f(x) = 1/(1-x) ,c=2 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=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) = 1/(1-x) centered at c=2, we list f^n(x) using the  Power rule for differentiation: d/(dx) u^n= n *u^(n-1) *(du)/(dx)  and basic differentiation property: d/(dx) c* f(x)= c * d/(dx) f(x) .
f(x)= 1/(1-x)
Let u=1-x then (du)/(dx)= -1 .
The derivative of f(x) will be:
d/(dx) (1/(1-x)) =d/(dx) (1-x)^(-1)
                  = (-1)*(1-x)^(-1-1)*(-1)
                  =(1-x)^-2 or 1/(1-x)^2
Then, we list the derivatives of f(x) as:
f'(x) = d/(dx) (1/(1-x))
           =(1-x)^-2 or 1/(1-x)^2
f^2(x)= d/(dx) (1-x)^(-2)
            =-2*((1-x)^(-2-1))*(-1)
           =2(1-x)^(-3) or 2/(1-x)^3
f^3(x)= d/(dx) 2(1-x)^(-3)
            =2*d/(dx) (1-x)^(-3)
            =2* (-3*(1-x)^(-3-1))*(-1)
            =6(1-x)^(-4) or 6/(1-x)^4
f^4(x)= d/(dx)6(1-x)^(-4)
            =6*d/(dx) (1-x)^(-4)
            =6* (-4*(1-x)^(-4-1))*(-1)
            =24(1-x)^(-5) or 24/(1-x)^5
Plug-in x=2 , we get:
f(2)=1/(1-2)
        =1/(-1)
        =-1
f'(2)=1/(1-2)^2
          =1/(-1)^2
         =1/1
         =1
f^2(2)=2/(1-2)^3
           =2/(-1)^3
         =2/(-1)
        =-2
f^3(2)=6/(1-2)^4
          =6/(-1)^4
          =6/1
          =6
f^4(2)=24/(1-2)^5
           =24/(-1)^5
          =24/(-1)
          =-24
Plug-in the values on the formula for Taylor series, we get:
1/(1-x) =sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n
=f(2)+f'(2)(x-2) +(f^2(2))/(2!)(x-2)^2 +(f^3(2))/(3!)(x-2)^3 +(f^4(2))/(4!)(x-2)^4 +...
=-1+1*(x-2) + (-2)/(2!)(x-2)^2 +6/(3!)(x-2)^3 + (-24)/(4!)(x-2)^4 +...
=-1+(x-2) -2/2(x-2)^2 +6/6(x-2)^3 -24/24(x-2)^4 +...
=-1+(x-2) -(x-2)^2 + (x-2)^3 -(x-2)^4 +...
The Taylor series for the given function f(x)=1/(1-x) centered at c=2 will be:
1/(1-x)=-1+(x-2) -(x-2)^2 + (x-2)^3 -(x-2)^4 +...
or 
1/(1-x) = sum_(n=0)^oo (-1)^(n+1)(x-2)^n