A power series with center c is an infinite series where x is a variable. For example, is a power series with center c = 2.

There is a surprisingly simple way to describe the set of values x at which a power series F(x) converges. According to our next theorem, either F(x) converges absolutely for all values of x or there is a radius of convergence R such that Many functions that arise in applications can be represented as power series. This includes not only the familiar trigonometric, exponential, logarithm, and root functions, but also the host of “special functions” of physics and engineering such as Bessel functions and elliptic functions.

This means that F(x) converges for x in an interval of convergence consisting of the open interval (c − R, c + R) and possibly at one or both of the endpoints c − R and c + R (Figure 1). Note that F (x) automatically converges at x = c because We set R = 0 if F (x) converges only for x = c, and we set R = if F (x) converges for all values of x. Interval of convergence of a power series.

THEOREM 1 Radius of Convergence Every power series has a radius of convergence R, which is either a nonnegative number (R ≥ 0) or infinity (R = ∞). If R is finite, F (x) converges absolutely when |x − c| < R and diverges when |x − c| > R. If R = ∞, then F (x) converges absolutely for all x. We see that there are two steps in determining the interval of convergence of F (x): Step 1. Find the radius of convergence R (using the Ratio Test, in most cases). Step 2. Check convergence at the endpoints

Using the Ratio Test Where does converge? Step 1. Find the radius of convergence. Let and compute the ratio ρ of the Ratio Test: We find that Thus F(x) converges if |x| < 2. Similarly, Thus F(x) converges if |x| > 2. Therefore, the radius of convergence is R = 2. Ratio Test

THEOREM 1 Ratio Test Assume that the following limit exists: (i) If ρ < 1, then converges absolutely. (ii) If ρ > 1, then diverges. (iii) If ρ = 1, the test is inconclusive (the series may converge or diverge). Example

Using the Ratio Test Where does converge? Step 2. Check the endpoints. The Ratio Test is inconclusive for x = ±2, so we must check these cases directly: Both series diverge. We conclude that F(x) converges only for |x| < 2 (Figure 2).

Where does converge? We compute ρ with We find that Thus F (x) converges absolutely on the open interval (1, 9) of radius 4 with center c = 5. In other words, the radius of convergence is R = 4. Next, we check the endpoints:

THEOREM 2 Leibniz Test for Alternating Series Assume that {an} is a positive sequence that is decreasing and converges to 0: Then the following alternating series converges:

We conclude that F (x) converges for x in the half-open interval (1, 9] shown in Figure 3.

An Even Power Series Where does converge? Some power series contain only even (or odd) powers of x. The Ratio Test can still be used to find the radius of convergence. An Even Power Series Where does converge? Recursion Relation Furthermore, (2n + 2)! = (2n + 2)(2n + 1)(2n)!, so THEOREM 1 Ratio Test Assume that the following limit exists: Thus ρ = 0 for all x, and F (x) converges for all x. The radius of convergence is R = (i) If ρ < 1, then converges absolutely. (ii) If ρ > 1, then diverges. (iii) If ρ = 1, the test is inconclusive (the series may converge or diverge).

Geometric series are important examples of power series Geometric series are important examples of power series. Recall the formula valid for |r| < 1. Writing x in place of r, we obtain a power series expansion with radius of convergence R = 1: The next two examples show that we can modify this formula to find the power series expansions of other functions.

Geometric Series Prove that Substitute 2x for x:

Find a power series expansion with center c = 0 for and find the interval of convergence. We need to rewrite f (x) so we can use . We have

Our next theorem tells us that within the interval of convergence, we can treat a power series as though it were a polynomial; that is, we can differentiate and integrate term by term.

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 = Furthermore, we can integrate and differentiate term by term. For x (c − R, c + R), Proof These series have the same radius of convergence R. 2nd Proof

Differentiating a Power Series Prove that for −1 < x < 1, THEOREM 2 Sum of a Geometric Series Let c 0. If |r| < 1, then The geometric series has radius of convergence R = 1: By Theorem 2, we can differentiate term by term for |x| < 1 to obtain THM 2

Power Series for Arctangent Prove that for −1 < x < 1, Integral Formulas (7.8) THM 2

Setting x = 0, we obtain A = tan −1 0 = 0.

GRAPHICAL INSIGHT Let’s examine the expansion of the previous example graphically. The partial sums of the power series for f(x) = tan−1 x are For large N we can expect SN(x) to provide a good approximation to f (x) = tan−1 x on the interval (−1, 1), where the power series expansion is valid. Figure 4 confirms this expectation: The graphs of S50(x) and S51(x) are nearly indistinguishable from the graph of tan−1 x on (−1, 1). Thus we may use the partial sums to approximate the arctangent. For example, tan−1 (0.3) is approximated by Since the power series is an alternating series, the error is less than the first omitted term: The situation changes drastically in the region |x| > 1, where the power series diverges and the partial sums SN(x) deviate sharply from tan−1 x.

S50(x) and S51(x) are nearly indistinguishable from tan−1 x on (−1, 1).

Power Series Solutions of Differential Equations Power series are a basic tool in the study of differential equations. To illustrate, consider the differential equation with initial condition We know that f (x) = ex is the unique solution, but let’s try to find a power series that satisfies this initial value problem. We have a0 = a1, a1 = 2a2, a2 = 3a3, a3 = 4a4,… In other words,

An equation of this type is called a recursion relation An equation of this type is called a recursion relation. It enables us to determine all of the coefficients an successively from the first coefficient a0, which may be chosen arbitrarily. For example, To obtain a general formula for an, apply the recursion relation n times: We conclude that

In , we showed that this power series has radius of convergence R = Example 3 Moreover, F (0) = a0, so the initial condition y (0) = 1 is satisfied with a0 = 1. What we have shown is that f (x) = ex and F (x) with a0 = 1 are both solutions of the initial value problem. They must be equal because the solution is unique. This proves that for all x, In this example, we knew in advance that y = ex is a solution of but suppose we are given a differential equation whose solution is unknown. We can try to find a solution in the form of a In favorable cases, the differential equation leads to a recursion relation that enables us to determine the coefficients an.

Find a power series solution to the initial value problem Assume this equation has a power series solution Then

Now substitute the series for y, y , and y into the given differential equation to determine the recursion relation satisfied by the coefficients an: We combine the first three series into a single series using n (n − 1) + n − 1 = n2 − 1 and we shift the fourth series to begin at n = 2 rather than n = 0.

The differential equation is satisfied if The first few terms on each side of this equation are Matching up the coefficients of xn, we find that In general, and this yields the recursion relation Note that a0 = 0. The recursion relation forces all of the even coefficients a2, a4, a6,…to be zero:

As for the odd coefficients, a1 may be chosen arbitrarily As for the odd coefficients, a1 may be chosen arbitrarily. Because F (0) = a1, we set a1 = 1 to obtain a solution y = F (x) satisfying F (0) = 1. Now apply This shows the general pattern of coefficients. To express the coefficients in a compact form, let n = 2k + 1. Then the denominator in the recursion relation can be written n2 − 1 = (2k + 1)2 − 1 = 4k2 + 4k = 4k (k + 1) and

Applying this recursion relation k times, we obtain the closed formula Thus we obtain a power series representation of our solution: A straightforward application of the Ratio Test shows that F (x) has an infinite radius of convergence. Therefore, F (x) is a solution of the initial value problem for all x.