Lecture 34: Recall Defn: The n-th Taylor polynomial for a function f at a is: n f j (a) j! + f n (a)

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1 Lecture 34: Recall Defn: The n-th Taylor polynomial for a function f at a is: n f j (a) P n (x) = (x a) j. j! j=0 = f(a)+(f (a))(x a)+(1/2)(f (a))(x a) 2 +(1/3!)(f (a))(x a) f n (a) (x a) n n! Recall that P n (x) is the unique polynomial of degree n whose derivatives of order up to n match those of f. Recall that for f(x) = e x, a = 0, P n (x) = 1 + x + x2 2! + x3 3! xn n!. For x 0, you can show that e x P n (x), i.e., the error in the Taylor polynomial approximation is always 0. Taylor polynomials for f(x) = cos(x) at a = 0. f(x) = cos(x), f (x) = sin(x), f (x) = cos(x), f (x) = sin(x), f (4) (x) = cos(x) and then repeats. So, f n (x) = { ( 1) (n+1)/2 sin(x) n is odd ( 1) n/2 cos(x) n is eve n Can rigorously prove by induction, as we did earlier for sin(x). f(0) = 1, f (0) = 0, f (0) = 1, f (0) = 0, f (4) (0) = 1 and then repeats. { } f n 0 n is odd (0) = ( 1) n/2 n is even } 1

2 So, first few Taylor polynomials are P 0 (x) = P 1 (x) = 1, P 2 (x) = P 3 (x) = 1 x2 2, P 4(x) = P 5 (x) = 1 x2 2 +x4 4! General Taylor polynomial for cos(x): P 2n (x) = P 2n+1 (x) = 1 x2 2 + x4 4! ( 1)n x2n (2n)! Draw graphs of first few Taylor polynomials. Recall that we proved earlier that cos(x) > 1 x 2 /2. Can show inductively, using the MVT, that the errors in the Taylor polynomial approximations for cos(x) alternate in sign. Note that Taylor polynomial approximations can be very poor if x is very far away from a. Note that degree of P n (x) may be strictly less than n. Example: Taylor polynomials for f(x) = 1/x at a = 1. f(x) = 1/x, f (x) = 1/x 2, f (x) = 2/x 3, f (x) = 6/x 4, f (4) (x) = 4!/x 4. f (n) (x) = ( 1) n n!/x n+1. f (n) (1) = ( 1) n n!. P 0 (x) = 1, P 1 (x) = 1 (x 1), P 2 (x) = 1 (x 1) + (x 1) 2. P n (x) = 1 (x 1) + (x 1) 2 (x 1) ( 1) n (x 1) n. Note that this is just the formula for the first terms of the sum of a geometric series: 1 x = 1 1 (1 x) = 1 + (1 x) + (1 x)2 + (1 x) Taylor polynomials for other elementary functions: p

3 Taylor polynomials with a = 0 are often called MacLaurin polynomials. Define: E n (x) = f(x) P n (x), Theorem 12: (Taylor s formula with Lagrange remainder) If f (n+1) (t) exists for all t in an interval containing a and x, then there exists some s in between a and x s.t. E n (x) = f (n+1) (s) (n + 1)! (x a)n+1. Note that there are other formulas for the remainder. Proof: The case n = 1 was Theorem 11, which gave a formula for the error in the linearization approximation. The case n = 0 says that f(x) P 0 (x) = f(x) f(a) = f (s)(x a). which is just the MVT. Prove by induction. So assume E k 1 (x) = f (k) (s) (x a) k. k! for all f to which the theorem applies, i.e., f (k) (t) exists for all t in an interval that contains a and x. The idea is to apply the theorem to f and transform the E k 1 result for f to the E k result for f. We mimic proof of Theorem 11. By GMVT, for some u in (a, x), we have E k (x) (x a) = E k (x) E k (a) k+1 (x a) k+1 (a a) = E k (u) k+1 (k + 1)(u a) k 3

4 E k(t) = f (t) ( f(a) + (f (a))(t a) + (1/2)(f (a))((t a) 2 ) + (1/3!)(f (a))((t a) 3 ) f (k) (a) ((t a) k ). k! = f (t) ( f (a) + (f (a))(t a) + (1/2)(f (a))(t a) f (k) (a) (k 1)! ((t a)k 1 ) = F k 1 (t) where F k 1 is the error in the (k 1)-th Taylor polynomial for f instead of f. So, By the induction hypothesis, F k 1 (t) = (f ) (k) (s) (t a) k. k! E k (x) (x a) = E k (u) k+1 (k + 1)(u a) = F k 1 (u) k (k + 1)(u a) = k (f ) (k) (s)(u a) k (k + 1)(k!)(u a) = f (k+1) (s) k (k + 1)! 4

5 Lecture 35: Recall Theorem 12: (Taylor s formula with Lagrange remainder) E n (x) := f(x) P n (x) If f (n+1) (t) exists for all t in an interval containing a and x, then there exists some s in between a and x s.t. E n (x) = f (n+1) (s) (n + 1)! (x a)n+1. Recall that the case n = 0 was the MVT and the case n = 1 was the the error in the linearization (Theorem 11). Rewrite the conclusion: f(x) = P n (x) + E n (x) = f(a) + (f (a))(x a) + f (a) (x a) f (n) (a) (x a) n + f (n+1) (s) n! (n + 1)! (x a)n+1. where the error E n (x) is regarded as a remainder. Apply to f(x) = e x, a = 0. P n (x) = 1 + x + x 2 /2 + x 3 /3! x n /n! E n (x) = where s is between 0 and x. e s (n + 1)! xn+1 So, if x > 0, then s > 0 and E n (x) > 0 and so e x > P n (x). Apply to f(x) = cos(x), a = 0. Recall: f (n) (x) = { ( 1) (n+1)/2 sin(x) n is odd ( 1) n/2 cos(x) n is even 5 }

6 P 2n (x) = P 2n+1 (x) = 1 x2 2 + x4 4! (note that degree(p k (x)) can be less than k) ( 1)n x2n (2n)! E 2n (x) = E 2n+1 (x) = f (2n+2) (s) (2n + 2)! x2n+2 = ( 1) n+1 cos(s (2n + 2)! x2n+2 So, for all π/2 x π/2, error will alternate in sign: 1 x2 2 x2 cos(x) x4 4! As I claimed earlier it can be shown that the error will alternate in sign for all x (using MVT). Application: Use Taylor polynomial approximations for f(x) = x at a = 25 to get a better approximation for 26. Recall that using the linearization, we got and using the error in the linearization, we got < 26 < 5.1 f (x) = (1/2)x 1/2, f (x) = ( 1/4)x 3/2, f (3) (x) = (3/8)x 5/2 P 1 (x) = f(25) + f (25)(x 25) = (x 25) P 2 (x) = f(25) + f (25)(x 25) + f (25) (x 25) 2 2 = (x 25) (x 25) P 1 (26) = 5.1 P 2 (26) =

7 E 2 (26) = f (3) (s) 3! (26 25) 3 for some 25 < s < 26. Since f (3) (s) > 0, we have 26 = f(26) > P 2 (26) = So So, 0 f (3) (s) = (3/8) 1 s 5/2 < (3/8) 1 (25) 5/2 = (3/8) = E 2 (26) ! which is an improvement. (26 25) 3 = = < 26 < < 26 < Defn: f(x) = O(u(x)) as x a if there exists K > 0 and an open interval I containing a s.t. for all x I f(x) < K u(x) Defn: f(x) = g(x)+o(u(x)) as x a if f(x) g(x) = O(u(x)). Examples: 1. x 2 = O(x) as x 0 because for x < 1, we have x 2 < x. 2. Similarly, if n > m, then x n = O(x m ) as x sin(x) = O(x) because sin(x) x. = 1: for suffi- x 4. x = O(sin(x)) as x 0 because lim x 0 sin(x) ciently small x, x 1 < 1/2 sin(x) x sin(x) < 1/2 sin(x) x = x sin(x) + sin(x) x sin(x) + sin(x) < 3/2 sin(x) 7

8 In fact, for any ɛ, one can replace 3/2 by 1 + ɛ by requiring that x < δ, where δ corresponds to ɛ in the definition of the limit. 8

9 Lecture 36: Midterm 2: Friday, November 20 in class. Midterm Coverage: Sections : exp, log, growth and decay problems, inverse trig functions, Sections , : Indeterminate forms, extreme values, concavity, graph sketching, linear approximation, Taylor polynomials. No related rates or extreme value word problems. No HW for next week. Big O: Recall Defn: f(x) = O(u(x)) as x a if there exists K > 0 and an open interval I containing a s.t. for all x I f(x) < K u(x) Recall Defn: f(x) = g(x) + O(u(x)) as x a if f(x) g(x) = O(u(x)). Simple properties: 1. If f(x) = O(u(x)) as x a, then Cf(x) = O((u(x)) for any C. 2. If f(x) = O(u(x)) and g(x) = O(u(x)) as x a, then f(x) ± g(x) = O(u(x)) as x a (use triangle inequality). 3. If f(x) = O((x a) k u(x)) as x a, then f(x) as x a. f(x) (x a) k = O(u(x)) f(x) (x a) k = (note that technically need not be defined at a; so (x a) k O(u(x)) should be taken to mean that there is a constant K > 0 s.t. for all x a in an open interval containing a, f(x) < K u(x). (x a) k 9

10 4. If f(x) = O(u(x)) as x a and u(x) = O(w(x)) as x a, then f(x) = O(w(x)). Proposition: If f (n+1) (t) is continuous at t = a, then and so E n (x) = O((x a) n+1 ). f(x) = P n (x) + O((x a) n+1 ). Proof: Since f n+1) (t) is continuous at t = a, it is defined on an open interval I containing a. By Taylor s theorem with remainder, for all x I for some s in between a and x. E n (x) = f (n+1) (s) (x a)n+1 (n + 1)! For a perhaps smaller open interval I containing a and all t I Thus, for all x I, f (n+1) (t) f (n+1) (a) + 1 := C E n (x) C x a n+1 (n + 1)! Note: in the text the proposition is stated only informally near the bottom of p. 277 and the assumption is weaker: merely that f n+1) (t) exists in an open interval containing a. This turns out to be OK, by the proof of Taylor s theorem but not by the mere statement of Taylor s theorem (as asserted in the text). Example: e x = 1 + x + x2 2 + O(x3 ) 10

11 cos(x) = 1 x 2 /2 + x 4 /4! + O(x 6 ) (since for f(x) = cos(x), a = 0, P 4 (x) = P 5 (x)). Note that P n (x) can have degree < n. We claim that P n (x) is the unique polynomial of degree at most n that approximates f(x) with error on the order of x a n+1 : Theorem 13: If f(x) = Q(x) + O((x a) n+1 (x) for some polynomial Q(x) of degree at most n, then Q(x) = P n (x). Proof: Since f(x) = P n (x) + O((x a) n+1 ), by property 2 above, we have R(x) := P n (x) Q(x) = (P n (x) f(x)) (Q(x) f(x)) = O((x a) n+1 ). We want to show that R(x) = 0. R(x) has degree at most n. Recall that any polynomial in x can be expressed as a polynomial in (x a) of the same degree. Write R(x) = c 0 + c 1 (x a) c n (x a) n. The idea is that for x near a, the terms in R(x) are too big to be dominated by K x a n+1 for x near a. Suppose R(x) is not 0. Precisely, let c k be the smallest non-zero coefficient. Then, R n (x) = (x a) k (c k + c k+1 (x a) +... c n (x a) n k Then But since lim x a R(x) (x a) k = c k 0. R(x) := O((x a) n+1 ), 11

12 we have by property 3 above, and n + 1 k > 0. So, Contradiction R(x) (x a) k = O((x a)n+1 k ) lim x a R(x) (x a) k = 0. 12