Taylor Series

Lecture 37 18.01 Fall 2006 Lecture 37: Taylor Series General Power Series What is cos x anyway? Recall: geometric series 1 + a + a2 + · · · = 1 1−a for | a |< 1 General power series is an infinite sum: f (x) = a0 + a1 x + a2 x2 + a3 x3 + · · · represents f when |x| < R where R = radius of convergence. This means that for |x| < R, |an xn | → 0 as n → ∞ (“geometrically”). On the other hand, if |x| > R, then |an xn | does not tend to 0. For 1 1 example, in the case of the geometric series, if |a| = , then |an | = n . Since the higher-order terms 2 2 get increasingly small if |a| < 1, the “tail” of the series is negligible. Example 1. If a = −1, |an | = 1 does not tend to 0. 1 − 1 + 1 − 1 + ··· The sum bounces back and forth between 0 and 1. Therefore it does not approach 0. Outside the interval −1 < a < 1, the series diverges. Basic Tools Rules of polynomials apply to series within the radius of convergence. Substitution/Algebra 1 = 1 + x + x2 + · · · 1−x Example 2. x = -u. 1 = 1 − u + u2 − u3 + · · · 1+u Example 3. x = −v 2 . 1 = 1 − v2 + v4 − v6 + · · · 1 + v2 1 Lecture 37 18.01 Fall 2006 Example 4. � 1 1−x �� 1 1−x � = (1 + x + x2 + · · · )(1 + x + x2 + · · · ) Term-by-term multiplication gives: 1 + 2x + 3x2 + · · · 1 Remember, here x is some number like . As you take higher and higher powers of x, the result 2 gets smaller and smaller. Differentiation (term by term) � � � d 1 d � = 1 + x + x2 + x3 + · · · dx 1 − x dx 1 = 0 + 1 + 2x + 3x2 + · · · (1 − x)2 where 1 is a0 , 2 is a1 and 3 is a2 Same answer as Example 4, but using a new method. Integration (term by term) � � � a1 a2 f (x) dx = c + a0 + x2 + x3 + · · · 2 3 where f (x) = a0 + a1 x + a2 x2 + · · · � Example 5. du 1+u � 1 = 1 − u + u2 − u3 + · · · 1+u � du u2 u3 u4 =c+u− + − + ··· 1+u 2 3 4 � x du x2 x3 x4 ln(1 + x) = =x− + + 3 4 2 0 1+u � So now we know the series expansion of ln(1 + x). Example 6. Integrate Example 3. 1 = 1 − v2 + v4 − v6 + · · · 1 + v2 � � � dv v3 v5 v7 =c+ v− + − + ··· 1 + v2 3 5 7 � x dv x3 x5 x7 − tan−1 x = =x− + + ··· 2 5 7 3 0 1+v 2 Lecture 37 18.01 Fall 2006 Taylor’s Series and Taylor’s Formula If f (x) = a0 + a1 x + a2 x2 + · · · , we want to figure out what all these coefficients are. Differentiating, f � (x) = a1 + 2a2 x + 3a3 x2 + · · · f �� (x) = (2)(1)a2 + (3)(2)a3 x + (4)(3)a4 x2 + · · · f ��� (x) = (3)(2)(1)a3 + (4)(3)(2)a4 x + · · · Let’s plug in x = 0 to all of these equations. f (0) = a0 ; f � (0) = a1 ; f �� (0) = 2a2 ; f ��� (0) = (3!)a3 Taylor’s Formula tells us what the coefficients are: f (n) (0) = (n!)an Remember, n! = n(n − 1)(n − 2) · · · (2)(1) and 0! = 1. Coefficients an are given by: � an = 1 n! � f (n) (0) Example 7. f (x) = ex . f � (x) = ex f �� (x) = ex f (n) (x) = ex f (n) (0) = e0 = 1 Therefore, by Taylor’s Formula an = ex = 1 and n! 1 1 1 1 + x + x2 + x3 + · · · 0! 1! 2! 3! Or in compact form, ex = ∞ � xn n! n=0 Now, we can calculate e to any accuracy: e=1+1+ 1 1 1 1 + + + + ··· 2 3! 4! 5! Example 7. f (x) = cos x. f � (x) = − sin x f �� (x) = − cos x 3 Lecture 37 18.01 Fall 2006 f ��� (x) = sin x f (4) (x) = cos x f (0) = cos(0) = 1 f � (0) = − sin(0) = 0 f �� (0) = − cos(0) = −1 f ��� (0) = sin(0) = 0 Only even coefficients are non-zero, and their signs alternate. Therefore, 1 1 1 1 cos x = 1 − x2 + x4 − x6 + x8 + · · · 2 4! 8! 6! Note: cos(x) is an even function. So is this power series — as it contains only even powers of x. There are two ways of finding the Taylor Series for sin x. Take derivative of cos x, or use Taylor’s formula. We will take the derivative: � � d 1 4 6 8 − sin x = cos x = 0 − 2 x + x3 − x5 + x7 + · · · dx 2 4! 6! 8! = −x + sin(x) = x − x3 x5 x7 − + + ··· 3! 5! 7! x3 x5 x7 + − + ··· 3! 5! 7! Compare with quadratic approximation from earlier in the term: 1 cos x ≈ 1 − x2 2 sin x ≈ x We can also write: cos x = ∞ � x2k x0 x2 1 (−1)k = (−1)0 + (−1)2 + · · · = 1 − x2 + · · · (2k)! 0! 2! 2 k=0 sin x = ∞ � x2k+1 (−1)k ← n = 2k + 1 (2k + 1)! k=0 Example 8: Binomial Expansion. f (x) = (1 + x)a a a(a − 1) 2 a(a − 1)(a − 2) 3 (1 + x)a = 1 + x + x + x + ··· 1 2! 3! 4 Lecture 37 18.01 Fall 2006 Taylor Series with Another Base Point A Taylor series with its base point at a (instead of at 0) looks like: f (3) (b) f �� (b) (x − b)2 + (x − b)3 + ... 2 3! √ √ Taylor series for x. It’s a bad idea to expand using b = 0 because x is not differentiable at x = 0. Instead use b = 1. � �� � 1 1 −1 1 2 2 x1/2 = 1 + (x − 1) + (x − 1)2 + · · · 2 2! f (x) = f (b) + f � (b)(x − b) + 5