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