tailieunhanh - Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 97

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 97. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Taylor Series 941 TAYLOR SERIES Defining Taylor Series In many examples in this chapter we ve observed that the higher the degree of the Taylor polynomial generated by at x fe the better it approximates x for x near fe. For functions such as sin x and cos x the higher the degree of the Taylor polynomial the longer the interval over which the polynomial follows the undulations of the function s graph. Letting the degree of the polynomial increase without bound gives us the Taylor series for . Definition If a function has derivatives of all orders at x fe then the Taylor series of at or about x fe is defined to be f b f b x - b 2 f b 5 m _i_ x - b H------1----- x - b H----- that is k 0 f k bL .k x - b k. k We refer to this series as the Taylor expansion of about x fe or centered at x fe. In the special case where fe 0 the series i 0 7 0 k can be called the Maclaurin series for . From the work we ve done with Taylor polynomials we can easily find the Maclaurin series for ex sin x and cos x. EXAMPLE SOLUTION Find the Maclaurin series for x ex. All derivatives of ex are ex. When evaluated at x 0 ex is 1. Maclaurin series for ex EXAMPLE SOLUTION Find the Maclaurin series for f x sin x. Even order derivatives Odd order derivatives f x sin x f 0 0 f x cos x f 0 1 f x sin x f 0 0 f x cos x f 0 1 f 4 x sin x f 4 0 0 f 5 x cos x f 5 0 1 f 2k x -1 k sin x f 2k 0 0 f 2k 1 x -1 k cos x f 2k 1 0 -1 k Maclaurin series for sin x x3 x5 x 2k 1 xx x x---- -------- - 1 k----- 3 5 2k 1 œ x2k 1 V -1 k 0 2k 1 942 CHAPTER 30 Series EXAMPLE Find the Maclaurin series for x y . SOLUTION x 1 - x -1 0 1 x 1 - x -2 0 1 x 2 1 - x -3 0 2 x 3 2 1 - x -4 0 3 . . . fc x fc 1 - x - fc 1 fc 0 fc Maclaurin series for y y 2x2 3 x3 4 x4 fc xfc 1 x ---1----1-1--------------1-1- 2 3 4 fc 1 x x2 x3 - xfc - xfc fc 0 The Maclaurin series for should look familiar. fc 0 xfc is a geometric series with a 1 and x. Therefore we know that it converges to for x 1 and diverges for x 1. This