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Unformatted text preview: R.l Whenever we have a series 2 un(x) of functions, there are three fundamental questions we ask: (1) Given the series 2 un(x), for what values of x does the series converge? (2) Given 2 un(x), if it converges to a function f(xl,
what properties does f(x) have? Specifically: Is f contin
uous? Can you calculate [b f(x) by integrating the series a
term—by~term? Is f differentiable, and can you calculate f'(x) by differentiating the series term—by—term?
(3) Given a function f(x), under what conditions does it equal such a series, where the functions un(x) are func— tions of a particular type?' We shall answer these questions for a power series. This is a series of the form: Theorem 1. Given a power series 2 anxn, exactly one of the following holds: (a) The series converges only for x = O. (b) The series converges absolutely for all x. (c) There is a number r > 0 such that the series converges absolutely if le < r and diverges if [X] > r. (Nothing is said about what happens when x  1r.) . . . n
Proof. Step 1. We show that if the series 2 anx converges for x = X0 # 0, then it converges<§bsolutely)for any with le < Ixol. R.2 Forthis purpose, we write
n _ n' n
\anx I~ lanxol lx/de = cnIX/xol , n .
where cn = ‘anxo" Now the series Ejlx/xoln converges,
because it is a geometric series of the form 2 yn with
[yl < 1. Furthermore, the sequence cn approaches 0 . n
as nave, because the series i anxO converges (by
hypothesis). We can choose N so that [anxgis l for
nzN. Then [anxn‘S \x/xoln for n>, N. The comparison thest then implies that the series :[anxnl converges. Step 2. Let S be the set of all numbers x for
which the series ilanxn converges. If S consists of
0 alone, then (a) holds. Otherwise, there is at least
one number x0 different from O belonging to S. It
then follows that there is a positive number xl belonging
to S; indeed, if x1 Vis any positive number such that
x1 < [xol , thens Sllanxgl converges by Step 1, so that jlanx?
_converges and x1 belongs to S. We now consider two cases. Case 1. The set S is bounded above. In this case, we set r = sup S, and show that the series Zanxn
diverges if \xl) r and converges (absolutely) if \x[.< r.
Divergence if lxl > r is clear. For suppose lxl> r . n '
and the series ‘2 arx converges. If we choose x so
y 2 that r< x <.\xl, then Step 1 implies that the series 2
j:[a xn\ converges. Then the series 51a xn converges.
n 2 n 2 so that x2 r is an upper bound for S. belongs to S, contradicting the fact that R.3 Convergence if [XI < r is also clear. If IxI< r, 3
would be‘a smaller upper bound on S than r). Step 1 then implies that leanxn‘_ we can choose an element x3 of S such that [xl< x (otherwise [xl converges. Case 2. The set S is unbounded above. We Show the series Elanxn converges (absolutely) for all x. Given x, choose an element x4 of S such that IX‘< x4. This we can do because 3 is unbounded above. Then i‘anxn‘ coverges, by Step 1. D Definition . The number r constructed in (c) of the preceding theorem is called the radius 2; convergence of the series. In case (a) we say that r = 0; and in case (b). we say that r = 63. ,ll. .w_ ~~ *..—._..—_ .— .4,. _ Theorem 2. Su pose 2 anxn has radius of convergence
2 _ _ um... __ ____________ r > 0. (Kg allow r = m.) Let & f(x) = 2 anxn
n=0 (a) f is continuous for [X] < r. (b) For [x] < r, we have x m Xn+1
Io f(t) dt = 211:0 an El— . (c) For Ix! < r, we have (d) The series in (b) and (c) have radii g: convergence precisely r. R.4 Proof.‘ In general, let — .0. m
pm(x) — a0 + alx + + amx . It is a polynomial of degree' m, and it is the mEE partial sum of our power series. We are going to prove parts (a) and (b) for the fixed
point b in (—r,r). So as a preliminary, let us choose R
so that [bl < R < r. Steg 1. Given a > 0, there is an integer N such that the inequality [f(X) ' Pm(X)[ < 8 holds for all m > N and all x with x < R. The proof is easy. Since 2 [aanl converges, we can
choose N sufficiently large that
00 la Rn] < s.
X=N+1 n It follows that if '[x[ < R and m > N, we have G3 ”I on
N N
[axn[< aR< aR <5.
E=m+l n n=g+l I n n=§+l I n I Then for m 9 N and x < R, f(x)p (x) = a x? < a xnl < a.
I m I Izn=m+l n I Zn=m+1 I n R.5 Step 2. We show that f is continuous at b. This
proves (a). Given a > 0, choose N as in Step 1. We have
f(x)  pN(x)[ < a,
for any x in the interval [—R,R]. In particular,f(b)  pN(b)l <. 6;. Now we use continuity of the polynomial pN(x) to choose 6 so that whenever Ix—bl < 6, then x is in [R,R], and IpN(x) — pN(b)l < 5. Adding these three inequalities and using the triangle inequality, we see that whenever Ixbl <'6, we have f(x) ~ f(b)[ < 35.
Step 3. We Show that 2 anbn+l/(n+l) converges to
b .
I f(x) dx. This proves (bL
0 Given a > 0, choose N as in Step 1. Then whenever m > N, the inequality s < f(x)  pm(x) < E holds for all x. in the interval [—R,R]. The comparison property of integrals tells us that
b .
[[0 <f(X)pm(X)) dx < elbI. This says that m+l b
[I f(x) dx  (a0b+alb2/2 +...+amb /(m+l))[ < eb{ O for all m 2 N. It follows that Z anbn+l/(n+l) converges to b I f(x) dx. 0 Step 4. ,We show that the power series has radius of convergence at least r. For this purpose, it suffices to show that if c is any number with O < c < r, then X nancn_l converges. In fact, it suffices to show that EZnancn converges, since multiplying
the:series by <3 dbes not afﬁan:conwaxﬁmce. This isvdwmzwe shall gxnn First,<ﬂmnse <1 smﬂlthat ci<<i< r. Thenvmitetme<¥xeraltenn ofcmu:series:h1the ﬁonm n ‘ — I
nanc  na (d) d? . We note that the series 2 and“ converges mcause d < r. It follows that the nth bean andn anxxraches 0 as r: baxmes large. Choose N sufficiently large that Iandnl< l for ma N. rIhen for ma N, we have n c n
nanc < ma) \ 2mg)“ converges by the ratio test, since 0< c/d< 1. Therefore the series Now the series 2 nancr1 converges, by the wnparison test. Step 5. We prove part (c). Let g(X) for Ix! < r. Part (b) of the theorem tells us that for lxl < r, we have x 0° 11
I g (t) dt
0 n=1 H
M
m
X II Part (a) of the theorem tells us that g(x) is continuous for [X] < r. Then the first fundamental theorem of calculus applies; we conclude that g(x) = f‘(x). which is what we wanted to prove. R.8 Step 6; We prove part (d). If the series E.anxn+l/(n+1)
had radius of convergence q > r, then so would the differenti ated series 2 anxn, by Step 4. But it does not. Similarly, if the series 2 na xnl n had radius of convergence q > r, then n so would the integrated series 2 a x . ‘But it does not. B :1 Remark. It follows readily that all the results of Theorem 2 hold for general power series of the form a o n
f(x) — 2 =0 an(xa) . There is a number r (which may be 0 or a) such that the series converges absolutely for xa < r and diverges for [x~a > r. Furthermore for lxal < I, one has: (a) f(x) is continuous. x w (xa)n+1
(b) [a f(t) dt = Xn=0 an “‘E$I‘” .
(c) f'(x) é nan(xa)nl.
n=1 The proof is immediate; one merely substitutes (xa) for x in the theorem. Here is a theorem proving the uniqueness of a power series representation: Theorem 3. Suppose f(x) = Z an(xa)n = X n
b (x—a)
11:0 11:0 n 23 some open interVal I containing a. Then for all k, , (k)
= b _ f '(a) ak k ‘ “‘TE“‘ 12.9 Proof. We apply the preceding theorem. We write f(x) : Z an(xa)n. Differentiating, we have , _ _ nl
f (x) — Z nan(x a) . n=l Applying the theorem once again, we have ’ on 2 n(nl)an(xa)n‘2. n=2 f"(X) And so on. Differentiating k times, we have f(k)(x) = Z n(nl)~(n—k+l)anxn‘k.
n=k When we evaluate at x = 0, all the terms vanish except for the first term. Thus f(k)(x) . klak, as desired. The same argument applies to compute b D k. Definition. If f(x) equals a power series 2 an(xa)n
in some open interval containing a, we say f is analytic
(or sometimes "real analytic") near a. By the preceding theorem, this power series is uniquely determined by f; its partial sums must be the Taylor polynomials of f at a. For R.lO this reason, the series is sometimes called the Tailor series of f at a. Corollarz . The function f(x) is analytic near a if and only if both the following hold: —— ——————.—— (1) All derivatives of f exist in an open interval about" a. (2) The error term En(x) in Taylor's formula aggroaches 0 as n ——> m, for each x in that interval. Remark. We know that it is possible for us to have C f(x) = 2 unm
n=0
for all x in an interval [c.d], where each function
un(x) is.continuous, without it following that f(x) is
continuous, or that its integral can be obtained by
integrating the series termrbyterm. However, this unpleasant
situation does not occur if the analogue of the statement in of the proof of Theorem 2 Step lAholds. This fact leads to the following definition. Definition. The series 2 un(x) is said to converge uniformly to f(x) on the interval [c.d] if given a > 0, .——————‘——d there is an N such that m If(x) Z unﬁt)! < 8
n=0 for all m > N and all x in [c,d]. R.ll Theorem 5. The series X un(x) converges uniformly on [c.d] if there is 5 Convergent Series 2 Mn of con— stants such that lun(x)l < Mn for all x in [c,d].
(The proof is just like that of Step 1. There the series of constants was the series 2 [aan!.) Under the hypothesis of uniform convergence, the
analogues of (a) and (b) of Theorem 2 hold: Theorem 6. Suppose X un(x) converges uniformly to
f(x) on [c,d]. if the functions un(x) are continuous, so is f(x), and furthermore the series “3 X
E (I un(t)dt) n=0 c
x
converges uniformlv to I f(t)dt 93 [c.d].
———————— c _ The proof is just like the ones given in Steps 2 and 3.
Remark. Part (c) of the theorem, about differentiating
a power series termhysterm, does not carry over to more general uniformly convergent series. For instance, the series 2 (sin nx)/n2 n=1 converges uniformly on any.interval. by comparison with the series of constants z l/nz, but the differentiated series 2 (cos nx)/n does not even converge at x = 0. If however the differentiated series does converge uniformly on [c.d], then f'(x) does exist and equals this differentiated series. The rproof is similar to that of (c). R.11 Exercises ll. Prove Theorem 3.
2. Prove Theorem 4. 3. Prove the following theorem about termbyterm differentiation: Suppose that the functions u$(x) are continuous, that no the series Znl u$(x) converges uniformly on [c,d], and that
Em i nh(x) converges for at least one x in [c,d]. Then:
n= (a) X:l un(x) converges uniformly on [c,d], say to
f(x). (b) f'(x) exists and equals 2 uA(x). [Hintz Integrate the series 2 uﬁ(x).] ...
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 Spring '09
 BRUBAKER

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