Section 8.7 Taylor and MacLaurin Series (1) Definitions, (2) Common Maclaurin Series, (3) Taylor Polynomials, (4) Applications. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 1 / 10
Math 126- Calculus Fall-2016 Quiz 5 (1) If the nth partial sum of the series is S n = 2n2 +2, then find the 3n 2 n sum of the series a n. (2) Determine if the following series converges or diverges (i) 4 n+1 π n (ii) arctan(n 2 +1) (iii) π n e n (3) Find the values of x for which the series converges. Find the sum of the series for those values of x. (4x 3) n 5 n MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 2 / 10
Theorem If f (x) has a power series representation centered at a, then f (x) = f (n) (a) n! (x a) n = f (a) + f (a) 1! (x a) + f (a) (x a) 2 +... 2! MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 3 / 10
Theorem If f (x) has a power series representation centered at a, then f (x) = f (n) (a) n! (x a) n = f (a) + f (a) 1! (x a) + f (a) (x a) 2 +... 2! This power series is called the Taylor Series of f (x) at x = a. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 3 / 10
Theorem If f (x) has a power series representation centered at a, then f (x) = f (n) (a) n! (x a) n = f (a) + f (a) 1! (x a) + f (a) (x a) 2 +... 2! This power series is called the Taylor Series of f (x) at x = a. If a = 0, then the series is called the MacLaurin Series of f (x). (1) f (x) = 1 x at a = 2 (2) f (x) = 8x x 3 at a = 2 Practice Problem: MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 3 / 10
Common MacLaurin Series 1 1 x MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
1 1 x = Common MacLaurin Series x n with R = 1. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
1 1 x = e x = x n n! Common MacLaurin Series x n with R = 1. with R =. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
Common MacLaurin Series 1 1 x = e x = sin(x) = x n n! x n with R = 1. with R =. ( 1) n x 2n+1 (2n + 1)! with R =. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
1 1 x = e x = sin(x) = cos(x) = x n n! Common MacLaurin Series x n with R = 1. with R =. ( 1) n x 2n+1 with R =. (2n + 1)! ( 1) n x 2n with R =. (2n)! MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
1 1 x = e x = sin(x) = cos(x) = x n n! Common MacLaurin Series x n with R = 1. with R =. ( 1) n x 2n+1 with R =. (2n + 1)! ( 1) n x 2n with R =. (2n)! arctan(x) = ( 1) n x 2n+1 with R = 1. 2n + 1 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 4 / 10
More Practice Problems (1) Use a Maclaurin series in the table below to obtain the Maclaurin series for the given function. (3) Use series to evaluate the limit. f (x) = 6e x + e 6x sin(2x) 2x + 4 3 lim x 3 x 0 x 5 2 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 5 / 10
Taylor s Inequality If f (n+1) (x) M for x a d, then for x a d R n (x) M x a n+1 (n + 1)! Use the Maclaurin series for sin(x) to compute 7 sin(1 ) correct to five decimal places. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 6 / 10
Alternating Series Test for Convergence If an alternating series ( 1) n b n satisfies (i) b n decreases, that is b n+1 b n for all n. (ii) lim b n = 0. n then the series converges. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 7 / 10
Alternating Series Test for Convergence If an alternating series ( 1) n b n satisfies (i) b n decreases, that is b n+1 b n for all n. (ii) lim b n = 0. n then the series converges. what can we say about the error term R k = s s k? MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 7 / 10
Alternating Series Test for Convergence If an alternating series ( 1) n b n satisfies (i) b n decreases, that is b n+1 b n for all n. (ii) lim b n = 0. n then the series converges. what can we say about the error term R k = s s k? R k = s s k b k+1 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 7 / 10
Alternating Series Test for Convergence If an alternating series ( 1) n b n satisfies (i) b n decreases, that is b n+1 b n for all n. (ii) lim b n = 0. n then the series converges. what can we say about the error term R k = s s k? R k = s s k b k+1 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 7 / 10
Practice Problem (1) (2) (3) ( 1) n 5n 2 3n + 2 ( 1) n+1 8n 4 error <.00005 n 1 3n ( 1) n 6 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 8 / 10
Ratio Test for Convergence Assume that L = lim a n+1 n a n. (i) If L < 1 then a n converges. (ii) If L > 1 or L = then a n diverges. (iii) If L = 1 the Ratio Test fails. MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 9 / 10
Ratio Test for Convergence Assume that L = lim a n+1 n a n. (i) If L < 1 then a n converges. (ii) If L > 1 or L = then a n diverges. (iii) If L = 1 the Ratio Test fails. Practice Problem: (i) 3 7 + 3.7 7.9 + 3.7.11 7.9.11 + 3.7.11.15 7.9.11.13 (ii) a 1 = 2 and a n+1 = 7n+1 3n+9 a n (iii) Assume that lim a n+1 n a n = 1/3.Determine if the following series converge or diverge, justify your answer. 4n a n MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 9 / 10
Homework Problems Determine whether the following series converge or diverge using the convergence tests. Clearly note which convergence test is used and then show the work required by the test. 1 j=1 j 2 (2j + 1)! 2 3 4 i=0 r=2 2 i i 3 i i π n n n r 3 r 7 + 2r 2 + 1 MATH 126 (Section 8.7) Taylor and MacLaurin Series The University of Kansas 10 / 10