{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Taylor Series - 4.5 Introduction to Taylor Series 4.5...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series 4.5 Introduction to Taylor Series Given a function f ( x ) that can be represented by a power series f ( x ) = n =0 a n ( x - a ) n centered at x = a , we have seen that f ( x ) has derivatives of all orders at x = a and that a n = f ( n ) ( a ) n ! . So, in fact, f ( x ) = n =0 f ( n ) ( a ) n ! ( x - a ) n . Definition. The series n =0 f ( n ) ( a ) n ! ( x - a ) n is called the Taylor Series for f ( x ) centered at x = a . We write f ( x ) n =0 f ( n ) ( a ) n ! ( x - a ) n . In the special case where a = 0, the series is often refered to as the Maclaurin Series for f ( x ). Up until now, we have started with a function that was represented by a power series. However, suppose that f ( x ) is any function for which f ( n ) ( a ) exists for each n . Then we can build the power series n =0 f ( n ) ( a ) n ! ( x - a ) n and investigate its properties. In particular, we want to know 1. For which values of x does the series n =0 f ( n ) ( a ) n ! ( x - a ) n converge? 2. If the series above converges at x 0 , is it true that f ( x 0 ) = n =0 f ( n ) ( a ) n ! ( x 0 - a ) n ? DE Math 128 310 (B. Forrest) 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series We know the answer to the first problem since we have developed a method for finding the interval of convergence of a power series. The second problem seems intuitively to be true. However, a closer look at what this says reveals why it may not be so. Essentially what we are asking is to be able to rebuild exactly a function over an interval that could very well be the whole real line by using only the information provided by the function at one single point. In this respect, it would seem that the fact that we can use information about e x at x = 0 to get that e x = n =0 x n n ! for all x seems quite remarkable and indeed it is! To further illustrate why e x is such a special function, consider g ( x ) = 1 e if x < - 1 e x if - 1 x 1 e if x > 1 On the interval [ - 1 , 1], g ( x ) behaves exactly like e x . In particular, g (0) = e 0 = 1 and g ( n ) (0) = e 0 = 1 for every n . This means that the Taylor series centered at x = 0 for g ( x ) is n =0 x n n ! which is exactly the same Taylor Series as for e x . However at x = 2, g (2) = e while n =0 2 n n ! = e 2 = g (2) . Example. Find the Taylor series centered at x = 0 for f ( x ) = cos( x ). We have that f ( x ) = sin( x ) = f (0) = - sin(0) = 0 f ( x ) = - cos( x ) = f (0) = - cos(0) = - 1 f ( x ) = sin( x ) = f (0) = sin(0) = 0 f (4) ( x ) = cos( x ) = f (4) (0) = cos(0) = 1 f (5) ( x ) = - sin( x ) = f (5) (0) = - sin(0) = 0 f (6) ( x ) = - cos( x ) = f (6) (0) = - cos(0) = - 1 f (7) ( x ) = sin( x ) = f (7) (0) = sin(0) = 0 f (8) ( x ) = cos( x ) = f (8) (0) = cos(0) = 1 . . . DE Math 128 311 (B. Forrest) 2
Background image of page 2
4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series with the cycle repeating itself every four derivatives. This gives f (4 k ) ( x ) = cos( x ) = f (4 k ) (0) = cos(0) = 1 f (4 k +1) ( x ) = - sin( x ) = f (4 k +1) (0) = - sin(0) = 0 f (4 k +2) ( x ) = - cos( x ) = f (4 k +2) (0) = - cos(0) = - 1 f (4 k +3) ( x ) = sin( x ) = f (4 k +3) (0) = sin(0) = 0 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}