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

a. T_3(x)
b. R_3(x) (remainder)

(The next one is optional)
c. What if the interval was 1\lex\le3?

1 Answer
May 31, 2018

a) T_3(x)=1-3(x-1)+6(x-1)^2-10(x-1)^3
b) R_3(x)<=0.9599

Explanation:

a) The nth degree Taylor Polynomial of a function f(x) 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(x) for f(x)=x^-3 at a=1, we need the f(1) and the first three derivatives each evaluated at 1.

f(1)=1^-3=1
f'(x)=-3x^-4, f'(1)=-3
f''(x)=12x^-5, f''(1)=12
f^((3))(x)=-60x^-6, f^((3))(1)=-60

Thus,

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

T_3(x)=1-3(x-1)+12(x-1)^2/(2)-60(x-1)^3/(6)

T_3(x)=1-3(x-1)+6(x-1)^2-10(x-1)^3

b) We can find R_3(x) 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 (n+1)th derivative on the interval 2.5<=x<=3.5 .

That is, C>=|f^((n+1))(x)| on 2.5<=x<=3.5

We're looking to maximize f^((n+1))(x) on 2.5<=x<=3.5.

Well, we have n=3, n+1=4, f^((4))(x)=360x^-7

f^((4))(x)=360/x^7

On 2.5<=x<=3.5, |360/x^7|=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^((4))(2.5)|=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<=x<=3.5

So,

2.5-1<=x-1<=3.5-1

1.5<=x-1<=2.5

We see |x-1|<=2.5, so |x-1|^4<=(2.5)^4=39.0625

Then,

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

R_3(x)<=0.9599