f(x) = secx , n=2 Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at a=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)x)/(1!)+(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=2 for the given function f(x)=sec(x) , we may apply the formula for Maclaurin series.
We list f^n(x) as:
f(x)=sec(x)
f'(x) =tan(x)sec(x)
f^2(x)=2sec^3(x)-sec(x)
Plug-in x=0 , we get:
f(0)=sec(0)
        =1
f'(0)=tan(0)sec(0)
          = 0 *1
          =0
f^2(0)=2sec^3(0)-sec(0)
        = 2*1 -1
        =1
Applying the formula for Maclaurin series, we get:
f(x)=sum_(n=0)^2 (f^n(0))/(n!) x^n
        =1+0/(1!)+1/(2!)x^2
       =1+0/1+1/2x^2
         =1+1/2x^2 or 1 +x^2/2
Note: 1! =1 and 2! =1*2 =2.
The 2nd degree Maclaurin polynomial for  the given function f(x)= sec(x) will be:
sec(x) =1+x^2/2
or P_2(x)=1+x^2/2