`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`