Power Series, Taylor and Maclaurin Polynomials and Series Power Series The Basics Definition 1 (Power Series). A power series [centered at x = 0] is a series of the form 1 X n 2 3 4 cnx = c0 + c1x + c2x + c3x + c4x + ::: n=0 (Informally, we can think of a power series as an “infinite polynomial.") A power series centered at x = a has the form 1 X n 2 3 4 cn(x − a) = c0 + c1(x − a) + c2(x − a) + c3(x − a) + c4(x − a) + ::: n=0 As we go through this chapter we'll be looking at at least two different ways to create the power series for a function. But for now, given a power series (however we came by it) we want to know for which values of x the power series converges (i.e., which values of x will yield a convergent series when we plug them into the power series?). 1 Example 1. P xn = 1 + x + x2 + x3 + ::: . n=0 In general we'll find the interval of convergence (which I'll abbreviate as IOC from here on) by using the Ratio Test (abbr. RT) on the power series and see which values of x will make the limit < 1, and then check convergence at the endpoints using some other convergence test. The test we use to check the endpoints will depend on what the series looks like, but we know for sure that the Ratio Test will not give us an answer since the endpoints are where the limit in the Ratio Test is = 1. 1 Example 2. P xn n=0 In general we have the following result: 1 1 P n Theorem 2. Given a power series cn(x − a) there are three possibilities for the IOC and ROC: n=0 1. The power series converges only when x = a (we'll get a limit > 1 otherwise in the RT). In this case IOC = fag (just a single point) and R = 0. 2. The power series converges for all values of x (we'll get a limit = 0 in the RT irrespective of the value that x takes on). In this case IOC = (−∞; 1) and R = 1. 3. There is a positive, finite number R (the radius of convergence) such that the series converges if jx − aj < R and diverges of jx − aj > R. We then need to use other convergence tests to see what happens at the endpoints x = a ± R. More Examples 1 1. P n!(x − 3)n n=0 1 (−1)nx2n+1 2. P n=0 (2n + 1)! 1 n+1 P (x − 2) 3. n+1 n=0 (n + 1)3 1 n P n(x − 4) 4. 3 n=0 n + 1 1 5. P n!(2x + 1)n n=0 1 xn 6. P n=0 n! 1 7. P n(x + 2)n n=0 1 (−1)nx2n 8. P n=0 (2n)! 1 n+1 n P (−1) (x − 5) 9. n n=0 n5 2 The Language of Convergence We say that . • . a series converges to a number, i.e., the sum of the series is what it converges to. • . a power series converges at a particular value of x if plugging that x into the power series yields a convergent series. • . a power series converges on an interval if it converges at every x in that interval. Taylor and Maclaurin Polynomials We now turn to the process of creating power series. To begin, we would like a polynomial (centered at some x = a) which approximates a function, at least around x = a: 2 N TN (x) = c0 + c1(x − a) + c2(x − a) + ··· + cN (x − a) (The degree is N.) We would like this function and its derivatives to match the original function and its derivative exactly when we plug in x = a. So, we construct the coefficients as follows: f(a) f(a) want= c + c (a − a) + c (a − a)2 + c (a − a)3 + ··· + c (a − a)N ) c = f(a) = 0 1 2 3 N 0 0! f 0(a) f 0(a) want= c + 2c (a − a) + 3c (a − a)2 + ··· + Nc (a − a)N−1 ) c = f 0(a) = 1 2 3 N 1 1! f 00(a) f 00(a) f 00(a) want= 2c + 3 · 2 · c (a − a) + ··· + N(N − 1)c (a − a)N−2 ) c = = 2 3 N 2 2 2! f 000(a) f 000(a) want= 3 · 2 · c + ··· + N(N − 1)(N − 2)c (a − a)N−3 ) c = 3 N 3 3! . f (N)(a) f (N)(a) want= n!c ) c = N N n! This leads to the following definition: Definition 3 (Taylor Polynomial). The N th Taylor polynomial of f(x) centered at x = a is f 00(a) f 000(a) f (N)(a) T (x) = f(a) + f 0(a)(x − a) + (x − a)2 + (x − a)3 + ··· + (x − a)N N 2! 3! N! N X f (n)(a) = (x − a)n n! n=0 If a Taylor polynomial is centered at a = 0 then we call it a Maclaurin polynomial: Definition 4 (Maclaurin Polynomial). The N th Maclaurin polynomial of f(x) is f 00(0) f 000(0) f (N)(0) M (x) = f(0) + f 0(0)x + x2 + x3 + ··· + xN N 2! 3! N! N X f (n)(0) = xn n! n=0 3 Examples 1 1. Find the third Maclaurin polynomial of f(x) = 1−x . 2. Find the first, second, and third Maclaurin polynomials of f(x) = ex and use them to approximate e0:05. 3. Find the fourth Taylor polynomial of f(x) = ln(x) centered at x = 1. 4. Find the fourth Taylor polynomial of f(x) = x2 + 3x + 7 centered at x = −1. 5. Find the seventh Maclaurin polynomial of f(x) = sin(x). Taylor and Maclaurin Series Once we have a Taylor or Maclaurin polynomial we can then extend it to a series: Definition 5. • The Taylor series of f(x) centered at x = a is 1 X f (n)(a) (x − a)n: n! n=0 • The Maclaurin series of f(x) is 1 X f (n)(0) xn: n! n=0 (Note: if f can be represented as a power series then it is identically equal to its Taylor/Maclaurin series within the IOC.) In general we would find a Taylor/Maclaurin series by writing out the first several terms until we can determine what pattern (if any) they follow and find a formula describing the terms. Important Examples 1 1 X 1. = 1 + x + x2 + x3 + ··· = xn 1 − x n=0 1 x2 x3 X xn 2. ex = 1 + x + + + ··· = 2! 3! n! n=0 1 x3 x5 x7 X (−1)nx2n+1 3. sin(x) = x − + − + ··· = 3! 5! 7! (2n + 1)! n=0 1 x2 x4 x6 X (−1)nx2n 4. cos(x) = 1 − + − + ··· = 2! 4! 6! (2n)! n=0 4 However, once we have a Taylor series for one function we can find the Taylor series for other, related functions by performing the following operations: 1. Substitution (i.e., composition): we replace x with some other polynomial in terms of x, usually just a single power of x or a constant times a single power of x. 2. Multiplication: we can multiply a power series by constants or [positive] powers of x. While it is possible to multiply two power series together, we won't be doing that this semester, and there will be another way to solve a problem that looks like it requires the multiplication of two power series. Also note that you should never need to divide by a power of x (or multiply by x to a negative power) { there will be another way to solve the problem. 3. Differentiation. When differentiating a power series, if the first term is a constant then you'll need to increase the starting index by 1 in the derivative's power series. This throws out the index corresponding to the term that became 0 in the derivative (so we don't need it in the sum anymore) and avoids potential issues with the derivative formula being undefined at that the starting index. (If the first term in the original series is not constant then the starting index remains the same after differentiating.) 4. Integration. To find the \+C" constant of integration, plug the value at which the power series is centered into both the function and the power series. This should make the power series become just 0's except for perhaps the first term and then you can solve for C. Note: Your final answer should always be of the form 1 X cnx n= i.e., you need to separate any constants from x and simplify the exponent. It's also helpful to get the power series into this form before differentiating or integrating. (The form above assumes the power series is centered at 0, which will be true for the examples that we're looking at since they'll all be based on the four \Important Examples" above which are all centered at 0.) Examples Find the Maclaurin series for the following functions using known Maclaurin series: x 1. f(x) = 1 − x 1 2. f(x) = 1 + 2x 2 3. f(x) = e−x 2 4. f(x) = −2xe−x 5. f(x) = x cos x3 5 6. f(x) = ln jx − 2j 1 7. f(x) = (1 − x)3 8. f(x) = xe2x 9. f(x) = arctan(x) x2 10.