Mathematical Methods for Physics and Engineering - Matematica.NET

4.6 TAYLOR SERIESWe may follow a similar procedure to obtain a Taylor series about an arbitrarypoint x = a.◮Expand f(x) =cosx as a Taylor series about x = π/3.As in the above example, it is easily shown that the nth derivative of f(x) isgivenby(f (n) (x) =cos x + nπ ).2Therefore the remainder after expanding f(x) asan(n − 1)th-order polynomial aboutx = π/3 isgivenbyR n (x) =(x − π/3)nn!(cos ξ + nπ ),2where ξ lies in the range [π/3,x]. The modulus of the cosine term is always less than orequal to unity, and so |R n (x)| < |(x−π/3) n |/n!. As in the previous example, lim n→∞ R n (x) =0 for any particular value of x, andsocosx can be represented by an infinite Taylor seriesabout x = π/3.Evaluating the function and its derivatives at x = π/3 weobtainf(π/3) = cos(π/3) = 1/2,f ′ (π/3) = cos(5π/6) = − √ 3/2,f ′′ (π/3) = cos(4π/3) = −1/2,and so on. Thus the Taylor series expansion of cos x about x = π/3 isgivenbycos x = 1 2 − √32(x − π/3)−12( ) 2x − π/3+ ··· . ◭2!4.6.2 Approximation errors in Taylor seriesIn the previous subsection we saw how to represent a function f(x) by an infinitepower series, which is exactly equal to f(x) for all x within the interval ofconvergence of the series. However, in physical problems we usually do not wantto have to sum an infinite number of terms, but prefer to use only a finite numberof terms in the Taylor series to approximate the function in some given rangeof x. In this case it is desirable to know what is the maximum possible errorassociated with the approximation.As given in (4.18), a function f(x) can be represented by a finite (n − 1)th-orderpower series together with a remainder term such thatf(x) =f(a)+(x − a)f ′ (a)+(x − a)2f ′′ (x − a)n−1(a)+···+2!(n − 1)! f(n−1) (a)+R n (x),where(x − a)nR n (x) = f (n) (ξ)n!and ξ lies in the range [a, x]. R n (x) is the remainder term, and represents the errorin approximating f(x) bytheabove(n − 1)th-order power series. Since the exact139