`f(x) = 1/(1-x) ,c=2` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Expert Answers
marizi eNotes educator| Certified Educator

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`