and Maclaurin Series
In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we find such representations?
We start by supposing that is any function that can be represented by a power series:
Let's try to determine what the coefficients must be in terms of . To begin, notice that if we put in Equation 1, then all terms after the first one are and we get
By Theorem 8.6.2, we can differentiate the series in Equation 1 term by term:
and substitution of in Equation 2 gives Now we differentiate both sides of Equation 2 and obtain
Again we put in Equation 3. The result is
Let's apply the procedure one more time. Differentiation of the series in Equation 3 gives
and substitution of in Equation 4 gives
By now you can see the pattern. If we continue to differentiate and substitute , we obtain
Solving this equation for the th coefficient , we get
This formula remains valid even for if we adopt the conventions that and .
Thus we have proved the following theorem.
THEOREM: If has a power series representation (expansion) at , that is, if
Substituting this formula for back into the series, we see that if has a power series expansion at , then it must be of the following form.
The series in Equation 6 is called the Taylor series of the function at (or about or centered at ). For the special case the Taylor series becomes
This case arises frequently enough that is is given the special name Maclaurin series.
NOTE: We have shown that if can be represented as a power series about , then is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. For example, one can show that the function defined by
is not equal to its Maclaurin series.
EXAMPLE 1: Find the Maclaurin series of the function and its radius of convergence.
Solution: If , then , so for all . Therefore, the Taylor series for at (that is, the Maclaurin series) is
To find the radius of convergence we let . Then
so, by the Ratio Test, the series converges for all and the radius of convergence is .
The conclusion we can draw from (5) and Example 1 is that if has a power series expansion at , then
So how can we determine whether does have a power series representation?
Let's investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if has derivatives of all orders, when is it true that
As with any convergent series, this means that is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are
Notice that is a polynomial of degree called the th-degree Taylor polynomial of at . For instance, for the exponential function , the result of Example 1 shows that the Taylor polynomials at (or Maclaurin polynomials) with , and 3 are
In general, is the sum of its Taylor series if
If we let
then is called the remainder of the Taylor series. If we can somehow show that , then it follows that
We have therefore proved the following.
THEOREM: If , where is the th-degree Taylor polynomial of at and
for , then is equal to the sum of its Taylor series on the interval .
In trying to show that for a specific function , we usually use the expression in the next theorem.
THEOREM (TAYLOR'S FORMULA): If has derivatives in an interval that contains the number , then for in there is a number strictly between and such that the remainder term in the Taylor series can be expressed as
NOTE 1: For the special case , if we put and in Taylor's Formula, we get
which is the Mean Value Theorem. In fact, Theorem 9 can be proved by a method similar to the proof of the Mean Value Theorem.
NOTE 2: Notice that the remainder term
is very similar to the terms in the Taylor series except that is evaluated at instead of at . All we say about the number is that it lies somewhere between and . The expression for in Equation 10 is known as Lagrange's form of the remainder term.
NOTE 3: In Section 8.8 we will explore the use of Taylor's Formula in approximating functions. Our immediate use of it is in conjunction with Theorem 8. In applying Theorems 8 and 9 it is often helpful to make use of the following fact:
This is true because we know from Example 1 that the series converges for all and so its th term approaches 0.
EXAMPLE 2: Prove that is equal to the sum of its Taylor series with (Maclaurin series).
Proof: If , then , so the remainder term in Taylor's Formula is
where lies between 0 and . (Note, however, that depends on .) If , then , so . Therefore
by Equation 11, so as by the Squeeze Theorem. If , then , so and
Again . Thus, by (8), is equal to the sum of its Maclaurin series.
From Example 2 it follows that
In particular, if we put in Equation 12, we obtain
EXAMPLE 3: Find the Taylor series for at .
Solution: We have and so, putting in the definition of a Taylor series (6), we get
Again it can be verified, as in Example 1, that the radius of convergence is . As in Example 2 we can verify that , so
We have two power series expansions for , the Maclaurin series in Equation 12 and the Taylor series in Equation 14. The first is better if we are interested in values of near and the second is better if is near 2.
EXAMPLE 4: Find the Maclaurin series for and prove that it represents for all .
Solution: We arrange our computation in two columns as follows:
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows:
Using the remainder term (10) with , we have
where and lies between 0 and . But is or . In any case, and so
By Equation 11 the right side of this inequality approaches as , so by the Squeeze Theorem. It follows that as , so is equal to the sum of its Maclaurin series by Theorem 8. Thus
EXAMPLE 5: Find the Maclaurin series for .
Solution 1: We arrange our computation in two columns as follows:
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows:
Solution 2: We differentiate the Maclaurin series for given by Equation 16:
Since the Maclaurin series for converges for all , Theorem 8.6.2 tells us that the differentiated series for also converges for all . Thus
EXAMPLE 6: Find the Maclaurin series for .
Solution: Instead of computing derivatives and substituting in Equation 7, it's easier to multiply the series for (Equation 17) by :
REMARK: The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 8.6 are indeed the Taylor or Maclaurin series of the given functions because Theorem 5 asserts that, no matter how we obtain a power series representation
it is always true that . In other words, the coefficients are uniquely determined.
EXAMPLE 7: Find the Maclaurin series for , where is any real number.
Solution: Arranging our work in columns, we have
Therefore, the Maclaurin series of is
This series is called the binomial series. If its th term is , then
Thus by the Ratio Test the binomial series converges if and diverges if .
The traditional notation for the coefficients in the binomial series is
and these numbers are called the binomial coefficients. The following theorem states that is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term approaches , but that turns out to be quite difficult.
THEOREM (THE BINOMIAL SERIES): If is any real number and , then
Although the binomial series always converges when , the question of whether or not it converges at the endpoints, , depends on the value of . It turns out that the series converges at 1 if and at both endpoints if . Notice that if is a positive integer and , then the expression for contains a factor , so
for . This means that the series terminates and reduces to the ordinary Binomial Theorem when is a positive integer.
EXAMPLE 8: Find the Maclaurin series for and its radius of convergence.
Solution: We write in a form where we can use the binomial series:
Using the binomial series with and with replaced by , we have
We know from (18) that this series converges when , that is, , so the radius of convergence is .
We collect in the following table, for future reference, some important Maclaurin series that we have derived in this section and the preceding one.
One reason that Taylor series are important is that they enable us to integrate functions that we couldn't previously handle. In fact, the function can't be integrated by techniques discussed so far because its antiderivative is not an elementary function (see Section 6.4). In the following example we write as the Maclaurin series to integrate this function.
(a) Evaluate as an infinite series.
(b) Evaluate correct to within an error of 0.001.
(a) First we find the Maclaurin series for . Although it's possible to use the direct method, let's find it simply by replacing with in the series for given in the table of Maclaurin series. Thus, for all values of ,
Now we integrate term by term:
This series converges for all because the original series for converges for all .
(b) The Fundamental Theorem of Calculus gives
The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than
Another use of Taylor series is illustrated in the next example. The limit could be found with l'Hospital's Rule, but instead we use a series.
EXAMPLE 10: Evaluate .
Solution: Using the Maclaurin series for , we have
because power series are continuous functions.