# f(x)=x^2cosx, n=2, c=pi Find the n'th Taylor Polynomial centered at c

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+...

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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`

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