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BC MATH 153 - Approximation Using Taylor Polynomials

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10.11 Approximation Using Taylor Polynomials Contemporary Calculus 1 10.11 APPROXIMATION USING TAYLOR POLYNOMIALS The previous two sections focused on obtaining power series representations for functions, finding their intervals of convergence, and using those power series to approximate values of functions, limits, and integrals. In the cases where the power series resulted in an alternating numerical series, we were also able to use the Estimation Bound for Alternating Series (Section 10.6) to get a bound on the "error:" "error" = | {exact value} – {partial sum approximation} | < | next term in the series | . If the power series did not result in an alternating numerical series, we did not have a bound on the size of the error of the approximation. In this section we introduce Taylor Polynomials (partial sums of the Taylor Series) and obtain a bound on the approximation error, the value |{ exact value of f(x) } – { Taylor Polynomial approximation of f(x) }| . The bound we get is valid even if the Taylor series is not an alternating series, and the pattern for the error bound looks very much like the next term in the series, the first unused term in the partial sum of the Taylor series. In mathematics, this error bound is important for determining which functions are approximated by their Taylor series. In computer and calculator applications, the error bound is important to designers to ensure that their machines calculate enough digits of functions such as ex and sin(x) for various values of x. In general, knowing this error bound can help us work efficiently by allowing us to use only the number of terms we really need. We also examine graphically how well the Taylor Polynomials of f(x) approximate f(x) Taylor Polynomials If we add a finite number of terms of a power series, the result is a polynomial. Definition For a function f, the nth degree Taylor Polynomial (centered at c), written Pn(x), is the partial sum of the terms up to the nth degree of the Taylor Series for f: Pn(x) = ∑k=0n f(k)(c)k! (x–c) k = f(c) + f '(c)(x–c) + f ''(c)2! (x–c) 2 + f '''(c)3! (x–c) 3 + f(4)(c)4! (x–c) 4 + f (n)(c)n! (x–c) n10.11 Approximation Using Taylor Polynomials Contemporary Calculus 2 Example 1: Write the first four Taylor Polynomials , P0(x) to P3(x), centered at 0 for ex, and then graph them for –1 < x < 1. Solution: The Maclaurin series for ex is ex = 1 + x + x22! + x33! + x44! + x55! + ... = ∑n=0∞ xnn! so P0(x) = 1, P1(x) = 1 + x, P2(x) = 1 + x + x22 , and P3(x) = 1 + x + x22 + x36 . The graphs of ex and P1(x), P2(x), and P3(x) are shown in Fig. 1. Notice that P0(x) and ex agree in value when x = 0, P1(x) , ex , and their first derivatives agree in value when x = 0, P2(x) , ex , their first derivatives, and their second derivatives agree in value when x = 0. Practice 1: Write the Taylor Polynomials P0(x), P2(x), and P4(x) centered at 0 for cos(x), and then graph them for –π < x < π. Write the Taylor Polynomials P1(x) and P3(x). When we center the Taylor Polynomial at x = c ≠ 0, the Taylor Polynomials approximate the function and its derivatives well for x close to c. Example 2: Write the Taylor Polynomials P0(x), P2(x), and P4(x) centered at 3π/2 for sin(x), and then graph them for 2 < x < 8. Solution: The Taylor series, centered at 3π/2, for sin(x) is sin(x) = –1 + 12! (x – 3π/2) 2 – 14! (x – 3π/2) 4 + 16! (x – 3π/2) + ... = ∑n=0∞ (–1) n+1 1(2n)! (x – 3π/2) 2n . Then P0(x) = –1 , P2(x) = –1 + 12 (x – 3π/2) 2 , and P4(x) = –1 + 12 (x – 3π/2) 2 – 124 (x – 3π/2) 4 . Fig. 11 + x-3123x46yex1 + x +2!x2-3123x46yex-3123x46yex1 + x +2!x23!x3+P23P1P10.11 Approximation Using Taylor Polynomials Contemporary Calculus 3 The graphs of sin(x), P0(x), P2(x), and P4(x) are shown in Fig. 2. Practice 2: Write the Taylor Polynomials P0(x), P1(x), and P3(x) centered at π/2 for cos(x), and then graph them for –1 < x < 4. The Remainder Approximation formulas such as the Taylor Polynomials are useful by themselves, but in many applied situations we want to know how good the approximation is or how many terms of a series are required to obtain a needed level of accuracy. If 2 terms of a series give you the needed level of accuracy for your application, it is a waste of time and money to use 100 terms. On the other hand, sometimes even 100 terms may not give the accuracy you need. Fortunately, it is possible to obtain a guarantee on how close a particular Taylor Polynomial approximation is to the exact value. Then we can work efficiently and use the number of terms that we need. The next theorem gives a pattern for the amount of "error" in our Taylor Polynomial approximation and can be used to obtain a bound on the size of the "error." Taylor's Formula with Remainder If f has n+1 derivatives in an interval I containing c, and x is in I, then there is a number z , strictly between c and x, so that f(x) = Pn(x) + Rn(x) where Rn(x) = f(n+1)(z)(n+1)! (x–c) n+1 . This says that f(x) is equal to the nth degree Taylor Polynomial plus a Remainder, and the Remainder Rn(x) has the form given in the theorem. Notice that the pattern for Rn looks like the pattern for the (n+1)st term of the Taylor series for f(x) except that it contains f(n+1)( z ) instead of f(n+1)( c ) . This particular pattern for Rn(x) is called the Lagrange form of the remainder, and is named for the French–Italian mathematician and astronomer Joseph Lagrange (1736–1813). xsin(x)-11y28-11xy3!/20P (x) = –1sin(x)28-11y3!/2P (x)2Fig. 2P (x)428x3!/2sin(x)10.11 Approximation Using Taylor Polynomials Contemporary Calculus 4 The main idea of the proof of the Taylor's Formula with Remainder is straightforward, but the technical details are rather complicated. The main idea and the technical details are given in the Appendix. The pattern for the remainder, f(n+1)(z)(n+1)! (x–c) n+1 , contains three pieces, (n+1)! , (x–c)n+1 , and f(n+1)(z) for some z between x and c. The Taylor Remainder Formula is typically used in two ways: In one type of use, the Taylor Polynomial is given so x, c, and n are known, and we can evaluate (n+1)! and (x–c)n+1 exactly. That leaves the piece f(n+1)(z) for some z


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