Operations on Power Series Related to Taylor Series In this problem, we perform elementary operations on Taylor series – term by term diﬀeren tiation and integration – to obtain new examples of power series for which we know their sum. Suppose that a function f has a power series representation of the form: f (x) = a0 + a1 (x − c) + a2 (x − c)2 + · · · = ∞ � an (x − c)n n=0 convergent on the interval (c − R, c + R) for some R. The results we use in this example are: • (Diﬀerentiation) Given f as above, f � (x) has a power series expansion obtained by by diﬀer entiating each term in the expansion of f (x): f � (x) = a1 + a2 (x − c) + 2a3 (x − c) + · · · = ∞ � nan (x − c)n−1 n=1 � • (Integration) Given f as above, f (x) dx has a power series expansion obtained by by inte grating each term in the expansion of f (x): � f (x) dx = C + a0 (x − c) + ∞ � a1 a2 an (x − c)2 + (x − c)3 + · · · = C + (x − c)n+1 2 3 n + 1 n=0 for some constant C depending on the choice of antiderivative of f . Questions: 1. Find a power series representation for the function f (x) = arctan(5x). (Note: arctan x is the inverse function to tan x.) 2. Use power series to approximate � 1 sin(x2 ) dx 0 (Note: sin(x2 ) is a function whose antiderivative is not an elementary function.) Solution: 1 , which then has a power 1 + x2 , where we subsitute −x2 for x. Hence: For question (1), we know that arctan x has a simple derivative: series representation similar to that of 1 1−x ∞ ∞ � � d 5 2 n arctan(5x) = = 5 (−25x ) = (−1)n 52n+1 x2n , dx 1 + 25x2 n=0 n=0 1 . 1−x The last equality presents a cleaner ﬁnal form after straightforward algebraic simpliﬁcation. Thus where the second equality above follows from the familiar geometric series representation for to obtain a power series expression for arctan x we may integrate this power series expression term by term. This gives: ∞ � 53 3 52n+1 2n+1 arctan(5x) = C + 5x − x + · · · = C + (−1)n x , 3 2n + 1 n=0 and we may solve for C by comparing both sides of the equality for any value of x. Choosing x = 0, we see that arctan(x) = 0 and all non-constant terms of the power series are 0, hence C = 0 as well. For question (2), we have seen that sin(x) has a power series expansion: sin(x) = x − ∞ � x 3 x5 x2n+1 + − ··· = (−1)n . 3! 5! (2n + 1)! n=0 Using a change of variable (replacing x by x2 in the power series above), we have the power series expansion ∞ � x6 x10 x4n+2 sin(x2 ) = x2 − + − ··· = (−1)n . 3! 5! (2n + 1)! n=0 Now taking the indeﬁnite integral of both sides, we obtain a power series representation for the antiderivative of sin(x2 ): � ∞ � 1 1 x7 1 x10 (−1)n x4n+3 sin(x2 ) dx = x3 − + − ··· = . 3 7 3! 11 5! 4n + 3 (2n + 1)! n=0 The power series expression is valid for any real number x since the power series for sin(x), and hence sin(x2 ) converged for all x. To approximate the deﬁnite integral, we may use as many terms of the series as we like. For example, using only the ﬁrst non-zero term would give: � 1 0 1 �x=1 1 sin(x ) dx ≈ x3 �� = . 3 3 x=0 2 � The ﬁrst two non-zero terms gives: � 2 1 � 2 sin(x ) dx ≈ 1 3 1 x7 x − 3 7 3! � �x=1 � � � 1 1 13 � = − = . � � 3 42 42 x=0 Using a numerical integration on a computer-algebra system, we ﬁnd that the answer is approxi mately .31026... while 13/42 = .309524. We can improve this estimate by using more terms in the power series.