The interval and radius of convergence of a power series

Transcription

1 The iterval ad radius of covergece of a power series Bro. David E. Brow, BYU Idaho Dept. of Mathematics. All rights reserved. Versio 1.1, of April, 014 Cotets 1 Itroductio 1 The ratio ad root tests 3 Examples 3 4 Exercises 6 5 Aswers 7 1 Itroductio Suppose give a power series 0 a (x x 0 ). Recall that the ceter of the power series is x 0, ad that we say that the series is cetered at x 0. Also recall that we say that the series coverges if it adds up to a real umber; 1 otherwise, we say that the series diverges. Of course, the series might coverge at some values of x ad diverge at others. The followig defiitio expresses this more formally. Defiitio 1.1. The above series coverges at x if ad oly if lim N N 0 a (x x 0 ) exists ad is fiite. At each x for which this limit exists, we make the defiitio a (x x 0 ) lim 0 N 0 N a (x x 0 ). If, for some x, lim N N 0 a (x x 0 ) does ot exist, we say that the series diverges at x. 1 is ot a real umber. 1

2 Every power series 0 a (x x 0 ) coverges at its ceter. (To see why, substitute x 0 for x i the formula of the series, ad simplify: 0 a (x 0 x 0 ) 0 a ) But this leaves us with the questio of whether the series coverges at some other x-value. If so, there s a whole iterval of x-values o which the series coverges. This fact is celebrated i the ext defiitio, which is here to remid you of the meaigs of the terms iterval of covergece ad radius of covergece: Defiitio 1.. Let 0 a (x x 0 ) be give. 1. Suppose the series oly coverges at x x 0. I this case, the iterval of covergece is {x 0 }, ad the radius of covergece is R 0.. Suppose series coverges at every x. I this case, the iterval of covergece is the etire umber lie, ad the radius of covergece is R. 3. Suppose the series coverges at each x i some iterval (a, b) that is ot the etire real lie. I this case, the iterval of covergece is (a, b) 3 ad the radius of covergece is R b a. This documet gives a method for calculatig the iterval ad radius of covergece of a power series. Sadly, o method ca do this for every power series. However, the method give here works well for the kids of series foud i a typical udergraduate differetial equatios course. The ratio ad root tests This sectio gives the promised method for fidig the iterval of covergece of a power series. For reasos that will become clear later, let A (x) a (x x 0 ), so that the power series is 0 A (x). The method ca make use of ay of a umber of tools, two of which are ow give. Tool 1: Ratio Test Theorem.1. Let ρ(x) lim A +1 (x) A (x), if the limit exists. The series coverges for all values of x for which ρ(x) < 1. If ρ(x) 1, the series may coverge or diverge at x. The series diverges for all values of x for which ρ(x) > 1. (Note that ρ is a fuctio of x, because both A +1 (x) ad A (x) are; after you take the limit, o loger appears, but x still ca. 4 ) Tool : The Root Test A Theorem.. Let ρ(x) lim (x), if the limit exists. The series coverges for all values of x for which ρ(x) < 1. The umber lie is kow variously as the real lie, the iterval (, ), the real umbers, or (most elegatly), R. 3 Note: Some authors defie the iterval of covergece a little differetly tha this. You see, the series may or may ot coverge at x 0 R or at x 0 + R, so some people like to isist that you iclude i your iterval of covergece ay edpoit at which the series coverges. We do t eed to bother with this. This semester we eed power series for differetial equatios, ad our iterest will be i ope itervals oly. So we re just usig the iterior of the iterval of covergece, which is (x 0 R, x 0 +R), whether the iterval of covergece actually icludes ay of its edpoits or ot. 4 You might be cocered about possibly dividig by 0. If ay of the terms i the series are 0, we ca throw them out ad reame the rest of the terms. Doig so will ot chage whether the series coverges, ad if the series coverges, throwig out the terms that equal zero will ot chage the sum. Page

3 If ρ(x) 1, the series may coverge or diverge at x. The series diverges for all values of x for which ρ(x) > 1. (Agai, ρ is a fuctio of x, because A (x) is; after you take the limit, o loger appears, but x still ca.) Note that the bulleted items are the same for the root test as for the ratio test. The Method: The method cosists of the followig steps: 1. Calculate ρ(x), whether by usig the Root Test or the Ratio Test.. Solve the iequality ρ(x) < 1 for x, ad you have the iterval of covergece. 3. Determie the radius of covergece, based o the iterval of covergece: If the iterval of covergece is {x 0 }, the radius of covergece is R 0. If the iterval of covergece is the etire real lie, the radius of covergece is R. Otherwise, the iterval of covergece is the iterval (a, b). I this case, the radius of covergece is r b a. The use of this method is illustrated i the ext sectio. There are may other tools for fidig the iterval ad radius of covergece of a power series. You ca fid some of them i just about ay advaced calculus text. Those give above are two of the most basic oes, but they are versatile eough for our eeds this semester. 3 Examples Example 3.1. Calculate the iterval ad radius of covergece of 0 (x 3). Usig the ratio test: The geeral term of the series is A (x 3), so A +1 (x) (x 3) +1. The ρ(x) lim A +1 (x) A (x) lim (x 3) +1 (x 3) lim (x 3) (x 3) (x 3) lim x 3 x 3. If we set ρ(x) < 1, we get x 3 < 1. Solvig this iequality 5 yields < x < 4, so the iterval of covergece is (, 4). The radius of covergece is R 4 1. Usig the root test: Agai, A (x) (x 3), for all. So ρ(x) lim A (x) lim (x 3) lim x 3 lim x 3 x 3, with the same results as before: You ll ed up solvig x 3 < 1 ad gettig (, 4) for the iterval of covergece ad 1 for the radius of covergece. Commet: I this example, we eded up solvig the same iequality whe we used the root test as whe we used the ratio test. Sometimes, you may thik the ratio test gives you a differet iequality to solve tha the root test. Not to worry: Whe both tests work, they do have to give you the same iterval of covergece (ad therefore the same radius of covergece). 5 Do t remember how to solve such a iequality? Well, recall that x 3 is the distace from 0 to x 3 o the umber lie. Therefore, to say x 3 < 1 is to say that x 3 is less that oe uit from 0 o the umber lie. This meas that x 3 has to be betwee 1 ad 1, that is, 1 < x 3 < 1. Now solve, by addig 3 throughout, to get that < x < 4. Page 3

4 Example 3.. Calculate the iterval ad radius of covergece of x 0!. Usig the ratio test: This time, A (x) x!, so A +1(x) x(+1) ( + 1)!, ad A +1 (x) A (x) x (+1) ( + 1)! x!x+ ( + 1)!x 1 3 ( 1) x x x 1 3 ( 1) ( + 1) x + 1.! x Therefore, ρ(x) lim 0. Um, we re supposed to solve ρ(x) < 1, which is 0 < 1, which is + 1 true o matter what the value of x is! I other words, the series coverges for all x. That meas the iterval of covergece is (, ). Naturally, the radius of covergece is R. A Usig the root test: (x) x 1/ x! which looks challegig. What could! possibly! tur out to be? Aswer: A mess. Do t use the root test whe you have factorials i your series! Example 3.3. Calculate the iterval ad radius of covergece of 0 (x) ( ) +1 Usig the ratio test: ρ(x) lim ( + 1)x (x) lim ( + 1) +1 x who-kows-what!? Hmm... If you thik of ( + 1) +1 as a polyomial i, its degree ought to be + 1, which ought to imply ( + 1) +1 as. The algebra we d have to do to verify this is messy ad complicated. This is a example for which the ratio test is ot the best tool to use. Usig the root test: ρ(x) (x) x lim lim lim x, which is 0 if x 0 ad otherwise. So the iterval of covergece of this series is {0}, with radius of covergece R 0. Example 3.4. Calculate the iterval ad radius of covergece of 1 ( ) x ( Usig the ratio test: A (x) 1 + ) 1 ( ) + 1 x x ; after the previous item, the ratio test does t seem very attractive here. Let s try the root test, istead. ( ) Usig the root test: + 1 x x, so ρ(x) lim x x. Solvig x < 1 tells us that the iterval of covergece is ( 1, 1), so the radius of covergece must be 1. ( (Commet: It may be of iterest to you to kow that lim e.7188, the base of the ) atural logarithm fuctio.) Example 3.5. Calculate the iterval ad radius of covergece of (3x) 0! Usig the ratio test: A (x) (3x), so A +1(x)! A (x) (3x) +1 ( + 1)! (3x) 3x 0 as, with or without + 1! the absolute value bars. Oce agai, there is o restrictio o x; the iterval of covergece is all of R, ad R. Page 4

5 Usig the root test: Actually, do t use the root test o this oe, because doig so would ivolve dealig with!, which is otrivial to do. Example 3.6. This example is a little more advaced ad ivolves somewhat more complicated algebra tha the other examples. You may skip it, the first time you read this documet. Calculate the iterval ad radius of covergece of 1 ( 1)+1 x (x 1) +1. (Note that this is t actually a power series. However, the method for fidig the iterval ad radius of covergece still works, whether you use the root test or the ratio test.) Usig the ratio test: (The algebra is a little hairy, so I ll put plety of steps i. but you eed to get to the poit where you ca skip some of these steps.) A (x) ( 1) +1 x (x 1) +1, so Therefore, ρ(x) lim A +1 (x) A (x) ( 1) + x+1 (x 1) + ( + 1) +1 ( 1) +1 x (x 1) +1 ( 1)+ ( 1) +1 x +1 (x 1) + ( + 1) +1 x (x 1) +1 x+1 x x x x x(x 1) ( + 1) lim (x 1)+ (x 1) (x 1)(x 1)+1 (x 1) +1 x (x 1) x(x 1). ( + 1) ( + 1) x(x 1) 1 x(x 1) lim + 1 x(x 1). x(x 1) We eed to solve the iequality < 1. Do a little algebra to get < x x < ad ote that this is really a system of two quadratic iequalities to solve simultaeously: { x x + > 0 x x < 0. Let s solve this system by solvig each iequality separately, ad takig the x-values that solve both iequalities. The usual recommedatio for solvig a polyomial iequality is based o the followig: Thik of the polyomial as beig the formula for a fuctio, ad cosider its graph i the x,y-plae. The zeros of the polyomial are boudaries betwee itervals o which the y-values of the polyomial are positive (above the x-axis) ad itervals o which they are egative (below the x-axis). You fid the zeros of your polyomial, ad the use test poits (x-values that are ot zeros of the polyomial) to determie the sigs of the polyomials all alog the umber lie. Let s follow this course, startig with, say, the secod iequality i our system above. The zeros of x x are 1 ad. A coveiet test poit is x 0; it s betwee the zeros of the polyomial, ad it s easy to work with. Whe x 0, y. Now, we all kow what the graph of a parabola looks like. If the y-values are egative betwee the two zeros the they are positive everywhere else. So the solutio set of the iequality x x < 0 is ( 1, ). Page 5

6 Now for the first iequality i our little system: The zeros of x x+ are complex. Complex umbers do ot appear aywhere o the x-axis, so the graph of this polyomial is either always above the x-axis or always below the x-axis. We ote with satisfactio that the y-itercept (which is ) is positive, so x x + > 0 for all x; i.e., every real umber is a solutio of x x + > 0, our first iequality. Fially, we eed the set of x-values for which both iequalities are true. These are the x s i the iterval ( 1, ). This iterval is the iterval of covergece of the give series, ad R ( ( 1) ) / 3/. Usig the root test: ( 1)+1 x (x 1) +1 x x 1 (+1)/. As, this quatity goes to x x x(x 1) x x, which is therefore ρ(x). We eed to solve the iequality < 1, which we did for the first half of this example. The rest of this half of this example goes exactly the same way as i the first half. Example 3.7. Calculate the iterval ad radius of covergece of 0 (si )x (si( + 1))x +1 Usig the ratio test: ρ(x) lim (si )x lim si( + 1) x, which is problematic. You si see, as, there are may times whe si is close to 0. At these times, si( + 1) is ot close to 0, so the fractio si( + 1)/ si is very large. This prevets the limit from existig, whe x 0. Of course, if x 0, the sum is 0 ad therefore coverges. So the iterval of covergece cotais oly the poit 0, ad R 0. (si Usig the root test: ρ(x) lim )x lim si x. Agai, this limit does ot exist for ozero x, this time because si x oscillates forever. The values of si vary ifiitely may times from just uder 1 to just over 0, as gets larger ad larger. So, agai, the iterval of covergece is {0} ad the radius of covergece is R 0. 4 Exercises Calculate the iterval ad radius of covergece of each of the followig series. Exercise (x) Exercise 4.. x +1 0 ( + 1)! Exercise (x + ) 0 Exercise 4.4. (x) 1 3 Exercise 4.5. ( x) 0! Exercise ( 1) x 4+3 (4 + 3)( + 1)! Page 6

7 5 Aswers If you have questios about these exercises or their aswers, please cotact me. Exercise 4.1: Use either the ratio test or the root test. The iterval of covergece is ( 1/, 1/), ad R 1/. Exercise 4.: Use the ratio test. The iterval of covergece is R, ad R. Exercise 4.3: Use either the ratio test or the root test. The iterval of covergece is (5/3, 7/3), ad R 1/3. Exercise 4.4: Use the root test. The iterval of covergece is {0}, ad R 0. Exercise 4.5: Use the ratio test. The iterval of covergece is R, ad R. Exercise 4.6: Use the ratio test. The iterval of covergece is R, ad R. Page 7