# For f(x)=x^-3, a=1, and 2.5\lex\le3.5, find the following?

## a. ${T}_{3} \left(x\right)$ b. ${R}_{3} \left(x\right)$ (remainder) (The next one is optional) c. What if the interval was $1 \setminus \le x \setminus \le 3$?

May 31, 2018

a) ${T}_{3} \left(x\right) = 1 - 3 \left(x - 1\right) + 6 {\left(x - 1\right)}^{2} - 10 {\left(x - 1\right)}^{3}$
b) ${R}_{3} \left(x\right) \le 0.9599$

#### Explanation:

a) The $n t h$ degree Taylor Polynomial of a function $f \left(x\right)$ about $x = a$ is defined as

T_n(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2/(2!) +f^((3))(a)(x-a)^3/(3!)+...+f^((n))(a)(x-a)^n/(n!)

So, to find ${T}_{3} \left(x\right)$ for $f \left(x\right) = {x}^{-} 3$ at $a = 1 ,$ we need the $f \left(1\right)$ and the first three derivatives each evaluated at $1$.

$f \left(1\right) = {1}^{-} 3 = 1$
$f ' \left(x\right) = - 3 {x}^{-} 4 , f ' \left(1\right) = - 3$
$f ' ' \left(x\right) = 12 {x}^{-} 5 , f ' ' \left(1\right) = 12$
${f}^{\left(3\right)} \left(x\right) = - 60 {x}^{-} 6 , {f}^{\left(3\right)} \left(1\right) = - 60$

Thus,

T_3(x)=1-3(x-1)+12(x-1)^2/(2!)-60(x-1)^3/(3!)

${T}_{3} \left(x\right) = 1 - 3 \left(x - 1\right) + 12 {\left(x - 1\right)}^{2} / \left(2\right) - 60 {\left(x - 1\right)}^{3} / \left(6\right)$

${T}_{3} \left(x\right) = 1 - 3 \left(x - 1\right) + 6 {\left(x - 1\right)}^{2} - 10 {\left(x - 1\right)}^{3}$

b) We can find ${R}_{3} \left(x\right)$ using Taylor's Inequality, which tells us that

|R_n(x)|<=C|x-a|^(n+1)/((n+1)!)

Where $C$ is the bounding value for the $\left(n + 1\right) t h$ derivative on the interval $2.5 \le x \le 3.5$ .

That is, $C \ge | {f}^{\left(n + 1\right)} \left(x\right) |$ on $2.5 \le x \le 3.5$

We're looking to maximize ${f}^{\left(n + 1\right)} \left(x\right)$ on $2.5 \le x \le 3.5$.

Well, we have $n = 3 , n + 1 = 4 , {f}^{\left(4\right)} \left(x\right) = 360 {x}^{-} 7$

${f}^{\left(4\right)} \left(x\right) = \frac{360}{x} ^ 7$

On $2.5 \le x \le 3.5 , | \frac{360}{x} ^ 7 | = \frac{360}{x} ^ 7$ is maximized for $x = 2.5 ,$ which yields the smallest possible denominator on this interval and therefore the largest value overall.

So, $C = | {f}^{\left(4\right)} \left(2.5\right) | = \frac{360}{2.5} ^ 7 = 0.589824$

Thus,

|R_3(x)|<=C|x-1|^(3+1)/((3+1)!)

|R_3(x)|<=0.589824|x-1|^(4)/(4!)

We know

$2.5 \le x \le 3.5$

So,

$2.5 - 1 \le x - 1 \le 3.5 - 1$

$1.5 \le x - 1 \le 2.5$

We see $| x - 1 | \le 2.5$, so $| x - 1 {|}^{4} \le {\left(2.5\right)}^{4} = 39.0625$

Then,

|R_3(x)|<=0.589824*39.0625/(4!)=0.9599

${R}_{3} \left(x\right) \le 0.9599$