11.8 Power Series. Recall the geometric series. (1) x n = 1+x+x 2 + +x n +

Transcription

1 Power Series Recall the geometric series (1) x n 1+x+x 2 + +x n + n As we saw in section 11.2, the series (1) diverges if the common ratio x > 1 and converges if x < 1. In fact, for all x ( 1,1) this series has the closed form representation (2) n x n 1 1 x, 1 < x < 1 Also, the series is clearly divergent if x 1 since Finally, for x 1 we have (3) ( 1) n+1 + which is also divergent since the terms do not approach. We ll return to this case later.

2 Power Series and Convergence Equation (1) is an example of a power series. Formally, we have Definition. Power Series, Center, and Coefficients A power series about x a is a series of the form (4) c n (x a) n c +c 1 (x a)+c 2 (x a) 2 + +c n (x a) n + n The center a and coefficients c,c 1,...,c n,... are constants. Remark. (i) For many examples the center is chosen to be. (ii) Notice that every power series converges (trivially) at its center. The question is, for what other x-values does the series in (4) converge?. For example, the series in (1) is a power series centered at x and the coefficients are c 1,c 1 1,...,c n 1,... That is, x n 1 (x ) n n n

3 Example 1. Testing for Convergence For which values of x does the following series converge? (2x) n 2 n x n n n Notice that the center is and the coefficients are c n 2 n. We try the Ratio Test (Actually, the Root Test is a better choice here!). Let a n (2x) n. Then ρ lim n lim n 2x a n+1 a n (2x) n+1 (2x) n It follows that the series converges absolutely if 2x < 1 Notice that, by the Ratio Test, this series diverges for all x > 1/2. In general, the end points must be always be explicitly checked. In this example, the series also diverges at ±1/2 as one can easily verify. The interval ( 1/2, 1/2) is called the interval of convergence.

4 Example 2. Testing for Convergence (cont.) For which values of x does the following series converge? (3x 5) n n n1 In this example the center is a 5/3 and c n 3 n / n. Again we try the Ratio Test. Let a n (3x 5) n / n. Then ρ lim a n+1 n a n lim (3x 5) n+1 n n n+1 (3x 5) n 3x 5 It follows that the series converges absolutely if ρ < 1, that is if 1 < 3x 5 < 1 x (4/3,2) or x (a 1/3,a+1/3) I Remark. Once again, I is called the interval of convergence. Also, the number, 1/3, is called the radius of convergence. It is not difficult to verify the the series diverges for x 2 and x < 4/3. Also, notice that the series converges conditionally at x 4/3 by Leibniz s Theorem.

5 Theorem 1. The Convergence Theorem for Power Series If the power series a n x n converges for x c, then the series converges absolutely for all x with x < c. If the series diverges for some x d, then it diverges for all x with x > d. Corollary 2. Corollary to Theorem 1 The convergence of the series a n (x a) n has only one of three possibilities. 1. There is a positive number R (called the radius of convergence) such that the series diverges for all x with x a > R but converges absolutely for all x with x a < R. The series must be explicitly tested at the end points x a±r. 2. The series converges absolutely for all x. (In this case, R.) 3. The series converges at x a only and diverges elsewhere. (In this case, R.) Remark. If c n (x a) n converges for x (a R,a+R), R > then the power series defines a function f: (5) f(x) c n (x a) n, a R < x < a+r n

6 Example 3. Geometric Series Earlier we saw that the series n xn converged absolutely on the interval ( 1,1). So for all x ( 1,1) this series defines a function, say f. We have f(x) x n, 1 < x < 1 n In fact, f(x) has the closed form. f(x) 1 1 x, 1 < x < 1 Example 4. Find the interval of convergence for the power series (6) 1 x+x 2 x 3 + ( 1) n x n It is easy to see (by the ratio test) that the series in (6) has the same interval of convergence as the series in the previous example. Also 1 1+x 1 1 ( x) n 1 x+x 2 x 3 +x 4 + ( 1) n x n n for x ( 1,1). It follows that the series in (6) defines a function g with (7) g(x) ( 1) n x n 1 1+x, x ( 1,1) n

7 It is worth noting that 1 lim lim x 1 g(x) x 1 1+x 1/2 ( 1) n n Theorem 3. Term-by-Term Differentiation Theorem Suppose that (5) holds. That is, suppose f(x) c n (x a) n, a R < x < a+r n Then f has derivatives of all orders inside the interval of convergence. In fact, we differentiate term-by-term. That is, f (x) n c n (x a) n 1 f (x) n1 n(n 1) c n (x a) n 2, n2 and so on. Each of the derived series converging at each point in (a R,a+R). Example 5. Find the power series expansion of each of the following about a. What is the interval of convergence? (a) 1 1+x 2

8 (b) x (1+x 2 ) 2 As we saw in Example 4, 1 1+x 1 x+x2 x 3 +x 4 + The substitution x x 2 to gives 1 1+x 2 1 x2 +(x 2 ) 2 (x 2 ) 3 +(x 2 ) x 2 +x 4 x 6 +x 8 + ( 1) n+1 x 2n n and this series converges for all 1 < x 2 < 1. That is, for 1 < x < 1. For part (b) we let f(x) 1/(1+x 2 ). Then So by part (a) f (x) 2x (1+x 2 ) 2 x (1+x 2 ) f (x) d ( 1 x 2 +x 4 x 6 +x + ) 8 dx ( 2x+4x 3 6x 5 +8x + ) 7

9 by Theorem 3. Since the power series in (a) converges for all 1 < x < 1, the series in (b) must have the same interval of convergence. Example 6. Let (8) h(x) ( 1) n+1 nx n n1 (a) Find the radius and interval of convergence. In other words, find the domain of h It follows from a straight-forward application of the ratio test that the series converges (absolutely) for all x < 1. Hence the series in (8) defines a differentiable function on ( 1, 1). (b) Show that lim x 1 h(x) 1/4. Let g(x) ( 1) n+1 x n n 1+x x 2 +x x, 1 < x < 1

10 So by Theorem 3 Now observe that g (x) 1 2x+3x 2 4x 3 + h(x) x ( 1) n+1 nx n 1 n1 1 (1+x) 2, 1 < x < 1 ( 1) n+1 nx n n1 ( 1) n+1 nx n 1 n1 xg (x) x (1+x) 2, 1 < x < 1 It follows that (9) lim lim x 1 h(x) x 1 x (1+x) 2 1/4, (Why?) Remark. In section 11.6 we noted that the alternating series ( 1) n+1 n n1 diverged by the n th term test. It follows that h(1) ( 1) n+1 n(1) n ( 1) n+1 n n1 n1

11 does not exist. So h is not (left) continuous at x 1 even though it has a (left-hand) limit there. To elaborate further, let k(x) x/(1+x) 2. Then k(x) is defined for all real x 1 but, the function h(x), given in (8), is defined only for x ( 1,1). In particular, h(x) k(x). On the other hand, if we restrict ourselves to x ( 1,1), then the two functions are equal. We used this fact to evaluate the limit in (9). Summability Theory The previous example touches on a subject called Summability Theory. A series n a n is said to be (Abel) summable (to L) if (i) The power series n a nx n converges for all x < 1 and, (ii) f(x) n a nx n L as x 1. In the last example we showed that the divergent series ( 1) n+1 n is Abel summable to 1/4. (Compare with Example ) The fact the convergent series are necessarily (Abel) summable was proven by N. H. Abel in Theorem 4. (Abel) Suppose that n a n converges to a real number, say L. Then the (power) series n a nx n converges for all x ( 1,1) and (1) lim a n x n a n L x 1 n n

12 Before proceeding with the proof, it is useful to rewrite the Abel sum in a more convenient form. As usual, let s n n j a j. Observe that a n x n a + a n x n n s + s + s + n1 (s n s n 1 ) x n }{{} a n s n x n s n 1 x n n1 n1 s n x n n1 s n x n n n1 s n x n+1 n s n x n+1 n s n (x n x n+1 ) n (1 x) s n x n n Proof. We leave it as an exercise to show that (11) f(x) a n x n converges for all x ( 1,1). n To prove (ii) above, we proceed in much the same way as we did in the proof of Cesro s Theorem (section 11.2).

13 Let ε >. Now choose N so large that s n L < ε whenever n N. Then f(x) L (1 x) s n x n L n (1 x) s n x n L(1 x) x n n (1 x) s n x n (1 x) (1 x) (1 x) < (1 x)k < (1 x)k n s n L x n n N 1 n s n L }{{} bounded N 1 n N 1 n n Lx n n x n +(1 x) x n +(1 x)ε 1+(1 x)ε ε (1 x)kn +(1 x) 1 x (1 x)kn +ε Now let x 1 and the result follows. nn n x n x n nn s n L }{{} <ε x n

14 It is interesting to compare the Abel sum with the Cesro sum (from section 11.2). Let p(j) 1 j/n, j,1,2...n 1. Given a (formal) series n a n, its Cesro sum is defined by σ n n 1 j ( 1 j n a p()+ ) n 1 j1 a j a j p(j) n 1 s p()+ (s j s j 1 )p(j) s p()+ 1 n n 1 j n 1 j n 1 j j1 n 1 j1 s j p(j) s j p(j) n 1 j s j (p(j) p(j +1)) s j ( 1 j n n 1 j s j 1 n 1 j 1 n 1 (s j 1) j n 1 j1 s j 1 p(j) s j p(j +1), (since p(n) ) ) j n

15 And its Abel sum is given by a n x n (1 x) n s n x n n 1 n xn n s n x n Comparing the final form of both sums, we see that Cesro and Abel sums represent a sort of averaging process. Tauberian Theory So we have an interesting collection of results concerning summability. Start with a convergent series, say n a n. Then immediately we can conclude that this series is both Cesro and Abel summable (by Theorems and 1.7.4). Tauberian Theory poses the converse question. We ll focus on Abel summable series. Suppose that n a n is Abel summable to L. That is, suppose that a n x n converges for x ( 1,1) and n f(x) a n x n L as n x 1 Is it necessarily true that the series n a n is convergent. By Example 6 the answer must be No. For an easier example, recall that is Cesro (and hence Abel) summable to 1/2. But, as we saw in section 11.2, this series does not converge (to anything).

16 The Tauberian Question: Can we add conditions to the coefficients a n of a summable series n a n series that guarantee that the series actually converges? We have Theorem 5. (Tauber ) Suppose that the series a n is Abel summable and na n. Then an converges. and Theorem 6. (Hardy & Landau - 191) Suppose the series a n is Cesro summable and let C >. Then either the two-sided condition na n C (Hardy) or the one-sided condition na n C (Landau) imply that the series a n converges.

17 Theorem 7. Term-by-Term Integration Theorem Suppose that (5) holds. That is, suppose f(x) c n (x a) n n converges for a R < x < a+r. Then (x a) n+1 c n n+1 n converges for a R < x < a+r and f(x)dx for a R < x < a+r n c n (x a) n+1 n+1 Example 7. Find the power series expansion of f(x) ln(1+x). Also, find the interval of convergence. Observe that f (x) 1 1+x 1 x+x 2 x 3 +, 1 < x < 1 It follows by Theorem 7 that f(x) ln(1+x) C +x x2 2 + x3 3 x4 4 +, 1 < x < 1

18 The initial condition f() ln1 C. Hence ln(1+x) ( 1) n+1xn n, 1 < x < 1 n1 In section 1.6 we showed that the Alternating Harmonic Series converged, say to L. So, as an added benefit, we can apply Abel s Theorem (Theorem 4) to conclude that ( 1) n+1 L lim ( 1) n+1xn n x 1 n lim ln2 x 1 ln(1+x) n1 n1 Example 8. about x. Find the power series expansion of ˆ sinx x dx In the next section we will learn that the power series expansion for the sine function about x is It follows that sinx x x3 3! + x5 5! sinx x n +( 1)n x2n+1 (2n+1)! + ( 1) n x2n+1 (2n+1)!, x R 1 x2 3! + x4 5! +( 1)n x 2n (2n+1)! +

19 So by Theorem 7, ˆ sinx x dx x x3 3(3!) x5 5(5!) + x7 7(7!) +( 1) n x 2n+1 (2n+1)(2n+1)! + for n,1,2,... and the series converges for all x R.

20 Example 9. In section 8.7 we proved that (12) sint t dt ( sint t ) 2 dt π 2 The proof was long (7 pages) and difficult. Now that you know a bit about Summability and Tauberian theory, we can try for an easier(?) proof. Let (13) F(s) ( ) 2 sint e st dt, s > t F(s) is called the Laplace Transform L of (sint/t) 2. Observe that the improper integral in (13) converges absolutely for all s >. In fact, (13) defines a (differentiable) function for all s >. The Laplace Transform will be studied in more detail in Math 235 where it will be used to transform certain differential equations into algebraic equations (which, in practice, are easier to solve). It turns out that (14) F(s) L{(sint/t) 2 } π 2 tan 1 s 2 s 4 ln ( 1+ 4 s 2 ), s > (See Proposition 1 below.) Now it is easy to see that (15) lim lim s +F(s) s + Exercise - Verify (15). { π s 2 tan 1 2 s (1+ 4 ln 4s )} 2 π/2

21 It seems logical to conclude that ( ) 2 sint ( ) 2 sint (16) dt ˆ e t() dt t t ( ) 2 sint (17)? lim e st dt s + t lim π/2 s +F(s) Unfortunately, we have been misled by such observations before. In Example 4, we saw that ( 1) n+1 ( 1) n+1 (1) n n n lim x 1 1/2 n ( 1) n+1 x n lim x x We need a condition that guarantees that the equality in (17) is legitimate. Fortunately, such a condition does exist. Theorem 8. Suppose that the improper integral e st f(t)dt converges for s > and that the function F(s) e st f(t)dt satisfies (18) lim s +F(s) A R (In such cases we say that f(t)dt as Abel summable to A.) Then the (Tauberian) condition n+1 (19) lim t f(t) dt n ˆ n

22 guarantees that the improper integral f(t)dt converges to A. That is, ˆ b (2) f(t)dt lim f(t)dt A b Now observe that for t > ( ) 2 sint t sin2 t 1 t t t It follows that ˆ n+1 ( ) 2 sint t dt t n ˆ n+1 n 1 t dt ln(n+1) lnn ln ( 1+ 1 ) n And the last expression approaches as n. So by the Squeeze Law, condition (19) is satisfied with f(t) (sint/t) 2. Now (13), (14) and (15) imply that (sint/t)2 dt is Abel summable to π/2. It follows by the above theorem that (sint/t) 2 dt π 2 Proof. (of the theorem) Let w(u) u f(t)dt. We will prove that (21) w(u) F(1/u) as u Notice that w(u) F(1/u) (22) ˆ u ˆ u f(t)dt (1 e t/u )f(t)dt e t/u f(t)dt u e t/u f(t)dt

23 In extra credit problem #2, we saw that 1+x e x for all real x. Thus for any t > and positive integer N, 1 e t/n t/n. Now let n+1 a n n t f(t) dt. Then ˆ N N ˆ n+1 (1 e t/u ) f(t) dt (1 e t/n ) f(t) dt n 1 N 1 N n N n N n ˆ n+1 n a n t f(t) dt But the last member goes to as N. To see this let ε >. By (19) we can choose N large so that a n a n < ε whenever n N. Thus ( N+m+1 1 N ) 1 N+m+1 a n a n + a n N +m+1 N +m+1 n < n 1 N +m+1 S N + nn+1 N+m+1 1 ε N +m+1 nn+1 1 N +m+1 S m N + N +m+1 ε 1 N +m+1 S N +ε Now the first member in the last expression vanishes as m. Thus (23) ˆ N (1 e t/u ) f(t) dt as N Now let N and a n be as defined above and observe that for t n N,

24 we have n/n t/n n/n t/n e n/n e t/n. Thus ˆ n+1 e t/n f(t) dt e t/n f(t) dt N nn 1 N 1 N < ε N n nn e n/nˆ n+1 n e n/n a n nn (e 1/N ) n nn t f(t) dt ε N 1 e 1/N ε 1/N e 1 e 1/N But the last expression approaches ε/e as N. Exercise - Verify this. Thus (24) Now by (22) N w(n) F(1/N) e 1 e t/n f(t) dt as ˆ N ˆ N (1 e t/u )f(t)dt (1 e t/u ) f(t) dt+ N N N e t/n f(t) dt e t/n f(t) dt which approaches as N as we saw in (23) and (24). Exercise - Prove that (21) is an immediate consequence.

25 It now follows that as desired. f(t)dt lim u w(u) lim u F(1/u) A

26 The following idea will be useful below (cf. section 7.7). Let f be a real-valued function. We define it s positive and negative parts by the rules (25) f + (t) f(t) +f(t) 2 and f (t) f(t) f(t) 2 Observe that f + (t) and f (t) for all t Dom(f), and (26) f + (t) f (t) f(t) +f(t) 2 f(t) f(t) 2 f(t) 1 y f + (t) y f (t) π Also, recall from section 8.2 (see Example 3) that two applications of integration by parts led to the convenient formula ˆ (27) e ax sinbxdx eax a 2 +b (asinbx bcosbx) 2 Similarly, we have ˆ (28) e ax cosbxdx eax a 2 +b (acosbx+bsinbx) 2

27 Differentiating Under the Integral Before we can prove (13) we need to introduce an old-fashioned technique. 1 Let f(s) sinstdt, s >. Then f is a well-defined function of s. In fact, f 1 (s) tcosstdt. To see this observe that by partial integration we have ˆ 1 ( cosst tcosstdt + tsinst ) t1 s 2 s t coss + sins 1 On the other hand, ˆs 2 s s 2 1 f(s) sinstdt cosst t1 s t Hence 1 s (1 coss) f (s) coss 1 s 2 + sins s Before we get too excited, suppose that g(s) sinst/tdt, s >. Then? g (s) cosstdt Clearly the right-hand side does not exist for any positive s. Apparently, there are some restrictions to this technique. Rather than develop a general theory, we will prove only what we need in the present context. Lemma 9. Suppose that the improper integrals e st f(t)dt, te st f(t)dt, and t 2 e st f(t)dt converge (absolutely) for each s >. Then F(s) e st f(t)dt is a differentiable function of s and (29) F (s) te st f(t)dt, s >

28 Proof. We need the following inequalities. 1+x e x, x R 1 x+ x2 2 e x, x > We have seen the first inequality before. We will prove the second in section Now let t >. Together the above inequalities imply (3) (31) t e ht 1 h t+ht 2 e ht 1 h Exercise - Verify (3) and (31). t+ht 2, h > t, h < Suppose first that f(t). Now if h > then (3) implies te st f(t)dt e st e ht 1 h e (s+h)t e st h f(t)dt f(t)dt ( t+ht 2 )e st f(t)dt te st f(t)dt+h t 2 e st f(t)dt But the last integral is finite (by assumption). Hence the Squeeze Law implies e (s+h)t e st lim f(t)dt te st f(t)dt h + h

29 For a general f the above results hold for its positive and negative parts. Now write f f + f. Then F(s+h) F(s) lim h + h lim h + lim h + lim h + lim h + e (s+h)t e st h e (s+h)t e st h e (s+h)t e st h te st f + (t)dt te st f(t)dt f(t)dt (f + (t) f (t))dt f + (t)dt e (s+h)t e st h f (t)dt te st f (t)dt The case for h < is proved in a similar manner and (29) is established. We can now prove (14). Proposition 1. (32) e st ( sint t ) 2 dt π s 2 tan 1 2 s (1+ 4 ln 4s ), s > 2 Proof. Let F(s) e st (sint/t) 2 dt. First observe that sin 2 t t 2 for t, hence (33) F(s) e st dt 1 s

30 So by the Squeeze Law F( ) def lim s F(s). Now by Lemma 9 we may differentiate under the integral to obtain (34) F (s) e st sin2 t t Since we are unable to evaluate this integral directly, we try differentiating again. First we note that F ( ). Now F (s) 1 2 e st sin 2 tdt e st dt 1 2 dt e st cos2tdt 1 2s e st s 2 +4 ( scos2t+2sin2t) t 1 2s 1 2 s 4+s 2 Effectively, we have transformed the problem into the following 2nd order differential equation. t (35) d 2 y ds 1 2 2s 1 2 s s 2 +4, s > subject to the boundary conditions, y( ) y ( ). Integrating both sides we obtain y (s) 1 2 lns 1 4 ln(4+s2 )+C 1 4 ln s 2 s C

31 And the second boundary condition implies C. Now by extra credit problem #3, y F(s) π 2 tan 1 s 2 s 4 ln ( 1+ 4 s 2 ) as desired. Example 1. Verify the following identity. (36) L{sint/t} st sint e dt π t 2 tan 1 s, s > Proof. The standard method for proving (36) is to differentiate under the integral sign. Instead we try another approach. This one involves an important technique that will be introduced in calculus III double (actually, iterated) integrals and Fubini s Theorem. First observe that for t > we have e tv dv 1 t e tv Then for s > st sint e dt t 1 t 1 t e st sint lim b (e tb e ) e tv dvdt e t(s+v) sintdtdv

32 Now we apply (27) with a (s+v) and b 1 to the inside integral to obtain e t(s+v) t (s+v) 2 +1 ( (s+v)sint cost) (s+v) 2 dv Now we apply a change of variables to conclude that for s >, st sint e dt t s du 1+u 2 π 2 tan 1 s t dv Exercise - Prove (36) by differentiating under the integral sign as we did above. Now (37) sint t dt? st sint lim e dt s + t lim s s +π/2 tan 1 π/2 Notice that Lemma 8 does not apply in this case. In order to prove that (37) is valid we need Theorem 11. Suppose that f is integrable on every finite interval, I [, ). Suppose also that the improper integral e st f(t)dt converges for s > and that the function F(s) e st f(t)dt satisfies (38) lim s +F(s) A R

33 Now let C R. Then the Tauberian condition (39) tf(t) C, for b < t < guarantees that the improper integral f(t)dt converges to A. Observe that sint/tdt is Abel summable to π/2 and u tf(t) sint 1 for all t. So by Theorem 11, sint/tdt π/2 as u. Unfortunately, the proof of Theorem 11 is much more difficult and involved than the proof of Theorem 8.