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1 Taylor-Maclaurin Series 1 Taylor-Maclaurin Series 2 00 2 Writing x = x0 + n4x; x1 = (n − 1)4x; ::, we get, (4y0) = y0 (4x) ::: and letting n ! 1, a leap of logic reduces the interpolation formula to: 1 y = y + (x − x )y0 + (x − x )2 y00 + ::: 0 0 0 0 2! 0 Definition 1.0.1. A function f is said to be an Cn function on (a; b) if f is n-times differentiable and the nth derivative f (n) is continuous on (a; b) and f is said to belong to C1 if every derivative of f exists (and is continuous) on (a; b). Taylor’s Formula: Suppose f belongs to C1 on (−R; R). Then for every n 2 N, and x 2 (−R; R) we have: 1 1 f(x) = f(0) + f 0(0)x + f (2)(0) x2 + ::: + f (n)(0) xn + R (x) 2! n! n f is said to be analytic at 0 if the remainder Rn(x) ! 0 as n ! 1. There are two standard forms for the remainder. 1. Integral form: Z x 1 (n+1) n Rn(x) = f (t)(x − t) dt n! 0 Proof. Integrating by parts, 1 Z x 1 1 Z x f (n+1)(t)(x − t)ndt = − f (n)(0)xn + f (n)(t)(x − t)n−1dt n! 0 n! (n − 1)! 0 Proof is completed by induction on n on observing that the result holds for n = 0 by the Funda- mental theorem of Calculus. 2. Lagrange’s form: There exists c 2 (0; x) such that: 1 R (x) = f (n+1)(c)xn+1 n (n + 1)! This form is easily derived from the integral form using intermediate value theorem of integral calculus. Lemma 1.0.2. If h(≥ 0) and g are continuous on [a; b], then there exists c 2 (a; b) such that Z b Z b h(t)g(t)dt = g(c) h(t)dt a a Proof. If m and M denote the min and max of g on [a; b], then h ≥ 0 implies (assuming h is not identically zero), Z b Z b m ≤ h(t)g(t)dt= h(t)dt ≤ M a a Now the intermediate value theorem for continuous functions assures the existence of c. 1 Examples: 1 (n+1) n+1 1. Let f(x) = sin(x). Then given x > 0, there exists ξ 2 (0; x) such that Rn(x) = (n+1)! f (ξ)x 1 n+1 and hence jRn(x)j ≤ (n+1)! x . Thus Rn(x) ! 0 as n ! 1 (uniformly on a finite subinterval 1 3 of R). Consequently, sin(x) = x − 3! x :::. Note that even if the series on the r.h.s. converges for every x, to show that it converges to sin(x), it is necessary to prove that Rn ! 0 as n ! 1. 2. Let f(x) = log(1 + x); x > −1. Then f (n)(x) = (−1)(n+1)(n − 1)!(1 + x)−n. Using the integral form (and assuming 0 < x < 1) we get: Z x j(x − t)j n 1 jRn(x)j ≤ ( ) dt 0 1 + t 1 + t Z x jxjn ≤ ( )dt 0 1 + t ≤ xn log(1 + x) (−1)n −n−1 Using Lagrange’s form, Rn(1) = n+1 1=(1 + ξ) . Thus Rn(x) ! 0 as n ! 1 if x 2 (0; 1]. x3 x5 log(1 + x) = x − 3 + 5 ::: for x 2 (−1; 1]. 3. Let f(x) = (1 + x)α where α 2 RnN; α 6= 0 and jxj < 1. Then 1 α f (k)(x) = (1 + x)α−k; k = 1; 2::: k! k Thus, Z x 1 x − t n α−1 Rn(x) = [α(α − 1)::(α − n)] ( ) (1 + t) dt n! 0 1 + t 1 If rn = n! [α(α − 1)::(α − n)], then jrn+1=rnj ! 1 as n ! 1. n n Now if jxj < 1, choose % such that %jxj < 1 and observe that jrnjjxj ≤ c(%jxj) ! 0 as n ! 1. x−t Consequently, since j 1+t j ≤ jxj for t 2 (0; x), as n ! 1, Z x n α−1 jRn(x)j ≤ rnjxj j (1 + t) dtj ) jRn(x)j ! 0 0 Thus, 1 X α (1 + x)α = 1 + xk k k=1 1 Example: Let α = − 2 . Then for jxj < 1, − 1 (2k)! 2 = (−1)k k 22k:(k!)2 Hence 1 X (2k)! (1 + x)−1=2 = 1 + (−1)k xk 22k:(k!)2 k=1 2 2 Power Series 2.1 Prerequisites Definition 2.1.1. (a)If ffng is a sequence of functions defined on a common domain A, we say that ffng converges to a function f defined on A if for every x 2 A; fn(x) ! f(x) as n ! 1. (b) We say that fn converges uniformly to f on A if sup fjfn(x) − f(x)j : x 2 Ag ! 0asn ! 1: What are the advantages of having uniform convergence? Recall that g is said to be continuous at x if g(x + 4x) ! g(x) as 4x ! 0. Theorem 2.1.2. Suppose fn ! f uniformly on A ⊆ R. The following hold. 1. If each fn is continuous on A, then so is f. R R 2. If A = [a; b], then A fn ! A f. Proof. 1. Fix x0 2 A. Given " > 0; choose N such that jf(x) − fN (x)j < "=3 for every x 2 A. In particular, jf(x0) − fN (x0)j < "=3 and jf(x0 + 4x0) − fN (x0 + 4x0)j < "=3. By hypothesis, fN is continuous, so there exists δ such that if j4x0j < δ, then jfN (x0 + 4x0) − fN (x0)j < "=3 Now, if j4x0j < δ jf(x0 + 4x0) − f(x0)j ≤ jf(x0) − fN (x0)j + jfN (x0 + 4x0) − fN (x0)j + jfN (x0 + 4x0) − fN (x0)j < "=3 + "=3 + "=3 i.e. f is continuous at x0. 2. Simply observe that Z b j fn − fdxj ≤ (b − a)maxfjfn(x) − f(x)j : a ≤ x ≤ bg ! 0 a Examples: n 1. Let A = [0; 1],and fn(x) = x . If f(1) = 1 and f(x) = 0 on [0; 1), then fn ! f, but not uniformly. 2. Let A = [0; 1) fn be the (isosceles) triangular graph with base [0; n] and height 2=n at x = n=2. Then fn(x) ! 0 uniformly on A. 3. Let A = [0; 1) fn be the (isosceles) triangular graph with base [0; 1=n] and height 1 at x = 1=n. Then fn(x) ! 0 on A, but not uniformly. 3 2.2 Power Series P1 n A power seriescentered at a is an infinite series of the form 0 cn(x − a) where fcng ⊆ R. For our purpose, it suffices to set a = 0 though the results are true for any a 2 R. There exists a number R, so that the series converges whenever jxj < R and diverges whenever jxj > R. For clear reasons, R is called the radius of convergence. If cn+1 ' 1=R or equivalently jc j1=n ' 1=R, then cn n jxj jc xij ' ( )i i R P k If jxj < R, then by comparison to the geometric series, ckx is convergent. Let 2 f(x) = c0 + c1x + c2x + ::: What makes power series very useful is that they are (almost) as easy to manipulate as polynomi- als. The principal reason is the following. Pn i Theorem If [a; b] ⊆ (−R; R), and x 2 [a; b], then sn(x) = 0 cix converges uniformly to f(x). i i Proof: As before, jcix j ≤ jb=Rj and jb=Rj < 1. The following corollary is now immediate. Corollary If [a; b] ⊆ (−R; R), then, Z b 1 Z b X i f(x)dx = c + ci x dx a 0 a 1=n 1=n P1 i−1 Also, jncnj ' jcnj ' 1=R and hence for jxj < R; i=n+1 icix ! 0 as n ! 1. Hence Z Z 0 0 sn ! g ) sn ! g R 0 R 0 0 0 But sn = sn ! f and thus f = g or g = f . Thus, sn ! f . It follows by induction that f is in C1 on (−R; R). Discussing the behavior of the power series at x = ±R takes work. P1 n P Abel’s Theorem: If f(x) = 0 cnx is convergent on (−1; 1) and cn is a convergent series of numbers, then X lim f(x) = cn x!1− 1 2 4 Example: Recall that, 1+x2 = 1 − x + x ::: Integrating term-by-term we have: −1 x3 x5 tan (x) = x − 3 + 5 ::: if x < 1. 1 1 By alternating series theorem, we know that 1 − 3 + 5 ::: is convergent. −1 π So by Abel’s theorem, using the fact that tan (1) = 4 we get π 1 1 = 1 − + ::: 4 3 5 Question: Let f(x) = e−1=x if x > 0 and 0 otherwise. Show that f 2 C1 on R. Is f analytic at 0? 4 3 Applications 3.1 Differential Equations Consider the second order linear differential equation of second order. y00 + p(x)y0 + q(x)y = 0 Definition 3.1.1. A point x0 in R is said to be a regular point if p(x) and q(x) have power-series expansion in an interval around x0. Otherwise, it is a singular point. Example 1: y00 + 2xy0 + 2y = 0 P1 k We look for a solution of the form y = k=0 ckx for x 2 (−R; R). Differentiate and substitute, 1 1 0 X k−1 X k−2 y = kckx ; y" = k(k − 1)ckx k=1 k=2 Thus, 1 1 1 X k X k X k ck+2(k + 1)(k + 2)x + 2 kckx + 2 ckx = 0 k=0 k=0 k=0 Equivalently, 2 (k + 1)(k + 2)c + 2(k + 1)c = 0 ) c = − c ; k ≥ 0 k+2 k k+2 k + 2 k In general,for k ≥ 0, (−1)k (−1)k2k c = c andc = c 2k k! 0 2k+1 1:3:5::(2k + 1) 1 Thus the general solution is: 1 1 X (−1)k X (−1)k2k y = c x2k + c x2k+1 0 k! 1 1:3:5::(2k + 1) k=0 k=0 Example 2 xy00 + y0 + xy = 0 Here x0 = 0 is a singular point.
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