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

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` .

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