Taylor s Formula G. B. Folland

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1 Taylor s Formula G. B. Folland There s a lot more to be said about Taylor s formula than the brief disussion on pp of Apostol. Let me begin with a few definitions. Definitions. A funtion f defined on an interval I is alled k times differentiable on I if the derivatives f, f,..., f (k) exist and are finite on I, and f is said to be of lass C k on I if these derivatives are all ontinuous on I. (Note that if f is k times differentiable, the derivatives f,..., f (k 1) are neessarily ontinuous, by Theorem 5.3; the only question is the ontinuity of f (k).) If f is (at least) k times differentiable on an open interval I and I, its kth order Taylor polynomial about is the polynomial P k, (x) = k j=0 f (j) () (x ) j j! (where, of ourse, the zeroth derivative f (0) is f itself), and its kth order Taylor remainder is the differene R k, (x) = f(x) P k, (x). Remark 1. The kth order Taylor polynomial P k, (x) is a polynomial of degree at most k, but its degree may be less than k beause f (k) () might be zero. Remark 2. We have P k, () = f(), and by differentiating the formula for P k, (x) repeatedly and then setting x = we see that P (j) k, () = f (j) () for j k. That is, P k, is the polynomial of degree k whose whose derivatives of order k at agree with those of f. For future referene, here are a few frequently used examples of Taylor polynomials: f(x) = e x ; P k,0 (x) = 0 j k f(x) = os x; P k,0 (x) = 0 j k/2 f(x) = sin x; P k,0 (x) = f(x) = log x; 0 j<k/2 P k,1 (x) = 1 j k x j j! ( 1) j x 2j (2j)! ( 1) j x 2j+1 (2j + 1)! ( 1) j 1 (x 1) j Note that (for example) x2 is both the 2nd order and the 3rd order Taylor polynomial of os x, beause the ubi term in its Taylor expansion vanishes. (Also note that in higher mathematis the natural logarithm funtion is almost always alled log rather than ln.) For k = 1 we have P 1, (x) = f() + f ()(x ); this is the linear funtion whose graph is the tangent line to the graph of f at x =. Just as this tangent line is the straight line 1 j

2 that best approximates the graph of f near x =, we shall see that P k, (x) is the polynomial of degree k that best approximates f(x) near x =. To justify this assertion we need to see that the remainder R k, (x) is suitably small near x =, and there are several ways of making this preise. The first one is simply this: the remainder R k, (x) tends to zero as x faster than any nonzero term in the polynomial P k, (x), that is, faster than (x ) k. Here is the result: Theorem 1. Suppose f is k times differentiable in an open interval I ontaining the point. Then R k, (x) lim x (x ) = lim f(x) P k, (x) = 0. k x (x ) k Proof. Sine f and its derivatives up to order k agree with P k, and its derivatives up to order k at x =, the differene R k, and its derivatives up to order k vanish at x =. Moreover, (x ) k and its derivatives up to order k 1 also vanish at x =, so we an apply l Hôpital s rule k times to obtain lim x R k, (x) (x ) k = lim x R (k) k, (x) k(k 1) 1(x ) 0 = 0 k! = 0. There is a onvenient notation to desribe the situation in Theorem 1: we say that R k, (x) = o((x ) k ) as x, meaning that R k, (x) is of smaller order than (x ) k as x. More generally, if g and h are two funtions, we say that h(x) = o(g(x)) as x (where might be ± ) if h(x)/g(x) 0 as x. The symbol o(g(x)) is pronouned little oh of g of x ; it does not denote any partiular funtion, but rather is a shorthand way of desribing any funtion that is of smaller order than g(x) as x. For example, Corollary 1 of l Hôpital s rule (see the notes on l Hôpital s rule) says that for any a > 0, x a = o(e x ) and log x = o(x a ) as x, and log x = o(x a ) as x 0+. Another example: saying that h(x) = o(1) as x simply means that lim x h(x) = 0. In order to simplify notation, in the following disussion we shall assume that = 0 and write P k instead of P k,. (The Taylor polynomial P k = P k,0 is often alled the kth order Malaurin polynomial of f.) There is no loss of generality in doing this, as one an always redue to the ase = 0 by making the hange of variable x = x and regarding all funtions in question as funtions of x rather than x. The onlusion of Theorem 1, that f(x) P k (x) = o(x k ), atually haraterizes the Taylor polynomial P k, ompletely: Theorem 2. Suppose f is k times differentiable on an open interval I ontaining 0. If Q is a polynomial of degree k suh that f(x) Q(x) = o(x k ) as x 0, then Q = P k. 2

3 Proof. Sine f Q and f P k are both of smaller order than x k, so is their differene P k Q. Let P k (x) = k 0 a jx j (of ourse a j = f (j) (0)/j!) and Q(x) = k 0 b jx j. Then (a 0 b 0 ) + (a 1 b 1 )x + + (a k b k )x k = P k (x) Q(x) = o(x k ). Letting x 0, we see that a 0 b 0 = 0. This being the ase, we have (a 1 b 1 ) + (a 2 b 2 )x + + (a k b k )x k 1 = P k(x) Q(x) x Letting x 0 here, we see that a 1 b 1 = 0. But then = o(x k 1 ). (a 2 b 2 ) + (a 3 b 3 )x + + (a k b k )x k 2 = P k(x) Q(x) x 2 = o(x k 2 ), whih likewise gives a 2 b 2 = 0. Proeeding indutively, we find that a j = b j for all j and hene P k = Q. Theorem 2 is very useful for alulating Taylor polynomials. It shows that using the formula a k = f (k) (0)/k! is not the only way to alulate P k ; rather, if by any means we an find a polynomial Q of degree k suh that f(x) = Q(x) + o(x k ), then Q must be P k. Here are two important appliations of this fat. Taylor Polynomials of Produts. Let P f k and P g k be the kth order Taylor polynomials of f and g, respetively. Then f(x)g(x) = [ P f k (x) + o(xk ) ][ P g k (x) + o(xk ) ] = [ terms of degree k in P f k (x)p g k (x)] + o(x k ). Thus, to find the kth order Taylor polynomial of fg, simply multiply the kth Taylor polynomials of f and g together, disarding all terms of degree > k. Example 1. What is the 6th order Taylor polynomial of x 3 e x? Solution: [ ] x 3 e x = x x + x2 2 + x3 6 + o(x3 ) = x 3 + x 4 + x5 2 + x6 6 + o(x6 ), so the answer is x 3 + x x x6. Example 2 What is the 5th order Taylor polynomial of e x sin 2x? Solution: ] ] e x sin 2x = [1 + x + x2 2 + x3 6 + x x o(x5 ) [2x (2x)3 + (2x) o(x5 ) [ 2 = 2x + 2x 2 + x ] [ 2 + x ] [ 2 + x ] + o(x 5 ), 120 so the answer is 2x + 2x x3 x x5. 3

4 Taylor Polynomials of Compositions. If f and g have derivatives up to order k, and g(0) = 0, we an find the kth Taylor polynomial of f g by substituting the Taylor expansion of g into the Taylor expansion of f, retaining only the terms of degree k. That is, suppose f(x) = a 0 + a 1 x + + a k x k + o(x k ). Sine g(0) = 0 and g is differentiable, we have g(x) g (0)x for x near 0, so anything that is o(g(x) k ) is also o(x k ) as x 0. Hene, f(g(x)) = a 0 + a 1 g(x) + + a k g(x) k + o(x k ). Now plug in the Taylor expansion of g on the right and multiply it out, disarding terms of degree > k. Example 3. What is the 16th order Taylor polynomial of e x6? Solution: e x = 1 + x + x2 2 + x3 6 + o(x3 ) = e x6 = 1 + x 6 + x x o(x18 ). The last two terms are both o(x 16 ), so the answer is 1 + x x12. Example 4. What is the 4th order Taylor polynomial of e sin x? Solution: e sin x = 1 + sin x + sin2 x 2 + sin3 x 6 + sin4 x 24 + o(x 4 ) sine sin x x. Now substitute x 1 6 x3 + o(x 4 ) for sin x on the right (yes, the error term is o(x 4 ) beause the 4th degree term in the Taylor expansion of sin x vanishes) and multiply out, throwing all terms of degree > 4 into the o(x 4 ) trash an: e sin x = 1 + [x x3 6 ] [x 2 x4 3 ] + x3 6 + x o(x4 ), so the answer is 1+x+ 1 2 x2 1 8 x4. (To appreiate how easy this is, try finding this polynomial by omputing the first four derivatives of e sin x.) Taylor Polynomials and l Hôpital s Rule. Taylor polynomials an often be used effetively in omputing limits of the form 0/0. Indeed, suppose f, g, and their first k 1 derivatives vanish at x = 0, but their kth derivatives do not both vanish. The Taylor expansions of f and g then look like f(x) = f (k) (0) k! Taking the quotient and anelling out x k /k!, we get x k + o(x k ), g(x) = g(k) (0) x k + o(x k ). k! f(x) g(x) = f (k) (0) + o(1) g (k) (0) + o(1) f (k) (0) g (k) (0) as x 0. This is in aordane with l Hôpital s rule, but the devies disussed above for omputing Taylor polynomials may lead to the answer more quikly than a diret appliation of l Hôpital. 4

5 Example 5. What is lim x 0 (x 2 sin 2 x)/x 2 sin 2 x? Solution: ] 2 sin 2 x = [x x3 6 + o(x4 ) = x 2 x4 3 + o(x4 ), so x 2 sin 2 x = x 4 + o(x 4 ), and x 2 sin 2 1 x x 2 sin 2 x = 3 x4 + o(x 4 1 ) x 4 + o(x 4 ) = + o(1) o(1)) 1 3. (Again, to appreiate how easy this is, try doing it by l Hôpital s rule.) Example 6. Evaluate [ 1 lim x 1 log x + x ]. x 1 Solution: Here we need to expand in powers of x 1. First of all, 1 log x x x 1 = x 1 x log x (x 1) log x (x 1) (x 1) log x log x =. (x 1) log x Next, log x = (x 1) 1 2 (x 1)2 + o((x 1) 2 ), and plugging this into the numerator and denominator gives (x 1) (x 1) 2 [ (x 1) 1 2 (x 1)2] + o((x 1) 2 ) (x 1) 2 + o((x 1) 2 ) = o(1) 1 + o(1) 1 2. Theorem 1 tells us a lot about the remainder R k, (x) = f(x) P k, (x) for small x, but sometimes one wants a more preise quantitative estimate of it. The most ommon ways of obtaining suh an estimate involve slightly stronger onditions on f; namely, instead of just being k times differentiable we assume that it is k + 1 times differentiable, or perhaps of lass C k+1, and the estimates we obtain involve bounds on the derivative f (k+1). There are several formulas for R k, (x) that lead to suh estimates; we shall present the two that are most often enountered. The first one is the one presented in Apostol. (It s Theorem 5.19, with the hange of variable k = n 1. Apostol states the hypotheses in a slightly more general, but also more ompliated, form; the version below usually suffies.) Theorem 3 (Lagrange s Form of the Remainder). Suppose f is k + 1 times differentiable on an open interval I and I. For eah x I there is a point x 1 between and x suh that R k, (x) = f (k+1) (x 1 ) (x ) k+1. (1) For the proof of Theorem 3 we refer to Apostol. The other popular form of the remainder requires a slightly stronger hypothesis, that f (k+1) not only exists but is ontinuous. (Atually, it s enough for it to be Riemann integrable, but these minor variations in the assumptions are usually of little importane.) I suspet the reason that Apostol doesn t mention it is that it involves an integral, and he doesn t want to disuss integrals until later. 5

6 Theorem 4 (Integral Form of the Remainder). Suppose f is of lass C k+1 on an open interval I and I. If x I, then R k, (x) = 1 Proof. Realling the definition of R k,, we an restate (2) as (x t) k f (k+1) (t) dt. (2) f(x) = k j=0 f (j) () (x ) j 1 + j! (x t) k f (k+1) (t) dt. (3) For k = 0, this simply says that f(x) = f() + f (t) dt, (4) whih is true by the fundamental theorem of alulus. Next, we integrate (4) by parts, taking u = f (t), du = f (t) dt; dv = dt, v = t x. Notie the twist: normally if dv = dt we would simply take v = t, but we are free to add a onstant of integration, and we take that onstant to be x. (The number x, like, is fixed in this disussion; the variable of integration is t.) The result is f(x) = f() + (t x)f (t) x (t x)f (t) dt = f() + (x )f () + whih is (3) with k = 1. Another integration by parts, with (x t)f (t) dt, u = f (t), du = f (t) dt; dv = (x t) dt, v = 1 2 (x t)2, (again, instead of taking v = xt 1 2 t2 we take v = 1 2 (x t)2 = 1 2 x2 + xt 1 2 t2 ) gives f(x) = f() + (x )f () 1 2 (x t)2 f (t) x + = f() + (x )f () + (x )2 f () + 1 2! 2! 1 (x 2 t)2 f (t) dt (x t) 2 f (t) dt, whih is (3) with k = 2. The pattern should now be lear: a k-fold integration by parts starting from (4) yields (3). The formal indutive proof is left to the reader. Let s be lear about the signifiane of Theorems 3 and 4. They are almost never used to find the exat value of the remainder term (whih amounts to knowing the exat value of the original f(x)); one doesn t know just where the point x 1 in (1) is, and the integral in 6

7 (2) is usually hard to evaluate. Instead, the philosophy is that Taylor polynomials P k, are used as (simpler) approximations to (ompliated) funtions f near, and the remainders R k, are regarded as junk to be disregarded. For this to work one needs some assurane that R k, (x) is small enough that one an safely neglet it or an estimate of the magnitude of the error one makes in doing so. The main purpose of Theorems 3 and 4 is to provide suh information via the following result. Corollary 1. Suppose f is k+1 times differentiable on an interval I and that f (k+1) (x) C for x I. Then for any x, I we have x k+1 R k, (x) C. (5) Proof. The estimate (5) is learly an immediate onsequene of (1). It also follows easily from (2): if x >, R k, (x) C (x t) k dt = C (x t) k+1 x (x )k+1 k! k! k + 1 = C, and if x <, R k, (x) C k! (x t) k dt = C k! x (t x) k dt = C ( x) k+1 k! k + 1 x k+1 = C. Observe that Corollary 1 is a more preise and quantitative version of Theorem 1 (under slightly stronger hypotheses on f): Theorem 1 says that R k, (x) vanishes faster than (x ) k as x ; Corollary 1 says that it vanishes at least at a rate proportional to (x ) k+1 and gives a good estimate for the proportionality onstant. The best estimate is obtained by taking C to be the least upper bound for f (k+1) on I, but it is usually not ruial to ompute this optimal value for C. What is ruial, however, and what some people find easy to forget, is the use of absolute values. It s the size of R k, (x) that matters. A typial use of Taylor polynomials is to evaluate integrals of funtions that don t have an elementary antiderivative. Here s an example. Example 7. The funtion f(x) = e x2 has no elementary antiderivative. However, we an do a Taylor approximation of e x2 and integrate the resulting polynomial. The effiient way to proeed is to onsider the Taylor approximations of e y (easier to ompute with!) and then set y = x 2. Sine (d/dy) j e y = ( 1) j e y 1 for y 0, the estimate (5) shows that e y = 1 y + y2 2 Setting y = x 2 yields + ( 1)k yk k! + R k,0(y), where R k,0 (y) yk+1 for y 0. e x2 = 1 x 2 + x4 2! + ( 1)k x2k k! + R k,0 (x 2 ), where R k,0 (x 2 ) x2k+2. 7

8 Therefore, where 0 e t2 dt = x x3 3 + x5 5 2! + ( 1)k x 2k+1 (2k + 1) k! + error, error For instane, if x = 1, we an take k = 4 and obtain 1 0 e t2 dt = ! ! 0 t 2k+2 dt = x 2k+3 (2k + 3). = with error less than A Few Conluding Remarks. Although Theorems 3 and 4 are most ommonly used through Corollary 1, there are other things that an be done with them. There s a nie appliation of Theorem 3 on p.376 of Apostol, whih we ll disuss toward the end of the quarter. For an extra twist on Theorem 4 that yields more estimates, as well as a sharper form of Theorem 1, see my paper Remainder estimates in Taylor s theorem, Amerian Mathematial Monthly 97 (1990), (available online through the UW Libraries site). 8