Answer:
See the explanation for the answer.
Step-by-step explanation:
Given function:
[tex]f(x) = x^{1/4}[/tex]
The n-th order Taylor polynomial for function f with its center at a is:
[tex]p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}[/tex]
As n = 3 So,
[tex]p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}[/tex]
[tex]p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}[/tex]
[tex]p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}[/tex]
[tex]p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}[/tex]
[tex]p_{3} (x)[/tex] = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6
(0.0000018522752) (x-81)³
[tex]p_{3} (x)[/tex] = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254
(x-81)³ + 2.25
Hence approximation at given quantity i.e.
x = 94
Putting x = 94
[tex]p_{3} (94)[/tex] = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +
0.00000030871254 (94-81)³ + 2.25
= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +
2.25
= 0.87037 03742 - 0.000042866941 (169) +
0.00000030871254(2197) + 2.25
= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25
[tex]p_{3} (94)[/tex] = 3.113804102621
Compute the absolute error in the approximation assuming the exact value is given by a calculator.
Compute [tex]\sqrt[4]{94}[/tex] as [tex]94^{1/4}[/tex] using calculator
Exact value:
[tex]E_{a}[/tex](94) = 3.113737258478
Compute absolute error:
Err = | 3.113804102621 - 3.113737258478 |
Err (94) = 0.000066844143
If you round off the values then you get error as:
|3.11380 - 3.113737| = 0.000063
Err (94) = 0.000063
If you round off the values up to 4 decimal places then you get error as:
|3.1138 - 3.1137| = 0.0001
Err (94) = 0.0001
