`f(x)=1/x ,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) = 1/x` centered at `c=1` , we list `f^n(x)` using the  Power rule for differentiation: `d/(dx) x^n= n *x^(n-1)` and basic differentiation property: `d/(dx) c* f(x)= c * d/(dx) f(x)` .

`f(x) =1/x`

`f'(x) = d/(dx) 1/x`

       `= d/(dx) x^(-1)`

       `=-1 *x^(-1-1)`

       `=-x^(-2) or -1/x^2`

`f^2(x)= d/(dx) -x^(-2)`

            `=-1 *d/(dx) x^(-2)`

            `=-1 *(-2x^(-2-1))`

           `=2x^(-3) or 2/x^3`

`f^3(x)= d/(dx) 2x^(-3)`

           `=2 *d/(dx) x^(-3)`

          `=2 *(-3x^(-3-1))`

          `=-6x^(-4) or -6/x^4`

`f^4(x)= d/(dx) -6x^(-4)`

            `=-6 *d/(dx) x^(-4)`

            `=-6 *(-4x^(-4-1))`

           `=24x^(-5) or 24/x^5`

Plug-in `x=1` , we get:

`f(1)=1/1 =1`

`f'(1)=-1/1^2 = -1`

`f^2(1)=2/1^3 =2`

`f^3(1)=-6/1^4 = -6`

`f^4(1)=24/1^5 = 24`

Plug-in the values on the formula for Taylor series, we get:

`1/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 +...`

` =1+(-1)*(x-1) +2/(2!)(x-1)^2 +(-6)/(3!)(x-1)^3 +24/(4!)(x-1)^4 +...`

` =1-(x-1) +2/2(x-1)^2 -6/6(x-1)^3 +24/24(x-1)^4 +...`

` = 1-(x-1)+ (x-1)^2 -(x-1)^3 + (x-1)^4 +...`

The Taylor series for the given function `f(x)=1/x` centered at `c=1` will be:

`1/x=1-(x-1)+ (x-1)^2 -(x-1)^3 + (x-1)^4 +...`

or

`1/x =sum_(n=0)^oo(-1)^n(x-1)^n`