# `f(x)=e^(-x) , n=5` 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=5` for the given function `f(x)=e^(-x)` , we may apply the formula for Maclaurin series..

To list `f^n(x)` , we may apply derivative formula for exponential function: `d/(dx) e^u = e^u * (du)/(dx)` .

Let `u =-x` then `(du)/(dx)= -1`

Applying the values on the derivative formula for exponential function, we get:

`d/(dx) e^(-x) = e^(-x) *(-1)`

`= -e^(-x)`

Applying `d/(dx) e^(-x)= -e^(-x)`  for each `f^n(x)` , we get:

`f'(x) = d/(dx) e^(-x)`

`=-e^(-x)`

`f^2(x) = d/(dx) (- e^(-x))`

`=-1 *d/(dx) e^(-x)`

`=-1 *(-e^(-x))`

`=e^(-x)`

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

`=-e^(-x)`

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

`=-1 *d/(dx) e^(-x)`

`=-1 *(-e^(-x))`

`=e^(-x)`

`f^5(x) = d/(dx) e^(-x)`

`=-e^(-x)`

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

`f(0) =e^(-0) =1`

`f'(0) =-e^(-0)=-1`

`f^2(0) =e^(-0)=1`

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

`f^4(0) =e^(-0)=1`

`f^5(0) =-e^(-0)=-1`

Note: `e^(-0)=e^0 =1` .

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

`f(x)=sum_(n=0)^5 (f^n(0))/(n!) x^n`

`= 1+(-1)/(1!)x+1/(2!)x^2+(-1)/(3!)x^3+1/(4!)x^4+(-1)/(5!)x^5`

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

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

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

`P_5(x)=1-x+x^2/2-x^3/6+x^4/24 -x^5/120`