8 Approximations, Taylor Polynomials, andTaylor SeriesPolynomials are the nicest possible functions. They are easy to differentiateand integrate, which is also true of the basic trigonometric functions, butmore importantly, polynomials can be evaluated at any point, which is nottrue for general functions. So what one does in practice is to approximate anyfunction f of interest by polynomials. When the approximation is done bylinear polynomials, then it is called a linear approximation, which pictoriallycorresponds to linearizing the graph of f. It turns out that the more timesone can differentiate f, the higher is the degree of the polynomial one canapproximate it with, and more importantly, the better the approximationbecomes, as one sees it intuitively. There is only one main theorem here,due to Taylor, but it is omnipresent in all the mathematical sciences, with anumber of ramifications, and should be understood precisely.8.1 Taylor polynomialsSuppose f is an N-times differentiable function on an open interval I. Fixany point a in I. Then for any non-negative integer n ≤ N, the nth Taylorpolynomial of f at x = a is given by(8.1.1) pn(f(x); a) =n∑j=0f(j)(a)j!(x −a)j,where f(j)(a) denotes the jth derivative of f at a. By convention, f(0)(a)just denotes f(a). (f is the 0th derivative of itself!)The coefficientsf(j)(a)j!are called the Taylor coefficients of f at a.The definition has been rigged so that the following holds:Lemma 8.1 Suppose f is itself a polynomial, i.e.,f(x) = a0+ a1x + . . . + amxm,1for some integer m ≥ 0. Then f is infinitely differentiable (which means itcan be differentiated any number of times), andpn(f(x); 0) ={a0+ a1x + . . . + anxn, if n < ma0+ a1x + . . . + amxm, if n ≥ mProof. Clearly, f is differentiable any number of times and moreover,f(n)(x) vanishes if n > m. So we have only to show that for n ≤ m,(8.1.2) f(n)(0) = n!an.When m = 0 this is clear. So let m > 0 and assume by induction that (8.1.2)holds for all polynomials of degree m−1 and n ≤ m−1. Define a polynomialg(x) by the formula(8.1.3) f(x) = a0+ xg(x).Theng(x) =m−1∑j=0aj+1xjand by the inductive hypothesis,(8.1.4) g(n)(0) = n!an+1for all non-negative n ≤ m −1. But by the product rule,f′(x) = g(x) + xg′(x), f′′(x) = 2g′(x) + xg′′(x), . . .By induction, we getf(n)(x) = ng(n−1)(x) + xg(n)(x),so that(8.1.5) f(n)(0) = ng(n−1)(0) ∀n ≤ m, n ≥ 1.The identity (8.1.2), and hence the Lemma, now follow by combining (8.1.4)and (8.1.5).22Lemma 8.2 (Linearity) Let f, g be n-times differentiable at a, and letα, β be arbitrary scalars. Thenpn(αf( x) + βg(x); a) = αpn(f(x); a) + βpn(g(x); a).This is easy to prove because the derivative is linear. In particular, wehave(αf + βg)(j)(a)j!= αf(j)(a)j!+ βg(j)(a)j!.It is helpful to look at some examples:(1): Letf(x) = sin x,which is infinitely differentiable, withf′(x) = cos x, f′′(x) = −sin x = −f(x).Thus(8.1.6) f(n)(x) ={(−1)ksin x, if n = 2k(−1)kcos x, if n = 2k + 1Since sin 0 = 0 and cos 0 = 1, the Taylor polynomials of sin x are given byp0(sin x; 0) = 0, p1(sin x; 0) = p2(sin x; 0) = x, p3(sin x; 0) = p4(sin x; 0) = x−x36, . . .More generally, for any positive integer k,(8.1.7) p2k−1(sin x; 0) = p2k(sin x; 0) = x −x33!+x55!−. . .−(−1)kx2k−1(2k − 1)!.(2): Putf(x) = log x.This function is not defined at 0, so we need to choose another point toevaluate the derivatives, and the easiest one isa = 1.3We havef′(x) =1x, f′′(x) = −1x2, f′′′(x) =2!x3, . . .By induction, we have for any n ≥ 1,f(n)(x) = (−1)n+1(n − 1)!xn.So the nth Taylor coefficient isf(n)(1)n!= (−1)n+11n,where we have used the simple fact that n! is n times (n −1)!. Consequently,since log 1 = 0, the nth Taylor p olynomial of log x is given by(8.1.8) pn(log x; 1) = x −x22+ . . . + (−1)n+1xnn.(3): Considerg(x) =1x.One has, for every n ≥ 0,g(n)(x) = f(n+1)(x),where f (x) is log x. Thus for any a > 0,(8.1.9)g(n)(1)n!= (n + 1)f(n+1)(a)(n + 1)!.As a consequence the Taylor polynomials of g at a = 1 are determinable fromthose of f. Let us make this idea precise.Lemma 8.3 Let f be a which is n times differentiable around a, withpn(f(x); a)) = a0+ a1(x − a) + . . . + an(x − a)n.Thenpn−1(f′(x); a) = a1+ 2a2x + . . . + nan(x − a)n−1.Moreover, if ϕ is a primitive of f around a,pn(ϕ(x); a) = ϕ(a) + a0(x − a) + a1(x − a)22+ . . . + an−1(x − a)nn.4The proof is immediate from the definition of Taylor polynomials.For a general f, even for such a simple function like11+x2, it is painful towork out the Taylor polynomials from scratch. One needs a better way tofind them, and this will b e accomplished in the next section.8.2 Approximation to order nDefinition 8.4 Let f, g be n times differentiable functions at a. We will saythat f and g agree up to order n at a iff we havelimx→af(x) − g(x)(x − a)n= 0.If g is a polynomial agreeing with f (or equalling f, as some would say)up to order n, then we would call g a polynomial approximation of f(x)to order n at x = a. The immediate question which arises is whetherthe nth Taylor polynomial of f is a polynomial approximation to order n.The answer turns out to be Yes, but even more importantly, the Taylorpolynomial is the only one which has this property. Here is the completestatement!Proposition 1 Let f be n times differentiable at a. Then(i) pn(f(x); a) is a polynomial approximation of f to order n;(ii) If q(x) is any polynomial in (x − a) of degree ≤ n which agrees with fup to order n, then q(x) = pn(f(x); a).Proof. (i): Put(8.2.1) g(x) = pn−1(f(x); a) and h(x) = (x − a)n.Then by definition,(8.2.2) pn(f(x); a) = g(x) +f(n)(a)n!h(x).5Hence(8.2.3)f(x) − pn(f(x); a)(x − a)n=f(x) − g(x)h(x)−f(n)(a)n!.So it suffices to prove the following:(8.2.4) limx→af(x) − g(x)h(x)=f(n)(a)n!.Applying Lemma 8.1, we get(8.2.5) g(j)(a) = f(j)(a), ∀j < n.Since g is a polynomial of degree ≤ n−1, its (n−1)th derivative is a constant;so(8.2.6) g(n−1)(x) = g(n−1)(a).Also,(8.2.7) h(j)(x) =n!(x − a)n−j(n − j)!.It follows from (8.2.5) and (8.2.7) that for every j < n −1,(8.2.8) limx→af(j)(x) − g(j)(x) =f(j)(a) − g(j)(a)h(j)(a)= 0andlimx→ah(j)(x) = h(j)(a) = 0.On the other hand, by (8.2.6) and (8.2.7),(8.2.9) limx→af(n−1)(x) − g(n−1)(x)h(n−1)(x)= limx→af(n−1)(x) − f(n−1)(a)n!(x − a)=f(n)(a)n!.In view of (8.2.8) and (8.2.9), we may apply L’Hopital’s rule as f and g aren-times