`f(x)=cos(pix) , n=4` Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at `c=0` . The expansion of the function about 0 follows the formula:

`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`

 or

`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the Maclaurin polynomial of degree `n=4` for the given function `f(x)=cos(pix)` , we may apply the formula for Maclaurin series.

To list `f^n(x)` up to `n=4` , we may apply the derivative formula for trigonometric functions: `d/(dx) sin(u) = cos(u) *(du)/(dx)`  and `d/(dx) cos(u) = -sin(u) *(du)/(dx)` .

Let `u =pix` then `(du)/(dx) =pi` .

`f(x) =cos(pix)`

`f'(x) = d/(dx) cos(pix)`

           `= -sin(pix) *pi`

           `=-pisin(pix)`

`f^2(x) = d/(dx)-pisin(pix)`

            `=-pi*d/(dx) sin(pix)`

            `= -pi * (cos(pi)* pi)`

            `= -pi^2cos(pix)`

`f^3(x) = d/(dx)-pi^2cos(pix)`

            `=-pi^2*d/(dx) cos(pix)`

            `= -pi^2 * (-sin(pix)*pi)`

            `= pi^3sin(pix)`

`f^4(x) = d/(dx)pi^3sin(pix)`

              `= pi^3d/(dx) sin(pix)`

             `= pi^3(cos(pix) *pi) `

             `=pi^4cos(pix)`

Plug-in` x=0` on each `f^n(x)` , we get:

`f(0)= cos(pi*0) =1`

`f'(0)= -pisin(pi*0) =0`

`f^2(0)= -pi^2cos(pi*0)=-pi^2`

`f^3(0)= pi^3sin(pi*0)=0`

`f^4(0) =pi^4cos(pi*0) =pi^4`

Note: `cos(pi*0) = cos(0)=1` and `sin(pi*0)=sin(0)=0` .

Plug-in the values on the formula for Maclaurin series, we get:

`sum_(n=0)^4 (f^n(0))/(n!)x^n`

        ` =f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4`

        ` =1+0/(1!)x+(-pi^2)/(2!)x^2+0/(3!)x^3+(pi^4)/(4!)x^4`

        `=1+0/1x-pi^2/2x^2+0/6x^3+pi^4/24x^4`

        `=1-pi^2/2x^2+pi^4/24x^4`

The Maclaurin polynomial of degree `n=4` for the given function `f(x)=cos(pix)` will be:

`P(x)=1-pi^2/2x^2+pi^4/24x^4`