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Taylor polynomial remainder term

  • #1
1,444
2
1. Homework Statement
Consider the followign function [itex] f(x) = x^-5 [/itex]
[itex] a=1 [/itex]
[itex] n=2 [/itex]
[itex]0.8 \leq x \leq 1.2 [/itex]

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 [itex] f(x) ≈ T_{n}(x) [/itex] when x lies in the interval

2. Homework Equations
[tex] f(x) = f(a) + f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 [/tex]
[tex] R_k(x) =\frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1} [/tex]

3. The Attempt at a Solution

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

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

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

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

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

and the result of the above is 1.6689

Is the above correct? Thanks for your help!
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
10
Since [tex] f^{(3)} (x) = -210x^{-8} [/tex]
and this is decreasing, we use x = 0.8 and we use [itex] M = -210(0.8)^{-8} [/itex]
Technially,
[itex] f^{(3)} (x) = -210x^{-8} [/itex]
is increasing on the interval. But since we need
[itex]M = max \left| f^{(k+1)}(\xi_L) \right|[/itex]
x = 0.8 is the correct value of xi.

I also got 1.6689 as the estimated error.
 
  • #3
1,444
2
Technially,
[itex] f^{(3)} (x) = -210x^{-8} [/itex]
is increasing on the interval. But since we need
[itex]M = max \left| f^{(k+1)}(\xi_L) \right|[/itex]
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!
 

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