Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i some cases, what it coverges to. More geerally however, istead of cosidered a series as a ifiite sum of umbers, we could cosider a ifiite sum of a expressio i a variable. If we could determie covergece properties of such series, this would allow us to make coclusios about ifiitely may series with just umbers simultaeously (by pluggig umbers i for the variable). We start with a defiitio. Defiitio 1.1. A power series is a series of the form c x =0 where x is a variable ad the c s are called the coefficiets of the series (ote that this series starts from 0, so there is a costat term to this polyomial). More geerally, a series of the form c (x a) =0 is called a power series cetered at a. If we are give a power series, it does ot immediately make sese to talk about covergece because x is a variable. However, if we choose a value of x, say a, the the power series becomes a regular series c a ad we ca asks questios about its covergece etc. A power series may coverge for some values of x ad diverge for others, so it ca be viewed as a fuctio whose domai is the set of all umbers for which it coverges. This leads us to the followig two questios: Questio 1.2. For what values does a power series coverge? Questio 1.3. How do we fid what values a power series coverges for? Sice a power series is essetially a series, we ca use most of the same tests we developed for regular series - the oly differece is that we must treat x like a variable ad determie the values it coverges for. We cosider a couple of easy examples to illustrate. 1
2 Example 1.4. (i) For what values of x does the series x coverge? For ay fixed value of x, this sum is just a geometric series, ad we kow all the values for which this coverges. Specifically, x coverges if ad oly if x < 1. (ii) For what values of x does the series x! coverge? Applyig the ratio test, we have x +1 (+1)! x! = x +1 (!) x ( + 1)! = x +1 + 1 = 0 idepedet of the choice of x. Therefore, this series coverges for all values of x. (iii) For what values of x does the series!x coverge? Applyig the ratio test, we have ( + 1)!(x 2) (+1)!(x 2) = ( + 1)(x 2) which diverges if x 2. Thus this series coverges oly at x = 2. The last three examples suggest that there are three possibilities for the covergece of a power series - either all choices of x coverge, oly those o a certai iterval coverge or oly oe does. This is true i geeral as the followig result tells us. Result 1.5. For a give power series c (x a) there are oly three possibilities: (i) The series coverges oly whe x = a. (ii) The series coverges for all x. (iii) There is a positive umber R called the radius of covergece such that the series coverges if x a < R ad diverges if x a > R. If x a = R (so x = R a or x = a R), the the series may coverge or diverge.
If the radius of covergece is R ad the series is cetered aroud a, we say that the iterval (a R, a + R) is the iterval of covergece (where we iclude the edpoits if the series coverges at them). This leads us to the followig questio: Questio 1.6. How do we fid the iterval of covergece? I order to fid the iterval of covergece of a power series c (x a) we proceed as follows: (i) Observe that if we apply the ratio test, we are evaluatig the it c +1 (x a) +1 c x = (x a) c +1 c. Let L be this it. I order to coverge, the ratio test has to be less tha 1, so we eed to solve (x a) L < 1. Therefore, i order to coverge we must have (x a) 1. It follows that L the radius of covergece will be the reciprocal of the absolute values of the it (ii) This tells us that if the c +1. c L = c +1 c c (x a) coverges o the iterval (a 1/L, a+1/l). To fiish the problem off, we determie whether the series coverges at either of the edpoits, a 1/L ad a + 1/L. We illustrate with some examples: Example 1.7. Fid the radius ad iterval of covergece of the followig power series. (i) x 3 Applyig the ratio test we get x +1 (+1)3 +1 x 3 x = 3( + 1) = x 3 < 1, so x < 3. So the radius of covergece is 3. This meas that x 3 3
4 (ii) coverges i the iterval ( 3, 3). We ow eed to check the edpoits: x = 3 gives 3 3 = 1 which is the harmoic series, so diverges. x = 3 gives ( 3) = ( 1) 3 which is the alteratig harmoic series, so coverges. Therefore, the iterval of covergece is [ 3, 3). ( 2) (x + 3) Applyig the ratio test we get 2 +1 (x+3) +1 +1 2 (x+3) = 2 x + 3 = 2 x + 3 < 1, + 1 (iii) so x + 3 < 1/2. So the radius of covergece is 1/2. This meas that x 3 coverges i the iterval ( 3.5, 2.5). We ow eed to check the edpoits: x = 2.5 gives ( 2) ( 1 2 ) = ( 1) which coverges by the alteratig series test. x = 3.5 gives ( 2) ( 1 2 ) = 1 which diverges by the p series test. Thus the iterval of covergece is [ 2.5, 3.5). (x 4) 3 + 1 Applyig the ratio test we get (+1) x 4 +1 (+1) 3 +1 x 4 3 +1 = ( + 1)( 3 + 1) x + 4 (( + 1) 3 + 1) = x + 4 < 1.
So the radius of covergece is 1. This meas that (x 4) 3 + 1 coverges i the iterval (3, 5). We ow eed to check the edpoits: x = 3 gives ( 1) 3 + 1 which coverges by the alteratig series test. x = 5 gives 3 + 1 1 2 so coverges usig the p-series test ad the compariso test. Thus the iterval of covergece is [3, 5]. 5