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, April 9, 9.10 Chapter Review Assignment: Page p505: 29-42 Previous Assignment: Page 503 Problem 2 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12). -3x f(x) = e 2 3 4 (-3x) (-3x) (-3x) f(x) = 1 + (-3x) + ----- + ----- + -------- 2 3! 4! 2 3 4 f(x) = 1 - 3x + (9/2) x - (9/2) x + (27/8) x f(0.12) = 0.697724 <== Approximation e^(-0.36) = 0.697676 error = 0.0000475139 5 f (c) 5 243 5 Error <= ----- x <= ----- (0.12) = 0.0000503885 5! 120 ---------------------------------------------- Get["font.math"]; f[x_] := 1 - 3 x + (9/2) x^2 - (9/2) x^3 + (27/8) x^4 P1 = Plot[f[x],{x,-0.12,0.12},PlotStyle->{RGBColor[0,1,1],Thickness[0.004]}]; P2 = Plot[E^(-3x),{x,-0.12,0.12}]; P3 = Show[P1,P2,PlotLabel->"P503 P2 blue = S4; black = e^(-3x)"]; Display["p2x.ps",P3]; P4 = Plot[E^(-3x)-f[x],{x,-0.12,0.12},PlotRange->All]; P5 = ListPlot[{{0.12, E^(-0.36)-f[0.12]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P6 = Show[P4,P5,PlotLabel->"P503 P3 E^(-3x)-S4",PlotRange->All]; Display["p2y.ps",P6]; --------------------------------------------------------- Page 503 Problem 4 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12). f(x) = Tan[x] 0 / 2 f (x) = Sec [x] 2 = 1 + Tan [x] 1 // 2 f (x) = 2 Tan[x] Sec [x] 3 = 2 Tan[x] + 2 Tan [x] 0 /// 2 2 2 f (x) = 2 Sec [x] + 6 Tan [x} Sec [x] 2 2 4 = 2 + 2 Tan [x] + 6 Tan [x] + 6 Tan [x] 2 4 = 2 + 8 Tan [x] + 6 Tan [x] 2 (4) 2 3 2 f (x) = 16 Tan[x] Sec [x] + 24 Tan [x] Sec [x] 3 3 5 = 16 Tan[x] + 16 Tan [x] + 24 Tan [x] + 24 Tan [x] 3 5 = 16 Tan[x] + 40 Tan [x] + 24 Tan [x] (5) 2 2 2 4 2 f (x) = 16 Sec [x] + 120 Tan [x] Sec [x] + 120 Tan [x] Sec [x] 2 2 4 4 6 16+16 Tan [x]+120 Tan [x]+120 Tan [x]+120 Tan [x] + 120 Tan [x] 2 4 6 16 + 136 Tan [x] + 240 Tan [x] + 120 Tan [x] Error < 16 + 136 (Tan[0.12])^2 + 240 (Tan[0.12])^4 + 120 (Tan[0.12])^6 -------------------------------------------------------------- (0.12)^5 5! = 18.0285 5 -------- (0.12) 120 -6 = 3.73838 10 0 1 0 2 2 3 0 4 f(x) = ------ + ----- x + ------ x + ----- x + ------ x 0! 1! 2! 3! 4! ------------------------------------------------------------------ 3 f(x) = x + (1/3) x f[0.12] = 0.120576 -6 error = -3.33721 10 ======================================================= Get["font.math"]; f[x_] := x + (1/3) x^3; P1 = Plot[f[x],{x,-0.12,0.12},PlotStyle->{RGBColor[0,1,1],Thickness[0.004]}]; P2 = Plot[Tan[x],{x,-0.12,0.12}]; P3 = Show[P1,P2,PlotLabel->"P503 P4 blue = S4; black = Tan[x]"]; Display["p4x.ps",P3]; P4 = Plot[Tan[x]-f[x],{x,-0.12,0.12},PlotRange->All]; P5 = ListPlot[{{0.12,Tan[0.12]-f[0.12]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P6 = Show[P4,P5,PlotLabel->"P503 P4 Tan[x]-S4",PlotRange->All]; Display["p4y.ps",P6]; ====================================================================== Page 503 Problem 6 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12) f(x) = Sqrt[1+x] f(0) = 1 1 / / f (x) = (1/2)(1+x)^(-1/2) f (0) = 1/2 +1/2 x // // 2 f (x) = (1/2)(-1/2) (1+x)^(-3/2) f (0) = -1/4 -1/4 x /2 /// /// 3 f (x) = (1/2)(-1/2)(-3/2) (1+x)^(-5/2) f (0) = 3/8 +3/8 x /6 //// //// 4 f (x) = (1/2)(-1/2)(-3/2)(-5/2) (1+x)^(-7/2) f (x) = -(15/16) -15/16 x /24 ///// f (x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(1+x)^(-9/2) (6) (-9/2) f (x) = 105/32 (1+x) 2 3 4 f[x] = 1 + (1/2)x - (1/8) x +(3/48) x -15/384 x (5) f (c) 5 105/32 5 Error < ------- x < -------- (0.12) = 5! 1.00 5! because we have to maximize the value of the fifth derivative. ------------------------------------------------------------ Get["font.math"]; f[x_] := 1 + (1/2)x - (1/8) x^2 +(3/48) x^3; P1 = Plot[f[x],{x,-0.12,0.12},PlotStyle->{RGBColor[0,1,1],Thickness[0.005]}]; P2 = Plot[Sqrt[1+x],{x,-0.12,0.12}]; P3 = Show[P1,P2,PlotLabel->"P503 P6 blue = series; black = Sqrt[1+x]"]; Display["p6x.ps",P3]; P4 = Plot[ Sqrt[1+x]-f[x],{x,-0.24,0.26},PlotRange->All]; P5 = Plot[-0.000145135,{x,-0.24,0.26},PlotStyle->{RGBColor[0,0,1]}]; P6 = ListPlot[{{0.12,Sqrt[1.12]-f[0.12]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P7 = Show[P4,P5,P6,PlotLabel->"P503 P6 y = Sqrt[1+x] - S3",PlotRange->All]; Display["p6y.ps",P7]; ------------------------------------------------------------ Page 503 Problem 8 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12) f(x) = Sinh(x) f(0) = 0 / / f (x) = Cosh(x) f (0) = 1 x // // f (x) = Sinh(x) f (0) = 0 /// /// 3 f (x) = Cosh(x) f (0) = 1 x /3! //// //// f (x) = Sinh(x) f (0) = 0 ///// f (x) = Cosh(x) 3 f(x) = x + x/3!; 5 f (c) 5 Cosh(0.12) 5 Error = -------- x <= ------------- (0.12) = 2.08855 10^(-7) 5! 120 -7 The exact error is -2.07431 10 ----------------------------------------------------------------------- Get["font.math"]; f[x_] := x + x^3/6; P1 = Plot[f[x],{x,-0.12,0.12},PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Plot[Sinh[x],{x,-0.12,0.12},PlotStyle->{Thickness[0.005]}]; P3 = Show[P1,P2,PlotLabel->"P503 p8 y = Sinh[x]"]; Display["p8x.ps",P3]; P4 = Plot[ Sinh[x]-f[x],{x,-0.12,0.12},PlotRange->All]; P5 = Plot[ Cosh[0.12]/120 (0.12)^5,{x,-0.12,0.12},PlotStyle->{RGBColor[0,0,1]}]; P6 = Plot[-Cosh[0.12]/120 (0.12)^5,{x,-0.12,0.12},PlotStyle->{RGBColor[0,0,1]}]; P7 = ListPlot[{{0.12,Sinh[0.12]-f[0.12]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P8 = Show[P4,P5,P6,P7,PlotLabel->"P503 p8 y = Sinh[x] - S4",PlotRange->All]; Display["p8y.ps",P8]; ----------------------------------------------------------------------------- Page 503 Problem 50 In analogy with Problem 49 2 4 x x Cos[x] = 1 - ------ + ------- + R (x) 2 24 5 If 0 <= x <= 1, give a good bound for | R (x) | 5 x=1 Then use your result to approximate INT Cos[x] dx x=0 and give a bound on the error. (5) f 5 1 Error <= ---- x <= ----- 5! 120 _ _ x=1 x=1 | 3 5 | INT Cos[x] = | x - x /6 + x /120 | = 1 - 1/6 + 1/120 = 0.841667 x=0 |_ _|x=0 The error is at most 1/120 = 0.00833333 Actual error is Sin[1]-0.841667 = -0.000196015 The actual error is less than the bound on the error as expected. ================================================================== New Material 29. If the power series INF n SUM a (x-3) n=0 n converges at x = -1.1, then it also converges at x=7. True The Radius of convergence is at least 4.1 so the convergence extends up to but not necessarily including 7.1. Thus 7 is inside the interval of convergence. INF n 30 If SUM a x converges at x = -2, then it also converges at x=2. n=0 n False. It may be conditionally convergent at x = -2 and not convergent at all at x = 2. n x An example is Sum ----------- n n 2 INF n 31. If f(x) = SUM a x and if the series converges at x = 1.5 n=0 n a x=1 INF n then INT f(x) dx = SUM ------ x=0 n=0 n+1 Since the interval of convergence extends to x=1.5 at the very least, thus the series converges absolutely at x=1. | n+1 | x=1 a x=1 INF n | INF x | n n INT SUM a x = | SUM a ------ | = SUM ---- x=0 n=0 n | n=0 n n+1 | x=0 n=0 n+1 32. Every power series converges for at least two values of the variable. false. Some power series only converge at x=0. INF n For example SUM n! x converges only at x=0 by the ratio test. n=1 / // 33. If F(x), f (0), f (x), ... all exist, then the Maclaurin series for f(x) converges to f(x) in a neighborhood of x=0. False. A flat function can have all of the derivatives at x=0 exist and equal to zero, and the function is not the zero function. An example is _ _ | -1 | | ------ | | 2 | |_ x _| f(x) = e 2 3 34. The function f(x) = 1 + x + x + x + ... / 2 satisfies the differential equation y = y on the interval (-1,1). 1 True. f(x) = --------- 1-x / 1 2 f (x) = ---------- = f(x) 2 (1-x) n INF n x 35. The function f(x) = SUM (-1) ----- satisfies the n=0 n! / differential equation y + y = 0 on the whole real line. -x True. f(x) = e / -x f (x) = - e / f (x) + f(x) = 0 36. If P(x) is the Maclaurin polynomial of order 2 for f(x), / / // // then P(0) = f(0), P (0) = f (0), and P (0) = f (0). True. This is the definition of the Maclaurin polynomial. 37. The Taylor polynomial of order n based at a for f(x) is unique; that is, f(x) has only one such polynomial. True. It is unique by definition. 5/2 38. f(x) = x has a second-order Maclaurin polynomial / 3/2 True: f (x) = 5/2 x // 1/2 f (x) = 15/4 x 2 The second order Maclaurin polynomial is 0 + 0 x + 0 x /2 3 2 39. The Maclaurin polynomial of order 3 for f(x) = 2 x - x + 7x - 11 is an exact representation of f(x). True: For polynomials, they are their own Maclaurin series. 40. The Maclaurin polynomial of order 16 for Cos[x] involves only even powers of x. True. The Cos[x] only has even terms for any degree. / / 41. If f (0) exists for an even function, then f (0) = 0. True. f( dx ) - f(0) f(-dx) - f(0) LIMIT --------------- = LIMIT -------------- dx -> 0 dx dx->0 -dx f( dx ) f(dx) f(dx) LIMIT -------- = LIMIT -------- = - LIMIT ---- dx -> 0 dx -dx->0 -dx -dx -> 0 dx Thus the limit is its own negative. If it exists, it is zero. / 42. Taylor s Formula with Remainder contains the Mean Value Theorem for Derivatives as a special case. True: / f(x) = f(a) + f (c) (x-a)