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

Expert Answers

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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`


`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) =tan(x)sec(x)`


Plug-in `x=0` , we get:




          `= 0 *1`



        `= 2*1 -1`


Applying the formula for Maclaurin series, we get:

`f(x)=sum_(n=0)^2 (f^n(0))/(n!) x^n`



         `=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`

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