POWER SERIES JAMES KEESLING 1. Introduction In this summary we will cover the basic features of power series. Power series is an important way to analyze functions in various settings. When a function agrees with its power series it is said to be analytic. Sometimes one specifies the interval over which the function is analytic. The functions with power series representation take us beyond the polynomials. The analytic functions include polynomials, rational functions, the loga- rithm function, the exponential function, the trigonometric functions, among many others. Virtually every function with major applications is in this category. 2. The Geometric Series Of first importance is the geometric series. We will prove the convergence of this series in this section. The proof is simple, but it is fundamental in what will follow. Theorem 1. Suppose that jxj < 1, then 1 X 1 xn = : 1 − x n=0 1 P1 n Furthermore, if 0 < " < 2 , then n=0 x converges uniformly to its limit on the interval [−1 + "; 1 − "]. If jxj ≥ 1, then the series diverges. PN n Proof. Assume that jxj < 1. Let N be a positive integer and let SN = n=0 x . Then N N+1 X n+1 X n x · SN = x = x n=0 n=1 and N N+1 X n X n N+1 (1 − x) · SN = x − x = 1 − x n=0 n=1 which gives us 1 − xN+! S = : N 1 − x 1 2 JAMES KEESLING n 1 Since jxj < 1, limn!1 x = 0 and thus limn!1 SN = 1−x . This is precisely what we mean by 1 X 1 xn = : 1 − x n=0 1 For the second part of the proof, let δ > 0 and suppose that 0 < " < 2 . Let x 2 P1 n (1−")K [−1 + "; 1 − "]. Let N be such that for all K ≥ N, n=K (1 − ") = " < δ. Now for any 0 ≤ jxj < 1 − " and K ≥ N, 1 1 1 X xK X jxjK X (1 − ")K xn = ≤ jxjn = ≤ (1 − ")n = < δ: 1 − x 1 − jxj " n=K n=K n=K P1 n 1 Thus, n=0 x converges uniformly to 1−x on [−1 + "; 1 − "]. P1 n It is easy to show that if jxj ≥ 1, then n=0 x diverges. We will leave that to the reader. 3. General Power Series and the Radius of Convergence P1 n Suppose that n=0 anx is any power series centered at 0. Let pn lim sup janj = α n!1 1 where 0 ≤ α ≤ 1. Let R = α , 0 ≤ R ≤ 1. Then we have the following theorem. P1 n Theorem 2. The series n=0 anx converges for all jxj < R. Furthermore, it converges uniformly on any interval [−M; M] with 0 ≤ M < R. If jxj > R, then the series diverges. For x = R or x = −R, the series may converge or it may not converge. Proof. The case R = 0 is trivial. The proof for R finite can be modified for the case that R = 1. So, we assume that R is finite. Let jxj < R. Suppose that N is such that for pn 1 1 1 pn n jxj all n ≥ N, janj < jxj . Note that jxj > α = R . So, for all n ≥ N, janjjxj < R . Of course, this implies that n jxjn pn ja jjxjn < n R and thus that 1 1 1 n X n X X jxj pn ja jjxjn = ja jjxjn < n n R n=N n=N n=N P1 n converges. This implies that n=N anx converges for jxj < R. This implies that P1 n n=0 anx converges for jxj < R since this last sum adds only a finite number of terms to the series. POWER SERIES 3 The fact that this series converges uniformly on any interval [−M; M] with 0 ≤ M < R follows from the proof for that case in the proof of Theorem 1. The fact that the series diverges for jxj > R is easy to prove and we leave it as an exercise for the reader. There are examples where the series may converge or diverge at x = R or x = −R. We say that R as defined above is the radius of convergence for the power series P1 n n=0 anx . Note that the radius of convergence R can be characterized as the largest R such that the series converges for all x 2 (−R; R) and diverges for jxj > R. The formula pn 1 lim supn!1 janj = α with R = α gives us a way to calculate the value of R, but if we could determine a value M by any means such the series converges for all x 2 (−M; M) and diverges for jxj > M, then M would be the radius of convergence for the series. In x9 we will give another way which is sometimes useful in determining the radius of convergence. 4. Antiderivatives and Integration of Power Series P1 n Theorem 3. Let f(x) = n=0 anx and suppose that the radius of convergence for this P1 an n+1 series is R > 0. Let g(x) = n=0 n+1 x . Then the radius of convergence for g(x) is also R and for all jxj < R, g0(x) = f(x). P1 n Proof. Consider the power series n=0 anx with radius of convergence R. Note that if R = 0, then the series only converges for x = 0. So, for a meaningful discussion, we have P1 n assumed that 0 < R ≤ 1. Now f(x) = n=0 anx is well-defined on the interval (−R; R). P1 an n+1 Let g(x) = n=0 n+1 x . It is easy to check that g(x) is represented by a power series that also has radius of convergence R. Let jxj < R. Then note that Z x Z x 1 ! Z x N ! X n X n (1) f(t)dt = ant dt = lim ant dt N!1 0 0 n=0 0 n=0 Z x N ! X n (2) = lim ant dt N!1 0 n=0 N+1 X an+1 (3) = lim xn N!1 n + 1 n=0 1 X an+1 (4) = xn n + 1 n=0 = g(x)(5) P1 n Going from the first to the second step is because the convergence of n=0 ant is uniform on the interval over which the integration takes place. So, this proves that g(x) is an 0 antiderivative of f(x) and that g (x) = f(x). We can apply this theorem to several examples. We know from elementary Calculus d 1 P1 n that dx ln(1 + x) = 1+x = n=0(−x) . By integrating term by term as above we get 4 JAMES KEESLING 1 X xn ln(1 + x) + C = (−1)n+1 n n=1 for all jxj < 1 for some constant C. By evaluating both sides of the equation at x = 0 we obtain that C = 0. Thus, 1 X xn ln(1 + x) = (−1)n+1 : n n=1 d 1 P1 2 n Similarly, dx arctan(x) = 1+x2 = n=0(−x ) . Again, by integrating term by term we get 1 X x2n+1 arctan(x) + C = (−1)n+1 2n + 1 n=1 for some constant C for all jxj < 1. By evaluating both sides of the equation at x = 0 we get that C = 0. Thus, 1 X x2n+1 arctan(x) = (−1)n+1 : 2n + 1 n=1 5. Derivatives of Power Series P1 n Now just because the power series n=0 anx converges on (−R; R) and even uniformly on [−M; M] ⊂ (−R; R) does not mean that the series which is the term-by-term derivative of that series would necessarily be the derivative of the series. It turns out that this is the case, but it requires a proof. The proof that we give is indirect. P1 n Theorem 4. Let f(x) = n=0 anx and suppose that f(x) has radius of convergence P1 n R. Let g(x) = n=1(n + 1)an+1x . Then g(x) also has radius of convergence R and f 0(x) = g(x) for all x 2 (−R; R). P1 n P1 n Proof. Let f(x) = n=0 anx and let g(x) = n=1(n + 1)an+1x . Note that g(x) is the term by term derivative of f(x). Suppose that R is the radius of convergence for the series that defines f(x). Then 1 pn = lim sup janj = α: R n!1 However, it is easy to show that 1 pn = lim sup (n + 1)jan+1j = α R n!1 as well. So, the series representing f(x) and that representing g(x) both have radius of convergence R. Since f(x) is the term by term integral of g(x), it must be that it is the 0 antiderivative of g(x). Thus, f (x) = g(x) as was to be proved. POWER SERIES 5 1 P1 n We have examples of this as well. Since the Geometric Series 1−x = n=0 x has radius of convergence R = 1, 1 1 X = (n + 1)xn (1 − x)2 n=0 also has radius of convergence R = 1. 6. Power Series Centered at an Arbitrary Point We have been discussing power series centered at 0. We could have defined a power series centered at a for any real a. This would take the form 1 X n an(x − a) : n=0 We calculate the radius of convergence the same way as in x3 using the coefficients an of the series.