Taylor’s formula with n=1 and a=0 gives the linearization of a funct

ikavumacj

ikavumacj

Answered question

2021-11-16

Taylor’s formula with n=1 and a=0 gives the linearization of a function at x=0 With n=2 and n=3 we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions:
a) For what values of x can the function be replaced by each approximation with an error less than 102
b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and interval.
Step 1: Plot the function over the specified interval.
Step 2: Find the Taylor polynomials P1(x), P2(x) and P3(x) at x=0
Step 3: Calculate the (n+1) st derivative f(n+1)(c) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of c over the specified interval and estimate its maximum absolute value, M.
Step 4: Calculate the remainder Rn(x) for each polynomial. Using the estimate M from Step 3 in place of f(n+1)(c) plot Rn(x) over the specified interval. Then estimate the values of x that answer question (a).
Step 5: Compare your estimated error with the actual error En(x)=|f(x)Pn(x)| by plotting En(x) over the specified interval. This will help answer question (b).
Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5.
f(x)=(1+x)32, 12x2

Answer & Explanation

Louis Gregory

Louis Gregory

Beginner2021-11-17Added 14 answers

Step 1
The given function is
f(x)=(1+x)32, 12x2
The graph of the given function is shown below.
image

Step 2
Now we have to find the Taylor polynomials P1(x), P2(x), and P3(x) at x=0.
Recall: The Taylor polynomial of order n generated by f at x=a is given be
Pn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2++fn(a)n!(xa)n
Let us first calculate derivatives of f(x) at the point x=0
f(x)=32(1+x)12f(0)=32
f(x)=34(1+x)12f(0)=34
f(x)=38(1+x)32f(0)=38
fiv(x)=916(1+x)52fiv(0)=916
Step 3
Now the Taylor polynomial P1(x) at x=0 is given by
P1(x)=f(0)+f(0)(x0)
P1(x)=1+3x2
The the Taylor polynomial P2(x) at x=0 is given by
P2(x)=f

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