Explanation Given: The function is f ( x ) = x 11 2 and the value of a is 1. Concept used: The n th Taylor polynomial centered at x = a for the function is, T n ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f n ( a ) n ! ( x − a ) n ....... (1) If a = 0 , then T n is also called the n th Maclaurin polynomial. Now, if f ( n + 1 ) exists and is continuous, then the error bound is, | T n ( x ) − f ( x ) | ≤ K | x − a | n + 1 ( n + 1 ) ! ....... (2) Here, K is a number such that | f ( n + 1 ) ( u ) | ≤ K for all u between a and x . Calculation: Choose n = 4 and differentiate the function f ( x ) = x 11 2 up to degree 4 as shown below. The first derivative for the function f ( x ) = x 11 2 is calculated as, f ′ ( x ) = d d x ( x 11 2 ) = 11 2 x 11 2 − 1 = 11 2 x 9 2 The secondderivative for the function f ( x ) = x 11 2 is calculated as, f ″ ( x ) = d d x ( 11 2 x 9 2 ) = 11 2 d d x ( x 9 2 ) = 11 2 ⋅ 9 2 x 9 2 − 1 = 99 4 x 7 2 The thirdderivative for the function f ( x ) = x 11 2 is calculated as, f ‴ ( x ) = d d x ( 99 4 x 7 2 ) = 99 4 d d x ( x 7 2 ) = 99 4 ⋅ 7 2 x 7 2 − 1 = 693 8 x 5 2 The fourthderivative for the function f ( x ) = x 11 2 is calculated as, f ( 4 ) ( x ) = d d x ( 693 8 x 5 2 ) = 693 8 d d x ( x 5 2 ) = 693 8 ⋅ 5 2 x 5 2 − 1 = 3465 16 x 3 2 In order to find the Taylor polynomial of degree 4 , use the firstfour derivative of f ( x ) evaluated at x = 1 as shown in Table 1. Derivative of function f ( x ) Value of derivative at x = 1 f ( x ) = x 11 2 f ( 1 ) = 1 f ′ ( x ) = 11 2 x 9 2 f ′ ( 1 ) = 11 2 f ″ ( x ) = 99 4 x 7 2 f ″ ( 1 ) = 99 4 f ‴ ( x ) = 693 8 x 5 2 f ‴ ( 1 ) = 693 8 f ( 4 ) ( x ) = 3465 16 x 3 2 f ( 4 ) ( 1 ) = 3465 16 Table 1 The Taylor polynomial of degree 4 is, T 4 ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 + f ‴ ( a ) 3 ! ( x − a ) 3 + f ( 4 ) ( a ) 4 ! ( x − a ) 4 Substitute 1 for a in above equation. T 4 ( x ) = f ( 1 ) + f ′ ( 1 ) 1 ! ( x − 1 ) 1 + f ″ ( 1 ) 2 ! ( x − 1 ) 2 + f ‴ ( 1 ) 3 ! ( x − 1 ) 3 + f ( 4 ) ( 1 ) 4 ! ( x − 1 ) 4 Substitute 1 for f ( 1 ) , 11 2 for f ′ ( 1 ) , 99 4 for f ″ ( 1 ) , 693 8 for f ‴ ( 1 ) and 3465 16 for f ( 4 ) ( 1 ) in above equation

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Videos

Question

Chapter 11.7, Problem 40E

To determine

To find: The Taylor polynomial for the function f(x)=x112 and use error bound to obtain the maximum possible size of the error |f(1.2)−T4(1.2)| .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.7, Problem 39E

Chapter 11.7, Problem 41E

Students have asked these similar questions

a. Find the linear approximating polynomial for the following function centered at the given point a. b. Find the quadratic approximating polynomial for the following function centered at the given point a. c. Use the polynomials obtained in parts a. and b. to approximate the given quantity. f(x) = 8x³/2, a = 1; approximate 8(1.73/2) a. p₁(x) = b. p₂(x) = c. Using the linear approximating polynomial to estimate, 8 (1.73/2) is approximately (Simplify your answer.) Using the quadratic approximating polynomial to estimate, 8(1.73/2) is approximately. (Simplify your answer.)

The figure below shows the graph of the polynomial function f. For which value of x is it true that f"(x) < f'(x) < f(x)? а. а b. b С. C la d. d е. е Graph of f

Give an interpolating polynomial for f(x) = ²x at x = 0, 1 and 3. Evaluate it at x = 2 and give the absolute error under 4 digit chopping. Give the bound for the error of the interpolating polynomial: |f(x) - p₂(x)|.

Chapter 11 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

The Taylor polynomial for the function f ( x ) = x 11 2 and use error bound to obtain the maximum possible size of the error | f ( 1.2 ) − T 4 ( 1.2 ) | .

Solution Summary: The author explains the Taylor polynomial for the function f(x)=x112 and the error bound.

Videos

To find: The Taylor polynomial for the function f(x)=x112 and use error bound to obtain the maximum possible size of the error |f(1.2)−T4(1.2)| .

Want to see the full answer?

Chapter 11 Solutions