Chapter 8: Q. 6 (page 679)

Let f be a twice-differentiable function at a point $x_{0}$ . Explain why the sum
$f (x) + f' (x) (x - x_{0}) + \frac{f'' (x)}{2!} {(x - x_{0})}^{2}$
is not the second-order Taylor polynomial for f at $x_{0}$ .

Short Answer

Expert verified

$The given function is not a second - order Taylor polynomial for f at x_{0} . This is because the function f (x) and its derivatives, that is, f' (x) and f'' (x) are not evaluated at x = x_{0} .$

Step by step solution

Step 1. Given information is:

f is a twice-differentiable function at a point $x_{0}$

Step 2. Solving

$Consider g (x) = f (x) + f' (x) (x - x_{0}) + \frac{f'' (x)}{2!} {(x - x_{0})}^{2} The second order Taylor polynomial for f at x_{0} : f (x_{0}) + f' (x_{0}) (x - x_{0}) + \frac{f'' (x_{0})}{2!} {(x - x_{0})}^{2} The second order Taylor polynomial in terms of g is : g (x_{0}) = f (x_{0}) + f' (x_{0}) (x - x_{0}) + \frac{f'' (x_{0})}{2!} {(x - x_{0})}^{2} The function g (x) is not a second - order Taylor polynomial for f at x_{0} . This is because the function f (x) and its derivatives, that is, f' (x) and f'' (x) are not evaluated at x = x_{0} .$

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Let f be a twice-differentiable function at a point $x_{0}$ . Explain why the sum
$f (x) + f' (x) (x - x_{0}) + \frac{f'' (x)}{2!} {(x - x_{0})}^{2}$
is not the second-order Taylor polynomial for f at $x_{0}$ .

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Solving

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Let f be a twice-differentiable function at a point x0. Explain why the sumf(x)+f'(x)(x-x0)+f''(x)2!(x-x0)2is not the second-order Taylor polynomial for f at x0.

Short Answer

Step by step solution

Step 1. Given information is:

Step 2. Solving

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

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Study anywhere. Anytime. Across all devices.

Let f be a twice-differentiable function at a point $x_{0}$ . Explain why the sum
$f (x) + f' (x) (x - x_{0}) + \frac{f'' (x)}{2!} {(x - x_{0})}^{2}$
is not the second-order Taylor polynomial for f at $x_{0}$ .