Power series and Taylor series D. DeTurck University of Pennsylvania March 29, 2018 D. DeTurck Math 104 002 2018A: Series 1 / 42 Series First . a review of what we have done so far: 1 We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. 2 We developed tests for convergence of series of constants. 3 We considered power series, derived formulas and other tricks for finding them, and know them for a few functions. D. DeTurck Math 104 002 2018A: Series 2 / 42 1. Geometric and telescoping series The geometric series is 1 X a a r n = a + ar + ar 2 + ar 3 + ··· = n 1 − r n=0 provided jrj < 1 (when jrj ≥ 1 the series diverges). We often use partial fractions to detect telescoping series, for which we can calculate explicitly the partial sums Sn. D. DeTurck Math 104 002 2018A: Series 3 / 42 2. Tests for convergence of series of constants 1 Fundamental divergence test (nth term must go to zero for convergence to be possible) 2 Integral test 3 Comparison and limit comparison tests 4 Ratio test 5 Root test 6 Alternating series test D. DeTurck Math 104 002 2018A: Series 4 / 42 3. Power series f (n)(0) f (x) = a + a x + a x2 + a x3 + ··· where a = 0 1 2 3 n n! or 1 X f (n)(0) f (x) = xn n! n=0 1 and we know the series for ex , sin x and . 1 − x D. DeTurck Math 104 002 2018A: Series 5 / 42 Convergence of power series Before we get too excited about finding series, let's make sure that, at the very least, the series converge. Later, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they \came from". (Strictly speaking, this is not always true | but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.) Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test. D. DeTurck Math 104 002 2018A: Series 6 / 42 Ratio test review 1 X Recall that for a series of constants bn, we have that the series n=0 converges (absolutely) if bn+1 lim n!1 bn is less than one, diverges if the limit is greater than one, and the test is indeterminate if the limit equals one. To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge. D. DeTurck Math 104 002 2018A: Series 7 / 42 For the exponential function The power series for ex is 1 x2 x3 x4 X xn ex = 1 + x + + + + ··· = 2! 3! 4! n! n=0 Therefore n+1 bn+1 x n! jxj lim = lim n = lim = 0: n!1 bn n!1 (n + 1)! x n!1 n + 1 No matter what x is, the limit is 0, which is less than 1. So the series for the exponential function converges for all values of x. Your turn! For which values of x does the series for f (x) = sin x converge? D. DeTurck Math 104 002 2018A: Series 8 / 42 A more interesting example: 1 X xn For the series f (x) = , This time the ratio test gives n n=1 n+1 x n n lim = lim jxj = jxj: n!1 n + 1 xn n!1 n + 1 So the series converges if jxj < 1 and diverges if jxj > 1 (reminiscent of the geometric series). It remains to check the endpoints x = 1 and x = −1 1 X 1 For x = 1 the series is , the (divergent) harmonic series. n n=1 1 X (−1)n For x = −1 the series is , the alternating harmonic n n=1 series, which we know to be (conditionally) convergent. 1 X xn So converges if −1 ≤ x < 1 and diverges otherwise. n n=1 D. DeTurck Math 104 002 2018A: Series 9 / 42 OK, your turn. 1 X (2x)n For which values of x does the series converge? n2 n=1 1 1 A. −1 < x < 1 B. −2 < x < 2 C. − ≤ x < 2 2 1 1 D. −2 ≤ x ≤ 2 E. − ≤ x ≤ 2 2 1 X For which values of x does the series nxn converge? n=1 A. −1 < x < 1 B. −1 ≤ x < 1 C. −1 < x ≤ 1 D. −1 ≤ x ≤ 1 E. 0 ≤ x < 1 D. DeTurck Math 104 002 2018A: Series 10 / 42 From these examples. it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form [−a; a], (−a; a], [−a; a) or (−a; a) (where a might be either zero or infinity). The interval is called the interval of convergence and the number a is called the radius of convergence. D. DeTurck Math 104 002 2018A: Series 11 / 42 Let's go back to finding series for functions There are two ways: The standard way: f (n)(0) Use the formula a = to find the coefficients. We've found n n! series for ex and sin x this way: 1 x2 x3 x4 X xn ex = 1 + x + + + + ··· = 2! 3! 4! n! n=0 and 1 x3 x5 x7 X (−1)nx2n+1 sin x = x − + − + ··· = 3! 5! 7! (2n + 1)! n=0 We could calculate other series this way (and sometimes we do have to resort to this), but the other way is more fun: D. DeTurck Math 104 002 2018A: Series 12 / 42 The other way Magic tricks: Start from known series and use algebraic and/or analytic manipulation to get others: Substitute x2 for x everywhere in the series for ex to get: 2 2 2 3 2 4 2 [x ] [x ] [x ] ex = 1 + [x2] + + + + ··· 2! 3! 4! x4 x6 x8 = 1 + x2 + + + + ··· 2! 3! 4! 1 X x2n = n! n=0 D. DeTurck Math 104 002 2018A: Series 13 / 42 Another example Take the derivative of the series for sin x to get: d d x3 x5 x7 cos x = sin x = x − + − + ··· dx dx 3! 5! 7! 3x2 5x4 7x6 = 1 − + − + ··· 3! 5! 7! x2 x4 x6 = 1 − + − + ··· 2! 4! 6! 1 X (−1)nx2n = (2n)! n=0 D. DeTurck Math 104 002 2018A: Series 14 / 42 Yet another example Integrate both sides of the geometric series from 0 to x to get: x 1 x dt = (1 + t + t2 + t3 + ··· ) dt ˆ0 1 − t ˆ0 x2 x3 x4 − ln(1 − x) = x + + + + ··· 2 3 4 Negate both sides and replace x by −x everywhere to get: 1 x2 x3 x4 X (−1)n+1xn ln(1 + x) = x − + − + ··· = 2 3 4 n n=1 Put x = 1 to learn that the fourth series from before sums to ln 2. D. DeTurck Math 104 002 2018A: Series 15 / 42 Start from the geometric series again: and substitute −x2 for x everywhere it appears to get: 1 2n 1 X (−1)n = 1 − x2 + x4 − x6 + x8 − x10 + ··· = 1 + x2 x n=0 Now integrate both sides from 0 to x to get: 1 x3 x5 x7 x9 x11 X (−1)nx2n+1 arctan x = x − + − + − + ··· = : 3 5 7 9 11 2n + 1 n=0 Put x = 1 to show that the first series from before sums to π=4. 1 X 1 π2 We still have the challenge of showing that = . n2 6 n=1 D. DeTurck Math 104 002 2018A: Series 16 / 42 Applications of series I: Limits at x = 0. sin x We know for other reasons that lim = 1. x!0 x But we could prove this using series: 1 1 sin x 1 X (−1)nx2n+1 X (−1)nx2n lim = lim = lim x!0 x x!0 x (2n + 1)! x!0 (2n + 1)! n=0 n=0 x2 x4 x6 = lim 1 − + − + ··· = 1 + 0 + 0 + ··· = 1 x!0 3! 5! 7! You can do this for complicated limits at 0 | substitute the series for the functions and do algebra. x − sin x Calculate the limit: lim x!0 1 − e−x3 1 1 A. 0 B. C. 1 D. E. does not exist 6 12 D. DeTurck Math 104 002 2018A: Series 17 / 42 Applications of series II: Approximate evaluation of integrals Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are). 1 2 Calculate e−x dx to the nearest 0.001. ˆ0 We begin by substituting −x2 for x in the known series for ex , and then integrating it. This will give us a numerical series that converges to the answer: 1 1 4 6 2 x x 1 1 1 e−x dx = 1−x2+ − +··· dx = 1− + − +··· ˆ0 ˆ0 2! 3! 3 5 · 2! 7 · 3! D. DeTurck Math 104 002 2018A: Series 18 / 42 Error estimates So far we have 1 2 1 1 1 e−x dx = 1 − + − + ··· ˆ0 3 5 · 2! 7 · 3! The series on the right is alternating, so if we want the error to be 1 less than 0:001 = 1000 , we need to take all the terms before the first one that is less than that.