f(x)=e^(2x) ,c=0 Use the definition of Taylor series to find the Taylor series, centered at c for the function.

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=a . The general formula for Taylor series is:
f(x) = sum_(n=0)^oo (f^n(a))/(n!) (x-a)^n
or
f(x) =f(a)+f'(a)(x-a) +(f''(a))/(2!)(x-a)^2 +(f^3(a))/(3!)(x-a)^3 +(f'^4(a))/(4!)(x-a)^4 +...
To apply the definition of Taylor series for the given function f(x) = e^(2x) , we list f^n(x) using the derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx) .
Let u =2x then (du)/(dx)= 2
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(2x) = e^(2x) *2
               = 2e^(2x)
Applying d/(dx) e^(2x)= 2e^(2x)   for each f^n(x) , we get:
f'(x) = d/(dx) e^(2x)
           = 2e^(2x)
f^2(x) = 2 *d/(dx) e^(2x)
           = 2*2e^(2x)
           =4e^(2x)
f^3(x) = 4*d/(dx) e^(2x)
            = 4*2e^(2x)
f^4(x) = 8*d/(dx) e^(2x)
           = 8*2e^(2x) 
           =16e^(2x)
Plug-in x=0 , 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. 
e^(2x)= sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n
          = sum_(n=0)^oo (f^n(0))/(n!) x^n
          = 1+2x+4/(2!)x^2+8/(3!)x^3+16/(4!)x^4+...
          =1+2x+4/(1*2)x^2+8/(1*2*3)x^3+16/(1*2*3*4)x^4+...
          = 1+2x+4/2x^2+8/6x^3+16/24x^4+...
         = 1+2x+2x^2+4/3x^3+2/3x^4+...
The Taylor series for the given function f(x)=e^(2x) centered at a=0 will be:
e^(2x) =1+2x+2x^2+4/3x^3+2/3x^4+...