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# Functions and Continuity: Power Series Functions

Module by: Lawrence Baggett. E-mail the author

Summary: We introduce next a new kind of function. It is a natural generalization of a polynomial function. Among these will be the exponential function and the trigonometric functions. We begin by discussing functions of a complex varible, although totally analogous definitions and theorems hold for functions of a real variable.

The class of functions that we know are continuous includes, among others, the polynomials, the rational functions, and the nnth root functions. We can combine these functions in various ways, e.g., sums, products, quotients, and so on. We also can combine continuous functions using composition, so that we know that nnth roots of rational functions are also continuous. The set of all functions obtained in this manner is called the class of “algebraic functions.” Now that we also have developed a notion of limit, or infinite sum, we can construct other continuous functions.

We introduce next a new kind of function. It is a natural generalization of a polynomial function. Among these will be the exponential function and the trigonometric functions. We begin by discussing functions of a complex varible, although totally analogous definitions and theorems hold for functions of a real variable.

Definition 1:

Let {an}0{an}0 be a sequence of real or complex numbers. By the power series functionf(z)=n=0anznf(z)=n=0anzn we mean the function f:SCf:SC where the domain SS is the set of all zCzC for which the infinite series anznanzn converges, and where ff is the rule that assigns to such a zSzS the sum of the series.

The numbers {an}{an} defining a power series function are called the coefficients of the function.

We associate to a power series function f(z)=n=0anznf(z)=n=0anzn its sequence {SN}{SN} of partial sums. We write

S N ( z ) = n = 0 N a n z n . S N ( z ) = n = 0 N a n z n .
(1)

Notice that polynomial functions are very special cases of power series functions. They are the power series functions for which the coefficients {an}{an} are all 0 beyond some point. Note also that each partial sum SNSN for any power series function is itself a polynomial function of degree less than or equal to N.N. Moreover, if ff is a power series function, then for each zz in its domain we have f(z)=limNSN(z).f(z)=limNSN(z). Evidently, every power series function is a “limit” of a sequence of polynomials.

Obviously, the domain SSfSSf of a power series function ff depends on the coefficients {an}{an} determining the function. Our first goal is to describe this domain.

## Theorem 1

Let ff be a power series function: f(z)=n=0anznf(z)=n=0anzn with domain S.S. Then:

1.  0 belongs to S.S.
2. If a number tt belongs to S,S, then every number u,u, for which |u|<|t|,|u|<|t|, also belongs to S.S.
3.  SS is a disk of radius rr around 0 in CC (possibly open, possibly closed, possibly neither, possibly infinite). That is, SS consists of the disk Br(0)={z:|z|<r}Br(0)={z:|z|<r} possibly together with some of the points zz for which |z|=r.|z|=r.
4. The radius rr of the disk in part (3) is given by the Cauchy-Hadamard formula:
r=1lim sup|an|1/n,r=1lim sup|an|1/n,
(2)
which we interpret to imply that r=0r=0 if and only if the limsup on the right is infinite, and r=r= if and only if that limsup is 0.0.

### Proof

Part (1) is clear.

To see part 2, assume that tt belongs to SS and that |u|<|t|.|u|<|t|. We wish to show that the infinites series anunanun converges. In fact, we will show that |anun||anun| is convergent, i.e., that anunanun is absolutely convergent. We are given that the infinite series antnantn converges, which implies that the terms antnantn tend to 0. Hence, let BB be a number such that |anzn|B|anzn|B for all n,n, and set α=|u|/|t|.α=|u|/|t|. Then α<1,α<1, and therefore the infinite series BαnBαn is convergent. Finally, |anun|=|antn|αnBαn,|anun|=|antn|αnBαn, which, by the Comparison Test, implies that |anun||anun| is convergent, as desired.

Part (3) follows, with just a little thought, from part 2.

To prove part (4), note that lim sup|an|1/nlim sup|an|1/n either is finite or it is infinite. assume first that the sequence {|an|1/n}{|an|1/n} is not bounded; i.e., that lim sup|an|1/n=.lim sup|an|1/n=. Then, given any number p,p, there are infinitely many terms |an|1/n|an|1/n that are larger than p.p. So, for any z0,z0, there exist infinitely many terms |an|1/n|an|1/n that are larger than 1/|z|.1/|z|. But then |anzn|>1|anzn|>1 for all such terms. Therefore the infinite series anznanzn is not convergent, since limanznlimanzn is not zero. So no such z is in the domain S.S. This shows that if lim sup|an|1/n=,lim sup|an|1/n=, then r=0=1/lim sup|an|1/n.r=0=1/lim sup|an|1/n.

Now, suppose the sequence {|an|1/n}{|an|1/n} is bounded, and let LL denote its limsup. We must show that 1/r=L.1/r=L. We will show the following two claims: (a) if 1/|z|>L,1/|z|>L, then zS,zS, and (b) if 1/|z|<L,1/|z|<L, then zS.zS. (Why will these two claims complete the proof?) Thus, suppose that 1/|z|>L.1/|z|>L. Let ββ be a number satisfying L<β<1/|z|,L<β<1/|z|, and let α=β|z|.α=β|z|. Then 0<α<1.0<α<1. Now there exists a natural number NN so that |an|1/n<β|an|1/n<β for all nN,nN, or equivalently |an|βn|an|βn for all nN.nN. (See part (a) of Exercise 2.17. ) This means that for all nNnN we have |anzn|=|an/βn||βz|nαn.|anzn|=|an/βn||βz|nαn. This implies by the Comparison Test that the power series anznanzn is absolutely convergent, whence convergent. Hence, zS,zS, and this proves claim (a) above. Incidentally, note also that if L=0,L=0, this argument shows that r=,r=, as desired.

To verify claim (b), suppose that 1/|z|<L.1/|z|<L. Then there are infinitely many terms of the sequence {|an|1/n}{|an|1/n} that are greater than 1/|z|.1/|z|. (Why?) For each such term, we would then have |anzn|1.|anzn|1. This means that the infinite series anznanzn is not convergent and zS,zS, which shows claim b.

Hence, in all cases, we have that r=1/lim sup|an|1/n,r=1/lim sup|an|1/n, as desired.

Definition 2:

If ff is a power series function, the number rr of the preceding theorem is called the radius of convergence of the power series. The disk SS of radius rr around 0, denoted by Br(0),Br(0), is called the disk of convergence.

## Exercise 1

Compute directly the radii of convergence for the following power series functions, i.e., without using the Cauchy-Hadamard formula. Then, when possible, verify that the Cauchy-Hadamard formula agrees with your computation.

1. f(z)=zn.f(z)=zn.
2. f(z)=n2zn.f(z)=n2zn.
3. f(z)=(-1)n(1/(n+1))zn.f(z)=(-1)n(1/(n+1))zn.
4. f(z)=(1/(n+1))z3n+1f(z)=(1/(n+1))z3n+1.
5. f(z)=n=0zn/n!.f(z)=n=0zn/n!.

## Exercise 2

1. Use part (e) of (Reference) to show that a power series function pp is an even function if and only if its only nonzero coefficients are even ones, i.e., the a2ka2k's. Show also that a power series function is an odd function if and only if its only nonzero coefficients are odd ones, i.e., the a2k+1a2k+1's.
2. Suppose f(z)=k=0a2kz2kf(z)=k=0a2kz2k is a power series function that is an even function. Show that
f(iz)=k=0(-1)ka2kz2k=fa(z),f(iz)=k=0(-1)ka2kz2k=fa(z),
(3)
where fafa is the power series function obtained from ff by alternating the signs of its coefficients. We call this function fafa the alternating version of f.f.
3. If g(z)=k=0a2k+1z2k+1g(z)=k=0a2k+1z2k+1 is a power series function that is an odd function, show that
g(iz)=ik=0(-1)ka2k+1z2k+1=iga(z),g(iz)=ik=0(-1)ka2k+1z2k+1=iga(z),
(4)
where again gaga is the power series function obtained from gg by alternating the signs of its coefficients.
4. If ff is any power series function, show that
f(iz)=fe(iz)+fo(iz)=fea(z)+ifoa(z),f(iz)=fe(iz)+fo(iz)=fea(z)+ifoa(z),
(5)
and hence that fe(iz)=fea(z)fe(iz)=fea(z) and fo(iz)=ipoa(z).fo(iz)=ipoa(z).

The next theorem will not come as a shock, but its proof is not so simple.

## Theorem 2

Let f(z)=anznf(z)=anzn be a power series function with radius of convergence r.r. Then ff is continuous at each point in the open disk Br(0),Br(0), i.e., at each point zz for which |z|<r.|z|<r.

### Proof

Let zBr(0)zBr(0) be given. We must make some auxiliary constructions before we can show that ff is continuous at z.z. First, choose a z'z' such that |z|<|z'|<r.|z|<|z'|<r. Next, set bn=|nan|,bn=|nan|, and define g(z)=bnzn.g(z)=bnzn. By the Cauchy-Hadamard formula, we see that the power series function gg has the same radius of convergence as the power series function f.f. Indeed, lim sup|bn|1/n=lim supn1/n|an|=limn1/nlim sup|an|.lim sup|bn|1/n=lim supn1/n|an|=limn1/nlim sup|an|. Therefore, z'z' belongs to the domain of g.g. Let MM be a number such that each partial sum of the series g(z')=n=0Nbnz'ng(z')=n=0Nbnz'n is bounded by M.M.

Now, let ϵ>0ϵ>0 be given, and choose δδ to be the minimum of the two positive numbers ϵ|z'|/Mϵ|z'|/M and |z'|-|z|.|z'|-|z|. We consider any yy for which |y-z|<δ.|y-z|<δ. Then yBr(0),yBr(0),|y|<|z'|,|y|<|z'|, and

| f ( y ) - f ( z ) | = lim | S N ( y ) - S N ( z ) | = lim | n = 0 N a n ( y n - z n ) | lim N n = 0 N | a n | | y n - z n | = lim N n = 1 N | a n | | y - z | j = 0 n - 1 | y j | | z n - 1 - j | lim N n = 1 N | a n | | y - z | j = 0 n - 1 | z ' | n - 1 lim N | y - z | ( 1 / | z ' | ) n = 0 N n | a n | | z ' | n | y - z | lim N M | z ' | < δ lim N M | z ' | ϵ . | f ( y ) - f ( z ) | = lim | S N ( y ) - S N ( z ) | = lim | n = 0 N a n ( y n - z n ) | lim N n = 0 N | a n | | y n - z n | = lim N n = 1 N | a n | | y - z | j = 0 n - 1 | y j | | z n - 1 - j | lim N n = 1 N | a n | | y - z | j = 0 n - 1 | z ' | n - 1 lim N | y - z | ( 1 / | z ' | ) n = 0 N n | a n | | z ' | n | y - z | lim N M | z ' | < δ lim N M | z ' | ϵ .
(6)

This completes the proof.

## Exercise 3

1. Let f(z)=n=0anznf(z)=n=0anzn be a power series function, and let p(z)=k=0mbkzkp(z)=k=0mbkzk be a polynomial function. Prove that f+pf+p and fpfp are both power series functions. Express the coefficients for f+pf+p and fpfp in terms of the anan's and bkbk's.
2. Suppose ff and gg are power series functions. Prove that f+gf+g is a power series function. What is its radius of convergence? What about cf?cf? What about fg?fg? What about f/g?f/g? What about |f|?|f|?

## Exercise 4

1. Prove that every polynomial is a power series function with infinite radius of convergence.
2. Prove that 1/z1/z and (1/(z-1)(z+2))(1/(z-1)(z+2)) are not power series functions. (Their domains aren't right.)
3. Define f(z)=n=0(-1)nz2n+1.f(z)=n=0(-1)nz2n+1. Prove that the radius of convergence of this power series function is 1, and that f(z)=z1+z2f(z)=z1+z2 for all zB1(0).zB1(0). Conclude that the rational function z/(1+z2)z/(1+z2) agrees with a power series function on the disk B1(0).B1(0). But, they are not the same function. HINT: Use the infinite geometric series.

Theorem 2 and Exercise 3 and Exercise 4 raise a very interesting and subtle point. Suppose f(z)=anznf(z)=anzn is a power series function having finite radius of convergence r>0.r>0. Theorem 2 says that ff is continuous on the open disk, but it does not say anything about the continuity of ff at points on the boundary of this disk that are in the domain of f,f, i,e., at points z0z0 for which |z0|=r.|z0|=r. and anz0nanz0n converges. Suppose g(z)g(z) is a continuous function whose domain contains the open disk Br(0)Br(0) and also a point z0,z0, and assume that f(z)=g(z)f(z)=g(z) for all zBr(0).zBr(0). Does f(z0)f(z0) have to agree with g(z0)?g(z0)? It's worth some thought to understand just what this question means. It amounts to a question of the equality of two different kinds of limits. f(z0)f(z0) is the sum of an infinite series, the limit of a sequence of partial sums, while, because gg is continuous at z0,z0,g(z0=limzz0g(z).g(z0=limzz0g(z). At the end of this chapter, we include a theorem of Abel that answers this question.

The next theorem is the analog for power series functions of part (2) of (Reference) for polynomials. We call it the “Identity Theorem,” but it equally well could be known as the “Uniqueness of Coefficients Theorem,” for it implies that different coefficients mean different functions.

## Theorem 3: Identity Theorem

Let f(z)=anznf(z)=anzn be a power series function with positive radius of convergence r.r. Suppose {zk}{zk} is a sequence of nonzero distinct numbers in the domain of ff such that:

1.  limzk=0.limzk=0.
2.  f(zk)=0f(zk)=0 for all k.k.

Then ff is identically 0 (f(z)0f(z)0 for all zSzS). Moreover, each coefficient anan of ff equals 0.0.

### Proof

Arguing by induction on n,n, let us prove that all the coefficients anan are 0.0. First, since ff is continuous at 0,0, and since limzk=0,limzk=0, we have that a0,a0, which equals f(0),=limf(zk)=0.f(0),=limf(zk)=0.

Assume then that a0=a1=...=an-1=0.a0=a1=...=an-1=0. Then

f ( z ) = a n z n + a n + 1 z n + 1 + ... = z n j = 0 b j z j , f ( z ) = a n z n + a n + 1 z n + 1 + ... = z n j = 0 b j z j ,
(7)

where bj=an+j.bj=an+j. If gg is the power series function defined by g(z)=bjzj,g(z)=bjzj, then, by the Cauchy-Hadamard Formula, we have that the radius of convergence for gg is the same as that for f.f. (Why does lim sup|bj|1/j=lim sup|ak|1/k?lim sup|bj|1/j=lim sup|ak|1/k?) We have that f(z)=zng(z)f(z)=zng(z) for all zz in the common disk of convergence of these functions ff and g.g. Since, for each k,zk0k,zk0 and f(zk)=zkng(zk)=0,f(zk)=zkng(zk)=0, it follows that g(zk)=0g(zk)=0 for every k.k. Since gg is continuous at 0,0, it then follows as in the argument above that g(0)=0.g(0)=0. But, g(0)=b0=an.g(0)=b0=an. Hence an=0,an=0, and so by induction all the coefficients of the power series function ff are 0. Clearly this implies that f(z)f(z) is identically 0.

## Rule 1

Suppose ff and gg are two power series functions, that {zk}{zk} is a sequence of nonzero points that converges to 0, and that f(zk)=g(zk)f(zk)=g(zk) for all k.k. Then ff and gg have the same coefficients, the same radius of convergence, and hence f(z)=g(z)f(z)=g(z) for all zz in their common domain.

## Exercise 5

1. Prove the preceding corollary. (Compare with the proof of (Reference).)
2. Use the corollary, and the power series function g(z)=z,g(z)=z, to prove that f(z)=|z|f(z)=|z| is not a power series function.
3. Show that there are power series functions that are not polynomial functions.
4. Let f(z)=anznf(z)=anzn be a power series function with infinite radius of convergence, all of whose coefficients are positive. Show that there is no rational function r=p/qr=p/q for which f(z)=r(z)f(z)=r(z) for all complex numbers z.z. Conclude that the collection of power series functions provides some new functions. HINT: Use the fact that for any nn we have that f(x)>anxnf(x)>anxn for all positive x.x. Then, by choosing nn appropriately, derive a contradiction to the resulting fact that |p(x)/q(x)|>anxn|p(x)/q(x)|>anxn for all positive x.x. See part (b) of (Reference).

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