Let us take that as the definition of a function that

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Calculus
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Chapter 16 / Exercise 14
Calculus
Stewart
Expert Verified
Let us take that as the definition of a function that is concave up over I . Now if p I , Taylor’s theorem applied to the tangent line T 1 ( x ) = f ( p ) + f 0 ( p )( x - p ) tells us that f ( x ) = T 1 ( x ) + f 00 ( z ) 2 ( x - p ) 2 for some z between p and x. (6) If f 00 ( t ) > 0 for all t in I and x 6 = p , we see that f 00 ( z ) 2 ( x - p ) 2 > 0 , since z is some number also in I . Hence equation (6) tells us that f ( x ) > T 1 ( x ) for all x in I , other than x = p of course. Thus we have proven the standard concavity test for functions, based on a proper mathematical definition of functions that are concave up. Exercise 12. Does the tangent line to y=xlnx, at any positivep, over-estimate or under-estimatexlnx? 13. Does the tangent line to y=e-x2atp= 1over-estimate or under-estimatethe function whenxis close to1?11
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Calculus
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Chapter 16 / Exercise 14
Calculus
Stewart
Expert Verified
14. Does there exist a function that is always negative onRwhile its doublederivative is always positive? Suppose that f has continuous derivatives of order up to n +1 on some interval I containing p , and that for some constant B we have f ( n +1) ( t ) B for all t in I . Then | f ( x ) - T n ( x ) | ≤ B | x - p | n +1 ( n + 1)! for every x in I. Proof. According to Taylor’s theorem, if x I , then f ( x ) - T n ( x ) = f ( n +1) ( z ) ( n + 1)! ( x - p ) n +1 for some z between p and x. Since z is between p and x , this elusive z is also in I . Our assumption yields that | f ( n +1) ( z ) | ≤ B. From that we obtain | f ( x ) - T n ( x ) | = f ( n +1) ( z ) ( n + 1)! ( x - p ) n +1 B | x - p | n +1 ( n + 1)! , as desired. 12
The challenge from here on is to find the constant B over an appropriate interval I . Example 3. Let f ( x ) = e x over I = [ - 1 , 1] . The Taylor polynomial of f of degree up to 3 taken at 0 is T 3 ( x ) = 1 + x + x 2 2 + x 3 6 . Now for x in the interval [0 , 1] we can see that | f (4) ( x ) | = e x e 1 3 In this example, Taylor’s inequality becomes the statement that for any x [ - 1 , 1] , | e x - T 3 ( x ) | = e x - (1 + x + x 2 2 + x 3 6 ) 3 24 | x | 4 = 1 8 | x | 4 As an application of this fact, let us estimate e = e 1 / 2 using T 3 (1 / 2) , and also let us place a bound on the error made by our estimate. From the formula for T 3 ( x ) we see that T 3 (1 / 2) = 1 + 1 2 + 1 2 · 1 2 2 + 1 6 · 1 2 3 = 79 48 . Furthermore we know that the error in this approximation is no more than 1 8 · 1 2 4 = 1 128 . Those who trust calculators more than mathematics can compare e and 79 / 48 on their electronic devices, and confirm that our error prediction is valid.

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