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Taylor Series From Functions to Series and Back, again! Chapter 1 FROM FUNCTIONS TO SERIES Before: Given a power series can we find a formula for it? Sometimes we could . . . 1 f ( x) x for all x in 1,1 . 1 x n 0 n Sometimes not . . . 1 k ( x 3) k 2 2 k k 1 Chapter 1 FROM FUNCTIONS TO SERIES Now we want to turn this around: Given a particular function, can we find a power series for it? •The book suggests that sometimes we can (chart you used for homework!) •We also suggested something similar in our discussion of Taylor Polynomials! Approximating f ( x) ( x 1) e The third order Maclaurin Polynomial is the best 3rd degree polynomial approximation for f at x = 0. And, as we added more and more terms the approximation got better and better. x2 Approximating f ( x) ( x 1) e The hundredth Maclaurin Polynomial is the best 100th degree polynomial approximation for f at x = 0. What if we take this to its logical conclusion? We get Taylor Series! x2 Taylor Series for cos(x) 1. Find the Maclaurin series for f (x) = cos(x). 2. Use Taylor’s Theorem to show that this series converges to f for all values of x. Maclaurin Series for f (x) = cos(x) n f (n)(x) f (n)(0) 0 cos( x) cos 0 1 a0 1 1 sin( x) sin 0 0 a1 0 2 cos( x) cos 0 1 a2 3 sin( x) sin(0) 0 a3 0 cos( x) cos(0) 1 a4 4 an= f (n)(0)/n! 1 2! 1 4! Can we deduce the rest of the pattern? What happens, in general, with the coefficients? Maclaurin Series for f (x) = cos(x) n f (n)(x) f (n)(0) 0 cos( x) cos 0 1 a0 1 1 sin( x) sin 0 0 a1 0 2 cos( x) cos 0 1 a2 3 sin( x) sin(0) 0 a3 0 cos( x) cos(0) 1 a4 4 x 2 x 4 x6 1 2! 4! 6! an= f (n)(0)/n! (1)n x2 n 2n ! k 0 1 2! 1 4! Show that this converges to cos(x) We start with the general set-up for Taylor’s Theorem. K n1 n 1 cos( x) Pn ( x) x n 1! What is Kn+1? d n1 cos( x) 1 for all n and all x. n1 dx It follows that n 1 1 n 1 x cos( x) Pn ( x) x n 1! n 1! This quantity goes to 0 as n→∞! Show that this converges to cos(x) We start with the general set-up for Taylor’s Theorem. K n1 n 1 cos( x) Pn ( x) x n 1! What is Kn+1? d n1 cos( x) 1 for all n and all x. n1 dx It follows that n 1 1 n 1 x cos( x) Pn ( x) x n 1! n 1! So what can we conclude? Show that this converges to cos(x) For all real x, cos( x) Pn ( x) 0. n What (on Earth!) does this mean? •The Pn’s are the ? . Show that this converges to cos(x) For all real x, cos( x) Pn ( x) 0. n What (on Earth!) does this mean? •The Pn’s are the partial sums of the Maclaurin Series •So cos( x) Pn ( x) converges to cos(x)! 0 means that the series . Chapter 2 FROM SERIES TO FUNCTIONS What is the relationship between Taylor Series and other power series? Coefficients of a Power Series Suppose that we have function f given by a power series f ( x) a0 a1 ( x x0 ) a2 ( x x0 )2 a3 ( x x0 )3 What can we say about the relationship between f and the coefficients a0, a1, a2, a3, a4, a5, . . ? Answer: Quite a bit, and the reasoning should look somewhat familiar to you. Here’s how it goes . . . If f ( x) a0 a1 ( x x0 ) a2 ( x x0 )2 a3 ( x x0 )3 Then f ( x0 ) a0 a1 ( x0 x0 ) a2 ( x0 x0 )2 a3 ( x0 x0 )3 a0 Furthermore, Theorem 13 (pg. 591 in OZ), says that f ( x) a1 2a2 ( x x0 ) 3a3 ( x x0 )2 4a4 ( x x0 )3 Thus f ( x0 ) a1. Is this beginning to look familiar? It should remind you of the process by which we computed the coefficients of the Taylor polynomial approximations Continuing to take derivatives and evaluate at x0, we have . . . f ( x) 2a2 3 2a3 ( x x0 ) 4 3a4 ( x x0 )2 f ( x0 ) 2a2 f ( x0 ) a2 2 f ( x) 3!a3 4!a4 ( x x0 ) 5 4 3a5 ( x x0 ) 2 f ( x0 ) 3!a3 f ( x0 ) a3 3! In general, we have: (n) f ( x0 ) ( n) f ( x0 ) n!an which tells us that an n! In other words, if a function f is given by a power series that is centered at x0, that power series must be the Taylor series for f based at x0. If, we have f ( x) a0 a1 ( x x0 ) a2 ( x x0 )2 a3 ( x x0 )3 Then (3) f ( x0 ) f ( x0 ) f ( x) f ( x0 ) f ( x0 )( x x0 ) ( x x0 ) 2 ( x x0 )3 2 3! It is easy to see that Taylor series are just a special kind of power series. Our discovery tells us that they are really the only kind of power series there is. To reiterate: If a function f is given by a power series, that power series must be the Taylor Series for f at the same base point. Chapter 3 TAYLOR SERIES CAUTIONS Weird stuff can happen if we are not careful. Now where were we? If a function f is given by a power series, that power series must be the Taylor Series for f at the same base point. Notice what this does not say. •It does not say that every function is given by its Taylor Series. •It does not even say that every function that has a Taylor series is given by its Taylor Series. For a Function f, Some Questions Arise 1. If f has a Taylor Series, does the series converge? Answer: Often, but not always, and certainly not always on the whole domain of the function. Consider the familiar case of f ( x) 1 . 2 1 3x What is the Taylor Series for this function? What can we say about its convergence? Next Question . . . 2. If the Taylor Series for f converges, is it equal to the f on its interval of convergence? Answer: Often, but not always. Consider the absolute value function f ( x) x . We know that we cannot expand it in a Taylor series about x = 0. (Why?) But f (x) = | x | has derivatives of all orders at all other points. What if we consider a Taylor series expansion about x = 1? Taylor Series for f (x)= |x| based at x = 1. What about the derivatives of f at x = 1? What do we get for the Taylor series expansion at x = 1? The Taylor Series expansion for f (x) = | x | converges on the entire real line, but is equal to f only on the interval [0,∞)! Things can get really weird e 1 x2 f ( x) 0 if x 0 if x 0 Facts: • f is continuous and has derivatives of all orders at x = 0. • f (n)(0)=0 for all n. What does all this tell us about the Maclaurin Series for f ? The Maclaurin Series for f converges everywhere, but is equal to f only at x = 0! So where does this leave us? To Summarize: Even if we can compute the Taylor Series for a function, •the Taylor Series does not always converge. •If the Taylor Series converges, the Taylor Series is not necessarily equal to the function, even on its interval of convergence. We know how to determine whether (and where) the Taylor series converges---Ratio test! But how do we know if the Taylor Series of the function is equal to the function on the interval where it converges? The answer is . . . Taylor’s Theorem. Ostebee and Zorn assure us that Though a Taylor series for f “might converge at all x but perhaps to a limit other than f, . . . “Taylor’s theorem guarantees that this unfortunate event seldom occurs.” Chapter 4 TAYLOR SERIES TO THE RESCUE Most of our familiar calculus functions ARE given by their Taylor Series, and Taylor’s Theorem helps us to establish this. More examples. Taylor Series f ( x) sin( x) 1. Find the Taylor series for f that is based at x = p/4. 2. Show that this Taylor series converges to f for all values of x. Taylor Series for f (x) = sin(x) at 0 sin( x) 1 cos( x) 2 sin( x) 3 cos( x) 4 sin( x) an= f (n)( p 4)/n! f (n)( p 4 ) f (n)(x) n 4 2 cos p 1 4 2 sin p 1 4 2 cos p 1 4 2 sin p 1 4 2 sin p 1 /4 1 1 2 0! 1 1 a1 2 1! 1 1 a2 2 2! 1 1 a3 2 3! 1 1 a4 2 4! a0 1 2 1 2 2 3 4 5 1 1 1 1 1 1 1 1 1 1 p p p p p x x x x x 4 4 4 4 4 2 2 22 2 3! 2 4! 2 5! Show that this converges to sin(x) We start with the general set-up for Taylor’s Theorem. n 1 K n1 p sin( x) Pn ( x) x Notice that I didn’t have to 4 what P was in order to n 1! know n gather this information. (In other words, our second question is independent of our first.) What is Kn+1? It follows that sin( x) Pn ( x) x p n 1 4 for all x. n 1! What happens to this quantity As n→∞? Epilogue---Two points of view Power series as functions First a series . . . n a ( x x ) n 0 n 0 . . . Then a function First a function . . . f ( x) . . . Then a series f ( x) an ( x x0 ) Taylor Series n n 0 Guarantees that f is equal to the power series where the power series converges. n 0 f ( n ) ( x0 ) ( x x0 )n n! No a priori guarantee that f is equal to its Taylor series. Why the Taylor series, then? Power series as functions First a series . . . n a ( x x ) n 0 n 0 . . . Then a function First a function . . . f ( x) . . . Then a series f ( x) an ( x x0 ) Taylor Series n n 0 Guarantees that the power series we started with is, in fact, the TAYLOR SERIES FOR f . n 0 f ( n ) ( x0 ) ( x x0 )n n! If f is equal to any power series at all, that power series must be the Taylor series for f. That’swhere we look! Taylor’s Theorem helps!