Expanding in power series: Methods survey

Given a function f and a center a, we want to expand f into a power series with this center.

We do it by applying various transformations to series that we already know. In most cases we can do with series for the exponential, sine and cosine, and the geometric series, so you should memorize them.

Sometimes one can also use the binomial series and perhaps some series for logarithm.

The basic tool is substitution.

This is used very often, for this reason we used y in the above expansions. The validity of a new expansion is determined easily by applying the corresponding restriction on y to the substituted expression A⋅(x − a) (see Example below).

Another popular modification to the above expansions is multiplication by a power of (x − a), sometimes one can even divide by such a power (see Taylor series in Theory - Series of functions). It is also good to recall that creating a series for a polynomial is trivial. This means that when deciding what series should our expansion start with, we can disregard parts of the given expression that will not cause troubles, namely multiplication by polynomials and addition of polynomials.

It is obvious that if we want to obtain a power series with center a, then y has to feature exactly this term (x − a); we also saw above that a multiplicative constant does not hurt. This observation leads to the following algorithm.

How to expand.
Step 1. Identify which series will be the basis for your expansion, keeping in mind that substitution as above can be used as well as multiplication by a suitable power. Form a plan.
There are three typical situations.
•• The given function is essentially one of the functions whose expansion you already know, with minor modifications. Then it is just a question of rewriting the given function to a suitable form as explained in Step 2.
•• The given function consists of functions whose expansions you already know (with minor modifications) put together by operations. Unless some trick can be used, this means that you have to expand each of these functions and then put the resulting series together. This can be a problem, for typical situations see this note.
•• The given function has parts that cannot be handled by the above approaches. Then you have to look for some trick. Very often you can use properties of power series, namely facts about differentiating or integrating series. If you differentiate or integrate the given function (or its troublesome part), do you get something that you can already expand (with modifications as above)? If this is true, use this known expansion and then reverse the operation that got you there (see Example below). This is most often used to handle expansions of logarithm or expressions like 1/(A + x)², both lead to the geometric expansion.

Step 2. After Step 1 it is known which function will be used as the basis for expansion, but before expanding, rewrite the given function so that all x in it appear in the form (x − a).
Usually some unwanted parts are created during this procedure; get rid of them using algebra and identities so that when it comes to expanding, you can get to known expansions using substitution. Then apply the expansion(s).
Typical situations can be found in these examples which also serve to illustrate the next steps.

Students usually find expansions using geometric series to be the most troublesome. The above example shows how one should proceed. First, create the term with appropriate center. Then rewrite the expression so that the constant in the denominator is in front and there is minus after it. Finally, factor out this constant. Note also how we created the right center for the polynomial 2x in the first example.

Step 3. Polish up the answer so that at the end there is just one power series.
This usually means moving constants A from substitution out of terms (Ax − B). Other frequently used operations: multiplicative terms in front of series are moved inside, series are added together. Sometimes you also have to multiply, divide, differentiate or integrate series. Again, recall that when adding series, indexing and powers have to be fixed first so that one can add corresponding powers (see Manipulating (power) series in Methods Survey - Series of functions).

Step 4. Determine the region where the resulting expansion is valid.
This is determined in the step where a series appeared in our calculations. That step was valid only for certain values of y, from which one easily finds restriction on x.

We will now show an example, it will be a bit longer to showcase many possible procedures (some of them less pleasant). For other examples see Solved Problems - Series of functions.

Example: Expand into a series with center a = 1 the function

What is the plan? Obviously, we will have to expand logarithm into a series with terms (x + 1). Some people remember it, but some do not, so we will try to somehow deduce this expansion. There is no way to get logarithm out of the four functions above using algebraic operations, so this will not help. Logarithm is an inverse function for the exponential whose expansion we know, but the Lagrange inversion formula is so ugly that we will leave it as the last (desperate) resort. The real key is the observation that when we differentiate logarithm, we get 1/x, which is something that can be handled using the geometric expansion. From this expansion we can get back to logarithm by integration and we are done.

Having expanded the logarithm, it remains to somehow involve the division by x. There are essentially two possibilities. One is to take the given expression as a product of logarithm and 1/x. We can expand both and then use the Cauchy product. The other option is to expand x itself and then use the algorithm for dividing series. We will show both ways. Thus the strategy is formed and we may proceed.

First we will expand the logarithm. We start with expansion of its derivative.

There are two ways to get to logarithm from 1/x. We can rewrite the above expansion using t as variable and then apply to both sides the definite integral from, say, 1 to x. The other possibility is to integrate both sides using antiderivatives, but then there is the problem of the constant. Still, it is probably easier than the definite integral way.

We shifted index up by 1 (the "new" k is given by k + 1) to fix the exponent in the power. How do we determine the right C? That is simple, this equality is supposed to be true for all x from the indicated set, so we can substitute for instance x = 1 to both sides and find that C = 0.

Now it is time to factor in the division by x. First the product of series approach. We are actually lucky and save time, since we already know how to expand 1/x.

This is a typical result that comes out of the Cauchy product. Do we know some formula for the partial sum of the harmonic series, so that we could write the coefficients in the answer somehow better? Not really, so we leave it as it is. Now for the second approach.

The polynomial x has 1 + (x − 1) as its expansion with center 1. Moreover, we know that the given function is "nice" at 1 (we do not divide by zero), so the ratio should also have some expansion at 1, we write a general power series there. We get

Usually we would need to use the Cauchy product on the right, but here we can actually just simplify it and express it as one power series.

Now we have two series that are equal, so by uniqueness of expansion, their coefficients must match. We get the system of equations

Thus we obtain a₀ = 0, a₁ = 1, a₂ = −1 − 1/2, a₃ = 1 + 1/2 + 1/3, a₄ = −1 − 1/2 − 1/3 − 1/4. In this way we can find as many coefficients as we need. We are getting the same coefficients as above and the same problem, we are not able to express them using a nice formula. Sometimes we are lucky and we get a pattern that we can recognize, then we can make a guess for a general k and prove it by induction. Not this time, though.

Summing up function series
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