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Calculus 1120, Fall 2012

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Calculus 1120, Fall 2012 Calculus 1120, Fall 2012 Dan Barbasch November 13, 2012 Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 1 / 5 Absolute and Conditional Convergence We need to consider series with arbitrary terms not just positive (nonnegative) ones. For such series the partial sums sn = a1 + ··· + an are not necessarily increasing. So we cannot use the bounded monotone convergence theorem. We cannot use the root/ratio test or the integral test directly on such series. The notion of absolute convergence allows us to deal with the problem. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 2 / 5 Absolute and Conditional Convergence P Definition: A series an is said to converge absolutely, if the series P janj converges. For convergence without absolute values, we say plain convergent. For plain convergence, but the series with absolute values does not converge, we say conditionally convergent. Theorem: If a series is absolutely convergent, then it is convergent. Note the difference between if and only if and if then. Typically we apply the root/ratio integral test and comparison test for absolute convergence, and use the theorem. REMARK: NOTE the precise definitions of the terms and the differences. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 2 / 5 Alternating Series Test P1 n−1 Theorem: (Alternating Series Test) Suppose n=1(−1) an satisfies 1 an ≥ 0; (series is alternating) 2 an is decreasing 3 limn!1 an = 0: Then the series is (possibly only conditionally) convergent. Error Estimate: jS − snj ≤ an+1. P1 i−1 S − sn = i=n+1(−1) ai . an+1 − an+2 + an+3 − · · · =(an+1 − an+2) + (an+3 − an+4) + ··· = =an+1 − (an+2 − an+3) − (an+4 − an+5) − ::: So 0 ≤ jS − snj = an+1 − an+2 + an+3 − · · · ≤ an+1 The proof is in the text and in the previous slides (also every text on calculus). Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 3 / 5 Alternating Series Test Examples: Determine whether the following series converge absolutely or conditionally. p X n p 1 (−1) ( n + 1 − n). n=1 1 X n 1 2 (−1) . ln(ln(n + 2)) n=1 1 n X n 1 3 Compare the error estimate for (−1) with the actual 10 n=0 error. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 3 / 5 Power Series 1 X n an(x − a) . n=0 a is (sometimes) called the center. an are called the coefficients. Issues/Summary of what we will learn For what values of x does the power series converge? Answer: There is an open interval (centered at a)(a − R; a + R) so that the series converges absolutely for x in this interval. R is called the radius of convergence. At the endpoints the series may or may not converge absolutely or conditionally. The totality of values for which the series converges is called the interval of convergence. What are its uses? (Short) Answer: Within the radius of convergence a − R < x < a + R you can add/substract/divide/multiply, differentiate and integrate the series term by term as if it were a polynomial. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 4 / 5 Power Series Examples: 1 P1 n For what values of x does n=0 x converge? n 2 P1 n−1 P1 x Same for n=1 nx and n=1 n : 3 P1 n n Same for n=1(−1) (x − 3) : (x−1)n 4 P1 n For what values of x does n=1(−1) n converge? What is the value? n 5 P x For what values of x does n! converge? What is the sum? 6 P1 n For what values of x does n=0 n!x converge? What is the sum? 7 1 Find a power series expansion for 1−x about a = 2: 8 1 Find a power series expansion for (1−x)2 about a = 0: Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 4 / 5 Summary Theorem: (corrollary to theorem 18, page 596). The convergence of a P1 n power series n=0 an(x − a) is described by one of the following three cases: 1 There is a number R > 0 so that the series converges absolutely for jx − aj < R; and diverges for jx − aj > R: 2 The series converges for x = a only. 3 The series converges for all x. In case (2) we write R = 0: In case (3) we write R = 1: This number R is called the radius of convergence ja j 1 (you can see from the formulas that R is lim n or ). n!1 pn jan+1j janj This says nothing about x = a − R or x = a + R. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 5 / 5 Summary In practice, to find the radius of convergence, you apply the ratio/root test. The condition n+1 jan+1jjx − aj jan+1j lim n = jx − aj lim < 1 n!1 janjjx − aj n!1 janj pn n pn lim janjjx − aj = jx − aj lim janj < 1 n!1 n!1 should translate into an inequality jx − aj < R for some number R: The R is the radius of convergence. jx − aj < R ,− R < x − a < R , a − R < x < a + R: In the examples we drew this last interval; R is half its length. Dan Barbasch () Calculus 1120, Fall 2012 November 13, 2012 5 / 5.
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