Lecture 38 18.01 Fall 2006 Lecture 38: Final Review Review: Diﬀerentiating and Integrating Series. If f (x) = ∞ � an xn , then n=0 � f (x) = ∞ � nan x n−1 � f (x)dx = C + and n=1 ∞ � an xn+1 n+1 n=0 Example 1: Normal (or Gaussian) Distribution. � � x � x� 2 (−t2 )2 (−t2 )3 e−t dt = 1 − t2 + + + · · · dt 2! 3! 0 0 � � x� t4 t6 t8 2 = 1 − t + − + − ... dt 2! 3! 4! 0 =x− x � x3 1 x5 1 x7 + − + ... 3 2! 5 3! 7 2 e−t dt isn’t an elementary function, we can still compute it. Elementary functions Even though 0 are still a little bit better, though. For example: sin x = x − x3 x5 π π (π/2)3 (π/2)5 + − · · · =⇒ sin = − + − ··· 3! 5! 2 2 3! 5! But to compute sin(π/2) numerically is a waste of time. We know that the sum if something very simple, namely, π sin = 1 2 It’s not obvious from the series expansion that sin x deals with angles. Series are sometimes com plicated and unintuitive. π π Nevertheless, we can read this formula backwards to ﬁnd a formula for . Start with sin = 1. 2 2 Then, � 1 �1 dx π π � √ = sin−1 x� = sin−1 1 − sin−1 0 = − 0 = 2 2 2 0 1−x 0 We want to ﬁnd the series expansion for (1 − x2 )−1/2 , but let’s tackle a simpler case ﬁrst: � �� 1 1 1 1 1 � � − − −1 − −2 − − −1 1 2 2 2 2 2 (1 + u)−1/2 = 1 + − u+ u2 + u3 + · · · 1·2 1·2·3 2 1 1·3 2 1·3·5 3 =1− u+ u − u + ··· 2·4 2 2·4·6 Notice the pattern: odd numbers go on the top, even numbers go on the bottom, and the signs alternate. 1 Lecture 38 18.01 Fall 2006 Now, let u = −x2 . 1 1·3 4 1·3·5 6 x + x + ··· (1 − x2 )−1/2 = 1 + x2 + 2 2·4 2·4·6 � � � 1 x3 1 · 3 x5 1 · 3 · 5 x7 2 −1/2 (1 − x ) dx = C + x + + + + ··· 2 3 2·4 5 2·4·6 7 � � � �� 1 π 1 1 1·3 1 1·3·5 1 = (1 − x2 )−1/2 dx = 1 + + + + ··· 2 2 3 2·4 5 2·4·6 7 0 Here’s a hard (optional) extra credit problem: why does this series converge? L’Hôpital’s rule to ﬁnd out how quickly the terms decrease. � � � � � � = � � � � 2 Hint: use