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# lec17 - Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon...

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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon () Math 16b Autumn 2010 1 / 44 The derivative as the rst Taylor polynomial If f ( x ) is di erentiable at a , then the function p ( x ) = b + m ( x- a ) where b = f ( ) and m = f ( a ) is the best linear approximation to f near a . For x ≈ a we have f ( x ) ≈ p ( x ) . Note that f ( a ) = b = p ( a ) and f ( a ) = m = p ( a ) . Thomas Scanlon () Math 16b Autumn 2010 2 / 44 Linear approximation in an example If f ( x ) = sin ( x ) , then f ( ) = sin ( ) = and f ( ) = cos ( ) = 1 So L ( x ) = + 1 x = x is the best linear approximation to sin ( x ) near zero. Thomas Scanlon () Math 16b Autumn 2010 3 / 44 Linear approximation in an example at another point If f ( x ) = sin ( x ) , then f ( π/ 4 ) = sin ( π/ 4 ) = √ 2 / 2 and f ( π/ 4 ) = cos ( π/ 4 ) = √ 2 / 2 So Λ( x ) = √ 2 / 2 + √ 2 / 2 ( x- π/ 4 ) is the best linear approximation to sin ( x ) near π/ 4 . Thomas Scanlon () Math 16b Autumn 2010 4 / 44 Higher degree Taylor polynomials De nition If f ( x ) is a function which is n times di erentiable at a , then the n th Taylor polynomial of f at a is the polynomial P ( x ) of degree (at most n ) for which f ( i ) ( a ) = p ( i ) ( a ) for all i ≤ n . For n = 0, P is constant and P ≡ f ( a ) . For n = 1, P is a linear polynomial. Write P ( x ) = b + m ( x- a ) . Then f ( a ) = f ( ) ( a ) = P ( ) ( a ) = P ( ) = b and f ( a ) = f ( 1 ) ( a ) = P ( 1 ) ( a ) = P ( a ) = m . The n th Taylor polynomial of f at a is the best approximation to f near a using polynomials of degree at most n . Thomas Scanlon () Math 16b Autumn 2010 5 / 44 Example Compute the third Taylor polynomial of f ( x ) = e x at a = 0. Write P ( x ) = c + c 1 x + c 2 x 2 + c 3 x 3 We need to nd c , c 1 , c 2 , and c 3 so that P ( i ) ( ) = f ( i ) ( ) for i = 0, 1, 2, and 3 In our case f ( i ) ( x ) = e x for all i ≥ 0 and e = 1. So, f ( i ) ( ) = 1 for all i . Thomas Scanlon () Math 16b Autumn 2010 6 / 44 Solution, continued P ( x ) = c + c 1 x + c 2 x 2 + c 3 x 3 We compute P ( x ) = c 1 + 2 c 2 x + 3 c 3 x 2 P 00 ( x ) = 2 c 2 + 6 c 3 x P 000 ( x ) = 6 c 3 Thus, 1 = f ( ) ( ) = P ( ) ( ) = c , 1 = f ( 1 ) ( ) = P ( 1 ) ( ) = c 1 , 1 = f ( 2 ) ( ) = P ( 2 ) ( ) = 2 c 2 so that c 2 = 1 2 , and 1 = f ( 3 ) ( ) = P ( 3 ) ( ) = 6 c 3 so that c 3 = 1 6 . Thus, the third Taylor polynomial of f ( x ) = e x at a = 0 is P ( x ) = 1 6 x 3 + 1 2 x 2 + x + 1 Thomas Scanlon () Math 16b Autumn 2010 7 / 44 Graphs of Taylor polynomials of e x Thomas Scanlon () Math 16b Autumn 2010 8 / 44 Another example Problem Find the third Taylor polynomial of f ( x ) = ln ( x ) at a = 1. As before, we write P ( x ) = c + c 1 x + c 2 x 2 + c 3 x 3 We nd P ( x ) = c 1 + 2 c 2 x + 3 c 3 x 2 P 00 ( x ) = 2 c 2 + 6 c 3 x P 000 ( x ) = 6 c 3 Thomas Scanlon () Math 16b Autumn 2010 9 / 44 Solution, continued Di erentiating, f ( x ) = 1 x = x- 1 , f 00 ( x ) =- x- 2 , and f 000 ( x ) = 2 x- 3 ....
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