Taylor Series and Asymptotic Expansions

Transcription

1 Taylor Series and Asymptotic Epansions The importance of power series as a convenient representation, as an approimation tool, as a tool for solving differential equations and so on, is pretty obvious. What may not be so obvious is that power series can be very useful even when they diverge! Let us start by considering Taylor series. If f : [ a,a] has infinitely many continuous derivatives, Taylor s theorem says that for each n IN {}, f( P n (+R n ( n with P n ( m m! f(m ( m R n ( (n+! f(n+ ( n+ for some between and Since f (n+ is continuous, there eists a constant M n such that R n ( M n n+ for all [ a,a]. This says that as gets smaller and smaller, the polynomial P n ( becomes a better and better approimation to f(. The rate at which P n ( tends to f( as tends to zero is, at worst, proportional to n+. Thus the Taylor series m m! f(m ( m is an asymptotic epansion for f( where Definition. m a m m is said to be an asymptotic epansion for f(, denoted f( m a m m if, for each n IN {}, lim n [f( n a m m] m There are several important observations to make about this definition. (i The definition says that, for each fied n, n m a m m becomes a better and better approimation to f( as gets smaller. As, n m a m m approaches f( faster than n tends to zero. (ii The definition says nothing about what happens as n. There is no guarantee that for each fied, n m a m m tends to f( as n. (iii In particular, the series m a m m may have radius of convergence zero. In this case, m a m m is just a formal symbol. It does not have any meaning as a series. (iv We have seen that every infinitely differentiable function has an asymptotic epansion, regardless of whether its Taylor series converges or not. Now back to our Taylor series. There are three possibilities. (i The series m m! f(m ( m has radius of convergence zero. (ii The series m m! f(m ( m has radius of convergence r > and lim R n( for all r < < r. n In this case the Taylor series converges to f( for all ( r,r. Then f( is said to be analytic on ( r,r. (iii The series m m! f(m ( m has radius of convergence r > and the Taylor series converges to something other than f(. Here is an eample of each of these three types of behaviour. Eample (i For this eample, f( e t + t dt. Our work on interchanging the order of differentiation and integration yields that f( is infinitely differentiable and that the derivatives are given by d n d n f( n n ( e t + t dt c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions

2 Instead of evaluating the derivative directly, we will use a little trick to generate the Taylor series for f(. Recall that r n r rn+ r We just substitute r t to give and f( n n N + t n ( t n + ( t N+ + t ( t n e t ( dt+ t N+ + t e t dt ( n n t n e t dt+( N+ (N+ n ( n n +( N+ (N+ n t N+ + t e t dt t N+ + t e t dt since ( t n e t dt. To compute the m th derivative, f (m (, we choose N such that (N + > m and apply { d m d n if m n m ( if m n Applying dm d with m < (N + to ( N+ (N+ t N+ m + t e t dt gives a finite sum with every term containing a factor p with p >. When is set to, every such term is zero. So { d m d f( if m is odd m ( m!m! if m is even So the Taylor series epansion of f is m m! f(m ( m m m even ( m ( m! ( m m!m! m mn ( n n The high growth rate of makes the radius of convergence of this series zero. For eample, one formula for the radius of convergence of [ n a ]. [ ] n n is lim an+ n When an ( n, lim an+ n [ lim (n + ]. Replacing by gives that n n ( n n converges if and only if. Nonetheless, the error estimates N f( ( n n (N+ t N+ + t e t dt (N+ t N+ e t dt (N +! N+ n shows that if we fi N and make smaller and smaller, N n ( n n becomes a better and better approimation to f(. On the other hand, if we fi and let N grow, we see that the error grows dramatically for large N. Here is a graph of y (N+! N+ against N, for several different (fied values of. ( Denote γ n t n e t dt. Then γ e t dt and, by integration by parts with u t n, dv e t dt, du nt n dt and v e t, γ n nγ n for all n IN. So, by induction, γ n. c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions n

3 (N +! N N Eample (ii For this eample, f( e. By Taylor s theorem, f( n m m m! +R n( with R n ( e (n+! n+ for some between and. In this case, m m m! has radius of convergence infinity and the error Rn ( (n+! e n+ tends to zero both (a when n is fied and tends to zero and (b when is fied and n tends to infinity. Use R n ( (n+! e n+ to denote our bound on the error introduced by truncating the series at n. Then R n+ ( R n+ n (. So if, for eample, n, then increasing the inde n of R n ( by one reduces the value of R n ( by a factor of at least two. Eample (iii For this eample, f( { e / if if Of course by the usual differentiation rules, f( is infinitely differentiable, ecept possibly at. I shall show below that f is also infinitely differentiable at and f (n ( for every n IN {}. So in this case, the Taylor series is n n and has infinite radius of convergence. When we approimate f( by the Taylor polynomial, n m m, of degree n, the error, R n ( f(, is independent of n. The function f( agrees with its Taylor series only for. Similarly, the function { g( e / if > if agrees with its Taylor series if and only if. Here is the argument that f is infinitely differentiable at and f (n ( for every n IN {}. First note that, by the usual differentiation rules, when, every derivattive f (n ( is some polynomial in times e /. For any and any integer n e / + +! ( + + ( n ( + n Consequently, for any polynomial P(y, say of degree p, there is a constant C such that ( P e / P ( e C p / (p+!c +p (p+! (p+ c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions

4 for all. Hence, for any polynomial P(y, Applying ( with P(y y, we have lim P( e /. ( f f( f( ( lim lim e / f ( Applying ( with P(y y 4, we have f f ( lim ( f ( lim ( e / f ( { if e / if { if ( e / if 6 ( Indeed, for any n, if f (n ( P n e / for all, then applying ( with P(y yp n (y gives f (n f ( lim (n ( f (n ( lim P n ( e / and hence { if f (n ( [ P ( n ( + P n ] e / if Eample (iv Stirling formula ln ( n+ lnn n+ln π actually consists of the first few terms of an asymptotic epansion ln ( n+ lnn n+ln π+ n 6 n + If n >, the approimation ln ( n+ lnn n+ln π is accurate to within.6% and the eponentiated form n n+ πe n+ n is accurate to one part in,. But fiing n and taking many more terms in the epansion will in fact give you a much worse approimation to ln. Laplace s Method Stirling s formula may be generated by Laplace s method one of the more common ways of generating asymptotic epansions. Laplace s method provides good approimations to integrals of the form I(s e sf( d when s is very large and the function f is reasonably smooth and takes its minimum value at a unique point, m. The method is based on the following three observations. (i When s is verylargeand m, e sf( e sf(m e s[f( f(m] e sf(m, since f( f( m >. So, for large s, the integral will be dominated by values of near m. (ii For near m f( f( m + f ( m ( m since f ( m. (iii Integrals of the form e a( m d, or more generally e a( m ( m n d, can be computed eactly. Let s implement this strategy for the Γ function, defined in the lemma below. This will give Stirling s formula, because of part (d of the lemma. c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 4

5 Lemma. Define, for s >, the Gamma function Γ(s t s e t dt (a Γ( (b Γ ( π. (c For any s >, Γ(s+ sγ(s. (d For any n IN, Γ(n+. Proof: (a is a trivial integration. (b Make the change of variables t, dt d td. This gives Γ ( t e t dt We compute the square of this integral by switching e d e d Γ ( ( ( e d e y dy e ( +y ddy IR to polar coordinates. This standard trick gives Γ ( π dθ dr re r π dθ [ e r] π dθ π (c By by integration by parts with u t s, dv e t dt, du st s dt and v e t Γ(s+ t s e t dt t s ( e t ( e t st s dt s t s e t dt sγ(s (d is obvious by induction, using parts (a and (c. To derive Stirling s formula, using Laplace s method, We first manipulate Γ(s + into the form e sf( d. Γ(s+ t s e t dt s s+ s e s d s s+ e s e s( ln d where t s So f( ln and f (. Consequently f ( < for < and f ( > for >, so that f has a unique minimum at and increases monotonically as gets farther from. The minimum value of f is f(. (We multiplied and divided by e s in the last line above in order to arrange that the minimum value of f be eactly zero. c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 5

6 If ε > is any fied positive number, e sf( d +ε ε e sf( d will be agoodapproimationfor anysufficiently larges, because, when, f( > and lim s e sf(. We will derive, in an appendi, eplicit bounds on the error introduced by this approimation as well as by the other approimations we make in the course of this development. The remaining couple of steps in the procedure are motivated by the fact that integrals of the form e a( m ( m n d can be computed eactly. If we have chosen ε pretty small, Taylor s formula f( f(+f (( + f( (( +! f( (( + 4! f(4 (( 4 [ [ [ ] ( +! ] ( + ] 4! 4 ( 4 ( ( + 4 ( 4 will give a good approimation for all < ε, so that e sf( d +ε ε e s[ ( ( + 4 ( 4] d +ε ε e s ( e [s ( s 4 ( 4] d (We shall find the first two terms an epansion for Γ(s+. If we wanted to find more terms, we would keep more terms in the Taylor epansion of f(. The net step is to Taylor epand e [s ( s 4 ( 4] about, in order to give integrands of the form e s ( ( n. We could apply Taylor s formula directly, but it is easier to use the known epansion of the eponential. e sf( d +ε ε +ε ε e s ( { + [ s ( s 4 ( 4] + [ s ( s 4 ( 4] } d e s ( { + s ( s 4 ( 4 + s 8 ( 6} d e s ( { + s ( s 4 ( 4 + s 8 ( 6} d e s ( { s 4 ( 4 + s 8 ( 6} d since ( e s ( is odd about and hence integrates to zero. For n an even natural number e s ( ( n d ( s n+ ( s n+ ( s n+ ( s n+ Γ ( n+ e y y n dy where y e y y n dy e t t n dt s ( where t y, dt ydy tdy ( By the lemma at the beginning of this section, Γ ( ( π Γ Γ( ( π Γ 5 Γ( π Γ ( 7 5 Γ( 5 5 c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 6 π

7 so that e s( ln d ( s Γ ( ( π s s 5 4( Γ ( 5 s s 8 85 s s π ( + 4s + 5 6s + s 8( s 7 Γ ( and Γ(s+ s ( + s+ e s π + s So we have the first two terms in the asymptotic epansion for Γ. We can obviously get as many as we like by taking more and more terms in the Taylor epansion for f( and e y. So far, we have not attempted to keep track of the errors introduced by the various approimations. It is not hard to do so, but it is messy. So we have relegated the error estimates to the appendi. Summability We have now seen one way in which divergent epansions may be of some use. There is another possibility. Even if we cannot take the sum of the series n in the conventional sense, there may be some more general sense in which the series is summable. We will just briefly take a look at two such generalizations. Suppose that n is a (possibly divergent series. Let S N denote the series partial sum. Then the series is said to converge in the sense of Cesàro if the limit n S S C lim +S + +S N N N+ eists. That is, if the limit of the average of the partial sums eists. The series is said to converge in the sense of Borel if the power series g(t has radius of convergence and the integral S B n t n e t g(t dt converges. As the following theorem shows, these two definitions each etends our usual definition of series convergence. Theorem. If the series n converges, then S C and S B both eist and S C S B n Proof that S C n. Write S n and σ N S+S+ +SN N+ and observe that S σ N (S S +(S S + +(S S N N + c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 7

8 Pick any ε >. Since S lim S n there is an M such that S S m < ε n for all m M. Also since S lim S n, the sequence { } S n is bounded. That is, there is a K such that Sn K for all n. So if n N > ma { MK ε/,m} S σn S S + S S + + S S N N + S S + S S + + S S M + S S M + S S M+ + + S S N N + N + ε ε ε K + K + + K < + N + N + ε + ε ε MK N + Hence lim N σ N S and S C S. Proof that S B n. Write S n n The last step, namely the echange n + n n e t t n dt N M + N + g(t ε {}}{ t n e t dt n can be justified rigorously. If the radius of convergence of the series n α n is strictly greater than one, the justification is not very hard. But in general it is too involved to give here. That s why Proof is in quotation marks. The converse of this theorem is false of course. Consider, for eample, n ( rn with r. Then, as n, n S n ( r m ( rn+ converges to +r σ N N+ m if r < and diverges if r. But S n n +r + N+ +r r N+ (r+ + N+ As N, this converges to +r g(t n n +r [ N ( r n+] ( r N+ (r+ n N+ +r if r, r, and diverges if r >. And ( r n t n e rt e t g(t dt ] [N + ( r ( rn+ +r e (+rt dt +r if r > and diverges otherwise. Collecting together these three results, in the conventional sense if r (, ( r n +r in the sense of Cesàro if r (,] n in the sense of Borel if r (, c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 8

9 By way of conclusion, let me remark that these generalized modes of summation are not just mathematician s toys. The Fourier series of any continuous function converges in the sense of Cesàro. To get conventional convergence, one needs a certain amount of smoothness in addition of continuity. Also the calculation of the Lamb shift of the hydrogen atom spectrum, which provided (in addition to a Nobel prize spectacular numerical agreement with one of the most accurate measurements in science, consisted of adding up the first few terms of a series. The rigorous status of this series is not known. But umber of similar series (arising in other quantum field theories are known to diverge in the conventional sense but to converge in the sense of Borel. Appendi - Error Terms for Stirling s Formula In this appendi, we will treat the error terms arising in our development of Stirling s formula. A precise statement of that development is Step. Recall that f( ln. Write e sf( d and discard the last two terms. Step. Write +ε ε +ε ε e sf( d e sf( d+ +ε ε ε e sf( d+ e sf( d +ε e s( e s[f( ( ] d Apply e t +t+ t +e c! t with t s [ f( ( ] to give e s[f( ( ] s [ f( ( ] + s[ f( ( ] e c! s[ f( ( ] where c is between and s [ f( ( ]. Discard the last term. Step. Taylor epand f( ( ( + 4 ( 4 5 ( 5 + f(6 (c 6! ( 6 for some c [ ε,+ε], substitute this into s [ f( ( ] + s[ f( ( ] and discard any terms that contain f (6 (c and any terms of the form s m ( n with m n+ 5. (You ll see the reason for this condition shortly. This leaves +ε ε because the terms with n odd integrate to zero. Step 4. Replace +ε ε by The errors introduced in each step are as follows: Step. ε e sf( d and +ε e sf( d e s( { s 4 ( 4 + s 8 ( 6} d Step. +ε ε e s( e c! s[ f( ( ] d Step. A finite number of terms, each of the form umerical constant times an integral +ε ε e s( s m ( n d with m n+ 5. Here we have used that, since c [ ε,+ε], there is a constant M such that f (6 (c M. c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions 9

10 Step 4. A finite number of terms, each of the form umerical constant times an integral ε e s( s m ( n d. To treat these error terms, we use the following information about f( ln. (F f increases monotonically as increases. (F We have fied some small ε >. There is a δ > such that f( δ( for all +ε. To see that this is the case, we first observe that if δ ( d d f( δ( δ. So it suffices to pick δ sufficiently small that +ε δ The latter is always possible just because f(+ε >. (F For each [ ε,+ε] there is a c [ ε,+ε] such that Hence, there is a constant M such that f( (! f( (c ( f( ( M for all [ ε,+ε]. (F4 If [ ε,+ε] with ε small enough that εm 4, f( ( M 4 ( We can now get upper bounds on each of the error terms. For step, we have ε e sf( d F e sf( ε ε d e sf( ε and that f(+ε δε. and For step, we have +ε e sf( d F +ε e sδ( d δs e δεs +ε ε For step, we have e s( e c! s f( ( d F4 +ε ε e s ( s m n d! s +ε ε e s ( e s 4 ( f( ( d F +ε! M s e s 4 ( 9 d ε! M s e s 4 ( 9 d! M s ( s 4 5Γ(5 M s as in ( e s ( s m n d n+ s m( s M s 5 Γ ( n+ as in ( c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions

11 if m n+ 5. For step 4, we have e s ( s m n d ε ε e 4 ε s e 4 ε s e s 4 ( e s 4 ( s m n d ε e 4 εs s m( s 4 n+ M s m n+ e 4 ε s e s 4 ( s m n d e s 4 ( s m n d Γ ( n+ as in ( As promised these error terms all decay faster than πs ( + s as s. In particular, error terms introduced in Steps and 4, by ignoring contributions for ε, decay eponentially as s. Error terms introduced in Steps and, by dropping part of the Taylor series approimation, decay like s p for some p, depending on how much of the Taylor series we discarded. In particular, error terms introduced in Step decay like s 5, because of the condition m n+ 5. c Joel Feldman. 6. All rights reserved. December, 6 Taylor Series and Asymptotic Epansions