1、Series of Functions and Power Series1Overview2l Concepts of functional seriesl Power seriesl Tests for convergence of power seriesl Operations of power series Functional Series3A series of the form ,where is a function of x in a set A R,is called a series of functions(or functional series)defined in
2、 A.1()nnux ()nux if the series with constant terms converges,then we say that the series of functions is convergent at ,and is called a convergence point of the series of functions;if diverges,is called a divergence point.The set consisting of all the Convergence points is called the Convergence dom
3、ain 收敛域收敛域 of the series.0,xA 0()nux()nux 0 xx 0 x0()nux 0 xFunctional Series4()nux If D is the convergence domain of the series ,thenthe series has a sum .Hence the sum of the series is actually a function of x defined in D,called the sum function 和函数and denoted byIn this case,the series is said to
4、 exhibit everywhere convergence or pointwise convergence to in D.Generally,it is called simply convergence to in D.0 xD()S x1()(),nnuxS xxD ()nux 0()nux 0()S x()nux()S x()S xFunctional Series5Just as the series with constant terms,we can define the partial sum of series.Definition(Partial sum and Re
5、mainder)The quantity is called the partial sum部分和部分和 of the series ,while is called a remainder余项余项 of the series.1()()nnkkSxux 1()nnux 1()()()(),nnkk nRxS xSxuxxD It is easy to see that the necessary and sufficient condition for a series to be everywhere convergent in is that1()nnux Dlim()0lim()(),
6、nnnnRxSxS xxDorApplying the Definition6 Find the convergence domain of the series21sin.nnxn 0(,),x SolutionSince021sinnnxn is a convergence series,21sinnnxn the convergence domain of the series(,).isFinish.The Geometric Series7The sum function of the geometric series is The convergence domain of the
7、 given series is the interval ,centered at .1ax 11x 0 x Solution Discuss the convergence of the geometric series20,nnnaxaaxaxax where is a constant.Find its convergence domain and sum function.0a Finish.Applying the Definition8 Find the convergence domain and sum function of the series2321()()().nnx
8、xxxxxx 21(),nnnnSxxxxxxx SolutionSincefor lim()0,1;nnSxxwe have thatfor lim(),1;nnSxx for lim()1,1;nnSxx()nSxand the limit of does not exist for1.x Hence,the convergence domain of the given series is(1,1,01;()11.xS xx function is the sumFinish.Power Series9Definition (Power Series 幂级数幂级数)An expressi
9、on of the formis a power series centered at .An expression of the formis a power series centered at .The term is the nth term;the number a is the center.20120()()()()nnnnncxacc xacxacxa xa 20120nnnnnc xcc xc xc x 0 x ()nncxa A series whose general term is a power function is called a power series.Po
10、wer Series10Note:This is typical behavior,as we soon see.A power series converges for all x,converges on a finite interval with the same center as the series,or converges only at the center itself.The geometric seriesis a power series centered at .201nnnxxxx 0 x 11x 11x 0 x The geometric series conv
11、erges to on the interval ,also centered at .Power Series11 The power series21111(2)(2)(2)242nnxxx (1)is centered at with coefficients 02x 012c1,1/2,1/4,.,(1/2)nnccc 212x The series converges for or .04xThe sum is112,2112xrx so22(2)(2)11(2),04242nnxxxxx This is a geometric series with first term 1 an
12、d ratio .22xr Power Series1201222()11()1(2)222113()1(2)(2)32424P xxP xxxxP xxx Series(1)generates useful polynomial approximations of for values of x near 2.()2/f xx 21111(2)(2)(2)242nnxxx (1)Convergence of Power Series13 the sequence is bounded,that is,such that1nnc x0MProof of Part(1)Since converg
13、es,.11nnnc x 1lim0nnnc x Thus 1|N.nnc xMn for allConsider the power series .(1)If it converges at ,then it must converge absolutely in the interval ;(2)If it diverges at ,then it must diverge in .0nnnc x 1x10 x 1|xx 2x20 x 2|xx Convergence of Power Series14Proof of Part(1)(continued)Thus,111|,nnnnnn
14、xxc xc xMxx while the geometric series converges provided 11nnxx 1|xx because11.xx According to the comparison test,the power series converges for1|.xx Part(2)can be proved by contradiction.We do not mention here.The Radius and Interval of Convergence151.There is a positive R such that the series di
15、verges forbut converges for .The series may or may not converge at either of the endpoints and .2.The series converges for every .3.The series converges at and diverges elsewhere .|xaR|xaRxaRxaR()x Rxa(0)R 0()nnncxa There are three possibilities for with respect to convergence.The Radius and Interva
16、l of Convergence16The number R is the radius of convergence收敛半径收敛半径,the open interval(a-R,a+R)is called the convergence interval收敛区间收敛区间,and the set of all values of x for which the series converges is the convergence domain收敛域收敛域.The radius of convergence completely determines the domain of converg
17、ence if R is either zero or infinite.For ,however,there remains the question of what happensat the endpoints of the interval.0R The next example illustrates how to find the domain of convergence.DefinitionNote:Finding the Domain of Convergence17 For what values of x do the following power series con
18、verge?(d)231!12!3!nnn xxxx (a)2311(1)23nnnxxxxn (b)213511(1)2135nnnxxxxn (c)2311!2!3!nnxxxxn Finding the Domain of Convergence18At ,we 1x get the alternating harmonic series ,1111234At ,we get ,the negative of the harmonic1111234 1x Solution Apply the Ratio Test to the series ,where|nu nuis the nth
19、term of the series in question.(a)2311(1)23nnnxxxxn 1|.1nnunxxun|1x It diverges if The series converges absolutely for .|1x because the nth term does not converge to zero.series;it diverges.Series(a)converges for and diverges elsewhere.11x which converges.Finding the Domain of Convergence19Solution(
20、continued)22121|.21nnunxxun 21x It diverges if The series converges absolutely for .21x because the nth term does not converge to zero.At ,we get1x At ,we get ,it is again convergent.1111357 1x the series ,which converges by the1111357Series(b)converges for and diverges elsewhere.11x (b)213511(1)213
21、5nnnxxxxn Leibnizs Theorem.x1 1OFinding the Domain of Convergence20Solution(continued)(c)2311!2!3!nnxxxxn 11!|0(1)!1nnnnuxnxxunxn for everyThis series converges absolutely for all x.xOFinding the Domain of Convergence 21Solution(continued)11(1)!(1)|0!nnnnunxnxxun x for everyThis series diverges for
22、all values of x except x=0.xO(d)231!12!3!nnn xxxx Steps for Finding the Domain of Convergence22Step 1:Use the Ratio Test(or nth-Root Test)to find the intervalwhere the series converges absolutely.Ordinarily,the is an open interval|.xaRaRxaRorStep 2:If the interval of absolute convergence is finite,t
23、est forconvergence or divergence at,as in previous example(a)and(b).Use a Comparison Test,the Integral Test,or the Alternating Series Test.Step 3:If the interval of absolute convergence is the series diverges for (it does not even converge conditionally),because the nth term does not approach zero f
24、or those value of x.,aRxaR|xaRFinding the Domain of Convergence23Theorem(Page 318)If1limnnncc 1,nnc c 0(),nnncxa 10;0;0.R whereis the coefficients of the power seriesthen the radius of convergence is given by Finding the Domain of Convergence24 Find the radius of convergence,convergence interval and
25、 convergence domain of the following power series:(a)2301(1)121 262056nnnxxxxn (b)2213(2)13(1)19(1)(1)(1)23nnnnxxxxn (c)224622232014 44(1)4344nnnxxxxn Finding the Domain of Convergence1limnnncc 25Solution(a)2301(1)121262056nnnxxxxn Let 1(1),(21)2nnncn 1111(1)(23)2limlim1(1)(21)2nnnnnnnncncn 21lim2(2
26、3)nnn 1.2 and the convergenceSo,the radius of convergence of the series(a)is R=2At x=2,the series becomes (2,2).interval of the series(a)iswhich converges by the Leibnizs01(1),21nnn Hence,the convergence domain of the given series(a)is(2,2.Theorem.At x=-2,the series becomes it is divergent.01,21nn F
27、inding the Domain of Convergence26Solution(continued)1limnnncc (b)13(2)(1)nnnnxn Let 3(2),nnncn we have that 23(2)3lim3.2(1)13nnnnn 1113(2)1limlim3(2)nnnnnnnncncn So,the radius of convergence of the series(b)is R1,3 and the convergence or 12 41(,).33 3x interval of the series(a)isFinding the Domain
28、of Convergence2702(1)3.nnnn At the series becomes 2,3x Solution(continued)(b)13(2)(1)nnnnxn The series0213,nnn At the series becomes 4,3x 0(1)nnn Thus,the series(b)converges at 2.3x and the series02()3nnn are convergent.it is divergent.Hence,the convergence domain of the given series(b)is24,).33Find
29、ing the Domain of Convergence28Solution(continued)(c)2204(1)nnnxn The series contains only even order powers.20.4(1)nnntn Let 2,tx The series(c)becomesLet 2(),4(1)nnntu tn The series converges absolutely for 1()()lim.()4nnntuttu t and diverges for 4t Therefore,4.t the series(c)converges absolutely f
30、or and diverges for 2x 2.x and the convergenceradius of convergence of the series(a)is R2,(2,2).interval of the series(c)isThe201,(1)nn At the series become 2,x it is convergent.Hence,the convergence domain of the given series(c)is 2,2.Operations of Power Series29If and converge absolutely for and ,
31、respectively,and .0()nnnB xb x 0()nnnA xa x (1)The linear combination converges in the common convergence interval and2|xR 1|xR 12min,RR R 000()()()(,R).nnnnnnnnnna xb xabxA xB x (,)R R 00nnnnnna xb x(2)The product series also converges in and0111 100.nnnnnnkn kkca ba baba ba b (,)R R 000,nnnnnnnnna
32、 xb xc x where Operations of Power Series12min,RRR 30The radius of consequence of the sum and product of two powerseries may be larger than the radii and ,where .R2R1R12RR Note:It is easy to see that the series and areboth converge with ,but the radii of their sumis .0(12)nnnx 0(12)nnnx 1212RR02nnx
33、1R In this case,of course,the Addition Theorem holds.ExampleApplying the Multiplication Theorem31 Multiply the geometric series2011|1,1nnnxxxxxx forby itself to get a power series for ,for .21(1)x|1x SolutionLet20201()111()11nnnnnnnnA xa xxxxxB xb xxxxx and01101 termsnnnkn knnca ba ba ba b 1 ones111
34、1.nn Applying the Multiplication Theorem32Then,by the Series Multiplication Theorem,0()()nnnA xB xc x 0(1)nnnx 231234(1)nxxxnx is the series of .21(1)x The series all converge absolutely for|1.x Solution(continued)Finish.Multiply the geometric series2011|1,1nnnxxxxxx forby itself to get a power seri
35、es for ,for .21(1)x|1x Operations of Power Series33 Suppose converges for|x|0)and its sum function is S(x).Then(1)S(x)is continuous in the convergence interval(R,R).(2)S(x)is differential in the convergence interval(R,R),and the derivative be obtained term by term,that is,(1)The power series(1)has t
36、he same radius of convergence as the original series.(3)S(x)is integrable in the convergence interval(R,R),and the integration can be obtained term by term,that is,x (R,R),we have (2)The power series(2)has the same radius of convergence as the original series.0nnnc x 101()nnnnnnS xc xnc x 10000()1nx
37、tnnnnnxS t dtc t dtcn Applying Term-by-Term Differentiation0(1),11nnnxx 34 Find series for and if()fx()fx 1()1f xx 2341nxxxxx ,11.x Solution2232()2612(1)(1)nfxxxn nxx 2121()123(1)nfxxxnxx 22(1),11nnn nxx Finish.A series for arctan x,-1x135 Identify the function35(),11.35xxf xxx Solution We differentiate the original series term by term and get246()1,11.fxxxxx This is a geometric series with first term 1 and ratio ,so2x We can now integrate to get21()1fxx 2211().1()1fxxx 2()()arctan.1dxf xfx dxxCx The series for is zero when ,so .()f x0 x 0C 35arctan,11.35xxxxxHenceFinish.
