Section 11.10 Taylor Series Given a function f(x), we would like to be able to find a power series that represents the function. For example, in the last section we noted that we can represent ex by the power series x2 x3 x4 ex = 1 + x + + + + :::; 2! 3! 4! and the power series converges to ex for any value of x. In this section we will show how to find such a power series representation. If the function f(x) can be written as a power series on an interval I, then the power series is of the form X1 n f(x) = an(x − a) n=0 2 3 = a0 + a1(x − a) + a2(x − a) + a3(x − a) + ::: The only unknowns above are the coefficients ai; if we can determine their form, then we will know the precise form of the power series for f(x). The way to determine the coefficients is by using the derivatives of f; assuming that the power series for f converges on I, we know that its derivatives converge on I as well. We calculate the derivatives below: 2 3 4 f(x) = a0 + a1(x − a) + a2(x − a) + a3(x − a) + a4(x − a) + ::: 0 2 3 n−1 f (x) = a1 + 2a2(x − a) + 3a3(x − a) + 4a4(x − a) + ::: + nan(x − a) + :::; 00 2 n−2 f (x) = 2a2 + 2 · 3a3(x − a) + 3 · 4a4(x − a) + ::: + (n − 1) · nan(x − a) + :::; 000 n−2 f (x) = 2 · 3a3 + 2 · 3 · 4a4(x − a) + ::: + (n − 2) · (n − 1) · nan(x − a) + :::; and in general, (n) f (x) = n! an + :::; Each of these is a power series centered at x = a, so each one converges for x = a. That means that we can evaluate each of the power series at x = a: f(a) = a0; 0 f (a) = a1; 00 f (a) = 2a2; 000 f (a) = 2 · 3a3; and in general, (n) f (a) = n! an; since x − a = 0. This tells us the values for the coefficients: 1 Section 11.10 a0 = f(a); 0 a1 = f (a); f 00(a) a = ; 2 2! f 000(a) a = ; 3 3! and in general f (n)(a) a = : n n! Definition 0.0.1. Let f be a function with derivatives of all orders throughout some interval containing a as an interior point. Then the Taylor series generated by f at x = a is given by 1 X f (k)(a) f 00(a) (x − a)k = f(a) + f 0(a)(x − a) + (x − a)2 + ::: k! 2! k=0 where f (k)(a) is the kth derivative of f evaluated at x = a. The Taylor series generated by f at x = 0, given by 1 X f (k)(0) f 00(0) xk = f(0) + f 0(0)x + x2 + ::: k! 2! k=0 is generally called the Maclaurin series generated by f. We must be careful about our interpretation of the above theorem; it is not always the case that the Taylor series of a function f(x) actually converges to f(x) on its entire interval of convergence. In fact, the function f(x) might not be equal to its Taylor series. We will discuss this problem in more detail in the next section. Example: Find the Taylor series generated by f(x) = ex at x = 0. Since the Taylor series for f(x) at x = 0 has form 1 X f (k)(0) f 00(0) xk = f(0) + f 0(0)x + x2 + :::; k! 2! k=0 we will need to calculate derivatives of ex and evaluate them at x = 0: 2 Section 11.10 f 0(x) = ex f 0(0) = e0 = 1 f 00(x) = ex f 00(0) = e0 = 1 . f (n)(x) = ex f (n)(0) = e0 = 1 So the Taylor series for f(x) = ex at x = 0 is given by 1 X 1 1 xk = 1 + x + x2 + ::: k! 2! k=0 1 X 1 Again, at this point we do not know if ex actually equals xk; we will show that the Taylor k! k=0 series does actually equal ex in the next section. In Calculus I, we learned how about linearizations of a function. A linearization of a function f(x) atp the point x = a is the line tangent to the function at x = a. For example, the function 3 − x − 2 f(x) = x 4 and its linearization at x = 5, given by L(x) = 3 3 , are graphed below: We like linearizations because they give us a way to approximate a difficult function byp using a linear function, which is quite simple to work with. Close to x = 5, the function f(x) = 3 x − 4 x − 2 and its linearization 3 3 are nearly indistinguishable, and we may use L(x) as an approximation of f(x). In order to improve the quality of the approximations, we extend this idea to polynomials of higher degree. Definition 0.0.2. Let f be a function with derivatives of order k for k = 1; 2;:::;N in some interval containing a as an interior point. Then for any integer n from 0 to N, the Taylor polynomial of order n generated by f at x = a is the polynomial f 00(a) f (n)(a) P (x) = f(a) + f 0(a)(x − a) + (x − a)2 + ::: + (x − a)n: n 2! n! 3 Section 11.10 Notice that this polynomial (which has a finite number of terms) is simply a truncation of the Taylor series for f(x); we get the kth Taylor polynomial by erasing all terms from the Taylor series so that n > k. As we will see in the next example, the Taylor polynomials of f(x) become better and better approximations of f(x) as we increase the number of terms in the polynomial. Example: Given f(x) = cos x, find its Taylor polynomials P0(x), P1(x), P2(x), and P3(x) and its Taylor π series centered at x = 2 . The desired Taylor polynomials are given by π P = f( ) 0 2 π π π P = f( ) + f 0( )(x − ) 1 2 2 2 π π π f 00( π ) π P = f( ) + f 0( )(x − ) + 2 (x − )2 2 2 2 2 2! 2 π π π f 00( π ) π f 000( π ) π P = f( ) + f 0( )(x − ) + 2 (x − )2 + 2 (x − )3 3 2 2 2 2! 2 3! 2 π So we need to calculate the derivatives of f at 2 : π f(x) = cos x f( ) = 0 2 π f 0(x) = − sin x f 0( ) = −1 2 π f 00(x) = − cos x f 00( ) = 0 2 π f 000(x) = sin x f 000( ) = 1: 2 So the Taylor polynomials are given by π P (x) = f( ) = 0 0 2 π π π π P (x) = f( ) + f 0( )(x − ) = −(x − ) 1 2 2 2 2 π π π f 00( π ) π π π P (x) = f( ) + f 0( )(x − ) + 2 (x − )2 = −(x − ) + 0 = −(x − ) 2 2 2 2 2! 2 2 2 π π π f 00( π ) π f 000( π ) π π 1 π P (x) = f( ) + f 0( )(x − ) + 2 (x − )2 + 2 (x − )3 = −(x − ) + (x − )3: 3 2 2 2 2! 2 3! 2 2 3! 2 π Below is a graph of f(x) = cos x and P1(x) (which is just the linearization of cos x at x = 2 ): 4 Section 11.10 The graph below shows f(x) and P3(x): Notice that, as we add more terms, the Taylor polynomial seems to become a better and better approximation of the function. In order to find the Taylor series, we need to find 1 X f (k)( π ) π π π π f 00( π ) π 2 (x − )k = f( ) + f 0( )(x − ) + 2 (x − )2 + ::: k! 2 2 2 2 2! 2 k=0 We end up with π π π f 00( π ) π π 1 π f( ) + f 0( )(x − ) + 2 (x − )2 + ::: = 0 + −(x − ) + 0 + (x − )3 + ::: 2 2 2 2! 2 2 3! 2 π 1 π 1 π = −(x − ) + (x − )3 − (x − )5 + ::: ; 2 3! 2 5! 2 so the Taylor series may be written as 1 X (−1)n+1 π (x − )2n+1: (2n + 1)! 2 n=0 Convergence of Taylor Series 5 Section 11.10 We have one question left to answer: given a function f(x) and the Taylor series generated by f at x = a, does the Taylor series converge to f(x)? In other words, does the equal sign in 1 X f (k)(a) f(x) = (x − a)k k! k=0 make sense? We will see how to answer that question in this section. Theorem 0.0.3. Taylor's Formula If f(x) has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, f 00(a) f 000(a) f (n)(a) f(x) = f(a) + f 0(a)(x − a) + (x − a)2 + (x − a)3 + ::: + (x − a)n + R (x) 2! 3! n! n where f (n+1)(c) R (x) = (x − a)n+1 n (n + 1)! for some c between a and x. Notice that the theorem rewrites f(x) as f(x) = Pn(x) + Rn(x), the nth Taylor polynomial of f plus the remainder function Rn(x). In effect, the theorem says that plugging x into f is almost the same as plugging x into the nth Taylor polynomial; the remainder function is the difference between f(x) and Pn(x).