Student Question

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

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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`


`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)`


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)`



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

           `=2*0*1 +0*1`


`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`


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

            ` =12*1 -8*0*1 +0*1`


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:


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