Chapter Eleve Taylor Series 111 Power Series Now that we are kowledgeable about series, we ca retur to the problem of ivestigatig the approximatio of fuctios by Taylor polyomials of higher ad higher degree We begi with the idea of a so-called power series A power series is a series of the form A power series is thus a sequece of special polyomials: each term is obtaied from the previous oe by addig a costat multiple of the ext higher power of (x a) Clearly the questio of covergece will deped o x, as will the limit where there is oe The k th term of the series is (x a) k so the Ratio Test calculatio looks like r(x) = lim k +1 (x a) k +1 (x a) k = x a lim k +1 Recall that our series coverges for r(x) < 1 ad diverges for r(x) > 1 Thus this series coverges absolutely for all values of x if the umber lim k +1 = 0 Otherwise, we c have absolute covergece for x a < lim k ad divergece for k +1 c x a > lim k The umber k +1 R = lim k +1 is called the radius of covergece, ad the iterval x a < R is called the iterval of covergece There are thus exactly three possibilities for the covergece of our power series (i)the series coverges for o value of x except x = a ; or : 111
(ii)the series coverges for all values of x ; or (iii)there is a positive umber R so that the series coverges for x a < R ad diverges for x a > R Note that the Ratio Test tells us othig about the covergece or divergece of the series at the two poits where x a = R Example Cosider the series x k The R = lim k +1 = lim k (k +1)! = lim 1 k k +1 = 0 Thus this series coverges oly whe x = 0 Aother Example Now look at the series 3 k (x 1) k Here R = lim k +1 3 k 1 = lim = lim k k +1 3 k 3 = 1 3 Thus, this oe coverges for x 1 < 1 3 ad diverges for x 1 > 1 3 Exercises Fid the iterval of covergece for each of the followig power series: 1 (x + 5) k 2 1 (x 1)k k 112
3 k (x 4)k 3k + 1 4 3 k (x +1)k 5 k (x 9) 7(k 2 +1) 112 Limit of a Power Series If the iterval of covergece of the power series is x a < R, the, of course, the limit of the series defies a fuctio f : f (x) = (x a) k, for x a < R It is kow that this fuctio has a derivative, ad this derivative is the limit of the derivative of the series Moreover, the differetiated series has the same iterval of covergece as that of the series defiig f Thus for all x i the iterval of covergece, we have f '(x) = k (x a) k 1 k = 1 We ca ow apply this result to the power series for the derivative ad coclude that f has all derivatives, ad they are give by f (p ) (x) = k(k 1)K(k p + 1) (x a) k p k = p Example 113
We kow that 1 1 x = x k for x < 1 It follows that 1 (1 x) = kx k 1 = 1+ 2x + 3x 2 + 4x 3 +K 2 k =1 for x < 1 It is, miraculously eough, also true that the limit of a power series ca be itegrated, ad the itegral of the limit is the limit of the itegral Oce agai, the iterval of covergece of the itegrated series remais the same as that of the origial series: x f (t)dt = (x a) k +1 k +1 a Example We may simply itegrate the Geometric series to get x k +1 log(1 x) =, for 1 < x < 1, or 0 < 1 x < 2 k + 1 It is also valid to perform all the usual arithmetic operatios o power series Thus if f (x) = x k ad g(x) = d k x k for x < r, the f (x) ± g(x) = ( ± d k )x k, for x < r k =0 Also, f (x)g(x) = k c i d k i x k, for x < r i= 0 The essece of the story is that power series behave as if they were ifiite degree polyomials the limits of power series are just about the icest fuctios i the world 114
Exercises 6 What is the limit of the series x 2k? What is its iterval of covergece? 7 What is the limit of the series 2( 1) k kx k =1 2k 1? What is its iterval of covergece? 8 Fid a power series that coverges to ta 1 x o some otrivial iterval 9 Suppose f (x) = (x a) k What is f (p ) (a)? k=0 113 Taylor Series Our major iterest i fidig a power series that coverges to a give fuctio The obvious cadidate for such a series is simply the sequece of Taylor polyomials of icreasig degree Thus if f is a give fuctio, ad a is a poit i the iterior of the domai of f, the Taylor Series for f at a is the series f (k ) (a) The Taylor Series is thus a ifiite degree Taylor Polyomial> I geeral, the Taylor series for a fuctio may ot coverge o ay otrivial iterval to f, but, mercifully, for may sufficietly ice fuctios it does I such cases, we are provided with the ice aswer to the questio proposed back i Chapter Nie: Ca we approximate the fuctio f as well as we like by a Taylor Polyomial for sufficietly large degree? 115
Example The Taylor series for f (x) = si x at x = a is simply ( 1) k x 2k + 1 A easy (2k +1)! calculatio shows us that the radius of covergece is ifiite, or i other words, this power series coverges for all x But is the limit si x? That s easy to decide From Sectio 93, we kow that si x ( 1) k x 2 k+ 1 x 2 + 3 (2k +1)! (2 + 3)!, ad we kow that x 2+3 lim (2 + 3)! = 0, o mater what x is Thus we have k =0 si x = ( 1) k x 2 k+ 1, for all x (2k +1)! Exercises 10 Fid the Taylor Series at a = 0 for f (x) = e x Fid the iterval of covergece ad show that the series coverges to f o this iterval 11 Fid the Taylor Series at a = 0 for f (x) = cos x Fid the iterval of covergece ad show that the series coverges to f o this iterval 12 Fid the derivative of the cosie fuctio by differetiatig the Taylor Series you foud i Problem #11 13 Fid the Taylor Series at a = 1 for f (x) = logx Fid the iterval of covergece ad show that the series coverges to f o this iterval 116
14 Let the fuctio f be defied by 0, for x = 0 f (x) = e 1/ x 2, for x 0 Fid the Taylor Series at a = 0 for f Fid the iterval of covergece ad the limit of the series 117