sambellamy
Junior Member
- Joined
- Oct 21, 2014
- Messages
- 53
Firstly, this forum has been a huge help with understanding this chapter - thanks everyone. My final question from this set:
I am trying to estimate the accuracy of the approximation T2 when 0.5 ≤ x ≤ 1.5 with Taylor's inequality.
I have f(x) = x4/3. I am also given a=1.
I have determined:
f'(x) = (4/3)x1/3
f''(x) = (4/9)x2/3
T2 = 1 + (4/3)(x-1) + (8/9)(x-1)2, and f'''(x) = (-8/27)x-5/3
I believe all of the above to be true. I know that Taylor's inequality gives |Rn(x)| ≤ M/(n+1)!|x-a|n+1, for |x-a| ≤ d
given that M ≥ f(n+1)(x), and f'''(x) = (-8/27)x-5/3, I determined the largest value in this interval to be x=1.5, which gives M=-0.15
|Rn(x)| ≤ (-0.15/6) |x-1|3
|x-1| must also be ≤ d, which is 1, so x must be ≤ 2.
so for the largest value of |Rn(x)|, i calculated (-0.15/6)*13, which came out to about 0.025. Can anyone tell me that what I did is correct or incorrect? I fudged my way through this and am not sure if it's OK.
I am trying to estimate the accuracy of the approximation T2 when 0.5 ≤ x ≤ 1.5 with Taylor's inequality.
I have f(x) = x4/3. I am also given a=1.
I have determined:
f'(x) = (4/3)x1/3
f''(x) = (4/9)x2/3
T2 = 1 + (4/3)(x-1) + (8/9)(x-1)2, and f'''(x) = (-8/27)x-5/3
I believe all of the above to be true. I know that Taylor's inequality gives |Rn(x)| ≤ M/(n+1)!|x-a|n+1, for |x-a| ≤ d
given that M ≥ f(n+1)(x), and f'''(x) = (-8/27)x-5/3, I determined the largest value in this interval to be x=1.5, which gives M=-0.15
|Rn(x)| ≤ (-0.15/6) |x-1|3
|x-1| must also be ≤ d, which is 1, so x must be ≤ 2.
so for the largest value of |Rn(x)|, i calculated (-0.15/6)*13, which came out to about 0.025. Can anyone tell me that what I did is correct or incorrect? I fudged my way through this and am not sure if it's OK.