Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio cosists of all real umbers x such that the power series is coverget, i.e., ( ) D (f) x : (x a) is coverget : Example 4.. Some examples of power series: f (x) (!) x + x + (2!) x 2 + (!) x + (4!) x 4 + :::!; a 0 (x ) g (x) (x ) ; a (x + ) h (x) + (x + ) +!! ; a : Iterval of Covergece + (x )2 2 + (x + )2 2! (x ) + + ::: (x + )! + ::: We ow use Ratio Test to study domai of power series (). To this ed, we write, usig the stadard series otatio a (x a)
with ad compute ad So a (x a) a + + (x a) + a + a + (x a) + (x a) + jx aj : a +! a +! jx aj +! jx aj : De e a umber R by (so that R +!! + R ):! + : (2) We call R the Radius of Covergece for the power series. It follows that a +! a + jx aj! jx aj R : By the Ratio Test, the power series is absolutely coverget if a +! a jx aj < or jx aj < R; R ad it is diverget if a +! a jx aj R > or jx aj > R: I other words, if we draw a circle cetered at a with radius R;the the power series is absolutely coverget iside iterval (a R; a + R) ad diverget outside the iterval. 2
absolutely coverget iside (a R, a+r) diverget outside a R a a+r ucertai at edpoits: xa R ad xa+r at each edpoit a R or a+r, it could be coverget or diverget However, at this poit we kow othig about what would happe at the edpoits x a R ad x a + R: Example 4.2. Fid radius of covergece for the series i Example 4.: (a) f (x) (!) x ;!; a 0 (b) g (x) (c) h (x) (x ) ; ; a (x + ) ;!! ; a : Solutio: (a)!; + ( + )!;so R! + (b) ; + + ; R!! + (c)! ; c + ( + )! ; R! +!! ( + )!!! ( + )!! ( + ) + 0: :! ( + ) :
We summarize the above aalysis as follows. Theorem. Cosider the power series f (x) (x a) ad its Radius of Covergece, if exists, R! + There are three possibilities: () If R 0; the the power series is coverget oly at x a; (2) If R ; the the power series is absolutely coverget for all x;ad thus f (x) is de ed everywhere; () If 0 < R < ; the the power series is absolutely coverget for jx aj < R;i.e., a R < x < a + R; ad is diverget for jx aj > R: It is ot clear what happe whe jx aj R or x a R: I short, the domai of the power series () is a iterval with edpoits x a R. This iterval could be a ope iterval, a closed iterval, or half ope half closed iterval, ad is called Iterval of Covergece. Fidig Iterval of Covergece: Step #: d the radius of covergece R; ad the write dow a iterval of the form fa R; a + Rg Step #2: check covergece of the series f (a R) ad f (a + R) at two edpoits x a R ad x a + R Example 4.. Fid iterval of covergece for P (a) f (x) x p ; + P (b) g (x) (x + 2) : 5 Solutio: (a) We kow that a 0; ad we rst d its radius of covergece. To do so, write dow p + ; + : (+) p ( + ) + ; 4
ad calculate R! + p + 2 p +! (+)! We the write dow the iterval of the form fa R; a + Rg ; p + 2 p + : that oly idicates the iterval of covergece is a iterval with edpoits ad : It could be of ay type: ope iterval, or closed iterval, or a half ope iterval. Thus, iterval of covergece has the form ; ; where the type of iterval eeds to be determied. edpoits. At the left edpoit x ; f p + ( ) p + Next, we check each ( ) p + : By Alteratig Series Test, we d it coverget (but ot absolutely coverget). At the right edpoit x ; f p + p + This is a diverget p-series. We thus coclude that iterval of covergece [ (b) We follow the same process to study (x + 2) g (x) : 5 5 ; ): p :
I this case, a 2; R! + So the iterval of covergece has the form 5 ; + + 5 ; +! 5 5+ + 5! + 5: fa R; a + Rg f 7; g : We ext check covergece at edpoits. At x 7; ( g ( 7) 7 + 2) 5 ( 5) 5 ( ) 5 5 ( ) : This series is diverget (by Divergece Test). At x ; ( + 2) g () 5 (5) 5 is obviously diverget. We coclude that iterval of covergece ( 7; ): Derivatives ad Itegrals Suppose that a power series f (x) (x a) has the radius of covergece R: The Theorem. Iside the iterval of covergece, i.e., jx aj < R we have df (x) dx X ( (x a) ) 0 (x a) ; d2 f (x) dx 2 6 ( ) (x a) 2 ; ::: 2
ad f (x) dx (x a) dx C+ + (x a)+ ; C is a arbitrary costat. If c ad d are iside the iterval of covergece, i.e., jc aj ; jd aj < R; the d d f (x) dx (x a) dx: c c Example 4.4. For each series, d the itegral of covergece ad derivative ad ati-derivative. Fid also a explicit formula. (a) f (x) x ; (b) g (x) 5 + x Solutio: (a) Sice ; the radius of covergece R : + Withi the iterval of covergece, i.e., ( ; ) ; we have f 0 (x) f (x) dx (x ) 0 x dx x x + + + C X x + C: Recall the geometric series r r i jrj < : Therefore ad f (x) x x iterval of covergece ( ; ) 7
(b) We rst calculate, withi the iterval of covergece, g 0 (x) g (x) dx 5 + x X 0 5 + x 5 + x dx We ext rewrite the series as g (x) 5 + x 5 (5) x X 5 where Thus g (x) y 5x : 5 + + x+ + C 5x 5 X 5 + x 5f (y) 5 y 5 5x ; it is coverget for jyj j5x j < : So jxj < p 5 ad Iterval of covergece p ; 5 p : 5 ad Thus, Example 4.5. Study the power series f (x) Solutio: Recall from Example 4.4 (a) X x dx f (x) x x : x + + + C X x x : x + C dx l j xj + C: x 8 y
Set x 0; we d 0 0 l j 0j + C C: We coclude f (x) x l j xj l ( x) for jxj < : Homework: P. Suppose that x coverges whe x 4 ad diverges whe x 6: What ca you say about the covergece or divergece of the followig series? (a) (b) (c) (d) P P 8 P ( ) P ( ) 9 2. Fid the radius of covergece ad iterval of covergece of the series. (a) (b) (c) (d) P p x P P P x ( ) x 4 l ( ) x 2 (2)! 9
(e) P ( 2) p (x + ). Fid derivatives ad ati-derivative. Fid also the explicit formula. P (a) f (x) ( ) 2 x P (b) g (x) + x 2 0