Question

What is the Taylor series expansion of f(x)= x^2-x+1 about the point x=-1?

(a) f(x) = -3-3(x+1)+(x+1)^2

(b) f(x) = -3-(x+1)+(x+1)^2

(c) f(x) = -3-3(x+1)+2(x+1)^2

(d) f(x) = -1-3(x+1)+2(x+1)^2

This question was posed to me during an interview.

I would like to ask this question from Generalized Mean Value Theorem in portion Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (a) f(x) = -3-3(x+1)+(x+1)^2

The best I can explain: Given, f(x)= x^2-x+1 ,         f(-1)=1+1+1=3

f'(x) = 2x-1,f'(-1)=2(-1)-1= -3

f”(x) = 2,           f”(-1)=2

f^n (x)=0,     for n > 2

The Taylor series expansion of f(x) about x=a is,

f(x) =  f(a)+(x-a) f'(a)+(x-a)^2 \(\frac{f”}{2!}\) (a)+⋯

Here a=-1,

f(x) = f(-1)+(x+1) f'(-1)+(x+1)^2 \(\frac{f”}{2!}\) (-1)  since,for n > 2, f^n (x) = 0.

f(x) = -3-3(x+1)+(x+1)^2