Question:a) Find the Taylor Polynomials Po,b) Plot Po, P, P,, P2, and g(x) in the same Cartesian PlanP, P,, and P, for g(x) = ln(x) centered at c = 11'2

let f(x)=g(x)=ln(x) centered at a=c=1 A Taylor series is given by f(x)=∑k=0∞f(k)(a)k!(x−a)k here…

Answered: Question: a) Find the Taylor…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Step 1

let f(x)=g(x)=ln(x) centered at a=c=1

A Taylor series is given by

$f (x) = \sum_{k = 0}^{\infty} \frac{f^{(k)} (a)}{k!} {(x - a)}^{k}$

here a=1

Step 2

$S o, w h a t w e n e e d t o d o t o g e t t h e d e s i r e d p o l y n o m i a l i s t o c a l c u l a t e t h e d e r i v a t i v e s, e v a l u a t e t h e m a t t h e g i v e n p o i n t, a n d p l u g t h e r e s u l t s i n t o t h e g i v e n f o r m u l a . f^{(0)} (x) = f (x) = \ln (x) E v a l u a t e t h e f u n c t i o n a t t h e p o i n t : f (1) = 0 1 .) F i n d t h e 1 s t d e r i v a t i v e : f^{(1)} (x) = {(f^{(0)} (x))}^{'} = {(\ln (x))}^{'} = \frac{1}{x} E v a l u a t e t h e 1 s t d e r i v a t i v e a t t h e g i v e n p o i n t : {(f (1))}^{'} = 1 2) . F i n d t h e 2 n d d e r i v a t i v e : f^{(2)} (x) = {(f^{(1)} (x))}^{'} = {(\frac{1}{x})}^{'} = - \frac{1}{x^{2}} E v a l u a t e t h e 2 n d d e r i v a t i v e a t t h e g i v e n p o i n t : {(f (1))}^{''} = - 1 3) . F i n d t h e 3 r d d e r i v a t i v e : f^{(3)} (x) = {(f^{(2)} (x))}^{'} = {(- \frac{1}{x^{2}})}^{'} = \frac{2}{x^{3}} E v a l u a t e t h e 3 r d d e r i v a t i v e a t t h e g i v e n p o i n t : {(f (1))}^{'''} = 2 4) . F i n d t h e 4 t h d e r i v a t i v e : f^{(4)} (x) = {(f^{(3)} (x))}^{'} = {(\frac{2}{x^{3}})}^{'} = - \frac{6}{x^{4}} E v a l u a t e t h e 4 t h d e r i v a t i v e a t t h e g i v e n p o i n t : {(f (1))}^{''''} = - 6 5) . F i n d t h e 5 t h d e r i v a t i v e : f^{(5)} (x) = {(f^{(4)} (x))}^{'} = {(- \frac{6}{x^{4}})}^{'} = \frac{24}{x^{5}} E v a l u a t e t h e 5 t h d e r i v a t i v e a t t h e g i v e n p o i n t : {(f (1))}^{(5)} = 24 N o w, u s e t h e c a l c u l a t e d v a l u e s t o g e t a p o l y n o m i a l : f (x) \approx \frac{0}{0!} {(x - (1))}^{0} + \frac{1}{1!} {(x - (1))}^{1} + \frac{- 1}{2!} {(x - (1))}^{2} + \frac{2}{3!} {(x - (1))}^{3} + \frac{- 6}{4!} {(x - (1))}^{4} + \frac{24}{5!} {(x - (1))}^{5} a f t e r s i m p l i f y i n g w e g e t f (x) \approx P (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} - \frac{1}{4} {(x - 1)}^{4} + \frac{1}{5} {(x - 1)}^{5}$

now,

$P_{0} (x) = 0 (p o l y n o m i a l d e g r e e o f o r d e r 0) p_{1} (x) = (x - 1) (p o l y n o m i a l d e g r e e o f o r d e r 1) p_{2} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} (p o l y n o m i a l d e g r e e o f o r d e r 2) p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3} (p o l y n o m i a l d e g r e e o f o r d e r 3)$

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