University:
California State University, Northridge
Course:
MATH 150B | Calculus II
Academic year:

2023
Views:

230

Pages:

4
Author:

Yung D

10.9 Convergence of Taylor Series 77 Th (Taylor's Theorem) (7) A ssume fifif...f are continous on [a, b] and f is differentiable on (a,b), then there exists a number CE(a,b) such that: (1) f(b) = f(a)+f(a) (b-a) + (a) (b-a).... (b-a)" + 21 (+1) fic) (b-a) + (n+1)! *Note that Taylor's Theorem is a generalization of the MVT. *If we change b by x, we get Taylors formulat " (n) f(x) = f(a)+f(a)(x-a) + (a)(x-2)²+...+ £2 (x-a)" + R/(x) 21 where P(x) Ru(x) is the Remainder of order nor the error term on an open interval I results from approximating f by Pn (x) I given by minsa R(x) = (1+1) n+1 f(c) (x-a) (n+1)! where ce (a,x) * Convergence of Taylor Series: If Lim Rn (x) = 0 for all XE I then the Taylor Series generated by f at x = a converges to f 5148 on I: f (x) = fla (x-a) K K (K) Exe show that the Taylor series generated by f(x)=e 78 at x=0 converges to f(x) for every real valve x. X f(x) = e is smooth on IR= (-00, 00). Thus, has all orders of all derivatives. Taylor's formula is ce(0,x) 2 21 3 (X)"++ + where in = 1 + x + 31 e e = 1+x+✗ ++ ?! X 31 X k! = Rn (x)= C e n+1 X Note that lim R(x) = e lim X 43 (h+i)! Th5 sec10 (n+1)! 10 1-300 Thus, the series converges to é for every x > 3 * Note that e = 00 as a series. 15:0 CE (0,1) e = 1+1++++ Ru (1), where PM(U= (+) (+) e < In (The Remainder Estimation Theorem) Given Taylor's formula: f(x) = P(x) + R₄(x). If If M (+) Rn(x)/ M for all +E(a,x), then the Remainder "+1 (nti), 1x-al Further, if * then the series converges to f(x). holds for every n, * Exp show that the Taylor series for sinx at x=0 Recall that Taylor's formula for sinx at x=0 sinx = x- X (1) 2h+1 (24+1)! + 24+2 is ce(a,x) = (a,x) R(X). Now apply The Remainder 294) Estimation Theorem: converges for all x. 31 51 R(x) |= 24+2 fro (2n+2)! X IX. (2n+2)! ne (2n+1)! But lim/RIX)/ lin 1 = x => The Maclaurin series for sinx converges ⇒sinx=x+ to sinx x. (241) Exp show that Taylor series for cosx at x=0 converges to cosx for every value of x. The Taylor's Formula for cosx 6 2 21 COSX = 1- - 41 at x=0 is 27 (2n)! +(- + R(x). h وز X + Since | cosx and it's all derivative/ <1 we apply the Remainder Estimation Theorem with M=1 24+1 (2n+1)! IR(x) ≤ Now for every x we have f(x)→ as 100 by Ths sabe Therefore, the series converges to cosx for every x. Thus, 00 iz X 2 in X 3 COSX = (24)! =1-20 ... Exp Find the first four nonzero terms in the Maclaurin series for the functions: 2 4 (2x+xcosx) = +( = +3 + ih 21 3 213 5 x+ 7 x+ X 3(41) 3(61) X 6 72 2160+... X 6 cosx = (1+x+++) (+) 2 X 3 + 6 3 5 2! 2!2! 2!3! = 5 +(+++) 2!५! 3 X =1+--+... 3 6 80 31 COS 2 X COS X = الله Σ 24 2 X 6 (-1) X (24)! x X = 1-2+पना 2! 61 6 (2x) (2x) ५! 61 66 24+... 6! 00 24 (-1) (2x) no (24)! 8 COS2X= (1) 2X (2n)! =1- (2x)+ 2 1-2x+2x 2! nad Exp For what valves of x with an error sin x = x - can of magnitude 3 5 X we no replace sinx by more 3 X 31 メー than 3x10? 315! Alternating Series Estimation Theorem: 5 Therefore the error 5 -4 < 3 x10 120 Using the The error after <1) 3 31 will be less than 51 +1=191 -4 3X10 1x1<√360×100.514. EXP Estimate the error if P(x)=x-is used to estimate the value of sinx at x=0.1 f(x)= sinx = P (4) P₂(x) + R3(x), where R₃(x) = f(c) (x-0)" (4) 4! M ५ Apply The Remainder Estimation. The with M=1 (c) Error = /Rs(x) < 1x = 10.10 24.2x1060) 4 -6 24