f(x)=e^x ,c=1 Use the definition of Taylor series to find the Taylor series, centered at c for the function.

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