M138.W09.assmt10

# M138.W09.assmt10 - MATH 138 Assignment 10 g2 pages Winter...

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Unformatted text preview: MATH 138 Assignment 10 g2 pages) Winter 2009 Important: Assignment 10 covers the last three lectures on Taylor series, Taylor polynomials, Taylor’s Remainder Theorem and Inequalities and their applications. While it is not for submission. 5-10% of the ﬁnal exam will be based on this material. Thus it is essential that you do as many of the problems as possible. Some have been suggested with an asterisk (they would make good exam problems). Solutions for problems with ‘ will be posted by April 8‘“. There will be an open tutorial for Assignment 10 on Monday, April 6. from 4:30-6:30 pm. in RCH 101. *1. a) Use Taylor’s Remainder Theorem to show that \$2 \$4 2:6 cost — 1— — + 5—— forall 126R 2! E 720 b) Find an approximation for cos(0.2), and state an upper bound for the error. e) Following the method of Example 6 on page 123 of your Course Notes, use the result from a) to 1r/G ﬁnd an approximate value for / cos(:r2)dzr, and derive an upper bound on the error. Compare 0 this upper bound with the error bound predicted by the corollary of Alternating Series Test. 2. *a) State Taylor‘s Remainder Theorem with n = 2 and use it to show that 1.3 S — for all 12 0. l 1 lv1+z—(l+—;r——sr2) 16 2 8 *b) Use the result in a) to approximate each quantity and find an upper bound on the error. 7 1 11' (i) «1“.0‘2‘ *(ii) \/§ (Rewrite as 5 1+ 5) (iii) / y/1+%t4dt *(iv) / \/4 + ‘si—ntdi 0 0 ’3. a) Suppose P2(:r) centred at a: = a = 1 for a certain function f(\$) is 1320:) = 2 + 4(1- — 1) + 30; —1)‘2. . 1 Find the Taylor polynomial T2(;r) centred at .1: = 1 for the function g(:1:) = :r b) Find a value of N which guarantees that ' 1.2 1.;\' .1? __ n '_. . . . ‘_ “l < a ._ (. (1+.L+2!+ +N!>l310 for ,ie[1,1]. 4. *a) Show that if f(:r) is an odd function, then any Taylor polynomial PN(.7:) centred at .1' = a. = 0 will contain only odd powers of 1'. b) Prove a similar result for even f (1r). 5. Use known Maclaurin series to evaluate each limit. . 1—cos:r5 . 2V1 ar—2—:r t' —. 1— \$4 a) hm —7(-) ‘b) lini —+—2—— c) lim Emma—T *d) lim ﬁe— 1:—~0 a: .r—0 :1: 1—0 a: r—'0 .732 sm(z2) 6. ‘3) You are about to calculate sin(36°) when the batteries die on your calculator. You roommate has an old calculator, but it has no ‘sin‘ key. Unperturbed, she enters 3.1415926. divides by 5, enters 2 :I: the result in memory (called hereafter). Then she calculates :r (I — and uses it for the value of sin(36°)i Explain what she was doing in terms of Taylor polynomials, [ind the answer she got. and give an upper bound on the error. MATH 138 ~ Winter 2009 Assignment #10 Page 2 of ‘2 b) The electrical potential energy V at a point P due to the charge on a disc of radius a with constant charge density a is R V(P)=27ra(\/m—R>, where R is the distance from P to the disc Use the substitution (1 = Rat and the ﬁrst two terms of the Maclaurin series for \/1 + 12 to show that for large R, (ie for a: = (1/12 near 0), V(P) a: 7ra2a/R. c) Taylor series appr0ximations can be used to ﬁnd approximate solutions to equations. A nice ' example of this is the equation sinx+b(1+cos2\$+cosz) =0. where b is a very small positive constant. (This equation arose in Einstein’s calculations pre- dicting the bending of light by the gravitational ﬁeld of the sun.) i) Explain how you know there is a solution near (I: = 0‘ ii) Expand the equation about :3 = 0, using only the linear terms in :1: (Le. disregard terms of order x2 or higher), and solve for m (in terms of b). 7. Recall that Newton‘s method for solving for a root r of f (at) = 0 uses the recursion formula (4:) fl(\$n) 8.) Write down Taylor’s Remainder Theorem with N = l and a = 1:” by completing the RHS of the following statement: 33114-1 2 xn _ ftp) - _ — 1'11) = b) Substitute the root 2: = r in your statement in a); and then solve the resulting equation for ﬂat”) c) Substitute your result from b) for f(a:n) in the recursion (*), and heme prove that, if |f"| S 2M and |f’| > K for all x in an interval I containing x", \$7,“, and r, then M l\$n+1‘7'lS I; i-Tn " riz- (This explains why Newton’s method roughly doubles the number of decimal places of acuracy at each step.) ...
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