# Ch 8 - 8 Approximations, Taylor Polynomials, and Taylor...

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Unformatted text preview: 8 Approximations, Taylor Polynomials, and Taylor Series Polynomials are the nicest possible functions. They are easy to differentiate and integrate, which is also true of the basic trigonometric functions, but more importantly, polynomials can be evaluated at any point, which is not true for general functions. So what one does in practice is to approximate any function f of interest by polynomials. When the approximation is done by linear polynomials , then it is called a linear approximation , which pictorially corresponds to linearizing the graph of f . It turns out that the more times one can differentiate f , the higher is the degree of the polynomial one can approximate it with, and more importantly, the better the approximation becomes, as one sees it intuitively. There is only one main theorem here, due to Taylor, but it is omnipresent in all the mathematical sciences, with a number of ramifications, and should be understood precisely. 8.1 Taylor polynomials Suppose f is an N-times differentiable function on an open interval I . Fix any point a in I . Then for any non-negative integer n N , the n th Taylor polynomial of f at x = a is given by (8 . 1 . 1) p n ( f ( x ); a ) = n j =0 f ( j ) ( a ) j ! ( x- a ) j , where f ( j ) ( a ) denotes the j th derivative of f at a . By convention, f (0) ( a ) just denotes f ( a ). ( f is the 0th derivative of itself!) The coecients f ( j ) ( a ) j ! are called the Taylor coecients of f at a . The definition has been rigged so that the following holds: Lemma 8.1 Suppose f is itself a polynomial, i.e., f ( x ) = a + a 1 x + . . . + a m x m , 1 for some integer m . Then f is infinitely differentiable (which means it can be differentiated any number of times), and p n ( f ( x ); 0) = { a + a 1 x + . . . + a n x n , if n &lt; m a + a 1 x + . . . + a m x m , if n m Proof . Clearly, f is differentiable any number of times and moreover, f ( n ) ( x ) vanishes if n &gt; m . So we have only to show that for n m , (8 . 1 . 2) f ( n ) (0) = n ! a n . When m = 0 this is clear. So let m &gt; 0 and assume by induction that (8.1.2) holds for all polynomials of degree m- 1 and n m- 1. Define a polynomial g ( x ) by the formula (8 . 1 . 3) f ( x ) = a + xg ( x ) . Then g ( x ) = m 1 j =0 a j +1 x j and by the inductive hypothesis, (8 . 1 . 4) g ( n ) (0) = n ! a n +1 for all non-negative n m- 1. But by the product rule , f ( x ) = g ( x ) + xg ( x ) , f ( x ) = 2 g ( x ) + xg ( x ) , . . . By induction, we get f ( n ) ( x ) = ng ( n 1) ( x ) + xg ( n ) ( x ) , so that (8 . 1 . 5) f ( n ) (0) = ng ( n 1) (0) n m, n 1 . The identity (8.1.2), and hence the Lemma, now follow by combining (8.1.4) and (8.1.5)....
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## This note was uploaded on 11/10/2011 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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Ch 8 - 8 Approximations, Taylor Polynomials, and Taylor...

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