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.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Wednesday, April 30 These last three days of classes will be over series. There are no hand-in-homework and no tests. This material is part of the AP Calculus curriculum. The syllabus is: Mon, Apr 28 10.1, 10.2, Sequence, Series, Harmonic Series, 10.3, 10.4 Geometric Series, Ratio Test, Root Test Comparison Test, Wed, Apr 30 10.5, 10.6 Power series, Radius of Convergence, Term by term Differentiation and Integration Fri, May 2 10.7 MacLaurin Series, Taylor Series, Error Bounds --------------------------------------------------------------- Main Idea: Polynomials of infinite degree Key Words: Power Series, Radius of Convergence, Term by term differentiation and Integration Goal: View Power Series as a very important technique ------------------------------------------------------------ 2 3 4 n a + a x + a x + a x + a x + ...... + a x + ... 0 1 2 3 4 n is called a power series. It certainly converges for x=0. It may converge for other values of x as well. Given a series, it is generally a lot of work just to see if the thing converges. Fortunately, the convergence behavior of power series is extremely well predicted. ------------------------------------------------------- Everything in a nutshell: When a function is defined by a power series as Infinity n f(x) = SUM a x n=0 n There is an interval (-R,R) on which f(x) and all its derivatives exist. If |x| > R, then neither f nor any of its derivatives converge. --------------(++++++++++++++++++++++++++++)------------- Diverge -R Converge R Diverge ----------------------------------------------------------- Theorem: Given a power series, there is a number R such that |x| < R means the series converges absolutely at x. If |x| > R the series diverges. If |x| = R, the series may converge or diverge. It may converge absolutely or conditionally. Proof: The proof simply shows that if a power series converges at x , then it converges whenever |x| < |x |. 0 0 Surprisingly, this implies the theorem, even though the value of R is never mentioned. Suppose that the power series converges at a particular th value x . Then the n term has to go to zero. Thus the terms 0 in the series are bounded. | n | If | a x | < M, then for |x| < |x | | n 0 | 0 n n | n | | n | | x | | x | | a x | < | a x | | -----| < M |----| . | n | | n 0 | | x | | x | 0 0 | n | Since each term of |a x | is less than the corresponding | n | term for a geometric series, n SUM a x converges absolutely when |x| < |x |. n 0 ------------------------------------------------------------ 1 2 3 Example: f(x) = ------- = 1 + x + x + x + ... + 1-x This has radius of convergence R=1. Obvious, one suspects that x=1 would cause problems because the original function is not defined at x=1. ----------------------------------------------------------- 1 2 4 6 8 Example: f(x) = --------- = 1 - x + x - x + x - ... 2 1 + x Notice that the function is well defined for any value of x, but the power series does not converge for |x| >= 1. ------------------------------------------------------ Theorem: Power series have derivatives. The derivative is computed by differentiating term by term. WOW. (1) The Radius of convergence of the power series and the series obtained by differentiating each term is the same. (2) The Power Series has a derivative. (3) The derivative is the power series obtained by differenting term by term. And thus a function defined by a power series has derivatives of all orders. And all of these derivatives have the same radius of convergence. -------------------------------------------------------------- Applications: Show ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ..... 1 --- = 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 ,,, 1+x ln(1+x) = x - x^2/2 + x^3/3 -x^4/4 + x^5/5 - x^6/6 ... ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ... (* One has to be careful because we are evaluating the function right at its radius of convergence. But at least you can see a good reason to suspect that the alternating power series converges to ln(2) *) ----------------------------------------------------------- Pi/4 = 1/3 - 1/5 + 1/7 - 1/9 + 1/11 - ... 1 ------- = 1 - x^2 + x^4 - x^6 + x^8 - x^10 ... 1+x^2 ArcTan[x] = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - x^11/11 ... Pi/4 = ArcTan[1] = 1 - 1/3 + 1/5 - 1/7 + 1/9 - x/11 ... --------------------------------------------------------------- How many ways can you associate a string of n letters. a 1 way ab 1 way abc 2 ways abcd 5 ways abcde 14 ways c(n) = c(1)c(n-1) + c(2)c(n-2) + c(3) c(n-3) + ... Consider the series f(x) = c(0) + c(1) x + c(2) x^2 + c(3) x^3 + ... + c(n) x^n + ... f(x) = x + x^2 + 2 x^3 + 5 x^4 + 14 x^5 + ... c(n) x^n + ... n f(x)*f(x) = x^2+ 2 x^3 + 4 x^4 + 14 x^5 + ...+SUM c(i) c(n-i) x^(n+1) i=1 c(n+1) x^(n+1) f(x)*f(x) = f(x) -x f(x)^2 - f(x) + x = 0 1 +/- Sqrt[1-4x] f(x) = ------------------ 2 1+/-Sqrt[1-4x] If we write out the power series for ---------------- 2 The coefficients will give us the formula for the number of ways to associate a string of n letters. ------------------------------------------------------------ What is the series for e^x ? Since the derivative of e^x is itself, we build a series which has that property. e^x = 1 + x + x^2/2 + x^3/3! + x^4/4! + ... + x^n/n! + ... ---------------------------------------------------------------- What is the series for Sin[x]? We build a series where the second derivative is the negative of itself. Sin[x] = x - x^3/3! + x^5/5! - x^7/7! + ... + ---------------------------------------------------------------- What is the series for Cos[x] ? Cos[x] = 1 - x^2/2! + x^4/4! - x^6/6! + ... ---------------------------------------------------------------- Proofs of the theorem: Infinity n let f(x) = SUM a x have radius of convergence R n=0 n Infinity n-1 let fp(x) = SUM n a x be the term by term differentiation n=0 n We first show that if |x| < R then fp(x) converges. What we do is insert a value x which has absolute value greater than x and 0 still less than R. (----------------------------------------) -R x x R 0 By construction 0 < |x| < |x | < R. 0 We work with the absolute value of the terms of fp[x]. | n-1 | 1 | n | | n a x | = --- | n a x | | n | |x| | n | n n | n | | x | = --- | a x | |-----| |x| | n 0 | | x | 0 n | x | <= M n |----- | | x | 0 The M comes because the sequence f[x ] converges and thus the 0 terms are bounded. Using the ratio test we know that n | x | SUM n |----| converges. | x | 0 Thus fp(x) converges absolutely. ----------------------------------------------------------------- The following is the really hard proof. I hesitate to give it because it is too difficult. But it is so wonderful that I hate to pass over the opportunity. We have to show that the derivative of f[x] actually is fp[x]. We do that with the fundamental definition of the derivative. ___ ___ | | | n n | f(x)-f(t) Infinity | x - t n-1 | --------- - fp(x) = SUM ai |--------- - n x | x-t n=1 | x - t | |___ ___| (..................0....................) x t R We now use the mean value theorem to express n n x - t n-1 ----------- = n X for some X between x and t. x - t n (..................0....................) x X t R n ___ ___ | | | | f(x)-f(t) Infinity | n-1 n-1 | --------- - fp[x] = SUM a | n X - n x | x-t n=1 n | n | |___ ___| We now use the mean value theorem again to collapse n-1 n-1 n-2 X - x = (n-1) Y (X - x ) n n n for some Y between x and X . n n (..................0....................) x Y X t R n n ___ ___ | | | | f(x)-f(t) Infinity | n-2 | --------- - fp(x) = SUM a | n(n-1) Y (X -x) | x-t n=1 n | n n | |___ ___| We now look at the absolute value of each of the terms for | f(x) - f(t) | | ----------- - fp(x) | = | x-t | | n-2 | | n-2 | | | | n(n-1) a Y (X - x) | <= |n(n-1) a x | | x-t | | n n n | | n 0 | | | Where x is still in the interval of convergence but is 0 larger in absolute value than |x| or |t|. Also by construction | | |X - x | < | x - t |. | n | n-2 Since n(n-1) a x is the second derivative of f[x] n 0 and we already know that it converges absolutely at x . 0 | f[x] - f[t] | So | ------------- - fp[x] | <= M |x-t|. | x - t | Thus Limit approaches 0 as t->x and so the derivative of f[x] is fp[x]. ------------------------------------------------------------- Theorem: A function can have at most one power series expansion. (n) f (0) Proof. The nth coefficient has to be ------- n!. Thus the function designates all of the coefficients exactly. ------------------------------------------------------------ Some functions cannot be written as a power series. 2 -(1/x) f(x) = e has no power series. All of the derivatives at x = 0 are 0. Thus the only possible power series which can represet f is the all zero power series. And that is not f. ---------------------------------------------------------- Write the power series for f(x) = ln(1+x) ix Show that e = Cos[x] + i Sin[x]. What is ln[i] ? Check it out with your calculator. 1 1 1 What is 1 + ----- + ------ + -------- + ... 2 3 3 4 4 5 ---