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 2, 9.7 Term by term Assignment: page 487: 1, 5,11,14 Main Idea: The obvious thing to do is to differentiate the series term by term. The marvelous thing is that this works. Key Words: (a) Term by term differentiation, (b) Term by term integration. Goal: First show that after differentiating term by term the series still converges. Then show that it converges to the actual derivative. -------------------------------------------------------- Previous assignment: p483: (2),(6),(12),(24) Page 483 Problem 2 Find the convergence set of the given power series. 1 2 3 4 n x x x x x --- + ------ + ------ + ------ + ... + ------ + ... 1 2 3 4 n 3 3 3 3 3 n x a = ------ n n 3 By ratio test; a n+1 n n+1 x 3 x ----- = ------ ------ = ----- a n+1 n 3 n 3 x The series converges by the ratio test, |x| < 3 at x = -1,the series diverges. at x = 3, the series diverges. / \ -----------------(######################################)---------------- diverges \ converges absolutely / diverges -3 3 /|\ /|\ | | diverges diverges .............................................................. The series converges to x/3 x -------------- = ------- 1 - x/3 3-x ============================================================ Get["font.math"]; k[x_] := x/(3-x); f[x_,n_] := x^n/3^n; g[n_] := Plot[Sum[f[x,i],{i,1,n}],{x,-3.5,2.9}]; a = Plot[Sum[f[x,i],{i,1,10}],{x,-3.2 , 2.9}, PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; b = Plot[k[x],{x,-3.5, 2.9}, PlotStyle->{RGBColor[0,0,1],Thickness[0.003]}]; c = Show[a,b,PlotLabel->"P483 P2; SUM x^n/3^n); red=S10; blue=x/(3-x) ", PlotRange->{{-3.5,2.9},{-1,3}}]; Display["p2.ps",c]; ============================================================ Page 483 Problem 6 Find the convergence set of the given power series. n Infinity n x SUM (-1) ----- n=1 n a n+1 n+1 x n n ----- = --------- * ----- = x ---- ---------> x a n+1 n n+1 n x The series converges if |x| < 1 by ratio test. r = |x| When x = 1 the series converges by alternating series test. When x = -1 the series diverges because harmonic series diverges The convergence set is (-1,1] --------------------------------------------------------------- 1 2 3 4 x x x x f(x) = - ---- + ---- - ---- + ---- ... 1 2 3 4 / 2 3 f (x) = -1 + x - x + x ..... / -1 f (x) = ---------- 1+x f(x) = - ln(1+x) ============================================================ Get["font.math"]; k[x_] := - Log[1+x]; f[x_,n_] := (-1)^n x^n/n g[n_] := Plot[Sum[f[x,i],{i,1,n}],{x,-1.2, 1.20}]; a = Plot[Sum[f[x,i],{i,1,10}],{x,-1.2 , 1.2 }, PlotStyle->{RGBColor[1,0,0],Thickness[0.010]}]; b = Plot[k[x],{x,-1.2, 1.2}, PlotStyle->{RGBColor[0,0,1],Thickness[0.003]}]; c = Plot[0,{x,-1,1}, PlotStyle->{RGBColor[1,0,1],Thickness[0.003]}]; d = Show[a,b,c,PlotLabel->"P483 P6; SUM (-x)^n/n; red=S10; blue=-ln(1+x) ", PlotRange->{{-1.2,1.2},{-2,5}}]; Display["p6.ps",d]; ===================================================================== Page 483 Problem 12 Find the convergence set of the given power series. 2 4 6 8 10 x x x x x 1 - ---- + --- - ---- + --- - ---- ..... 2! 4! 6! 8! 10! 2 n x a = --------- n (2n)! a 2n+2 2 n+1 x (2n)! x ----- = ----------- ----- = ----------- -----------> 0 a (2n+2)! 2n (2n+2)(2n+1) n x The series converges everywhere. ................................................................. Get["font.math"]; k[x_] := Cos[x]; f[x_,n_] := (-1)^(n) x^(2n)/(2n)! g[n_] := Plot[Sum[f[x,i],{i,0,n}],{x,-4 Pi, 4 Pi}]; a = Plot[Sum[f[x,i],{i,0,10}],{x,-2 4 Pi, 4 Pi }, PlotStyle->{RGBColor[1,0,0],Thickness[0.010]}]; b = Plot[k[x],{x,-4 Pi, 4 Pi}, PlotStyle->{RGBColor[0,0,1],Thickness[0.003]}]; c = Plot[0,{x,-4Pi,4Pi}, PlotStyle->{RGBColor[1,0,1],Thickness[0.002]}]; d = Show[a,b,c, PlotLabel->"P483 P12; SUM (-x)^(2n)/(2n)!; red=S10; blue=Cos[x] ", PlotRange->{{-4 Pi,4 Pi},{-1.2, 1.2}}]; Display["p12.ps",d]; ========================================================================== Page 483 Problem 24 Find the radius of convergence of the following: 2 3 4 (x+2) (x+2) (x+2) 1 + (x+2) + ----- + ------- + ----- + ...... 2! 3! 4! a n+1 n+1 (x+2) n! (x+2) ----- = --------- -------- = ------- ---------> 0 a (n+1)! n n+1 n (x+2) The series converges everywhere. ------------------------------------------------------------------------ Get["font.math"]; k[x_] := E^(x+2); f[x_,n_] := (x+2)^n/n! g[n_] := Plot[Sum[f[x,i],{i,0,n}],{x,-8, 3}]; a = Plot[Sum[f[x,i],{i,0,10}],{x,-10, 3 }, PlotStyle->{RGBColor[1,0,0],Thickness[0.010]}]; b = Plot[k[x],{x,-10, 3}, PlotStyle->{RGBColor[0,0,1],Thickness[0.003]}]; c = Plot[0,{x,-8, 3}, PlotStyle->{RGBColor[0,0,1],Thickness[0.001]}]; d = Show[a,b,c,PlotLabel->"P483 P24; SUM (x+2)^n/n!; red=S10; blue=e^(x+2)", PlotRange->{{-8,2},{0,9}}]; Display["p24.ps",d]; ============================================================================= New Material Infinity i f(x) = SUM a x i=0 i Since f(x) is a power series, we know it has a convergence pattern of this form. diverges / converges absolutely \ diverges ----------------(------------0------------)------------------ \ / -R R / Infinity i-1 f (x) = SUM i a x i=0 i converges diverges / absolutely \ diverges ----------------(--------0--------)------------------ -R'\ / R' / f (x) is also a power series, but we do not initially know that / f(x) and f (x) have the same radius of convergence. ----------------------------------------------------------- / Our first task is to show that f(x) and f (x) have the same radius of convergence. We must avoid working right at the endpoints of the interval of convergence. It may very well / happen at that endpoint, f(x) converges, but f (x) diverges. / / / Inside the interval (-R,R) = (-R ,R ) both f(x) and f (x) converge absolutely. -------------------------------------------------------- / We start by showing that R has to be at least as big as R because if the original series converges, then the derivative series will also converge. x b | | ---------------(-----------0-- * --*--)------------- R R ----------(----------------0-----------------)------ R' R' Suppose that |x| < R. This means that f(x) converges absolutely because x is strictly inside the interval of convergence. / We show that f (x) also converges. / Infinity i-1 f (x) = SUM i a x i=1 i / We will show that the f (x) converges absolutely. Then we will / know by the absolute convergence theorem that f (x) converges as well. i-1 Infinity | | | | SUM i | a | | x | i=1 | i | | | i-1 i-1 Infinity | || | | x | = SUM i | a || b | | --- | where b is just some number i=1 | i || | | b | |x| < b < R. i-1 Infinity | x | <= SUM i M | --- | because the series for f(x) is i=1 | b | i absolutely convergent at b and thus the terms a b -> 0 i i-1 Infinity | x | <= M SUM i | --- | i=1 | b | By the ratio test, r = |x|/|b| < 1, this series converges. / Therefore the original series for f (x) is absolutely convergent by the comparison test. ------------------------------------------------------------ We have just shown that when the f(x) series converges, / then so does the f (x) series. / This means that the radius of convergence of f (x) is not smaller than the radius of convergence of f(x). --------------------------------------------------------- / We now show that R cannot extend beyond R by showing that if the derivative series converges, then the original series converges. ---------------(----------------------)-*----------- R R | x | ----------(-----------------------------*----)------- R' R' / We now show that when f (x) converges absolutely, then so does f(x). / If the f (x) series converges absolutely, then Infinity i Infinity i SUM | a x | <= |a | + SUM |a i x | i=0 i 0 i=1 i / The second series is the absolute value series for x f (x) which converges. Therefore f(x) converges absolutely by the comparison test. Therefore the series f(x) converges. This establishes that the radius of convergence of the two series are the same. .................................................................. / We now know that f(x) and f (x) have the same radius of convergence. // /// Thus f (x) and f (x) also have the same radius of convergence and you can take as many derivatives as you want. The radius of convergence never changes. ?????????????????????????????????????????????????????????????????????? / But the question remains, does this f (x) really work as the derivative of f(x)? ===================================================================== TAKE A BREATH.....WE REALLY NOW ARE GOING TO PUSH THE SUMMATIONS AROUND ===================================================================== diverges / converges absolutely \ diverges -----------|--------------------0---------*-*--*---*---*--|------------------ \ | | | | | / a u t x b i i Infinity i f(x) = SUM a x i=0 i / Infinity i-1 f (x) = SUM i a x i=1 i i i a (x - a ) f(x)-f(a) / Infinity i i-1 --------- - f (a) = SUM ------------- - i a a x-a i=1 x-a i diverges / converges absolutely \ diverges -----------|--------------------0---------*-*--*----*--*--|------------------ \ | | | | | / a u t x b i i i-1 a i t (x-a) f(x)-f(a) / Infinity i i i-1 --------- - f (a) = SUM ------------- - i a a x-a i=1 x-a i f(x)-f(a) / Infinity i-1 i-1 --------- - f (a) = SUM a i t - i a a x-a i=1 i i i f(x)-f(a) / Infinity i-1 i-1 --------- - f (a) = SUM a i ( t - a ) x-a i=1 i i f(x)-f(a) / Infinity i-2 --------- - f (a) = SUM a i (i-1) u (t -a) x-a i=1 i i i i-2 | f(x)-f(a) / | Infinity | | | | | | | --------- - f (a) | <= SUM | a | i (i-1) | u | | t -a | | x-a | i=1 | i | | i | | i | i-2 | f(x)-f(a) / | Infinity | | | | | | | --------- - f (a) | <= SUM | a | i (i-1) | b | | x-a | | x-a | i=1 | i | | | | | _ _ | i-2 | | f(x)-f(a) / | | Infinity | | | | | | --------- - f (a) | <= | SUM | a | i (i-1) | b | || x-a | | x-a | | i=1 | i | | | | |_ _| <------------This Converges---------> | f(x)-f(a) / | | | | --------- - f (a) | <= M | x-a | | x-a | | | Therefore; | f(x)-f(a) / | Limit | --------- - f (a) | = 0. x->a | x-a | ---------------------------------------------------------------------------- 1 (a) What is the series for ------- ? 1-x x (b) What is the series for e ? (c) What is the series for Sin[x] ? (d) What is the series for Cos[x] ? (e) What is the series for ArcTan[x] } i x (f) What is the series for e ?