f(x)=lnx ,c=1 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) = ln(x) , we list f^n(x) as:
f(x) = ln(x)
f'(x) = d/(dx)ln(x) =1/x
Apply Power rule for derivative: d/(dx) x^n= n *x^(n-1)
f^2(x) = d/(dx) 1/x
            = d/(dx) x^(-1)
            =-1 *x^(-1-1)
            =-x^(-2) or -1/x^2
f^3(x) = d/(dx) -x^(-2)
            =-1 *d/(dx) x^(-2)
            =-1 *(-2x^(-2-1))
           =2x^(-3) or 2/x^3
f^4(x)= d/(dx) 2x^(-3)
             =2 *d/(dx) x^(-3)
            =2 *(-3x^(-3-1))
            =-6x^(-4) or -6/x^4
Plug-in x=1 , we get:
f(1) =ln(1) =0
f'(1)=1/1 =1
f^2(1)=-1/1^2 = -1
f^3(1)=2/1^3 =2
f^4(1)=-6/1^4 = -6
Plug-in the values on the formula for Taylor series, we get:
ln(x) =sum_(n=0)^oo (f^n(1))/(n!) (x-1)^n
=f(1)+f'(1)(x-1) +(f^2(1))/(2!)(x-1)^2 +(f^3(1))/(3!)(x-1)^3 +(f^4(1))/(4!)(x-1)^4 +...
=0+1*(x-1) +(-1)/(2!)(x-1)^2 +2/(3!)(x-1)^3 +(-6)/(4!)(x-1)^4 +...
=x-1 -1/2(x-1)^2 +1/3(x-1)^3 -1/4(x-1)^4 +...
The Taylor series for the given function f(x)=ln(x) centered at c=1 will be:
ln(x) =x-1 -1/2(x-1)^2 +1/3(x-1)^3 -1/4(x-1)^4 +...
 or
ln(x) = sum_(n=1)^oo (-1)^(n+1)(x-1)^n/n