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

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

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

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

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

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

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

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

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

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

All of the `f^n(x)` is represented by `e^x` .

Plug-in `x=1` , we get:

`f(1) =e^1 =e`

`f'(1) =e^1 =e`

`f^2(1) =e^1 =e`

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

`f^4(1) =e^1 =e`

Note: `e^1=e^1 =e` .

Each `f^n(x) `  centered at `c=1 `  has a value of "`e` ".

Plug-in the values on the formula for Maclaurin series. 

`e^x= sum_(n=0)^oo (f^n(1))/(n!) (x-1)^n`

     `=e+e*(x-1) +e/(2!)(x-1)^2+e/(3!)(x-1)^3+ e/(4!)(x-1)^4+...`

     `=e+e*(x-1) +e/2(x-1)^2+e/6(x-1)^3+ e/24(x-1)^4+...`

The Taylor series for the given function ` f(x)=e^x` centered at `c=1` will be:

`e^x =e+e*(x-1) +e/2(x-1)^2+e/6(x-1)^3+ e/24(x-1)^4+...`

or

`e^x =sum_(n=0)^oo e/(n!)(x-1)^n`