**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+...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

**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`