11.10a Taylor and Maclaurin Series

Transcription

1 11.10a a Taylor and Maclaurin Series Let y = f(x) be a differentiable function at x = a. In first semester calculus we saw that (1) f(x) f(a)+f (a)(x a), for all x near a The right-hand side of (1) was called the standard linear approximation (or linearization) of f at a and the point x = a was called the center of the approximation. For example, if f(x) = 1/(1 x) then f (x) = 1/(1 x) 2 and the linearization of f at x = 0 is f(x) f(0)+f (0)(x 0) = 1+x On the other hand, we saw that if x ( 1,1), then (2) f(x) = 1+x+x 2 +x 3 + +x n + So the linearization of f was the constant and linear terms from the power series representation of f, at least in this case.

2 11.10a 2 Series Representations In this section we ask the following question. Suppose that f has derivatives of all orders on an open interval I and suppose that a I. Can we find a power series representation of f in the sense of (2). Suppose that f(x) has such a representation. f(x) = a n (x a) n n=0 = a 0 +a 1 (x a)+a 2 (x a) 2 + +a n (x a) n + with a positive radius convergence. Then by the Term-by-Term Differentiation Theorem we may take derivatives of all orders with the interval of convergence I. Thus, f (x) = a 1 +2a 2 (x a)+3a 3 (x a) 2 + +na n (x a) n 1 + f (x) = 2 a a 3 (x a)+3 4a 4 (x a) 2 + f (x) = 2 3a a 4 (x a)+3 4 5a 5 (x a) 2 + Notice that each of these derivatives is trivial to evaluate at x = a. In fact, f(a) = a 0 f (a) = a 1 f (a) = 1 2a 2 f (a) = 1 2 3a 3. f (n) (a) = n!a n so that the nth coefficient of the power series expansion of f is a n = f(n) (a) n!

3 11.10a 3 Taylor and Maclaurin Series Definition. Taylor and Maclaurin Series Let f be a function with derivatives of all orders on some open interval I and suppose that a I. Then the Taylor Series generated by f at x = a is (3) k=0 f (k) (a) k! (x a) k = f(a)+f (a)(x a)+ f (a) 2! (x a) f(n) (a) n! (x a) n + The Maclaurin series generated by f is (4) k=0 f (k) (0) k! x k = f(0)+f (0)x+ f (0) 2! x So a Maclaurin series is just a Taylor series generated at x = 0. Remark. Observe that we make no claims about the equivalence of the function f with its Taylor (or Maclaurin) series in (3) or (4).

4 11.10a 4 Example 1. Taylor Series for Sine Find the Taylor series generated by f(x) = sinx at a = 0. f(x) = sinx = f(0) = 0 f (x) = cosx = f (0) = 1 f (x) = sinx = f (0) = 0 f (x) = cosx = f (0) = 1. So that for n = 0,1,2,... f (2n) (0) = 0 and f (2n+1) (0) = ( 1) n It follows that the Taylor series for the sine function is (5) x x3 3! + x5 5! + + ( 1)n x 2n+1 + = (2n+1)! n=0 ( 1) n x 2n+1 (2n+1)! Remark. You should probably take the time to memorize the Taylor series for the sine and cosine functions (c.f., Table 10.1 on page 602 of the text). We should also point out that, by the Ratio Test, the series (5) converges absolutely for all real numbers since ρ = lim a n+1 n a n = lim x 2n+3 (2n+1)! n (2n+3)! x 2n+1 = x 2 lim n 1 (2n+2)(2n+3) = 0 In other words, the Taylor Series for the sine function has an infinite radius of convergence.

5 11.10a 5 Example 2. Taylor Series for the Exponential Function Find the Taylor series for g(x) = e x at a = 0. Hence g (n) (x) = e x, for n = 0,1,2,... g (n) (0) = e 0 = 1, for n = 0,1,2,... It follows that the Taylor Series for the exponential function is (6) 1+x+ x2 2! + + xn n! + = n=0 x n n! A trivial application of the ratio test confirms that the radius of convergence is infinite. Hence the series in (6) converges for all x R. Note: We will see in section 10.9 that the Taylor Series for sine, cosine and the exponential function all converge to their generating functions.

6 11.10a 6 Taylor Polynomials As mentioned earlier, the linearization of an infinitely differentiable function f is just the constant and linear terms from the Taylor expansion of f. Suppose that we wish to consider more terms from the Taylor expansion. This leads to Definition. Taylor Polynomial of Order n Let f be a function with derivatives of order k for k = 1,2,3,...,N on some open interval I and suppose a I. Then for any integer n from 0 through N, the Taylor polynomial of order n generated by f at x = a is the polynomial (7) P n (x) = f(a)+f (a)(x a)+ f (a) 2! (x a) f(k) (a) k! (x a) k + + f(n) (a) n! (x a) n In other words, the nth Taylor Polynomial is just the first n+1 terms (including any zero terms) from the Taylor series.

7 11.10a 7 Example 3. Taylor Polynomials of Sine Find Taylor polynomial of orders 3 and 4 generated by f(x) = sinx at x = 0. From (5) we have (8) (9) P 3 (x) = 0+x+0 x3 3! P 4 (x) = 0+x+0 x3 3! +0 Remark. So the order doesn t count the number of (nonzero) terms of the polynomial nor does it refer to the degree of the polynomial. What does the order refer to? We plot a few of the Taylor Polynomials for the sine function below. Notice that the fifth order polynomial appears to be a very good approximation on the interval ( π/2, π/2). How might this be useful? y = sinx y = P 3 (x) y = P 1 (x) y = P 5 (x) π/2 π/2 y = P 7 (x)

8 11.10a 8 π/ π/2 y = sinx P 7 (x) The sketch above is an attempt to give a visual of the error using a Taylor polynomial of order 7 to estimate the sine function. It appears that the estimate is quite good for π/2 x π/2. It turns out that for x [ π/2,π/2], the error in the approximation is less than Specifically, sinx P 7 (x) (π/2)9, for π 9! 2 x π 2 We will have more to say about this in the next section. Example 4. Taylor Polynomial for the Radical Function Let g(x) = x. Find the Taylor Polynomial for g(x) of order 5 about x = 1. Using the built-in Mathematica function Series[Sqrt[x], {x,1,5}] we get (10) 1+ x (x 1) (x 1) (x 1) (x 1)5 +O ( (x 1) 6) }{{} Error Term

9 11.10a 9 Example 5. Taylor Polynomials of Polynomials Let h(x) = x 2. Find the Taylor Series of h(x) about x = 3. and for all n 3 It follows that the Taylor Series is h(x) = x 2 = h(3) = 9 h (x) = 2x = h (3) = 6 h (x) = 2 = h (3) = 2 h (n) 0 So the Taylor Series is a finite sum. But 9+6(x 3)+ 2 2! (x 3) }{{} P 2 (x) = 9+6(x 3)+(x 3) 2 = P 2 (x) P 2 (x) = 9+6(x 3)+(x 3) 2 = 9+6x 18+x 2 6x+9 = x 2 = h(x) In other words, the Taylor Series is actually equal to (converges to) its generating function. In the next section, we will discover which function properties guarantee that a Taylor Series will converge to its generating function.