In the given equation as follows, use Taylor’s Theorem to obtain an up

osi4a2nxk

osi4a2nxk

Answered question

2021-11-18

In the given equation as follows, use Taylor’s Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error|:-
see the equation as attached here:-
cos(0.3)1(0.3)22!+(0.3)44!

Answer & Explanation

Lounctirough

Lounctirough

Beginner2021-11-19Added 14 answers

Step 1
Approximation is a result that is near, not exactly, but almost accurate. To find the simpler value of f(x), which has approximately the same value as F(x) over a specified interval. the analysis is then carried out on the more accessible f(x), knowing the difference will be small.
Upper bound on the size |Rn(x)| lof the error since T0(x)=f(a) then R0(x)=f(x)f(a).
T1(x)=f(a)+f(a)(xa) is the tangent line to f at a the remainder R1(x) is the difference between f(x) and the tangent line approximation at f.
Step 2
The formula for upper bound error estimate in Taylor's series |Rn(x)|max|fn+1(c)|(n+1)!|xa|n+1 in the given problem n=4max|fn+1(c)| is the maximum value off fn+1(c) between c and a, where a=0,x=0.3 and f(x)=cos(x) to find the fifth derivative of f(x).
f(x)=cos(x)
f(x)=sin(x)
f(x)=cos(x)
f(x)=sin(x)
f4(x)=cos(x)
f5(x)=sin(x)
|Rn(x)|max|fn+1(c)|(n+1)!|xa|n+1
|R4(x)||sin(0.3)|5!|0.30|5
|R4(x)|sin(0.3)5!(0.3)56×106
Upper bound for the error of the approximation is 6×106.
Step 3
To find the exact value of the error, subtract the given Taylor theorem by cos(0.3).

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