# Power series

Power series are natural generalizations of Taylor polynomials. We know that Taylor polynomials give (if we get lucky) approximations of functions that get better as polynomials get longer, so one can expect that infinite polynomials give precisely the functions from which they arise. Before we get to this relationship, we will explore power series as such, that is, we will try to make some sense of "infinite polynomials"; they have interesting and useful properties.

Definition.
By a power series with center a we mean any series of the form

where ak are real numbers.

A very important and useful concept is complex power series, which is beyond the scope of Math Tutor; however, note that most conclusions of this section are valid also in the complex case. We will actually below sometimes refer to complex series, since some notions are somewhat more clear in more dimensional cases.

Note that an ordinary polynomial can be also considered a power series. For instance, the polynomial 1 − 2x can be written as the series ∑ akxk, where a0 = 1, a1 = −2, and ak = 0 for all k > 1. Thus we also have to keep in mind that when we talk about power series, we can actually have finite polynomials. Some things are then much easier, often trivial, for instance such a "finite" series always converges. By the way, when we have a polynomial, then by the result on uniqueness below there is no other way to express it as a power series, in particular it cannot be rewritten as a "truly infinite" power series.

Back to power series. The basic notion - the region of convergence - behaves very nicely for power series. While for a general series of functions the region of convergence may be weird, here we know for sure that this set is the nicest possible - an interval (a circle in complex plane, a ball in general). We start with the following observation.

Fact.
Consider a power series with center a. If there exists x0 such that the given series converges at x0, then this series converges absolutely at all x satisfying

|x − a| < |x0 − a|.

Here it is exactly the case where considering complex numbers (or points from the plane) might help in understanding what is happening. If we have convergence at some point x0, then we get absolute convergence on the interior of the disc with center a on whose circumference x0 lies.

To put it another way, the series necessarily converges on the neighborhood U(a) with radius |x0 − a|. In our real case these neighborhoods translate into intervals on the real line.

Note that every power series converges at its center a. The above Fact shows that if we can move further from a (away from the center) with convergence, then we also automatically extend absolute convergence. There are three possible situations. One is that there is no convergence apart from the center a. This can indeed happen. Another situation is that we can "extend" convergence as far as we want from a, that is, we get convergence everywhere, consequently also absolute convergence everywhere. The third situation is that we "extend" convergence as far as possible from a, but we cannot go arbitrarily far, there is some limit beyond which we cannot go further. Then we obtain a neighborhood (a disc in the complex or planar case, a symmetric interval about a in the real case) such that the series converges absolutely inside and diverges outside (if we had but one convergent point outside, the by the above Fact we can enlarge this neighborhood of convergence accordingly).

Thus we have the following statement.

Theorem.
Consider a power series with center a. There exists a number R ≥ 0 (including R = ∞) such that the given series converges absolutely at all x satisfying |x − a| < R and diverges at all x satisfying |x − a| > R.
This number can be found as

These formulas also include the cases 1/0 = ∞ (infinite radius means convergence everywhere) and 1/∞ = 0 (radius zero means convergence only at a). Note that the inequality |x − a| < R describes the neighborhood UR(a), which in our case is the interval (a − R,a + R). Note also that this theorem does not tell us what happens on the border of this neighborhood, in our case at points a ± R. There we cannot make any general statement, every series has to be handled individually.

Definition.
Consider a power series with center a. The number R from the above theorem is called the radius of convergence of this series.

The theorem (including the formulas) was obtained in the following way. We tested the given power series for absolute convergence using suitable tests and it turned out that convergence depends on the distance from the center |x − a|. Applying the Root test one gets the first formula for R, applying the Ratio test one gets the second formula for R. When investigating a particular series most people do not directly apply the above formulas, instead they repeat the process with testing convergence, since it may give a better insight into the series and it is not substantially longer. Recall that when investigating convergence we do not really need to know where indexing start.

Example: Investigate convergence of the series

We will use the Ratio test to investigate absolute convergence of this series.
Note: We will use Ak to denote terms of the series in the Ratio test, since the traditional ak is in the context of power series used just for coefficients, not for whole terms of series.

(Note that in the last step we did the limit with respect to k, x is some fixed parameter here.) The test now shows that the investigated series converges absolutely if |x − 1| < 1 and does not converge if |x − 1| > 1. Therefore the radius of convergence is R = 1, the series converges absolutely on U1(1), that is, on (0,2), and diverges outside of [0,2].

What can we say about convergence at the endpoints? This has to be investigated individually, note that the Ratio test above gives lambda equal to 1 when x is from the border of the region of convergence, so it does not help.

x = 0: Then we get the series

This is the harmonic series that is known to diverge.

x = 2: Then we get the series

This is the alternating version of the harmonic series and by the Alternating series test it converges, see Convergence of general series in Theory - Testing convergence.

Conclusion: This series converges on the region of convergence (0,2]. Its region of absolute convergence is (0,2).

Remark: How do we know that we do not have absolute convergence at 2? Note that if we take the series we obtained with x = 2 and apply absolute value to its terms, we get the series that we investigated at the endpoint x = 0 and thus we get divergence. This is no accident. The endpoints have coordinates a ± R, and when we substitute them to the given series and check for absolute convergence, we get the same expression for both:

∑ |ak(a ± R − a)k| = ∑ |ak|⋅|±R|k = ∑ |ak|⋅Rk.

This means that either we have absolute convergence at both endpoints (then we also must have convergence there), or we do not have absolute convergence at any of them. From practical point of view, unless we have convergence at both endpoints, we never include endpoints in the region of absolute convergence. We will return to this below.

Example: Investigate convergence of the series

We will use the Ratio test to investigate absolute convergence of this series.

The test says that the investigated series converges absolutely if lambda is less than 1, but we see that this happens always, regardless of x. Thus we have R = ∞ and the region of convergence is the whole real line (it is also the region of absolute convergence).

Example: Investigate convergence of the series

We will use the Root test to investigate absolute convergence of this series.

The test says that the investigated series converges absolutely if rho is less than 1, but we see that this happens only if 2x + 3 = 0. Thus R = 0 and the region of convergence consists of the center of the series, it is {−3/2}.

Remark: The given series does not satisfy the definition of power series, but it can be rewritten into one.

Now we see that its center is indeed −3/2. We should have applied the standard procedure to this "proper" form, but usually it is somewhat easier to work with the given form as we did it here and the results are the same.

We see that the three cases discussed above (R positive, R = 0, R = ∞) can indeed happen. Apart from the last one, a typical power series then creates a function on a disc (on a neighborhood) of the center a. We will now look at some properties of power series and their consequences for functions that they define. But before we get to it we will add one more statement concerning the region of convergence.

Recall the Fact above that essentially says that if we have convergence at some point, then we gain convergence everywhere inside the disc up to that point (see again the picture above for the complex case, it is very instructive). Using this we concluded that the region of convergence is shaped as a disc. Assume now that we have a series with a positive (in particular finite) radius of convergence. Then the Fact (or the Theorem on radius of convergence) does not allow us to say anything about convergence on the circumference of the disc of convergence. And with a good reason, typically we have convergence at some places and divergence at other places of the circumference, sometimes it is just divergence everywhere or convergence everywhere. We will now show that the Fact we discuss can be strengthened if convergence at that special point is "better".

Fact.
Consider a power series with center a. If there exists x0 such that the given series converges absolutely at x0, then this series converges absolutely at all x satisfying

|x − a| ≤ |x0 − a|.

In particular, if we actually have absolute convergence at some point on the circumference of the disc of convergence, then we have convergence on the whole circumference.

## Properties of power series

Algebraic operations with series are defined in the natural way, using the definition for series of numbers, but due to their form we can only join series with a common center. An important fact is that adding/multiplying convergent power series does not spoil convergence (for the Cauchy product it follows from absolute convergence which we do have automatically for power series). Moreover, the sums of such series are what they should be.

Theorem.
Consider power series

with radii of convergence R f  and Rg, respectively. Then the following are true.

(i) The sum of these series converges with radius of convergence R ≥ min(R f ,Rg) and

(ii) For a real number c, the c-multiple of a series converges with the same radius of convergence and

(iii) The Cauchy product of these series converges with radius of convergence R ≥ min(R f ,Rg) and

We note that if in the part (i) the radii Rf and Rg are distinct, then necessarily R is equal to their minimum, see this problem in Solved Problems - Series of functions. The formula for the Cauchy product is entirely natural, we just write it in the long way and all is clear, for simplicity we use the center 0.

[a0 + a1x + a2x2 + ...] ⋅ [b0 + b1x + b2x2 + ...] = (a0b0) + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x2 + ...

This theorem has an interesting interpretation. Fix some center a, consider the set S of all sequences {ak} that create series with center a and positive radius of convergence, and the set F of functions that can be expressed as sums of such series. The first two statements in the above theorem show that both sets are linear spaces and that operations in one space correspond to natural operations in the other space. Formally, if we define the transformation

then this transformation is linear. In other words, instead of functions (which are sometimes rather complicated structures to work with) we can work with sequences that represent them and the correspondence is best possible, it also includes the usual linear operations. Thus from the algebraic point of view these spaces are the same. (Here we actually also need to know that this transformation is 1-1, which is true due to the result on uniqueness below.)

How does the third statement fit in? If a function f is coded by the sequence {ak} and a function g is coded by the sequence {bk}, the according to the above statement the product fg is coded by the sequence {ck} that is exactly the convolution of {ak} and {bk} (see Operations with sequences in Sequences - Theory - Introduction). Thus if we want to extend the above identification between the sets S and F to multiplication, we would say that the set F with the usual multiplication of functions corresponds to the set S equipped with convolution. This suggests that there are situations when convolution is a natural replacement for multiplication. Recall that we noted before that this operation satisfies the usual laws that we are glad to have for multiplication.

The above theorem can be symbolically expressed as follows: If we introduce the correspondence f ∼ {ak}, then

cf ∼ c{ak},
f + g ∼ {ak} + {bk},
fg ∼ {ak}*{bk}.

One can also make formulas for some other things that we do with functions, for instance substitution (division is actually quite tricky, by the way), but that is more related to expansion of functions that is covered in the next section.

An important factor when working with series of functions is the question of uniform convergence. Here the situation is pretty much the best possible. We just need to avoid the border and we are fine.

Theorem.
Consider a power series with center a and radius of convergence R > 0. Then for every positive r < R this series is uniformly convergent on [a − r,a + r].

More generally, such a power series converges uniformly on any closed interval that is a subset of (a − R,a + R). Below (the Abel's convergence theorem) we actually show that this statement can be made somewhat stronger.

As usual, if we use the language of neighborhoods: "A power series converges uniformly on any closed neighborhood that is included in the region of convergence," we get a statement that is true also in other settings (complex series etc.).

This theorem is even better than it seems at the first sight. Note that we get uniform convergence as close to the border as we want. If we take any x0 from UR(a), then it does not lie on the border and therefore this particular x0 is inside. Thus we can choose some r which is smaller than R, but still large enough that Ur(a) already includes x0, and we get uniform convergence on a neighborhood of x0.

In short, we get uniform convergence around all points inside the region of convergence, so we can use all the nice properties that we had for uniform convergence (see the Series of functions) everywhere in the region of convergence. Thus we get the following extremely useful theorem.

Theorem.
Consider a power series with center a and radius of convergence R > 0. Let f be the function that it defines on its region of convergence.
(i) The function f is continuous on UR(a) = (a − R,a + R).
(ii) The function f is differentiable on UR(a) = (a − R,a + R), on this set

and the convergence of this series is uniform on [a − r,a + r] for any positive r < R.
(iii) The function f is integrable on UR(a) = (a − R,a + R), on this set it has an antiderivative

and the convergence of this series is uniform on [a − r,a + r] for any positive r < R.

One can also say it like this. Given a power series, we can differentiate it term by term and integrate it term by term and the radius of convergence stays the same, convergence stays nice and the results are logical: When we differentiate a series term by term, we get the same thing as if we differentiated the function that the series defines (analogous statement is also true for integration). The fact that power series can be easily and reliably differentiated and integrated is one of the major reasons why they are so useful and popular.

Note two things. First, in the last expression on the right in (ii) we actually dropped the first term in the indexing. It is not necessary, since when k = 0, the first terms becomes zero (which is no surprise, the derivative of a constant is zero); thus it does not make any difference whether we keep the corresponding index or drop it. In most cases it is not worth bothering and we can as well keep it, but there are situations when keeping it is quite misleading. It is up to you. I usually try to remove unnecessary clutter from my series just to keep things tidy, and a few times it paid off.

Second, the series that we get after differentiating and integrating are not in their proper forms, for that we would have to do a shift in indexing.

Again, depending on situation, sometimes this is done (for instance if you want to present the result in a nice polished up way), but in practice people often do not bother.

Remark: The above theorem allows us to differentiate and integrate power series only inside their regions of convergence - and with good reason. If a power series also converges at some border point of its region of convergence, then this convergence need not survive differentiation and integration. For an example consider the first example above, the series there was found to be convergent on the interval (0,2], in particular it converges at 2. If we differentiate it term by term, we get

It is easy to check that the resulting series does not converge at 2.

By iterating the above theorem we find that power series yield functions that have derivatives of all orders.

Corollary.
Consider a power series with center a and radius of convergence R > 0. Let f be the function that it defines on its region of convergence. Then for every natural number n, the function f has the n-th derivative on UR(a), on this set

and the convergence of this series is uniform on Ur(a) for any positive r < R.

Now imagine that you have two series with center a that converge and their sums are equal on some neighborhood of a. We write this equality of series and then differentiate n times on each side, we get on the left and on the right expressions as in the Corollary. When we substitute x = a into this equality, most of the series disappear and only one term is left on each side (try it). Thus we get the following statement.

Corollary (uniqueness).
Consider two power series with center a, a series ∑ ak(x − a)k and a series ∑ bk(x − a)k such that they both converge on some Ur(a) for a positive radius r.
If the sums of the two series are equal on this neighborhood, then the series are necessarily the same, that is, ak = bk for all k.

This in particular means that a power series is entirely given by its values on some neighborhood of a, even very small. Once we know the values of it on some tiny neighborhood of a, then also values at other points of convergence are determined, even if the series converges everywhere.

In particular, when we have a polynomial, then it is not possible to write it in another way, not even as a truly infinite power series (with the same center, of course; if we choose a different center, we get a different expression for the same function).

A note on endpoints: We remarked that convergence of a series at endpoints need not survive differentiation, so no general statement concerning endpoints was possible in the second and the third part in the above theorem. However, the first statement (the one about continuity) is not the best possible. It can be improved, which is something that we do not get automatically from uniform convergence, but we have a special theorem about it.

Theorem (Abel's convergence theorem).
Consider a power series with center a and radius of convergence R > 0. Let f be the function that this series defines on its region of convergence.
• If the series converges at a + R, then f is continuous there from the left and the series converges uniformly also on the interval [a,a + R].
• If the series converges at a − R, then f is continuous there from the right and the series converges uniformly also on the interval [a − R,a].

This can be summed up like this.

Assume that a power series with center a and radius of convergence R > 0 converges on an interval I (so the interval I surely contains (a − R,a + R), but it may also include one or both its endpoints). Then the resulting function f is continuous on I and the convergence of this series is uniform on any closed interval J that is a subset of I.

In particular, if I is closed, that is, if the given power series actually converges on [a − R,a + R], then it converges also uniformly on this whole set.