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.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. March 23 10.7 Term by term p466: (4),(7),11,(16),18,25 ,(33),35 Main Idea: The obvious thing to do is to differentiate the series term by term. The marvelous thing is that this works. Key Words: Term by term differentiation, 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: Page 461 Problem 2 Find the convergence set of the given power series. 2 3 4 5 x x x x x ----- - ------ + ------ - -------- + ------ + ... 1*2 2*3 3*4 4*5 5*6 n x an = ------- n(n+1) an+1 x^(n+1) n(n+1) n --- = ----------- * -------- = x ----- an (n+1)(n+2) x^n n+2 The limit of an+1/an = x so the series converges absolutely with |x| < 1. It diverges if x > 1 at |x| = 1 the series converges absolutely because it Infinity 1 Infinity 1 SUM ----------- < SUM -------- n=1 n(n+1) n=1 2 n This is a p series, p=2 and it converges. The case |x| =1 is absolutely convergent by the comparison test. -------------------------[-1 xxxxxxxxxxxxxxx 1]-------------- The interval of convergence It converges absolutely at the endpoints. -------------------------------------------------- f[x_,n_] := x^n/(n(n+1)); s[x_,n_] := Sum[f[x,i],{i,1,n}]; A = Table[ Plot[ s[x,10*n],{x,-1.1,1.1}],{n,1,10}]; a = Show[A,PlotLabel->"P461 P2 x^n/(n(n+1))"]; Display["2.ps",a]; ----------------------------------------------------- Page 461 Problem 8 Find the convergence set of the given power series. 2 3 4 5 x x x x 1 + x + -------- + -------- + ---------- + --------- + ... Sqrt[2] Sqrt[3] Sqrt[4] Sqrt[5] n x an = -------- Sqrt[n] n+1 an+1 x Sqrt[n] Sqrt[n] ----- = --------------- * --------- = x ---------- an Sqrt[n+1] n Sqrt[n+1] x The limit an+1/an = x n->Infinity The Series converges absolute if |x| < 1. The series diverges if |x| > 1. If x = 1 it is the p series, p = 1/2 < 1 which diverges. If x = -1 it is the alternating p series, p=1/2 which is conditionally convergent. -----------------[-1 xxxxxxxxxxxxxxxxxxxxxxxxxxxx +1)------------------ Conditionally convergent at x = -1 divergent at x = 1 -------------------------------------------------- f[x_,n_] := x^n/Sqrt[n]; s[x_,n_] := Sum[f[x,i],{i,1,n}]; A = Table[ Plot[ s[x,10*n],{x,-1.1,1.1}],{n,1,10}]; a = Show[A,PlotLabel->"P461 P8 x^n/Sqrt[n]"]; Display["8.ps",a]; ---------------------------------------------------- Page 461 Problem 14 Find the convergence set of the given power series. 2 3 4 5 x 2x 3x 4x 5x -- + ---- + ----- + ------ + ------ + ..... 2 3 4 5 6 n n x an = -------- n+1 n+1 2 an+1 (n+1)x n+1 (n+1) ---- = ---------------- * ----------- = x ------ an n+2 n n(n+2) n x The limit as n -> Infinity of an+1/an = x Thus the series is absolutely convergent if |x| < 1. The series is divergent if |x| >1. When |x| = 1 the series is divergent by the nth term test. ---------------(-1 xxxxxxxxxxxxxxxxxxxxxxx 1)-------------------- Divergent at both endpoints. -------------------------------------------------------- f[x_,n_] := n x^n/(n+1); s[x_,n_] := Sum[f[x,i],{i,1,n}]; A = Table[ Plot[ s[x,10*n],{x,-1.01,1.01}],{n,1,10}]; a = Show[A,PlotLabel->"P461 P14 n x^n/(n+1)"]; Display["14.ps",a]; ------------------------------------------------------- Page 461 Problem 27 Find the convergence set for each series. n Infinity (3x+1) (a) SUM ---------- n=1 n n 2 n Infinity n (2x-3) (b) SUM (-1) -------------- n=1 n 4 Sqrt[n] ------------------------------------------------------ n (3x+1) an = ----------- n n 2 n+1 n (3x+1) n 2 (3x+1) n an+1/an = ------------ * --------- = ------ ----- n+1 n 2 n+1 (n+1) 2 (3x+1) |3x+1| The series is absolutely convergent if ------- < 1 2 -2 < 3x+1 < 2 -3 < 3x < 1 -1 < x < 1/3 The series is absolutely convergent if -1 < x < 1/3. The series diverges if x < -1 or x > 1/3. n (-2) at x = -1 SUM ---------- n n 2 Is the alternating harmonic series which is conditionally convergent. n 2 at x = 1/3 SUM --------- n n 2 Is the harmonic series which is divergent. -------------------------------------------------------- f[x_,n_] := (3x+1)^n/(n 2^n); s[x_,n_] := Sum[f[x,i],{i,1,n}]; A = Table[ Plot[ s[x,10*n],{x,-1.05 ,0.34}],{n,1,10}]; a = Show[A,PlotLabel->"P461 P27a (3x+1)^n/(n 2^n) "]; Display["27a.ps",a]; ------------------------------------------------------- n Infinity n (2x-3) (b) SUM (-1) -------------- n=1 n 4 Sqrt[n] an+1 (2x-3)^(n+1) 4^n Sqrt[n] (2x-3) Sqrt[n] ---- = ------------------- * -------------- = -------- ----------- an 4^(n+1) Sqrt[n+1] (2x-3)^n 4 Sqrt[n+1] The series converges if |2x-3|/4 < 1 -4 < 2x-3 < 4 -1 < 2x < 7 -1/2 < x < 7/2 When x = -1/2 n Infinity n (-4) (b) SUM (-1) -------------- n=1 n 4 Sqrt[n] The series is the p series p = 1/2. The series is divergent. .......................................... when x = 7/2 n Infinity n (+4) (b) SUM (-1) -------------- n=1 n 4 Sqrt[n] The series is the alternating p series p = 1/2 and is conditionally convergent. ----------------------(----0-----------------]---------- -1/2 7/2 -------------------------------------------------------- f[x_,n_] := (-1)^n (2x-3)^n/(4^n Sqrt[n]); s[x_,n_] := Sum[f[x,i],{i,1,n}]; A = Table[ Plot[ s[x,10*n],{x,-0.51 ,3.59}],{n,1,10}]; a = Show[A,PlotLabel->"P461 P27b (-1)^n (2x-3)^n /(4^n Sqrt[n]) ",PlotRange->All]; Display["27b.ps",a]; ------------------------------------------------------- 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. -------------------------------------------------------- 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. Infinity i-1 SUM i |a | |x| i=1 i i-1 Infinity i-1 |x| = SUM i |a ||b| |-| were 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, p = |x|/|b| < 1, so this series converges. Therefore the original series for f'(x) is absolutely convergent by the comparison test. ------------------------------------------------------------ This shows that when the f(x) series converges, then so does the f'(x) series. This shows that the radius of convergence of f'(x) is not smaller than the radius of convergence of f(x). --------------------------------------------------------- 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-1 SUM | a x | <= |ao| + SUM |a i x | i=0 i i=1 i eventually, that is, when i > |x| The second series is the absolute value series for f'(x) which converges. Therefore f(x) converges absolutely by the comparison test. Therefore the series f(x) converges. ===================================================================== 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)? diverges / converges absolutely \ diverges ----------------|------------0---a--x---b-|------------------ \ / 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 i-1 a i t (x-a) f(x)-f(a) Infinity 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] ? i x (e) What is the series for e ?