[i+1]] = {i+1,f[ A[[i,2]] ]},{i,1,9}]; Return[ListPlot[A,PlotJoined->True]]); B = Table[g[t/5],{t,-15,25}]; P6 = Plot[(1+Sqrt[13])/2,{x,0,9},PlotStyle->{RGBColor[1,0,0]}]; P7 = Show[B,P3,P4,P6,PlotRange->All,PlotLabel->"P453 P37 un+1 = Sqrt[3+un]"]; Display["p37y.ps",P7]; ----------------------------------------------------------------- New Material. Zeno's Paradox. Achilles chases a tortoise. The tortoise has a 1 mile head start. When Achilles reaches the spot the tortoise was, the tortoise has moved 1/10 mile further. When Achilles reaches the new spot, the tortoise will have moved 1/100 mile further. When and where will Achilles catch the tortoise. ---------------------------------------------------------------- The hand of the clock are together at 12:00. When the hour hand gets to 1, the minute hand is on the 12. When the minute hand moves to the 1, the hour hand has moved further away. When the minute hand catches up the hour hand has moved ever so tiny much further. When will the minute hand catch the hour hand? ------------------------------------------------------------------ Sue needs a medicine in her body. During each day her metabolism reduces the amount of medicine in her body by 60 %. That is, only 40% of what is in her body today will be there tomorrow. Each pill has 10 mg of medicine. There must be 16 mg in her body for the medicine to be effective. How many doses till this level is reached? The medicine has moderate side effects when the dose in the body reaches 16.6 mg. How many doses till this is reached? The medicine has horrible side effects when the medicine in the body reaches 16.7 mg. How many doses till this is reached? -------------------------------------------------------- A ball bounced 90 % of the way back up. (1) If it is dropped from 16 feet, how far does it go before it stops? (2) How much time is spent before it stops? ------------------------------------------------------ Using puzzle peg jumps on a half filled checker board (a) Can you jump a marker to the king row? (b) Can you jump a marker to the corner of the king row? (c) Can you jump a marker off the board? (d) Suppose the bottom half of the checker board were infinite, could you jump a checker beyond the king row? +--+--+--+--+--+--+--+--+ | | | | | | | | | +--+--+--+--+--+--+--+--+ | | | | | | | | | +--+--+--+--+--+--+--+--+ | | | | | | | | | +--+--+--+--+--+--+--+--+ | | | | | | | | | +==+==+==+==+==+==+==+==+ |()|()|()|()|()|()|()|()| +--+--+--+--+--+--+--+--+ |()|()|()|()|()|()|()|()| +--+--+--+--+--+--+--+--+ |()|()|()|()|()|()|()|()| +--+--+--+--+--+--+--+--+ |()|()|()|()|()|()|()|()| +--+--+--+--+--+--+--+--+ Zeno's Paradox. Achilles chases a tortoise. The tortoise runs 1 m/h. Achilles runs 10 mph. The tortoise has a 1 mile head start. When Achilles reaches the spot the tortoise was, the tortoise has moved 1/10 mile further. When Achilles reaches the new spot, the tortoise will have moved 1/100 mile further. When and where will Achilles catch the tortoise. delta A delta T delta time 1 0.1 0.1 0.1 0.01 0.01 0.01 0.001 0.001 ----------- --------- -------------- 1.1111... 0.1111... 0.1111... The time limit is 1/9. Achilles runs 10/9 miles and the tortoise runs 1/9 miles. 10/9 - 1/9 = 9/9 = 1 which is the initial separation of Achilles and the tortoise. ---------------------------------------------------------------- The minute hand moves 12 times faster than the hour hand. If the minute hand moves x degrees, the hour hand move x/12 degrees. The hand of the clock are together at 12:00. When the hour hand gets to 1, the minute hand is on the 12. When the minute hand moves to the 1, the hour hand has moved further away. When the minute hand catches up the hour hand has moved ever so tiny much further. When will the minute hand catch the hour hand? delta Hour Hand delta Minute Hand delta time 1/12 0 1 2 (1/12) 1/12 1/12 3 2 2 (1/12) (1/12) (1/12) . . . . . . . . . 2 3 The time is 1 + (1/12) + (1/12) + (1/12) + ..... 1 12 = --------- = ----- hours. 1-1/12 11 The hour hand moves 12/11 of 30 degrees The minute hand moves 12/11 of 360 degrees 12/11 360 - 12/11 30 = 12/11(330) = 360 which is the extra distance the minute hand moved. Notice that the hands pass between each number except for between 11 and 12 and 12 and 1. Symmetrically placing the places, they cross ll times in 12 hours which puts the time between intersections as 12/11 hours. ------------------------------------------------------------------ Sue needs a medicine in her body. During each day her metabolism reduces the amount of medicine in her body by 60 %. That is, only 40% of what is in her body today will be there tomorrow. Each pill has 10 mg of medicine. There must be 16 mg in her body for the medicine to be effective. How many doses till this level is reached? The medicine has moderate side effects when the dose in the body reaches 16.6 mg. How many doses till this is reached? The medicine has horrible side effects when the medicine in the body reaches 16.7 mg. How many doses till this is reached? a = 10 0 a = 0.4 a + 10 n+1 n a = 10 0 a = 14. 1 a = 15.6 2 a = 16.24 <== Effective at start of day 3 or 4 doses. 3 a = 16.496 4 a = 16.5984 5 a = 16.6394 <== Moderate side effects at start of day 6. 6 10 10 100 -------- = ------ = ------ = 16 2/3 so horrible side effects will 1-0.40 0.6 6 never occur. ------------------------------------------------------------------ A ball bounced 90 % of the way back up. (1) If it is dropped from 16 feet, how far does it go before it stops? (2) How much time is spent before it stops? 2 3 d = 16 + 2 16 0.9 ( 1 + 0.9 + 0.9 + 0.9 + ..... ) .9 d = 16 + 2 16 ------------ 1-.9 d = 16 + 2 16 9 = 304 feet. 2 s = 16 t so the time do fall distance s is Sqrt[s/16]. call this f[s] t = f[16] + 2 f[16 0.9] + 2 f[ 16 0.9^2] + 2 f[16 0.9^3] + ... t = 1 + 2 ( Sqrt[0.9] + Sqrt[0.9^2] + Sqrt[0.9^3 + ..... ) Sqrt[0.9] Sqrt[0.9](1+Sqrt[0.9] t = 1 + 2 ------------ = 1 + 2 ----------------------- 1-Sqrt[0.9] 1-0.9 t = 1 + 20 (Sqrt[0.9] + 0.9) t = 1 + 20 Sqrt[0.9] + 18 = 19 + 20 Sqrt[0.9] = 37.9737 seconds. ------------------------------------------------------ Suppose the bottom half of the checker board were infinite, could you jump a checker beyond the king row? 0 e --------------------------------------------------------------------- | | | | 1 | | | | | | | | | | e | | | | | | -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | | | | 2 | | | | | | | | | | e | | | | | | -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | | | | 3 | | | | | | | | | | e | | | | | | -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | | | | 4 | | | | | | | | | | e | | | | | | -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 8 | 7 | 6 | 5 | 6 | 7 | 8 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 9 | 8 | 7 | 6 | 7 | 8 | 9 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 10 | 9 | 8 | 7 | 8 | 9 | 10 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 11 | 10 | 9 | 8 | 9 | 10 | 11 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 12 | 11 | 10 | 9 | 10 | 11 | 12 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 13 | 12 | 11 | 10 | 11 | 12 | 13 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | 14 | 13 | 12 | 11 | 12 | 13 | 14 | | | 0 | e | e | e | e | e | e | e | 0 | 0 | 0 -----+-----+-----+-----+-----+-----+-----+-----+-----+-----+--------- | | | | | | | | | | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Summing the columns: 8 7 6 5 6 7 8 e e e e e e e ... + --- + --- + --- + --- + --- + --- + --- + ... 1-e 1-e 1-e 1-e 1-e 1-e 1-e Summing left and right column sumns. 6 5 6 e e e ------ + --- + ----- 2 2 (1-e) (1-e) (1-e) 6 5 6 6 e + e - e + e ------------------------ 2 (1-e) 5 e (1+e) ----------------- 2 (1-e) 5 e (1+e) -------------- 4 e 2 e(1+e) = e+e = 1. 0 There is just enough total weight to put a checker at the e position. But it cannot be done in a finite number of moves. 1 2 3 4 5 6 7 8 ------------------------------------------------- | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 | | e | e | e | e | e | e | e | e | ------------------------------------------------- | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | | e | e | e | e | e | e | e | e | ------------------------------------------------- | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | | e | e | e | e | e | e | e | e | ------------------------------------------------- | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | | e | e | e | e | e | e | e | e | ------------------------------------------------- | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | | e | e | e | e | e | e | e | e | ------------------------------------------------- | 12 | 11 | 10 | 9 | 8 | 7 | 6 | 5 | | e | e |e | e | e | e | e | e | ------------------------------------------------- | 13 | 12 | 11 | 10 | 9 | 8 | 7 | 6 | | e | e |e |e | e | e | e | e | ------------------------------------------------- | 14 | 13 | 12 | 11 | 10 | 9 | 8 | 7 | | e | e |e |e |e | e | e | e | ------------------------------------------------- The value of each square is shown. It is a power of e = (-1+Sqrt[5])/2 = 0.618034. 2 e is a solution of e + e = 1. The weight of a partially filled board is the sum of the value of the squares containing checkers. If half of the board is filled, the weight is 4 5 6 7 8 9 10 11 12 13 14 ans = e + 2 e + 3 e + 4 e + 4 e + 4 e + 4 e + 4 e + 3 e + 2 e + e ans = 135 - 60 Sqrt[5] = 0.835921 Theorem: For each jump, the weight of the board either decreases or remains the same. There are two cases: Jumping "backwards" n n+1 n+2 e + e is replaced by e But the weight of the board changes by n+2 n+1 n n e - e - e = e (e^2 -e -1) n = e (1-e -e -1) n+1 = - 2 e < 0 Jumping "forwards" n+2 n+1 n e + e is replaced by e n n+1 n+2 n 2 e - e - e = e (1 - e - e ) = 0 Since the weight of the board starts less than one and o never increases, it is impossible to put a peg in the e position which would be a board position of weight 1.