Taylor series, expanding functions, summing up series

Here we address the main question. We have a function f and a number a and we would like to express this function as a sum of a power series with center a. We start with some theory and then look at expanding using properties. At the end we address the reverse process, to a given series we try to find its sum.

As we outlined in general in the section Systems of functions, there are two questions to be answered: What functions can be expressed in this way and how do we find the series for them. And the first question is indeed hard, we will just give a partial answer later. We start by a definition.

Definition.
Let f be a function and a some point from the interior of its domain.
We say that we expanded the function f in a power series (at a) if we find some power series such that

on some neighborhood of a.
This power series is then called an expansion of f.

From the results in the previous section it follows that every function that can be expanded in this way must have derivatives of all orders at a. This tells us that there is no point in trying to expand other functions. It also says that when we talk about expanding, we have all these derivatives available.

The starting point of our investigation is the Corollary at the end of the previous section. Assume that we have a function that can be expressed by a series at a. If we substitute this center a into the formula for the n-th derivative, we immediately get the following.

Theorem (uniqueness of expansion).
Let f be a function that can be expanded in a power series

on some neighborhood of a. Then for every k, the coefficients ak necessarily satisfy

In other words, there is only one way to expand a function into a power series (if it can be done at all). This unique power series deserves a name.

Definition.
Let f be a function and a some point from the interior of its domain. Assume that f has derivatives of all orders at a. Then we define its Taylor series at a by the formula

The formula for coefficients is the same as in the case of Taylor polynomial, so this is really something like an "infinite Taylor polynomial".

Note that the theorem above was an implication. This means that once we have a function expanded into a power series, then this series is necessarily the Taylor series. However, if we take a function with all derivatives at a and construct the Taylor series according to the above formula, then there is no guarantee that this series will be convergent also somewhere else than at a, and even if it converges, there is no guarantee that it actually converges to the original function f. For instance, in this note we show an example of a function that is "nice" by most standards (it has derivatives of all orders everywhere), its Taylor series converges to its sum uniformly on the whole real line, yet the sum of this Taylor series is equal to the original function only at one point, the center of the series (where it actually has to work for any function and series, so it has no choice). This situation where we have trouble with getting the series to go to the right place appears often when we work with systems of functions, and it inspires the following notation.

The tilde means that the series on the right was obtained from f using the above formulas (it is the Taylor series for f ), but it also means that the process of creating this Taylor series is purely formal, since at this moment we have no information whether the sum of this series actually has anything in common with f.

Obviously, our aim is to change this tilde into equality, that is, we would like to see that the Taylor series on the right actually converges to f at least somewhere (apart from a, where the convergence is automatic). How do we recognize this? By definition we should take the partial sums TN of T and check whether they converge to f.

Equivalently, we would like to find out how the difference f − TN| behaves, hoping that it would go to zero as N goes to infinity. However, this is actually something that we already saw before, since TN are exactly Taylor polynomials (see Derivatives - Theory - Applications), so we know quite a bit about this difference. In particular, we have the Lagrange estimate of it. Thus we can restate the question like this:

Assume that f has derivatives of all orders on some neighborhood U of a. Let x be such that the closed interval I with endpoints a and x lies in U. We want to know when it is true that

Note that in this formula, x is fixed and thus x − a is a constant. When we recall that cN/N! tends to zero at infinity, we see that all depends on how quickly the maximum grows with N. If it does not grow at all or grows very slowly, then the Lagrange estimate of the error goes to zero. Thus we get the following.

Theorem.
Let f be a function such that it has derivatives of all orders on some neighborhood U of a point a. Let T be its Taylor series at a.
If there is a constant M such that f (N)(t)| ≤ M for all positive integers N and all t from U, then T converges (uniformly) to f on U.

Thus we get one group of functions whose Taylor series is actually equal to them, namely functions with uniformly bounded derivatives. However, this condition is too restrictive, there are also other functions whose Taylor series converge to where they should. Finding the right characterization is a hard task, definitely beyond the scope of Math Tutor, and we leave it to professional mathematicians.

There are some functions for which we can easily find their expansions. By uniqueness of expansion, every power series is its own Taylor series (with the same center). Expanding with a different center than "their own" is not so simple any more, but it can still be done in case of "finite" series, that is, in case of polynomials. We just need to create the right center. For instance, if we want to find the Taylor series for x2 with center a = 1, we simply do

x2 = [(x − 1) + 1]2 = (x − 1)2 + 2(x − 1) + 1.

The latter is a power series with center 1 (its coefficients are zero for k > 2), and so it is the desired Taylor series. (We would get the same series if we used the formula from the definition, of course.) Trying this trick for an infinite series would not be nice any more.

Now we now look at six famous expansions that are the basis of most other expansions.

Theorem.
The following expansions are true on the indicated sets.

As these formulas suggest, the choice a = 0 for the center is very popular. In fact, it is so popular that some people use the traditional name McLaurin series for Taylor series with this center. However, it is not universal and you find it especially in older texts.

In all six cases we find the formula for Taylor polynomials easily from definition. How about their convergence? The second and third formula is clear, convergence follows from the previous theorem (sine and cosine have all derivatives bounded by 1 everywhere).

For the exponential we can use this theorem as well, but we cannot do it globally, since the exponential is bounded only on bounded sets. However, this is no problem. If we take any real number x, we can always consider for instance the open neighborhood U2|x|(0). This is a bounded set and all derivatives of ex are therefore uniformly bounded on it, so by the above theorem, the Taylor series converges to the exponential there; in particular it converges to the exponential at x, since x belongs to this neighborhood. For the sake of completeness we include a direct proof using the Lagrange estimate, see this note.

The expansion for logarithm can be actually obtained in several ways, we will return to it also below; concerning its convergence, the interval we stated is clearly the largest one can hope for, since in the previous section we proved that this interval is exactly the region of convergence of this Taylor series. Thus it remains to show that the sum of the series on this region is exactly ln(x). However, here one cannot use the above Theorem, since derivatives are not uniformly bounded. This case is therefore far trickier, for more details see this note.

The fifth formula is just the geometric series, so there is nothing new about it.

Probably the most interesting of the six is the last formula, also called the binomial series or binomial expansion. It is true for all c and A, which in particular shows, that the binomial numbers can be also defined for a general number A. It is done by the second formula with the understanding that when k = 0, we get 1. As an example we show how the series looks like when c = 1, which is a frequently used form. We also show a special formula for "minus A", it is sometimes useful when working with reciprocals. Both work for |x| < 1.

Note also that if A is a natural number then by definition, for k > A all the coefficients are zero. Thus the sum is finite, it always converges and it is in fact the familiar binomial formula that we know from high-school.

One reason why these formulas are important is this: We rarely expand functions by definition, the way we did it above. In most cases we use various tricks to somehow get new expansions out of those that we already know. The above six are the starting point for most such calculations. The tricks are coming up just now.

Expansion and operations

In order to get expansions of new functions from known expansions we use properties known to work for power series. The theorem on algebraic operations from the previous section gives this handy fact.

Theorem.
Assume that we have these expansions on some neighborhood of a.

Then on this neighborhood also

How about the division f /g? Obviously it is enough to know how to expand the function 1/f (assuming that f (a) ≠ 0).

Theorem.
Assume a function f can be expanded in a Taylor series at a and that f (a) ≠ 0. Then also the function 1/f can be expanded in a Taylor series at a and this series has a positive radius of convergence.

This theorem does not offer any formula for the new series - and with a good reason, there is no reasonable way to specify it. In practice we use the method of undetermined coefficients. We can actually apply it directly to f /g and take 1/f as the special case. Assume that both f and g are expanded into their Taylor series at a and that g(a) ≠ 0. By the above theorem, also their ratio can be expanded in a Taylor series with yet unknown coefficients ck. We can write

The resulting system of infinitely many linear equations with infinitely many unknown ck is usually impossible to solve explicitly, but it allows us to calculate as many coefficients ck as we want, which is very useful when we want to approximate f /g using just a finite part of the Taylor series. For an example we refer to Solved problems, namely this problem (it is unusually nice but illustrates the idea well) and this problem (it is more typical).

We also often relate functions to other functions using derivative or integral. Then we can use this theorem from the previous section.

Theorem.
Assume that we have the following expansion on some neighborhood of a.

Then on this neighborhood also

One can also use indefinite integral in the second statement, but then it is necessary to work out the right constant. We will show an example below. Finally, there are other operations that can help in expanding new functions.

Theorem (substitution in series).
Assume that we have the following expansion on some neighborhood of a.

Then also

The first statement holds for any A and the radius of convergence of the new series is the same as the original one, but its center (and region of convergence) are shifted accordingly.
The second statement holds for all non-zero A and the radius of convergence of the new series is Rf /|A|.

The third statement is a bit shaky, since the outcome is a power series only for very special g. The region of convergence then has to be investigated individually.

In fact, in the first two statements we actually know even the region of convergence of the new series. In the first statement, the new region is just the original one but shifted. The region of convergence in the second case can be obtained by shrinking the original one, for A negative one also has to flip it around a. That is, if the original series converges at the right endpoint, then the new one also converges at the right endpoint for A > 0 and it converges at the left endpoint for A < 0. It is all actually just a common sense and understanding substitution well.

Example: We will deduce the expansion for f (x) = ln(1 − x) at a = 0.

The key to this expansion is the observation that f ′(x) = −1/(1 − x) and this is exactly the sum of the geometric series that everybody knows. So we start by taking the formula for geometric series (we use variable t) and then we integrate both sides from 0 to x

This series is often used for logarithm. The original series converges for t satisfying −1 < t < 1, so the new one also converges at least on this set and its radius of convergence must stay the same (that is, 1). However, we may have gained convergence at some endpoints. When we put in x = −1 and x = 1 and use suitable tests, we find that the region of convergence of this series is [−1,1). How do we know that the sum of this series at x = −1 is equal to ln(2)? One possible argument is to use continuity, see Abel's convergence theorem in the previous section. Thus we have expansion valid on [−1,1).

If we write this result using y and then use the substitution y = −x, we get

The original series converges for y satisfying −1 ≤ y < 1, so when we substitute -x for y, we get the region of convergence (−1,1] for the new series.

On the other hand, if we substitute y = 1 − x, then we get exactly the series for ln(x) that we had in the above Theorem.

 

It is perhaps surprising in view of difficulties we had with finding series for 1/f that there is in fact an explicit way to expand inverse function.

Theorem (Lagrange inversion formula).
Assume that a function f can be expanded in a series on some neighborhood of a point a. Assume further that f ′(a) is not zero. Denote b = f (a).
Then there is a neighborhood of a on which the function f is invertible and a neighborhood of b on which this inverse f−1 can be expanded in a power series. Moreover, this series can be found as

We conclude this part with yet another useful statement.

Theorem.
Assume that we have the following expansion on some neighborhood of a.

Then on this neighborhood we also have the following expansions.

(i) For any positive integer n,

(ii) If f has a root of multiplicity n at a, then

The second statement makes sense for the following reason. If a is a root of multiplicity n for f, then the first n − 1 derivatives of f at a are zero (see this note). Therefore also the first n − 1 coefficients of the Taylor series are zero, which means that in fact in the series for f the indexing starts at n, not at 0. This explains why after shifting the index we get the second formula.

For more examples of using the above properties to expand functions see Methods Survey and Solved Problems - Series of functions.

We conclude this part on properties of Taylor series with a fact that one could already guess from the six expansions above.

Fact.
Let f be a function that has Taylor series with center at a = 0, let ak be its coefficients.
If f is odd, then ak = 0 for all even k.
If f is even, then ak = 0 for all odd k.

Summing up series

We mentioned several times that summing up series can be rather difficult. The above tricks for expanding a function can be also used to sum up a power series (assuming that we are lucky). The basic idea is to change the given series into one that we already know by applying transformations from the above theorems to it while keeping track of what it does with its sum. It is best explained on an example.

Example: Find the sum of

Is there a series that looks like this? There is a series that has terms yk/k, we can almost get there by denoting y = x + 1. It remains to fix the power so that it agrees with the number in the denominator, but that is not a problem. Here we go.

What about the value x = −1? This is simple, we just put it into the given series and find that f (−1) = 0.

We can also try another approach. Often we need to get rid of extra k in a series. In our series we have extra k + 1 in the denominator, so we would appreciate if k + 1 appeared also in the numerator. There is a way to get k in the numerator, namely by taking derivative of a series. We need k + 1, so before differentiating we need to create (x + 1)k + 1 there. Once there is no division by k + 1, we use geometric series.

It remains to fix the right C. The easiest way it to simply put some nice x into the last equality. Since on the left we substitute into f, that is, into the given series, we do not have much choice. The only value that we can really put in this series is x = −1, but unfortunately, that is the one value we are not allowed in the last line. What can we do?

It would seem that we could fix this problem by using the second last line. Looking closer we see this is not true. Note that if x = −1, then on the second line we multiply by zero, which is not allowed in the equation. Thus for this value all subsequent equalities are suspect. Fortunately for us, we can look still closer. What do we have on that second last line? We have one function on the left and another on the right and we know for sure that they are equal for all x from a certain interval except one point in its middle. Since these two functions are continuous, we also must have equality at this particular point. We are saved, we substitute −1 in the second last line and learn that C = 0. Since we used integration that influences endpoints, we should now check on convergence at −2 and at 0 and we would get the same conclusion as above.


Fourier series
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