Ma 530 Power Series II Please note that there is material on power series at Visual Calculus. Some of this material was used as part of the presentation of the topics that follow. Operations on Power Series Addition and Subtraction Theorem Addition: Subtraction: Let fx n0 c n x a n and gx d n x a n be power series centered at n0 x a and let R f and R g be their radii of convergence, respectively. Further, let R minr f,r g be the smaller of these two radii. Then the sum and difference of f and g may be computed term by term and the radius of convergence is at least R fx gx c n d n x a n n0 fx gx c n d n x a n n0 Find the power series expansion for 5x centered about x 0, and find its radius 6x x of convergence. Solution: We first factor the denominator and do a partial fraction expansion to get 5x 6x x 5x 3x x 3x (Lots of work here) x Each of these fractions is the sum of a geometric series with a different ratio: Subtracting these, we find 3x 3x n 3 n x n for 3x or x 3 n0 n0 x x n n x n for x or x n0 n0 5x 6x x 3 n x n n x n n0 n0 3 n n x n for x 3 n0 At this point, we know the radius of convergence is at least, but we need to find it explicitly using 3 the ratio test:
a L lim n n a n x lim n x lim n lim n 3 n n x n 3 n n x n 3 n n 3 n n 3 n 3 n n 3 n 3 n 3 n n 3 n n 3 0 x lim n n 3 x 0 The series converges when L 3 x or x 3. So the the radius of convergence is R 3 after all. 3 n 3 n When is the radius of convergence not the minimum? Exercise Find the power series expansion for 4x centered about x 0, and find its 4x radius of convergence......... Operations on Power Series Multiplication by Polynomials Let fx c n x a n be a power series centered at x a and whose radius of convergence is R. n0 Also let k be a constant and let p be a non-negative integer. Then: Muliplication by a Polynomial: kx a p fx n0 which also has radius of convergence R. kc n x a np kc ip x a i ip The following series was found above, but here is an easier derivation.: Find the power series expansion for 4x centered about x 0, and find its radius of 4x convergence. Solution: We first use substitution to write as a geometric series: 4x 4x n 4 n x n for 4x or x 4x Finally we multiply by 4x: n0 n0
Remark This agrees with the previous result 4x 4 n x n for x 4x n0 4x k x k 4x k0 Exercise Find the power series expansion for 3x centered about x 0, and find its 3 x radius of convergence...... Operations on Power Series Differentiation Let fx c n x a n be a power series centered at x a and whose radius of convergence is R. n0 Then the derivative of f may be computed by differentiating the terms of the series for f: Differentiation: f x n0 which also has radius of convergence R. nc n x a n nc n x a n n k c k x a k k0 Find the power series expansion for 3x centered about x 0, and find its radius of x 3 convergence. Solution: We first notice that 3x x 3 is the derivative of whose series is 3 x x 3 n x 3n for x x 3 n0 n0 So we differentiate this series term by term: 3x x 3 n0 Notice that we can drop the n 0 term because it is zero. 3nx 3n 3nx 3n for x n Exercise Find the power series expansion for centered about x 0, and find its x radius of convergence... 3
Operations on Power Series Integration Let fx c n x a n be a power series centered at x a and whose radius of convergence is R. n0 Then the integral of f may be computed by integrating the terms of the series for f and adding a constant of integration: Integration: fxdx c n x a n dx n0 x a n c n n n0 k ck k C x ak C which also has radius of convergence R. Remark If you use a specific antiderivative on the left, then you must evaluate the constant on the right usually by evaluating both sides at the center x a. Find the power series expansion for arctanx centered about x 0, and find its interval of convergence. Solution: We first notice that arctanx is the integral of whose series is x x n n x n for x or x x n0 n0 So we integrate this series term by term: (Don t forget the constant!) arctanx n0 n x n dx n0 To find the constant, we evaluate both sides at x 0. arctan 0 n0 n n xn C for x n n 0n C Recall arctan 0 0and0 n 0. Thus C 0. We substitute back to conclude arctanx n0 n n xn for x We check the convergence at the endpoints. At x, the series becomes n which converges n n0 by the Alternating Series Test. At x, the series becomes n n n n n n0 n0 which also converges by the Alternating Series Test. So the interval of convergence is x. 4
Find a power series representation for fx x. 3 We begin with the fact that x x x x x 3 x 4 Differentiating, we obtain: x 3x 4x 3 x and hence Differentiating again, yields Finally, dividing by, we obtain: x x 3 x 3 x 3x 4x 3 6x x 3x 6x Exercise Find the power series expansion for ln x centered about x 0, and find its interval of convergence...... Remark Mathematicians also use series to define new functions: : The Bessel Function The Bessel function of order 0 is defined by the series J 0 x n0 n 4 n xn Find the domain of J 0 x. Remark Notice that the series for J 0 x only has even order terms. This simply means that the coefficients of the odd order terms are all zero. Solution: We find the radius of convergence by applying the ratio test: a L lim n n a n lim x n 4 lim n n! x n 4 n 4 n n! x n lim x n 0 4n Since L 0, the series converges for all x and the domain of J 0 x is,. Remark Since J 0 x is given by an alternating series, the values of the function may be 5
approximated by taking a finite number of terms and the error will be bounded by the next term. Taylor and MacLaurin Series If fx has a power series expansion at a, itmustbeoftheform: fx fa f ax a f ax a! f ax a 3 3! n0 f n a x a n This series is called the Taylor Series of the function f at a. In the special case when a 0, the series is called the MacLaurin Series. In other words, fx f0 f 0x f 0 x! f 0 x3 3! n0 f n 0 x n is the MacLaurin Series for fx. Remark The nth partial sum associated with a Taylor series for a given function is called nth degree Taylor polynomial for fx at x a and is denoted by T n x. Thus n T n x k0 f k a k! x a k It will be useful to know the MacLaurin Series for e x,sinx, and cosx. e x x x! x3 3! n0 sin x x x3 3! x5 5! n0 cosx x! x4 n0 n n x n xn n! xn Find the first 5 terms of the Taylor Series for fx ln x at a. fx ln x f ln f x x f f x f x 4 Thus f x f x 3 4 f iv x 6 f iv 6 x 4 6 3 8 6
fx ln x ln x 4 x! 4 x 3 3! 3 8 x 4 Use known MacLaurin Series to find the MacLaurin Series of Solution: Since fx e 3x e u u u! u3 3! e 3x 3x 9x! 7x3 3! Solution: Use known MacLaurin Series to find the MacLaurin Series of fx sin x cosx so cosu u! u4 Then cosx 4x! 6x4 and therefore cosx 4x! 6x4 Finally Solution: Evaluate cosx 4x! sin x 4x! sinx x 6x4 6x4 dx sinx x dx x x x3 3! x5 5! dx x 3! x4 5! dx C x x 3 33! x 5 55! Let be a nonzero real number. Find the MacLaurin series for 7
Solution: fx x fx x f x x f x x f x x 3 f n x 3 n x n Thus f n 0 3 n So the MacLaurin series is given by x n0 f n 0 x Using the Ratio Test we have that L lim n lim n x n n! x 3 n n! nx x n 3 n 3! x 3 x n 3 n x n x n Thus this series converges for x. 8