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Chapter 8: Q- 33E (page 435)

Question: In Problems 29–34, determine the Taylor series about the point X0for the given functions and values of X0.

33. f (X)= x3+3x-4, x0= 1

Short Answer

Expert verified

The required expression is 6(x-1)+3(x-1)2+(x-1)3.

Step by step solution

01

Taylor series

For a function f(x) the Taylor series expansion about a point x0is given by,f(x-x0)=f(x0)+f'(x0).(x-x0)+f''(x0).(x-x0)22!+f'''(x0).(x-x0)33!+.....

02

Derivatives of function at x0.

We have to calculate the Taylor series expansion for, f (x) = x3+3x-4 at x0=1 .

Calculating the derivatives of function at x0

f(x) = x3+3x-4 then f (X0)=0

f'(x) = 3x2+3 then f'(x0)=6

f''(x) = 6x then f''(x0)=6

f'''(x) =6 thenf'''(x0)=6

f''''(x)=0 thenf''''(x0)=60

03

Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at ,

x0=1 .then,

x3+3x-4=0+6.(x-1)+6.x-122!+6.x-133!+0+0+....=6(x-1)+3(x-1)2+(x-1)3

Hence, the required expression is 6(x-1)+3(x-1)2+(x-1)3

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Most popular questions from this chapter

The solution to the initial value problem

xy''(x)+2y'(x)+xy(x)=0y(0)=1,   y'(0)=0

has derivatives of all orders atx=0(although this is far from obvious). Use L'Hôpital's rule to compute the Taylor polynomial of degree 2 approximating this solution.

In Problems 21-28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion about x=0 of a general solution to the given differential equation.

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Question: In Problems 29–34, determine the Taylor series about the point X0for the given functions and values of X0.

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