Chapter 9 AP Calculus BC 9.1 Power Series 1 1 1 1 ... 1 2 4 8 16 1 1 1 1 1 ... 2 3 4 5 111111... ??? If the sequence of partial sums has a limit S, as ninfinity, then we say the series Converges to S. Otherwise it Diverges. Infinite Series: ak a1 a2 a3 ... an ... k 1 a1, a2 are terms an is the nth term Partial Sums: S1 a1 S2 a1 a2 S3 a1 a2 a3 n Sn ak k 1 Geometric Series: a ar ar 2 ar3 ... ar n1 ... converges if r 1 and diverges if r 1 ar n1 n1 Interval of convergence : -1 < r < 1 Is the IOC for Geom. Series. converges to sum a 1 r Representing functions by series: If x 1, then the Geometric Series formula assures us that: 1 x x2 x3 x4 ... xn ... 1 1 x xn is a Power Series centered at x=0 and it converges on the interval (-1,1) n0 Definition of a Power Series: An expression of the form: cn xn c0 c1x1 c2 x2 ... cn xn ... is a Power Series centered at x=0. n0 cn (x a)n c0 c1(x a)1 c2(x a)2 ... cn(x a)n ... n0 Given: is a Power Series centered at x = a. Find a Power Series for: If x 1, 1 x x2 x3 x4 ... xn ... 1 1 x 1 on (1,1) 1 x x 1 x 1 1 2x Again, Given: If x 1, 1 x x2 x3 x4 ... xn ... 1 1 x Find a Power Series for: 1 (1 x)2 Now, write down the series for: and use it to write one for : Answer: nxn1 n1 1 1 x ln(1 x) (1)n xn1 Answer: n 1 n0 Copy down Theorem 1 p. 478 and Theorem 2 p. 479 9.2 Taylor Series P( x) a0 a1x a2 x2 a3x3 a4 x4 Given: P(0) 1 P '(0) 2 P ''(0) 3 P '''(0) 4 P IV (0) 5 Find the Taylor Polynomial….. P( x) 1 2x 3 x2 2 x3 5 x4 2 3 24 Find the 4th order Taylor Polynomial of ln(1 x) Construct a Power Series for: Taylor Series generated by f at x=0: Answer: sin x and cos x at x = 0 nth term answers: 2n1 sin x (1)n x (2n 1)! n0 2n cos x (1)n x (2n)! n0 f k (0) n (0) f ''(0) f '''(0) f 2 3 n f (0) f '(0) x x x ... x ... xk 2! 3! n! k 0 k ! 9.2 cont’d. p. 489 Taylor Series centered at x = a. Copy it down!!!! p. 491 5 most important series: Copy them down, know them!!! Occasionally you need 6, 7……. 9.3 Taylor’s Theorem Adequate substitution: a Taylor series that is off from the actual by less than 0.0001 Truncation Error: NEXT TERM!!!! If f has derivatives of all orders in an open interval, I, containing a, then for each positive integer, n, and for each x in the interval: n (a) f ''( a ) f 2 f ( x) f (a) f '(a)( x a) ( x a) ... ( x a)n Rn( x) 2! n! n1(c) f Where: Rn ( x) (n 1)! ( x a)n1 Largest value of derivative..f part If Rn ( x) 0 as n for all x in I, we say that the Taylor Series generated by f at x = a converges to f on I. Theorem 4 – Remainder Estimation Theorem. (do examples) 9.4 Radius of Convergence A convergent series is a number and may be treated as such…. Theorem 5 - Convergence Theorem for Power Series -There are 3 possibilities for cn( x a)n n0 1. There is a R such that the series diverges for x a R but converges for x a R. The series may or may not converge at the endpoints. 2. The series converges for every x. (R=) 3. The series converges at x = a, but diverges elsewhere. (R=0) R = radius of convergence and the set of x-values for which the series converges is called the Interval of Convergence. Theorem 6 – nth term test for divergence a 0. an diverges if nlim n n1 9.4 cont’d. Direct Comparison Test (DCT) (non-negative terms) Greatest Power Rules!!!!! Let an be a series with no negative terms... (a) an converges if there is a convergent series cn with an cn (b) an diverges if there is a divergent series dn with an dn Absolute Convergence - If the series an of absolute values converges, then an converges absolutely. Theorem 8….. Ratio Test – (Powers and Factorials) an1 Let an be a series with terms and with nlim L an Then: (a) the series converges if L < 1. (b) the series diverges if L > 1. Telescoping series: (c) the test is inconclusive if L = 1. p. 510…… 9.5 Testing Convergence at Endpoints Theorem 10 - Integral Test - Let {an} be a sequence of terms. Suppose that an f (n), where f is a cts., , decreasing function of x. Then the series an and the Integral N f ( x)dx either n N both converge or both diverge. 1 P-series test: n p converges if p > 1, diverges if p 1. n1 Limit Comparison Test (LCT) Suppose that an 0, bn 0: an c, 0 c , then a & b both converge or diverge (1) If nlim n n b n an 0, and b converges then a converges. (2) If nlim n n b n an , and b diverges then a diverges. (3) If nlim n n b n 9.5 cont’d. Alternating series Test (Liebniz’s Theorem) The series (1)n1un u1 u2 u3 u4 ... n1 converges if all three of the following conditions are met: (1) Each un is positive. (2) un un1 for all n... Error is next term sign (3) nlim u 0. n included………… Look at examples 4 -6 pp. 518 - 520 Absolute and Conditional Convergence……