Student Question

# `f(x)=x^2e^(-x) , n=4` Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at `c=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=4 for the given function `f(x)=x^2e^(-x)` , we may apply the formula for Maclaurin series.

To list `f^n(x) ` up to `n=4` , we may apply the following formula:

Product rule for differentiation: `d/(dx) (u*v) = u' *v +u*v' `

Derivative property: `d/(dx) (f+-g+-h) = d/(dx) f +-d/(dx) g+-d/(dx) h`

Power rule for differentiation: `d/(dx) x^n =n*x^(n-1)`

Derivative formula for exponential function: `d/(dx) e^u = e^u * (du)/(dx)`

`f(x)=x^2e^(-x)`

Let `u =x^2` then `u' = 2x`

`v = e^(-x)`  then `v' = e^x*(-1) =-e^(-x)`

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

`= 2xe^(-x) -x^2e^(-x)`

Let: `u =x` then `u' =1`

`v = e^(-x)`  then `v' =-e^(-x)`

Note: `c` = constant value.

`d/(dx) c*xe^(-x) = c*d/(dx) xe^(-x)`

`= c*[1*e^x +x * (-e^(-x))]`

` = c*[e^x-xe^(-x)]`

` = ce^x-cxe^(-x)`

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

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

`=-ce^(-x)`

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

`= 2xe^(-x) -x^2e^(-x)`

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

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

`= [2e^(-x)-2xe^(-x)] - [2xe^(-x) -x^2e^(-x)]`

`=2e^(-x)-2xe^(-x) - 2xe^(-x) +x^2e^(-x)`

`=2e^(-x)-4xe^(-x) +x^2e^(-x)`

`f^3(x) = d/(dx) (2e^(-x)-4xe^(-x) +x^2e^(-x))`

`=d/(dx) 2e^(-x) -d/(dx) 4xe^(-x)+ d/(dx) x^2e^(-x)`

`=[-2e^(-x)] -[4e^(-x)-4xe^(-x)]+ [2xe^(-x) -x^2e^(-x)]`

`=- 2e^(-x) -4e^(-x)+4xe^(-x)+ 2xe^(-x) -x^2e^(-x)`

`=- 6e^(-x)+6xe^(-x) -x^2e^(-x)`

`f^4(x) = d/(dx) ( - 6e^(-x)+6xe^(-x) -x^2e^(-x))`

`=d/(dx) (-6e^(-x)) + d/(dx) 6xe^(-x)- d/(dx) x^2e^(-x)`

`=[ 6e^(-x)]+[6e^(-x)-6xe^(-x)] -[2xe^(-x) -x^2e^(-x)]`

`=6e^(-x)+6e^(-x)-6xe^(-x) -2xe^(-x) +x^2e^(-x)`

`=12e^(-x)-8xe^(-x) +x^2e^(-x)`

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

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

`=0*1`

`=0`

`f'(0)=2*0*e^(-0) -0^2e^(-0)`

`=2*0*1 +0*1`

`=0`

`f^2(0)=2e^(-0)-4*0*e^(-0) +0^2e^(-0)`

`=2*1 -4*0*1 +0*1`

` =2`

`f^3(0)=- 6e^(-0)+6*0*e^(-0) -0^2e^(-0)`

`=-6*1 +6*0*1 +0*1`

`=-6`

`f^4(0)=12e^(-0)-8*0*e^(-0) +0^2e^(-0)`

` =12*1 -8*0*1 +0*1`

`=12`

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

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

`sum_(n=0)^4 (f^n(0))/(n!) x^n`

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

`= 0+0/(1!)x+2/(2!)x^2+(-6)/(3!)x^3+12/(4!)x^4`

`= 0+0/1x+2/2x^2-6/6x^3+12/24x^4`

`= 0+0+x^2-x^3+1/2x^4`

`= x^2-x^3+1/2x^4`

The Maclaurin polynomial of degree `n=4` for the given function `f(x)=x^2e^(-x)` will be:

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