`f(x)=e^(-2x)` Prove that the Maclaurin series for the function converges to the function for all x

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=5 for the given function `f(x)=e^(-2x)` , we may apply the formula for Maclaurin series..

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

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

Let `u =-2x` then `(du)/(dx)= -2 .`

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

                 `= -2e^(-2x)`

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

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

        `=-2e^(-2x)`

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

            `=-2 d/(dx) (e^(-2x)) `

            `=(-2)*(-2e^(-2x))`

            `=4e^(-2x)`

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

            `=4d/(dx) (e^(-2x)) `

            `=4*(-2e^(-2x))`

             `=-8e^(-2x)`

 `f^4(x) = d/(dx) (- 8e^(-2x))`

             ` =-8 d/(dx) (e^(-2x)) `

             `=(-8)*(-2e^(-2x))`

             `=16e^(-2x)`

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

 `f(0) =e^(-2*0) =1`

 `f'(0) =-2e^(-2*0)=-2`

 `f^2(0) =4e^(-2*0)=4`

 `f^3(0) =-8e^(-2*0)=-8`

 `f^4(0) =16e^(-2*0)=16`

Note: `e^(-2*0) = e^0 = 1.`

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

 `sum_(n=0)^oo (f^n(0))/(n!) x^n`

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

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

 ` =sum_(n=0)^oo (-2)^n/(n!)x^n`

 `=sum_(n=0)^oo (-2x)^n/(n!)`

To determine the interval of convergence for the Maclaurin series:` sum_(n=0)^oo (-2x)^n/(n!)` , we may apply Ratio Test.  

In Ratio test, we determine the limit as: `lim_(n-gtoo)|a_(n+1)/a_n| = L.`

The series converges absolutely when it satisfies `Llt1` .

For the  Maclaurin series: `sum_(n=0)^oo (-2x)^n/(n!)` , we have:

`a_n=(-2x)^n/(n!)`

Then,

`1/a_n= (n!)/(-2x)^n`

`a_(n+1)=(-2x)^(n+1)/((n+1)!)`

            ` =((-2x)^n*(-2x)^1)/((n+1)*(n!))`

             ` =((-2x)^n(-2x))/((n+1)*(n!))`

Applying the Ratio test, we set-up the limit as:

`lim_(n-gtoo)|a_(n+1)/a_n|=lim_(n-gtoo)|a_(n+1)*1/a_n|`

                         ` =lim_(n-gtoo)|((-2x)^n(-2x))/((n+1)*(n!))*(n!)/(-2x)^n|`

Cancel out common factors: `(-2x)^n`  and `(n!)` .

`lim_(n-gtoo)|(-2x)/(n+1)|`

Evaluate the limit.

`lim_(n-gtoo)|(-2x)/(n+1)|=|-2x| lim_(n-gtoo)|1/(n+1)|`

                       ` =|2x|lim_(n-gtoo)1/(n+1)`

                       ` =|2x|* 1/oo`

                        `= |2x|*0`

                        ` =0`

The `L=0` satisfies ` Llt1` for all `x` .

Thus, the Maclaurin series: `sum_(n=0)^oo (-2x)^n/(n!)` is absolutely converges for all `x` .

Interval of convergence: `-ooltxltoo`