Answer:
Step-by-step explanation: please go through the attached file for detailed explanation .
The Taylor series of f(x) of degree 2 is given by [tex]\rm f(x) = x^2-3x+3[/tex] and according to the remainder estimation theorem [tex]\rm |R_2(x)| \leq 7.716049383(|x-1|)^{3}[/tex].
Given :
Consider the following function-- f(x) = 2/x, a = 1, n = 2, 0.6 ≤ x ≤ 1.4.
a) The Taylor series is given by:
[tex]\rm f(x) = f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f"(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+....\\[/tex]
Now, at (a = 1) and (n = 2) the above series becomes:
[tex]\rm f(x) = 1-\dfrac{1}{a^2}}(x-a)+\dfrac{1}{2!}\times \dfrac{2}{a^3}\times (x-a)^2[/tex]
Substitute (a = 1) in the above series.
[tex]\rm f(x) = 1-\dfrac{1}{1^2}}(x-1)+\dfrac{1}{2!}\times \dfrac{2}{1^3}\times (x-1)^2[/tex]
[tex]\rm f(x) = 1-(x-1)+(x-1)^2[/tex]
[tex]\rm f(x) = x^2+1-2x-x+1+1[/tex]
[tex]\rm f(x) = x^2-3x+3[/tex]
b) According to remainder estimation theorem:
[tex]\rm |f^{n+1}(x)|\leq m[/tex]
[tex]\rm |R_n(x)| \leq \dfrac{m(|x-a|)^{n+1}}{(n+1)!}[/tex]
So, at (a = 1) and (n = 2) the above expression becomes:
[tex]\rm |R_2(x)| \leq \dfrac{m(|x-1|)^{2+1}}{(2+1)!}[/tex]
[tex]\rm |R_2(x)| \leq \dfrac{m(|x-1|)^{3}}{(3)!}[/tex] ---- (1)
where m is ( [tex]\rm |f^{n+1}(x)|\leq m[/tex] ).
[tex]\rm |f^{2+1}(x)|\leq m[/tex]
[tex]\rm |f^{3}(x)|\leq m[/tex]
[tex]\rm f'''(x)=-\dfrac{6}{x^4}[/tex]
m is maximum on [0.6,1.4]. So, if x = 0.6 then:
[tex]\rm f'''(0.6)=-\dfrac{6}{0.6^4} = -46.2962963[/tex]
So, [tex]\rm |f'''(0.6)|=46.2962963[/tex]
Now, put the value of m in equation (1).
[tex]\rm |R_2(x)| \leq \dfrac{46.2962963(|x-1|)^{3}}{6}[/tex]
[tex]\rm |R_2(x)| \leq 7.716049383(|x-1|)^{3}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/23044118
