`f(x)=tanx , n=3` 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`

 or

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

To determine the Maclaurin polynomial of degree n=3 for the given function `f(x)=tan(x)` , we may apply the formula for Maclaurin series.

We may list `f^n(x)` as:

`f(x) = tan(x)`

`f'(x)=d/(dx) tan(x) =sec^2(x)`

`f^2(x)=d/(dx)sec^2(x) =2tan(x)sec^2(x)`

`f^3(x)=d/(dx) 2tan(x)sec^2(x) =6sec^4(x)-4sec^2(x)`

Plug-in x=0, we get:

`f(0)=tan(0)`

         `=0`

`f'(0)=sec^2(0) or (sec(0))^2`

          `= 1^2`

          `=1`

`f^2(0)=2tan(0)sec^2(0) `

            `=2*0*1`

            `=0`

`f^3(0)=6sec^4(0)-4sec^2(0)`

           `=6(sec(0))^4-4(sec(0))^2`

           `=6*1^4 -4*1^2`

           `= 6-4`

           `=2`

Applying the formula for Maclaurin series, we get:

`sum_(n=0)^3 (f^n(0))/(n!) x^n =0+1/(1!)x+0/(2!)x^2+2/(3!)x^3`

                         `=0+1/1x+0/(1*2)x^2+2/(1*2*3)x^3`

                         `= 0+x+0/2x^2+2/6x^3`

                         `=x+2/6x^3`

                         `=x+x^3/3` 

The Maclaurin polynomial of degree n=3 for the given function `f(x)=tan(x)` will be:

`P_3(x)=x+x^3/3 `

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