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 21 10.6 Power series p461: (2),(8),(14),14,17-19,21,(27) Main Idea: Power series are polynomials of infinite degree. They behave like polynomials as long as you remember to avoid like the plague the places that they do not converge. Key Words: Radius of Convergence, Interval of Convergence, Power Series, Goal: For series of the form: 2 3 n f(x) = a + a x + a x + a x + ... + a x + ... 0 1 2 3 n It is quite easy to visualize where they converge. -------------------------------------------------------- Previous assignment: March 11 10.5 Alt Series p457: 1-3,(4),(14),15-17,21,(28),29,(40) Page 457 Problem 4 Show that each alternating series converges and then estimate the error made by using the partial sum S 9 as an approximation to the sum S of the series. Infinity n+1 n SUM (-1) --------- n=1 2 n + 1 n 1 Limit ---------- = Limit ----------- = 0 n->Infinity 2 n->Infinity 2n n +1 2 2 x x +1 - x(2x) 1-x f(x) = -------- so f'(x) = ---------------- = ----------------- 2 2 2 2 2 x +1 (x + 1) (x + 1) Since the derivative is negative for x > 1, the series is decreasing Since it is alternating and converges to zero, it must converge. 10 S is within -------- of the correct answer. 9 101 ---------------------------------------------- f[x_] := x/(x^2+1); A = Table[{n,Sum[ -(-1)^i f[i],{i,1,n}]},{n,1,200}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; b = Show[a,PlotLabel->"P457 P4 SUM n/(n^2+1)"]; Display["4.ps",b]; ---------------------------------------------- Page 457 Problem 14 Classify each series as absolutely convergent, conditionally convergent, or divergent. Infinity n+1 1 SUM (-1) -------- i=1 1.1 5 n The absolute valued series is a p-series p = 1.1 which converges. Therefore the original series is absolutely convergent. --------------------------------------------------- Page 457 Problem 28 Classify each series as absolutely convergent, conditionally convergent, or divergent. n+1 Infinity (-1) SUM ------------------ n=1 Sqrt[n+1] + Sqrt[n] This is an alternating series, it is decreasing to zero so it must converge. Infinity 1 Infinity 1 SUM ------------------ > SUM ----------- n=1 Sqrt[n+1] + Sqrt[n] n=1 2 Sqrt[n+1] This is a p series, p = 1/2 < 1 so it diverges. Therefore the original series is conditionally convergent. ------------------------------------------------------------ Page 457 Problem 40 Discuss the convergence or divergence of 1 1 1 1 1 1 --------- - ----------- + ---------- - --------- + --------- - --------- Sqrt[2]-1 Sqrt[2] + 1 Sqrt[3]-1 Sqrt[3]+1 Sqrt[4]-1 Sqrt[4]+1 It is alternating The terms go to zero But the terms are not decreasing to zero. Combining them in pairs gives 2 2 2 2 ---------- + ------------ + ------------ + .... + ----------- 1 2 3 n-1 The pairwise series is the harmonic series so diverges. ---------------------------------------------- f[n_] := 1/(Sqrt[n]-1); g[n_] := -1/(Sqrt[n]+1); A = Table[0,{i,1,101}]; Do[A[[2*i]] = f[i],{i,2,100}]; Do[A[[2*i+1]] = g[i],{i,1,100}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; b = Plot[0,{x,3,100}]; B = Table[{n,Sum[A[[i]],{i,1,n}]},{n,3,100}]; c = ListPlot[B,PlotStyle->{RGBColor[0,0,1]}]; d = Show[a,b,c,PlotLabel->"P457 P40 "]; Display["40.ps",d]; ---------------------------------------------------------------- New Material: In Euclidean geometry, you think that the angle of 20 degrees exists, and you can approximate it as closely as you want. But because you can not construct it exactly, you are not allowed to say that it exists. In calculus if you can tell how to approximate something, that means you know that number and can use it in any calculations you want. This concept is really used in power series. Even if we really do not know what a series converges to exactly, we can still take its derivatives and its integrals. These will again be power series. We can write down a series which is the solution to a problem and then we can claim that we have the answer because the answer will be what the solution converges to. ------------------------------------------------------------------ For a series 2 3 n f(x) = a + a x + a x + a x + ... + a x + .... 0 1 2 3 n (a) Where does it converge. -------------------|##############################|------------------- diverges /|\ Converges absolutely /|\ diverges | | | | | | | | At the interface the series may converge, may diverge, or may converge conditionally. (b) f'(x) is obtained by differentiating the power series term by term. (c) INT f(x) is obtained by integrating the power series term by term. Conclusion, If you allow yourself the luxury of working with power series, even the most complicated problem is reduced to polynomials. ------------------------------------------------------------------ Page 458 Example 1. 2 3 n f(x) = a + a x + a x + a x + .... + a x ..... What is its limit. Where does it converge. f(x) is a geometric series, p = x and it converges if |x| < 1. a f(x) = ------ 1-x a f(x) = -------- 1-x ----------------|#################################|------------------ -1 1 diverges diverges f[x_,n_] := x^n; g[n_] := Plot[Sum[f[x,i],{i,0,n}],{x,-1,1}]; H = Table[g[ n],{n,1,10}]; a = Plot[Sum[f[x,i],{i,0,200}],{x,-1,1},PlotStyle->{RGBColor[1,0,0]}]; b = Plot[1/(1-x),{x,-0.8,0.8},PlotStyle->{RGBColor[0,0,1],Thickness[0.02]}] c = Show[H,a,b,PlotLabel->"Page 458 Example 1"]; Display["1.ps",c]; ----------------------------------------------------------------- Page 459 Example 2. n 2 3 Infinity x x x x f(x) = SUM ------------ = 1 + -------- + --------- + --------- + ... n=0 n 2 3 (n+1) 2 2 2 3 2 4 2 n x a = ----------- n n (n+1) 2 a n+1 n n+1 x (n+1) 2 (n+1) x (a) ------- = -------------- * ------------ = --------------- =====> x/2 a n+1 n (n+2) 2 n (n+2) 2 x (b) The series converges by the ratio test when |x| < 2. diverges converges absolutely diverges ----------------|#################################|------------------ -1/2 1/2 conditional diverges convergence At x = -1/2 the series is the alternating harmonic series and converges conditionally. At x = 1/2 the series is the harmonic series and diverges. ------------------------------------------------------------------- f[x_,n_] := x^n/((n+1) 2^n); g[n_] := Plot[Sum[f[x,i],{i,1,n}],{x,-1,1}]; H = Table[g[2 n],{n,1,5}]; a = Plot[Sum[f[x,i],{i,1,200}],{x,-1,1},PlotStyle->{RGBColor[1,0,0]}]; b = Show[H,a,PlotLabel->"Page 459 Example 2"]; Display["2.ps",b]; ------------------------------------------------------------------- Page 459 Example 3 n Infinity x f(x) = SUM -------- n=0 n! n x a = -------- n n! | a | n+1 | n+1 | |x| n! |x| (a) | ----- |= -------- * ------ = ------------ ======> 0 | a | (n+1)! n n+1 | n | |x| (b) by the ratio test, p = 0, the series converges for all x. Converges absolutely <###########################################################################> ------------------------------------------------------------------------- The proof of convergence is very very very clever. We simply show that if the series converges at x = b, then it converges absolutely for |x| < |b|. Infinity n Proof: Suppose that SUM a b converges. n=1 n n Then Limit a b = 0 so there exists an N such that n > N n->Infinity n 0 0 | n | implies |a b | < 1. | n | _ _ n _ _ n Infinity n Infinity n | |x| | Infinity | |x| | (a) Now SUM |a ||x| = SUM |a b | | --- | < SUM | ----| n=1 n n=1 n |_|b|_| n=1 |_|b|_| for n > No (b) The last series is a geometric series with p = |x|/|b| which converges if |x| < |b|. (c) Therefore the original series converges by the comparison test. Write the power series for 1 (a) f(x) = ------ 1-x 1 (b) f(x) = ----- 1+x x (c) f(x) = ---------- 2 1+x (d) f(x) = ln(1+x) x (e) f(x) = e