**Taylor series** is an example of infinite series derived from the expansion of `f(x)` about a single point. It is represented by infinite sum of `f^n(x)` centered at `x=c` . The **general formula for Taylor series** is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...`

To apply the definition of Taylor series for the given function `f(x) = e^(-4x)` centered at `x=0` , we list `f^n(x)` using the derivative formula for exponential function: `d/(dx) e^u = e^u * (du)/(dx)` .

Let `u =-4x` then `(du)/(dx)= -4` .

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

`d/(dx) e^(-4x) = e^(-4x) *(-4)`

`= -4e^(-4x)`

Applying `d/(dx) e^(-4x)= -4e^(-4x)` and `d/(dx) c*f(x) = c d/(dx) f(x)` for each` f^n(x)` , we get:

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

`= -4e^(-4x)`

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

`= -4*(-4e^(-4x))`

`=16e^(-4x)`

`f^3(x) = 16*d/(dx) e^(-4x)`

`= 16*(-4e^(-4x))`

`=-64e^(-4x)`

`f^4(x) =- 64*d/(dx) e^(-4x)`

`= -64*(-4e^(-4x))`

`=256e^(-4x)`

Plug-in `x=0` , we get:

`f(0) =e^(-4*0) =1`

`f'(0) =-4e^(-4*0)=-4`

`f^2(0) =16e^(-4*0)=16`

`f^3(0) =-64e^(-4*0)=-64`

`f^4(0) =2564e^(-4*0)=256`

Note: e`^(-4*0)=e^0 =1` .

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

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

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

`= 1+(-4)/(1!)x+16/(2!)x^2+(-64)/(3!)x^3+256/(4!)x^4+...`

`=1- 4/1x +16/(1*2)x^2 - 64/(1*2*3)x^3 +256/(1*2*3*4)x^4 +...`

`=1- 4x + 16/2x^2 - 64/6x^3 +256/24x^4 +...`

`= 1-4x+ 8x^2 - 32/3x^3 + 32/3x^4+...`

The Taylor series for the given function `f(x)=e^(-4x)` centered at `c=0` will be:

`e^(-4x) =1-4x+ 8x^2 - 32/3x^3 + 32/3x^4+...`

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