`f(x) = e^(x/3) , n=4` Find the n'th Maclaurin polynomial for the function.

Expert Answers
marizi eNotes educator| Certified Educator

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 +... `

We may apply the formula for Maclaurin series to determine the Maclaurin polynomial of degree `n=4` for the given function `f(x)=e^(x/3)` .

Apply derivative formula for exponential function: `d/(dx) e^u = e^u * (du)/(dx)` to list `f^n(x)` as:

Let `u =x/3` then `(du)/(dx)= 1/3`

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

`d/(dx) e^(x/3) = e^(x/3) *(1/3)`

         `= e^(x/3)/3 or 1/3e^(x/3)`

Applying `d/(dx) e^(x/3)= 1/3e^(x/3)`  for each `f^n(x)` , we get:

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

          `=1/3e^(x/3)`

`f^2(x) = d/(dx) (1/3e^(x/3))`

          `=1/3 *d/(dx)e^(x/3)`

          `=1/3 *(1/3e^(x/3))`

          `=1/9e^(x/3)`

`f^3(x) = d/(dx) (1/9e^(x/3))`

           `=1/9 *d/(dx) e^(x/3)`

          `=1/9 *(1/3e^(x/3))`

          `=1/27e^(x/3)`

`f^4(x) = d/(dx) (1/27e^(x/3))`

           `=1/27 *d/(dx) e^(x/3)`

           `=1/27 *(1/3e^(x/3))`

           `=1/81e^(x/3)`

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

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

`f'(0)=1/3e^(0/3) = 1/3`

`f^2(0)=1/9e^(0/3)=1/9`

`f^3(0)=1/27e^(0/3)=1/27`

`f^4(0)=1/81e^(0/3)=1/81`

Note: `e ^(0/3) = 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+(1/3)/(1!)x+(1/9)/(2!)x^2+(1/27)/(3!)x^3+(1/81)/(4!)x^4`

   `=1+1/3x+1/18x^2+1/162x^3+1/1944x^4`

The Maclaurin polynomial of degree n=4 for the given function `f(x)=e^(x/3)` will be:

`P_4(x)=1+1/3x+1/18x^2+1/162x^3+1/1944x^4`