Lecture Notes 9-16

# Lecture Notes 9-16 - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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Lecture 9: Linear and Quadratic Approximations Unit 2: Applications of Diﬀerentiation Today, we’ll be using diﬀerentiation to make approximations. Linear Approximation y=f(x) y = b+a(x-x 0 ) y x b = f(x 0 ) ; x 0 ,f(x 0 ) ( ) a = f’(x 0 ) Figure 1: Tangent as a linear approximation to a curve The tangent line approximates f ( x ). It gives a good approximation near the tangent point x 0 . As you move away from x 0 , however, the approximation grows less accurate. f ( x ) f ( x 0 ) + f ( x 0 )( x x 0 ) Example 1. f ( x ) = ln x, x 0 = 1 (basepoint) 1 f (1) = ln 1 = 0; f (1) = = 1 x x =1 ln x Change the basepoint: Basepoint u 0 = x 0 1 = 0. f (1) + f (1)( x 1) = 0 + 1 · ( x 1) = x 1 x = 1 + u = u = x 1 ln(1 + u ) u 1 Lecture 9 18.01 Fall 2006
Basic list of linear approximations In this list, we always use base point x 0 = 0 and assume that | x | << 1 . 1. sin x x (if x 0) (see part a of Fig. 2) 2. cos x 1 (if x 0) (see part b of Fig. 2) x 3. e 1 + x (if x 0) 4. ln(1 + x ) x (if x 0) 5. (1 + x ) r 1 + rx (if x 0) Proofs Proof of 1: Take f ( x ) = sin x , then f ( x ) = cos x and f (0) = 0 f (0) = 1 ,f ( x ) f (0) + f (0)( x 0) = 0 + 1 .x So using basepoint x 0 = 0 ,f ( x ) = x . (The proofs of 2, 3 are similar. We already proved 4 above.) Proof of 5: f ( x ) = (1 + x ) r ; f (0) = 1 f (0) = d (1 + x ) r x =0 = r (1 + x ) r 1 x =0 = r dx | | f ( x ) = f (0) + f (0) x = 1 + rx y = x sin(x) y=1 cos(x) (a) (b) Figure 2: Linear approximation to (a) sin x (on left) and (b) cos x (on right). To Fnd them, apply f ( x ) f ( x 0 ) + f ( x 0 )( x x 0 ) ( x 0 = 0) e 2 x Example 2. Find the linear approximation of f ( x ) = near x = 0. 1 + x We could calculate f ( x ) and ±nd f (0). But instead, we will do this by combining basic approxi- mations algebraically. u e 2 x 1 + ( 2 x ) ( e 1 + u, where u = 2 x ) 2 Lecture 9 18.01 Fall 2006

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1 + x = (1 + x ) 1 / 2 1 + 1 x 2 Put these two approximations together to get e 2 x 1 2 x x (1 2 x )(1 + 1 x ) 1 1 + x 1 + 1 2 2 Moreover (1 + 1 x ) 1 1 1 x (using (1 + u ) 1 1 u with u = x/ 2). Thus 1 2 2 e 2 x 1 1 1 2 1 + x (1 2 x )(1 2 x ) = 1 2 x 2 x + 2( 2 ) x Now, we discard that last x 2 term, because we’ve already thrown out a number of other x 2 (and higher order) terms in making these approximations. Remember, we’re assuming that x << 1. 2 3 | | This means that x is very small, x is even smaller, etc. We can ignore these higher-order terms, because they are very, very small. This yields e 2 x 1 5 1 + x 1 2 x 2 x = 1 2 x Because f ( x ) 1 5 x , we can deduce f (0) = 1 and f (0) = 5 directly from our linear approxi- 2 2 mation,
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Lecture Notes 9-16 - MIT OpenCourseWare http/ocw.mit.edu...

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