Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 30

Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of f^n(x) centered at x=c . The general formula for Taylor series is:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) = f(c) + f'(c) (x-c)+ (f'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...
To determine the Taylor polynomial of degree n=2 centered at c=pi , we may apply the definition of the Taylor series by listing the f^n(x) up to n=2 .
f(x) = x^2cos(x)
Apply Product rule of differentiation: d/(dx) (u*v) = v*du + u*dv for each derivative.
f'(x) = d/(dx) (x^2cos(x))
Let u = x^2 then du =2x
v = cos(x) then dv = -sin(x)
f'(x) =cos(x) *(2x) + x^2*(-sin(x))
=2xcos(x)-x^2sin(x)
f^2= d/(dx)(2xcos(x)-x^2sin(x) )
=d/(dx)2xcos(x)- d/(dx) x^2sin(x)
For d/(dx)2xcos(x) , we let:
u = 2x then du =2
v = cos(x) then dv = -sin(x)
d/(dx)2xcos(x)= cos(x)*2 + 2x*(-sin(x))
=2cos(x) -2xsin(x)
For d/(dx)x^2sin(x) , we let:
u = x^2 then du =2x
v = sin(x) then dv = cos(x)
d/(dx)2xcos(x)= sin(x)*2x + x^2*cos(x)
=2xsin(x) +x^2cos(x)
Then,
d/(dx)2xcos(x)-d/(dx) x^2sin(x) = [2cos(x) -2xsin(x)] -[2xsin(x) +x^2cos(x)]
= 2cos(x) -2xsin(x) -2xsin(x) -x^2cos(x)
=2cos(x) -4xsin(x) -x^2cos(x)
Thus, f^2(x) =2cos(x) -4xsin(x) -x^2cos(x).
Plug-in x=pi , we get:
f(pi) =pi^2*cos(pi)
=pi^2*(-1)
=-pi^2
f'(pi)=2pi*cos(pi)-pi^2*sin(pi)
=2pi*(-1) -pi^2 *(0)
=-2pi
f^2(pi) =2cos(pi) -4*pi*sin(pi) -pi^2*cos(pi)
=2(-1) -4*pi*0 -pi^2*(-1)
=-2+pi^2 or -(2-pi^2)
Applying the formula for Taylor series centered at c=pi , we get:
sum_(n=0)^2 (f^n(pi))/(n!)(x-pi)^n
=f(pi) + f'(pi) (x-pi)+ (f'(pi))/(2!) (x-pi)^2
=(-pi^2) + (-2pi) (x-pi)+ (-(2-pi^2))/(2!) (x-pi)^2
= -pi^2 -2pi (x-pi)-(2-pi^2)/2 (x-pi)^2
The Taylor polynomial of degree n=2 for the given function f(x)=x^2cos(x) centered at c=pi will be:
P(x) =-pi^2 -2pi (x-pi)-(2-pi^2)/2 (x-pi)^2