**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))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the Maclaurin polynomial of degree n=5 for the given function `f(x)=sin(x)` , we may apply the formula for Maclaurin series.

To list `f^n(x)` , we may apply the derivative formula for trigonometric functions: `d/(dx) sin(x) = cos(x)` and `d/(dx) cos(x) = -sin(x)` .

`f(x) =sin(x)`

`f'(x) = d/(dx) sin(x)`

`= cos(x)`

`f^2(x) = d/(dx) cos(x)`

`= -sin(x)`

`f^3(x) = d/(dx) -sin(x)`

`=-1*d/(dx) sin(x)`

`= -1 * cos(x)`

`= -cos(x)`

`f^4(x) = d/(dx) -cos(x)`

`=-1*d/(dx) cos(x)`

`= -1 * (-sin(x))`

`= sin(x)`

`f^5(x) = d/(dx) sin(x)`

`= cos(x)`

Plug-in `x=0 ` on each `f^n(x)` , we get:

`f(0) =sin(0) =0`

`f'(0)= cos(0) =1`

`f^2(0)= -sin(0)=0`

`f^3(0)= -cos(0)=-1`

`f^4(0)= sin(0)=0`

`f^5(0)= cos(0)=1`

Plug-in the values on the formula for Maclaurin series, we get:

`sum_(n=0)^5 (f^n(0))/(n!) x^n= 0+1/(1!)x+0/(2!)x^2+(-1)/(3!)x^3+0/(4!)x^4+1/(5!)x^5`

`= 0+1/1x+0/2x^2-1/6x^3+0/24x^4+1/120x^5`

`= x-1/6x^3+1/120x^5`

The **Maclaurin polynomial of degree n=5** for the given
function `f(x)=sin(x)` will be:

`P_5(x)=x-1/6x^3+1/120x^5`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.