Student Question

`f(x)=e^(-x/2) , n=4` 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`


`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=4` for the given function `f(x)=e^(-x/2)` , 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/2` then `(du)/(dx)= -1/2`

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

`d/(dx) e^(-x/2) = e^(-x/2) *(-1/2)`

            ` = -e^(-x/2)/2 or -1/2e^(-x/2)`

Applying `d/(dx) e^(-x/2)= -e^(-x/2)/2`   for each `f^n(x)` , we get:

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


`f^2(x) = d/(dx) (-1/2e^(-x/2))`

          `=-1/2 *d/(dx) e^(-x/2)`

           `=-1/2 *(-1/2e^(-x/2))`


 `f^3(x) = d/(dx) (1/4e^(-x/2))`

            `=1/4 *d/(dx) e^(-x/2)`

            `=1/4 *(-1/2e^(-x/2))`


`f^4(x) = d/(dx) (-1/8e^(-x/2))`

           `=-1/8 *d/(dx) e^(-x/2)`

           `=-1/8 *(-1/2e^(-x/2))`


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

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

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




Note: ` e ^(-0/2) = e^0 =1` .

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

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

     `= 1+(-1/2)/(1!)x+(1/4)/(2!)x^2+(-1/8)/(3!)x^3+(1/16)/(4!)x^4`


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


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