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

Expert Answers
marizi eNotes educator| Certified Educator

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`