Answered: f(x) = coslx, n = 2, a 2 2.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.3: Algebraic Expressions

Problem 32E

See similar textbooks

Related questions

Question

Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a.

Expert Solution

Step 1

Given that f(x)=cos^-1(x) centered at a=0.5.

We need to find the Taylor polynomial for n=2.

Step 2

$T h e n t h o r d e r T a y l o r p o l y n o m i a l c e n t e r e d a t a i s g i v e n b y P_{n} (x) = f (a) + f' (a) (x - a) + \frac{f'' (a)}{2!} {(x - a)}^{2} + - - - + \frac{f^{n} (a)}{n!} {(x - a)}^{n} f o r n = 2 a n d a = 0.5 P_{2} (x) = f (0.5) + f' (0.5) (x - 0.5) + \frac{f'' (0.5)}{2!} {(x - 0.5)}^{2} S o f (x) = \cos^{- 1} (x) f (0.5) = \cos^{- 1} (0.5) = \frac{π}{3} = 1.0472 f' (x) = \frac{- 1}{\sqrt{1 - x^{2}}} f' (0.5) = \frac{- 1}{\sqrt{1 - 0 . 5^{2}}} = - 1.1547 f'' (x) = \frac{(\sqrt{1 - x^{2}} * 0) - [- 1 * \frac{- 2 x}{\sqrt{1 - x^{2}}}]}{{(\sqrt{1 - x^{2}})}^{2}} f'' (x) = \frac{- x}{{(\sqrt{1 - x^{2}})}^{2} (\sqrt{1 - x}}$ ²)f''(x)=-x(1-x2)32f''(0.5)=-0.5(1-0.52)32=-0.7698.

Step by step

Solved in 3 steps

Knowledge Booster

$Power Series$

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.

Similar questions