# Lec8.pdf - Calculus I Lecture 8 February 3 2018 Problem set...

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Unformatted text preview: Calculus I Lecture 8 February 3, 2018 Problem set 8 Section 11.9: 4, 10, 12, 40 Section 11.10: 16, 18, 20, 55, 68, 70, 72. Outline 1 Power series representations (11.9) 2 Taylor series (11.10) Power series representations 1 We have shown that 1−x has a power series expansion in powers of 1 1−x = ∞ X x n, x ∀x ∈ (−1, 1). n=0 From this we can derive other power series expansions by as follows. Substitute −x 2 for x in the series above, we obtain Substitution: 1 1 + x2 = ∞ X 5 + 3x = 1 2 + 3u This holds for x2 = 1 ∀x ∈ (−1, 1). n=0 1 To expand 5+3x in powers of 1 (−1)n x 2n x + 1, 1 2 1 + 3u 2 − 32 u ∈ (−1, 1) Example: 2x+3 in powers of = i.e. 1 2 let ∞ X n=0 u = x + 1. 3 (− u)n = 2 x ∈ ( −35 , −31 ). x − 1. Then ∞ X (−3)n n=0 2n+1 (x + 1)n . Integration: get 1 Integrating the power series for 1−x term by term, we − ln(1 − x) = C + ∞ X xj j=1 j ∀x ∈ [−1, 1). Let x = 0 we get C = 0. 1 Integrate the power series of 1+x 2 we get arctan x = ∞ X (−1)j x 2j+1 j=0 2j +1 ∀x ∈ (−1, 1]. (note the change in the interval of convergence) Dierentiation: Example: 1 (1 − x)2 = ∞ X j=0 (j + 1)x j ∀x ∈ (−1, 1). Taylor polynomials Recall that when x is close to a, f (x) ≈ f (a) + f 0 (a)(x − a). The RHS is called the linearization of f at a. To get better approximation we use higher order polynomials. Tn (x) = n X f (k) (a) k=0 Note that at the point a, f and Tn k! (x − a)k . agree to the (n) nth order derivatives: Tn (a) = f (a), Tn0 (a) = f 0 (a), . . . , Tn (a) = f (n) (a). These are the Taylor polynomials of f at a. Taylor remainder Using integration by parts and induction, we can prove that Z f (x) − Tn (x) = x (x − t)n dt. n! f (n+1) (t) a n = 0: (1) This is none other than the fundamental theorem of calculus Z x f 0 (t)dt. f (x) − f (a) = (2) a n = 1: Integrating by part, using Z x 0 Z f (t)dt = a u =f0 x udv = uv |xa a = f 0 (a)(x − a) + v = −(x − t) and x Z − Z x we have vdu a f 00 (t)(x − t)dt. a Putting this into formula (2) we get f (x) = f (a) + f 0 (a)(x − a) + Z a x f 00 (t)(x − t)dt (3) v = − 12 (x − t)2 we have Z x Z x x 00 x f (t)(x − t)dt = udv = uv |a − vdu a a a Z x (x − t)2 f 00 (a) 2 (x − a) + f 000 (t) dt. = n = 2: Z Integrating by part, using u = f 00 2 and 2 a Putting this into formula (3) we get 0 f (x) = f (a) + f (a)(x − a) + f 00 (a) 2 2 Z x (x − a) + f 000 (t) a Exercises: I Follow this procedure, prove the formula (1) when I Use induction to prove (1) for all n. n = 3. (x − t)2 2 dt Taylor Series The Taylor series centered at a of a function ∞ X f (n) (a) n=0 n! f is (x − a)n . For many, but not all, functions, this series converges to near a. (To check whether the Taylor series of f f (x) for x converges to f, one use the Taylor remainder formula.) When the Taylor series does converges to representation for f in powers of f (x) = (When a=0 we obtain a x −a ∞ X f (n) (a) n=0 f, n! (x − a)n . this is also known as Maclaurin series.) Examples: x e = sin x = (1 + x)k = ∞ X xk k=0 ∞ X k=0 ∞ X j=0 k! (−1)k x 2k+1 (2k + 1)! k(k − 1) . . . (k − j + 1) j x j! ∀x ∈ (−1, 1). (binomial series) (4) SOME EXAMPLES OF BINOMIAL Taylor series can also be used to compute limit. E.g., sin(x 2 ) − x 2 , x→0 1 − cos(x 3 ) lim 2 ex − 1 . x→0 ln(1 + x sin x) lim ...
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