Math 312: Lecture 32 John Roe Penn State University April 23, 2014 • Reading for today is sections 29–31. The topic is differentiation theorems and Taylor series. • Homework 11 late due date today. • SRTEs are available on ANGEL; please complete. • Final exam information: The final is Monday May 5th, 4:40-6:30 p.m., 119 Osmond. The final exam is comprehensive. • Practice final has been posted. First question on practice final is Quiz 10, available on ANGEL until today 11:55 p.m. • For review information contact Dr Nate Brown, 320 McAllister. Office hours (next week) MF 11:15 and T 3:30, other times by appt, open door policy for finals. [email protected]. Generalized Rolle’s theorem Notation: For a function f that is n times differentiable we use the notation f (n) to denote its nth derivative: that is, f (0)(x) = f (x) and, for n > 0, f (n) is the derivative of f (n−1). Theorem Let f be n + 1 times differentiable on an open interval J, and let p; x 2 J. Suppose that f (p) = f 0(p) = ··· = f (n)(p) = 0 and f (x) = 0. Then there exists y between p and x such that f (n+1)(y) = 0. Proof. The proof is by induction on n: let T (n) denote the statement of the theorem for given n. The base case T (0) is the ordinary version of Rolle’s theorem. Suppose that T (n − 1) is true. Let g = f 0 be the derivative of f . By Rolle’s theorem (the ordinary version) there exists x1 0 between p and x for which f (x1) = g(x1) = 0. Moreover, g(p) = ··· = g(n−1)(p) = 0. By T (n − 1) (applied to the function g on the interval from p to x1) we see there is a point y such (n+1) (n) that f (y) = g (y) = 0. Since y is between p and x1, it is certainly between p and x. This proves T (n) and completes the induction. Taylor’s theorem Theorem Let f be n times differentiable in some open interval J containing p. Then for each x 2 p there is some y between x and p such that n−1 X f (k)(p) f (n)(y) f (x) = (x − p)k + (x − p)n: k! n! k=0 The case n = 1 is the mean value theorem; the case n = 2 is the “second mean value theorem” that we proved last time. This general case is called Taylor’s theorem with Lagrange’s form of the remainder. Proof of Taylor’s theorem Define s to be the real number that makes n−1 X f (k)(p) s f (x) = (x − p)k + (x − p)n: k! n! k=0 Consider the auxiliary function n−1 ! X f (k)(p) s g(t) = f (t) − (t − p)k + (t − p)n : k! n! k=0 By construction g(p) = ··· = g(n−1)(p) = 0 and g(x) = 0. By our generalized Rolle’s theorem there exists y between p and x such that 0 = g(n)(y) = f (n)(y) − s. So s = f (n)(y) as required. Taylor series expansion A function is called smooth if it is differentiable infinitely many times (all the derivatives f (n) exist). Proposition Suppose that f is smooth on an open interval J = (p − R; p + R) containing p, and that there are constants M (n) n and c such that supfjf (x)j : x 2 Jg 6 Mc . Then the Taylor series n−1 X f (k)(p) (x − p)k k! k=0 converges to f (x) for all x 2 J. (In particular, the radius of convergence of the series is at least R.) Proof. Let Sn(x) denote the nth partial sum of the Taylor series. By Taylor’s theorem, f (n)(y)(x − p)n jSn(x) − f (x)j = for some y 2 J n! (Rc)n M ! 0 as n ! 1 6 n! Example: the binomial series This is the series 1 X α(α − 1) ··· (α − k + 1) (1 + x)α = 1 + xk k! k=1 for α 2 R. Theorem The binomial series converges to (1 + x)α for jxj < 1. For 0 6 x < 1 this can be proved using Lagrange’s form of the remainder, which we have just demonstrated. But for −1 < x < 0 the derivatives grow too fast for Lagrange’s form of the remainder to be useful. Another version (Cauchy’s form of the remainder) is used to prove convergence in this case..