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. Friday, March 7 9.1 Sequences p453: 2,22,36,37 Main Idea: The calculus of power series is beautiful and gives a real understanding to some bizarre results. Key Words: Series, Sequence, Squeeze theorem, Limit, Converge Diverge, Monotone, {a } n Goal: Learn about limits of sequences. -------------------------------------------------------- Previous assignment: p445: (8),(14),(16),(42), Page 445 Problem 8 x=-5 1 INT --------- dx x= 5 2/3 x x= 5 1 - INT --------- dx x=-5 2/3 x x= 0 1 x=5 1 - INT ----- dx - INT ---- dx x=-5 2/3 x=0 2/3 x x _ 1/3 _ /|\ _ 1/3 _ | x |x | 0 | x | x=5 -| ------ | - | ----- | | |_ 1/3 _|x=-5 |_ 1/3 _| x \|/ 0 _ _ _ _ | 1/3 | | 1/3 | -3 | 0-(-5) | - 3 | 5 - 0 | |_ _| |_ _| 1/3 1/3 -3 5 - 3 5 1/3 = -6 5 (* Answer is negative because the limits are upside down *) --------------------------------------------------------- Get["font.math"]; P1 = Plot[ 1/(x^2)(1/3),{x,-5,-0.2}]; P2 = Plot[ 1/(x^2)(1/3),{x,0.2, 5 }]; P3 = Show[P1,P2,PlotLabel->"P445 p8 y = x^(-2/3)",AspectRatio->Automatic]; Display["p8.ps",P3]; ---------------------------------------- Page 445 Problem 14 x=3 x INT ----------- dx x=0 2 Sqrt[9-x ] _ _ /|\ | 2 1/2 | x | 3 x=3 x | (9-x ) | INT ----------- dx = | ------------ | x=0 2 |_ (1/2)(-2) _| x=0 Sqrt[9-x ] = - ( 0 - 3) = 3. ------------------------------------------------- Integrate[ x/Sqrt[9-x^2],{x,0,3}]; Get["font.math"]; f[x_] := x/Sqrt[9-x^2]; g[x_] := 3-Sqrt[9-x^2]; P1 = Plot[ f[x],{x,0,2.95},PlotStyle->{Thickness[0.01]}]; P2 = ParametricPlot[{3,y},{y,0,f[2.95]},PlotStyle->{RGBColor[1,0,0]}]; P3 = Plot[g[x],{x,0,3},PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; P4 = Show[P1,P2,P3, PlotLabel->"P445 P14 black y = x/Sqrt[9-x^2; blue = Area function]", AspectRatio->Automatic]; Display["p14.ps",P4]; ----------------------------------------------------- Page 445 Problem 16 1 1 --- ----- x=3 dx x=3 3 -3 INT ----------- = INT ----- + ------ x=0 2 x=0 x-1 x+2 x + x - 2 / g (x) = 2x+1 x=1 x=-2 /|\ x | 1 dx x=3 dx INT ----------- + INT ------------- x=0 (x-1)(x+2) x | 1 (x-1)(x+2) \|/ /|\ x | 1 1 1 x=3 1 1 1/3 INT --- - ---- +1/3 INT ---- - ---- x=0 x-1 x+2 x | 1 x-1 x+2 \|/ /|\ x | 1 -1 1 x = 3 1 1 1/3 INT --- - ---- + 1/3 INT ---- - ---- x=0 1-x x+2 x | 1 x-1 x+2 \|/ _ _ /|\ _ _ | | x | 1 | | x = 3 1/3 | ln(1-x) - ln(x+2) | +1/3 | ln(x-1)-ln(x+2) | |_ _| x = 0 |_ _| x | 1 \|/ _ _ /|\ _ _ | 1-x | x | 1 | x-1 | x = 3 1/3 | ln( (------)| +1/3 | ln ( -------) | |_ x+2 _| x = 0 |_ x+2 _| x | 1 \|/ _ _ _ _ | | | | 1/3 | - Infinity + ln(2) | +1/3 | ln(2/5) - ( -Infinity) | |_ _| |_ _| = - Infinity + Infinity The integral does not exist. ----------------------------------------- Get["font.math"]; f[x_] := 1/(x^2 + x - 2); P1 = Plot[f[x],{x,0,0.9}]; P2 = Plot[f[x],{x,1.1, 3}]; P3 = ParametricPlot[{1,y},{y,f[0.9],f[1.1]},PlotStyle->{RGBColor[1,0,0]}]; P4 = Show[P1,P2,P3,AspectRatio->Automatic, PlotLabel->"P426 P16 y = 1/(x^2+x-2)"]; Display["p16.ps",P4]; ----------------------------------------------------- Page 445 Problem 42 Find the area of the region between the curves y = 1/x and y = 1/(x^3 + x) for 0 < x <= 1. ---------------------------------------------------------------- Get["font.math"]; f[x_] := 1/x; g[x_] := 1/(x^3+x); P1 = Plot[{f[x],g[x]},{x,0.2, 1.0}]; P2 = ParametricPlot[{1,y},{y,f[1],g[1]},PlotStyle->{RGBColor[1,0,0]}]; P3 = ParametricPlot[{0,y},{y,0,g[1]}]; P4 = Show[P1,P2,P3, PlotLabel->"P445 p42 y=1/(x^3+x); y=1/x", AspectRatio->Automatic,PlotRange->All]; Display["p42.ps",P4]; ---------------------------------------------------------------- x=1 1 1 INT --- - --------- dx x=0 x 3 x + x x=1 1 1 INT ------ - --------- x=0 x 2 x(x + 1) 2 x=1 x + 1 - 1 INT ----------------- dx x=0 2 x(x + 1) x=1 x INT ----------- dx x=0 2 x + 1 _ _ | 2 | x = 1 | 1/2 ln(x + 1) | | |_ _| x \|/ 0 1/2 ln(2) - 0 = 1/2 ln(2) ============================================================== New Material A SEQUENCE is an infinite stream of numbers separated by commas. A SERIES is an infinite stream of numbers separated by +/- signs. __ | | Think of the tail of the |__| in se"q"uence as the comma separator. | \ ------------------------------------------------------------------ The Cake problem. 1. Each day you eat 1/2 of the remainder of the cake. At the end of day n: (a) How much remains? (b) How much has been eaten. -------------------------------------------------------- Get["font.math"]; A = Table[ ListPlot[{{1/2^n,0},{1/2^n,2/2^n}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0]}],{n,1,8}]; B = Table[ ListPlot[{{0,1/2^n},{1/2^n,1/2^n}},PlotJoined->True, PlotStyle->{RGBColor[1,0,0]}],{n,1,8}]; P1 = ListPlot[{{0,1},{1,1},{1,0},{0,0},{0,1}}, PlotJoined->True,PlotStyle->{Thickness[0.01]}]; P2 = Show[A,B,P1,AspectRatio->Automatic,PlotRange->All, PlotLabel->"Eat 1/2 the Remainder"]; Display["halfcake.ps",P2] 2. The first day you eat 1/3 of the cake. On subsequent days, you eat 1/3 of what you ate the day before. At the end of day n, how much remains? How much has been eaten? ---------------------------------------------------------------- Get["font.math"]; day[p_,q_] := Graphics[ Line[ { {p,0},{p,1},{q,1},{q,0},{0,0}}]]; v[p_] := Graphics[{RGBColor[1,0,0],Line[ { {p,0},{p,1}}]}]; a1 = Show[day[0,1],v[1/3],v[2/3]]; a2 = Show[day[0,1],day[1/3,2/3],v[1/2-1/18],v[1/2+1/18]]; a3 = Show[day[0,1],day[1/3,2/3],day[1/2-1/18,1/2+1/18],v[1/2-1/54],v[1/2+1/54]]; b1 = Show[a1,PlotLabel->"First day, Whole Cake, eat left 1/3"]; b2 = Show[a2,PlotLabel->"Second day, 2/3 remain, eat 1/9"]; b3 = Show[a3,PlotLabel->"Third day, three, eat 1/27"]; Display["cake1.ps",b1]; Display["cake2.ps",b2]; Display["cake3.ps",b3]; -------------------------------------------------------------------------- Famous sequences. 1, 2, 3, 4, 5, ..... 2, 4, 6, 8, 10, ... 1, -1, 1, -1, 1, -1 ... 1/2, 1/3, 1/4, 1/5, ... 1.1, 1.11, 1.111, 1.1111, 1.11111, ... 1, 1, 1, 1, 1, 1, 1, ... 1, 1, 2, 3, 5, 8, 13, 21, ... 1/2, 2/3, 3/4, 4/5, 5/6, ... 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ... -------------------------------------------------------------------- The Limit of a sequence. The sequence {a } is said to "converge" to L n If for each positive number epsilon, there is a corresponding positive number N such that: | | n >= N implies | a - L | < epsilon | n | We indicate that { a } converges to L by writing: n Lim a = L n->Infinity n A sequence that fails to converge is said to "diverge". ------------------------------------------------------------------ 2 3 n Example 2 Page 451. Find Limit ---------- n->Infinity 2 7 n + 1 ----------------------------------------------------- Get["font.math"]; f[n_] := 3 n^2 / (7 n^2 + 1); A = Table[{n,f[n]},{n,1,40}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P2 = Plot[3/7,{x,1,40},PlotStyle->{RGBColor[0,1,0]}]; P3 = Plot[0,{x,0,40}]; P4 = Show[P1,P2,P3,PlotLabel->"P451 Ex 2, f[n] = 3 n^2/(7 n^2 + 1)"]; Display["ex2.ps",P4]; ----------------------------------------------------------------- ln(n) Example 3 Page 451. Does -------- Converge? n->Infinity n e Use L'Hopital's rule by "extending the definition to all n. ----------------------------------------------------- Get["font.math"]; f[n_] := Log[n]/E^n; A = Table[{n,f[n]},{n,1 ,20}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.03]}]; P2 = Plot[0,{x,0,20}]; P3 = Plot[f[x],{x,1,20},PlotStyle->{RGBColor[0,0,1]}]; P4 = Show[P1,P2,P3,PlotLabel->"P451 Ex 3, f[n] = ln(n)/e^n",PlotRange->All]; Display["ex3.ps",P4]; --------------------------------------------------------------- Theorem B Page 451: Squeeze Theorem (1) If {a }, {c } both converge to L, n n (2) and a <= b <= c for n >= K, (K a fixed integer), n n n (3) then {b } also converges to L. n ---------------------------------------------------------------- n Page 452 Example 5. Show that if -1 < r < 1 then Lim r = 0. n->Infinity 1 ------ = 1 + p for some small positive number p. |r| 1 n n(n-1) 2 n(n-1)(n-2) 3 ----- = (1+p) = 1 + n p + ------- p + ------------ p + (positive stuff) n 2 3 2 |r| 1 ----- > 1 + n p n |r| 1 n So ------------- > |r| 1+n p 1 Since 1+n p approaches infinity, ----- ----> 0 and so 1+n p n |r| ---> 0 by the squeeze theorem. ------------------------------------------------ Get["font.math"]; P1 = Table[ Plot[(x)^n,{x,-2^(1/n), 2^(1/n) }],{n,1,20}]; P2 = Show[P1,PlotLabel->"P452 Ex 5 y = |r|^n",PlotRange->All, AspectRatio->Automatic]; Display["ex.5.ps",P2]; ------------------------------------------------ Sin[n] Show Limit --------- = 0. n->Infinity n ---------------------------------------------------- Get["font.math"]; A = Table[{n,Sin[n]/n},{n,1,100}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P2 = Show[P1,PlotLabel->"P453 Ex 6 f[n] = Sin[n]/n",PlotRange->All]; Display["ex.6x.ps",P2]; P3 = Plot[Sin[x]/x,{x,1,100}]; P4 = Show[P1,P3,PlotLabel->"P452 Ex 6 f[x] = Sin[x]/x",PlotRange->All]; Display["ex6y.ps",P4]; ------------------------------------------------ A sequence is monotone increasing is a <= a n n+1 A sequence is monotone decreasing is a >= a n n+1 ---------------------------------------------------------------- Theorem D page 433. If U is an upper bound for a nondecreasing sequence {a } n then the sequence converges to a limit A that is less than or equal to U. Similarly, if L is a lower bound for a nonincreasing sequence {b }, then the sequence {b } converges to a limit B that is n n greater than or equal to L. ------------------------------------------------------------- Examples: 3n+2 a = -------- write first four terms and find the limit. n n+1 -------------------------------------- Get["font.math"]; A = Table[0,{i,1,10}]; Do[A[[n]] = (3n+2)/(n+1),{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P2 = Plot[0,{x,0,10}]; P3 = Plot[3,{x,0,10},PlotStyle->{RGBColor[0,1,0]}]; P4 = Show[P1,P2,P3,PlotLabel->"Example, an = (3n+2)/(n+1))"]; Display["exu.ps",P4]; ----------------------------------------------------------------- _ _ | 2 | a = 2, a = 1/2 |a + ---- | write first four terms and 1 n+1 | n a | |_ n _| show that it converges. ------------------------------------------------- Get["font.math"]; A = Table[0,{i,1,11}]; A[[1]] = 2.0; Do[A[[n+1]] = 1/2 (A[[n]] + 2/A[[n]]),{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P2 = Plot[0,{x,0,10}]; P3 = Plot[Sqrt[2],{x,0,10},PlotStyle->{RGBColor[0,1,0]}]; P4 = Show[P1,P2,P3,PlotLabel->"Example, a_(n+1) = 1/2 (a_n+2/a_n)"]; Display["exv.ps",P4]; --------------------------------------------------------- u = Sqrt[3] u = Sqrt[3+u ] Find the limit. 1 n+1 n ------------------------------------------------------------- ------------------------------------------------- Get["font.math"]; A = Table[0,{i,1,10}]; A[[1]] = Sqrt[3]; Do[A[[n+1]] = Sqrt[3+A[[n]]],{n,1,10}]; P1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P2 = Plot[0,{x,0,10}]; P3 = Plot[(1+Sqrt[13])/2,{x,0,10},PlotStyle->{RGBColor[0,1,0]}]; P4 = Show[P1,P2,P3,PlotLabel->"Example, a_(n+1) = Sqrt[3+a_n ] "]; Display["exw.ps",P4]; ---------------------------------------------------------