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Calculus / Calculus: Early Transcendentals 1 / Chapter 9.1 / Problem 52E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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QUESTION:

Estimating errors Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at 0. Estimates are not unique.

In 1.04; n = 3

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QUESTION:

Estimating errors Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at 0. Estimates are not unique.

In 1.04; n = 3

ANSWER:

Solution 52E

Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at 0. Estimates are not unique.

In 1.04; n = 3

Step 1:

Here let $f(x)=1+x$

We know that The taylor polynomial of ${n}^{th}\$ order is generally given by

$f(x)=\sum\limits_{n=0}^{\infty{}}\frac{{f}^{(n)}(a)}{n!}(x-a{)}^{n}$

Thus taylor series expansion of f(x) for n=3 at a=0 is

$f(x)=\sum\limits_{n=0}^{3}\frac{{f}^{(n)}(0)}{n!}(x{)}^{n}$ $\$

$\Rightarrow{}{f}^{(0)}(0)=ln(1+0)=0$

$\Rightarrow{}\frac{d}{dx}{f}^{\ }(0)=\frac{1}{1+0}=1$

$\Rightarrow{}\frac{{d}^{2}}{d{x}^{2}}{f}^{\ }(0)=\frac{-1}{(1+0{)}^{2}}=-1$

$\Rightarrow{}\frac{{d}^{3}}{d{x}^{3}}{f}^{\ }(0)=\frac{2}{(1+0{)}^{3}}=2$

Therefore the taylor series at n=3 is

$ln(1+x)=0+1\times{}x-1\times{}\frac{{x}^{2}}{2!}+2\times{}\frac{{x}^{3}}{3!}=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}$

$ln(1+x)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}$

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