`f(x)=3/(2x-1) ,c=2` Find a power series for the function, centered at c and determine the interval of convergence.

Expert Answers
marizi eNotes educator| Certified Educator

To determine the power series centered at c, we may apply the formula for Taylor series:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) =f(c)+f'(c)(x-c) +(f''(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f'^4(c))/(4!)(x-c)^4 +...`

To list the `f^n(x) ` for the given function ` f(x)=3/(2x-1)` centered at `c=2` , we may apply Law of Exponent: `1/x^n = x^-n`  and  Power rule for derivative: `d/(dx) x^n= n *x^(n-1)` .

`f(x) =3/(2x-1)`

          ` =3(2x-1)^(-1)`

Let `u =2x-1` then `(du)/(dx) = 2`

`d/(dx) c*(2x-1)^n = c *d/(dx) (2x-1)^n`

                           `= c *(n* (2x-1)^(n-1)*2`

                            ` = 2cn(2x-1)^(n-1)`

`f'(x) =d/(dx) 3(2x-1)^(-1)`

            `=2*3*(-1)(2x-1)^(-1-1)`

            `=-6(2x-1)^(-2) or 2/(2x-1)^2`

`f^2(x) =d/(dx) -6(2x-1)^(-2)`

            `=2*(-6)(-2)(2x-1)^(-2-1)`

            ` =24(2x-1)^(-3) or 24/(2x-1)^3`

`f^3(x) =d/(dx) 24(2x-1)^(-3)`

           `=2*(24)(-3)(2x-1)^(-3-1)`

          ` =-144(2x-1)^(-4) or -144/(2x-1)^4`

Plug-in x=2 for each f^n(x), we get:

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

         ` =3/ 3`

          `=1`

`f'(2)=-6/(2(2)-1)^2`

          `=-6/3^2`

          `= -2/3`

`f^2(2)=24/(2(2)-1)^3 `

            `=24/3^3`

            `=8/9`

`f^3(2)=-144/(2(2)-1)^4 `

            `=-144/3^4`

            `= -16/9`

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

`3/(2x-1) = sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n`

` = sum_(n=0)^oo (f^n(2))/(n!) (x-2)^n`

` =1+(-2/3)(x-2) +(8/9)/(2!)(x-2)^2 +(-16/9)/(3!)(x-2)^3 +...`

` =1-2/3(x-2) +(8/9)/2(x-2)^2 +(-16/9)/6(x-2)^3 +...`

`=1-2/3(x-2) +4/9(x-2)^2 +8/27(x-2)^3 +...`

`= sum_(n=0)^oo (-(2(x-2))/3)^n`

To determine the interval of convergence, we may apply geometric series test wherein the series `sum_(n=0)^oo a*r^n `  is convergent if `|r|lt1 or -1 ltrlt 1` . If `|r|gt=1` then the geometric series diverges.

By comparing  `sum_(n=0)^oo (-(2(x-2))/3)^n` with  `sum_(n=0)^oo a*r^n` , we determine: `r = -(2(x-2))/3` .

Apply the condition for convergence of geometric series: `|r|lt1` .

`|-(2(x-2))/3|lt1`

`|-1|*|(2(x-2))/3|lt1`

`1*|(2(x-2))/3|lt1`

`|(2(x-2))/3|lt1`

`|(2x-4)/3|lt1`

`-1lt(2x-4)/3lt1`

Multiply each sides by `3` :

`-1*3lt(2x-4)/3*3lt1*3`

`-3lt2x-4lt3`

Add 4 on each sides:

`-3+4lt2x-4+4lt3+4`

`1lt2xlt7`

Divide each side by `2` :

`1/2lt2x/2lt7/2`

`1/2ltxlt7/2`

Thus, the power series  of the function `f(x) =3/(2x-1) ` centered at `c=2` is `sum_(n=0)^oo (-(2(x-2))/3)^n`  and has an interval of convergence: `1/2ltxlt7/2` .