In this section we develop general methods for finding power series representations. Suppose that f (x) is represented by a power series centered at x = c on an interval (c − R, c + R) with R > 0: According to THEOREM 2 Term-by-Term Differentiation and Integration Assume that has radius of convergence R > 0. Then F (x) is differentiable on (c − R, c + R) [or for all x if R =differentiable Furthermore, we can integrate and differentiate term by term. For x (c − R, c + R),

we can compute the derivatives of f (x) by differentiating the series expansion term by term:derivatives In general, Setting x = c in each of these series, we find that We see that a k is the kth coefficient of the Taylor polynomial studied in Section 9.4Section 9.4

This proves the next theorem. THEOREM 1 Taylor Series Expansion If f (x) is represented by a power series centered at c in an interval |x − c| 0, then that power series is the Taylor series In the special case c = 0, T (x) is also called the Maclaurin series: We define the nth Taylor polynomial centered at x = a as follows: Therefore f (x) = T (x), where T (x) is the Taylor series of f (x) centered at x = c:

Find the Taylor series for f (x) = x −3 centered at c = 1.

we can write the coefficients of the Taylor series as: The Taylor series for f (x) = x −3 centered at c = 1 is

However, there is no guarantee that T(x) converges to f(x), even if T(x) converges. To study convergence, we consider the kth partial sum, which is the Taylor polynomial of degree k: Theorem 1 tells us that if we want to represent a function f (x) by a power series centered at c, then the only candidate for the job is the Taylor series: In Section 8.4, we defined the remainder asSection 8.4 Since T(x) is the limit of the partial sums T k (x), we see that the Taylor series converges to f(x)

THEOREM 2 Let I = (c − R, c + R), where R > 0. Suppose there exists K > 0 such that all derivatives of f are bounded by K on I: Then f (x) is represented by its Taylor series in I: Taylor expansions were studied throughout the seventeenth and eighteenth centuries by Gregory, Leibniz, Newton, Maclaurin, Taylor, Euler, and others. These developments were anticipated by the great Hindu mathematician Madhava (c. 1340–1425), who discovered the expansions of sine and cosine and many other results two centuries earlier. There is no general method for determining whether R k (x) tends to zero, but the following theorem can be applied in some important cases. Taylor Expansion

Expansions of Sine and Cosine Show that the following Maclaurin expansions are valid for all x. Recall that the derivatives of f (x) = sinx and their values at x = 0 form a repeating pattern of period 4:

For f (x) = cosx, the situation is reversed. The odd derivatives are zero and the even derivatives alternate in sign: f (2n) (0) = (−1) n cos0 = (−1) n. Therefore the nonzero Taylor coefficients for cosx are a 2n = (−1) n /(2n)!. In other words, the even derivatives are zero and the odd derivatives alternate in sign: f (2n+1) (0) = (−1) n. Therefore, the nonzero Taylor coefficients for sinx are We can apply Theorem 2 with K = 1 and any value of R because both sine and cosine satisfy |f (n) (x)| ≤ 1 for all x and n. The conclusion is that the Taylor series converges to f (x) for |x| < R. Since R is arbitrary, the Taylor expansions hold for all x. THM 2

We have f (n) (c) = e c for all x, and thus Taylor Expansion of f(x) = e x at x = c Find the Taylor series T(x) of f (x) = e x at x = c. Because e x is increasing for all R > 0 we have |f (k) (x)| ≤ e c+R for x (c − R, c + R). Applying Theorem 2 with K = e c+R, we conclude that T(x) converges to f(x) for all x (c − R, c + R). Since R is arbitrary, the Taylor expansion holds for all x. For c = 0, we obtain the standard Maclaurin series

Shortcuts to Finding Taylor Series There are several methods for generating new Taylor series from known ones. First of all, we can differentiate and integrate Taylor series term by term within its interval of convergence, by Theorem 2 of Section We can also multiply two Taylor series or substitute one Taylor series into another (we omit the proofs of these facts). Section 11.6 Find the Maclaurin series for f (x) = x 2 e x.

Substitution Find the Maclaurin series for Substitute −x 2 for x in the Maclaurin series for e x. The Taylor expansion of e x is valid for all x, so this expansion is also valid for all x.

Integration Integrate the geometric series with common ratio −x (valid for |x| < 1) and c = 1, to find the Maclaurin series for f (x) = ln(1 + x). The constant of integration is zero because ln(1 + x) = 0 for x = 0. This expansion is valid for |x| < 1. It also holds for x = 1 (see Exercise 84).Exercise 84

In many cases, there is no convenient general formula for the Taylor coefficients, but we can still compute as many coefficients as desired. Multiplying Taylor Series Write out the terms up to degree five in the Maclaurin series for f (x) = e x cosx. We multiply the fifth-order Taylor polynomials of e x and cos x together, dropping the terms of degree greater than 5: Distributing the term on the right (and ignoring terms of degree greater than 5), we obtain:

In the next example, we express the definite integral of sin(x 2 ) as an infinite series. This is useful because the integral cannot be evaluated explicitly. Figure 1 shows the graph of the Taylor polynomial T 12 (x) of the Taylor series expansion of the antiderivative.Figure 1 Graph of T 12 (x) for the power series expansion of the antiderivative (a) Express J as an infinite series. (b) Determine J to within an error less than 10 −4. We obtain an infinite series for J by integration:

The infinite series for J is an alternating series with decreasing terms, so the sum of the first N terms is accurate to within an error that is less than the (N + 1) st term. The absolute value of the fourth term (1/75,600) is smaller than 10 −4 so we obtain the desired accuracy using the first three terms of the series for J:absolute value The error satisfies The percentage error is less than 0.005% with just three terms. (b) Determine J to within an error less than 10 −4.

Binomial Series Isaac Newton discovered an important generalization of the Binomial Theorem around For any number a (integer or not) and integer n ≥ 0, we define the binomial coefficient:

Let The Binomial Theorem of algebra states that for any whole number a,whole number Setting r = 1 and s = x, we obtain the expansion of f (x):

We derive Newton’s generalization by computing the Maclaurin series of f (x) without assuming that a is a whole number. Observe that the derivatives follow a pattern: derivatives In general, f (n) (0) = a (a − 1)(a − 2)…(a − (n − 1)) and Hence the Maclaurin series for f (x) = (1 + x) a is the binomial series

THEOREM 3 The Binomial Series For any exponent a and for |x| < 1, Find the terms through degree four in the Maclaurin expansion of f (x) = (1 + x) 4/3

Find the Maclaurin series for First, let’s find the coefficients in the binomial series for (1 + x) −1/2 : The general pattern is

Thus, the following binomial expansion is valid for |x| < 1: If |x| < 1, then |x| 2 < 1, and we can substitute -x 2 for x to obtain Find the Maclaurin series for