Taylor polynomial remainder term

[stunner5000pt]

Homework Statement

Consider the followign function f(x) = x^-5
a=1
n=2
0.8 \leq x \leq 1.2

a) Approximate f with a tayloy polynomial of nth degree at the number a = 1
b) use taylor's inequality to estimate the accuracy of approximation f(x) ≈ T_{n}(x) when x lies in the interval

Homework Equations

f(x) = f(a) + f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2
R_k(x) =\frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1}

The Attempt at a Solution

Part a is simple:
T_2(x) = 1 - 5(x-1) + 15(x-1)^2

Since we have found the taylor polynomial at n = 2 the remainder:

R_2(x) \leq | \frac{M}{3!}(x-1)^3 |

Since f^{(3)} (x) = -210x^{-8}
and this is decreasing, we use x = 0.8 and we use M = -210(0.8)^{-8}

R_2(x) \leq |\frac{-210(0.8)^{-8}}{3!} (0.8-1)^3

and the result of the above is 1.6689

Is the above correct? Thanks for your help!

 

[eumyang]

Since f^{(3)} (x) = -210x^{-8}
and this is decreasing, we use x = 0.8 and we use M = -210(0.8)^{-8}

Technially,
f^{(3)} (x) = -210x^{-8}
is increasing on the interval. But since we need
M = max \left| f^{(k+1)}(\xi_L) \right|
x = 0.8 is the correct value of xi.

I also got 1.6689 as the estimated error.

 

[stunner5000pt]

Technially,
f^{(3)} (x) = -210x^{-8}
is increasing on the interval. But since we need
M = max \left| f^{(k+1)}(\xi_L) \right|
x = 0.8 is the correct value of xi.

I also got 1.6689 as the estimated error.

Ah yes, didn't consider the negative sign - thanks for your help!